12.3 BEHAVIOUR OF GASES where M is the mass of the gas containing N Properties of gases are easier to understand than molecules, M0 is the molar mass and NA the those of solids and liquids. This is mainly Avogadro’s number. Using Eqs. (12.4) and (12.3) because in a gas, molecules are far from each can also be written as other and their mutual interactions are PV = kB NT or P = kB nT negligible except when two molecules collide. Gases at low pressures and high temperatures much above that at which they liquefy (or solidify) approximately satisfy a simple relation ) between their pressure, temperature and volume –1 given by (see Chapter 10) K –1 PV = KT (12.1) mol Jfor a given sample of the gas. Here T is the ( temperature in kelvin or (absolute) scale. K is a T pV µ constant for the given sample but varies with the volume of the gas. If we now bring in the idea of atoms or molecules, then K is proportional to the number of molecules, (say) N in the sample. We can write K = N k . Observation tells P (atm) us that this k is same for all gases. It is called Fig.12.1 Real gases approach ideal gas behaviour at Boltzmann constant and is denoted by k B. low pressures and high temperatures. P1V1 P2 V2 where n is the number density, i.e. number ofAs = = constant = kB (12.2) N 1 T1 N 2 T2 molecules per unit volume. kB is the Boltzmann constant introduced above. Its value in SI unitsif P, V and T are same, then N is also same for all is 1.38 × 10–23 J K–1.gases. This is Avogadro’s hypothesis, that the Another useful form of Eq. (12.3) isnumber of molecules per unit volume is ρRT the same for all gases at a fixed temperature and P = (12.5) pressure. The number in 22.4 litres of any gas M 0 Reprint 2025-26 KINETIC THEORY 247 where ρ is the mass density of the gas. etc. in a vessel of volume V at temperature T and A gas that satisfies Eq. (12.3) exactly at all pressure P. It is then found that the equation of pressures and temperatures is defined to be an state of the mixture is : ideal gas. An ideal gas is a simple theoretical PV = ( µ1 + µ2 +… ) RT (12.7)model of a gas. No real gas is truly ideal. Fig. 12.1 shows departures from ideal gas RT RT i.e. P = µ1 + µ2 + ... (12.8)behaviour for a real gas at three different V V temperatures. Notice that all curves approach = P1 + P2 + … (12.9)the ideal gas behaviour for low pressures and high temperatures. Clearly P1 = µ1 R T/V is the pressure that At low pressures or high temperatures the gas 1 would exert at the same conditions of molecules are far apart and molecular volume and temperature if no other gases were interactions are negligible. Without interactions present. This is called the partial pressure of the the gas behaves like an ideal one. gas. Thus, the total pressure of a mixture of ideal If we fix µ and T in Eq. (12.3), we get gases is the sum of partial pressures. This is Dalton’s law of partial pressures. PV = constant (12.6) i.e., keeping temperature constant, pressure of a given mass of gas varies inversely with volume. This is the famous Boyle’s law. Fig. 12.2 shows comparison between experimental P-V curves and the theoretical curves predicted by Boyle’s law. Once again you see that the agreement is good at high temperatures and low pressures. Next, if you fix P, Eq. (12.1) shows that V ∝ T i.e., for a fixed pressure, the volume of a gas is proportional to its absolute temperature T (Charles’ law). See Fig. 12.3. Fig. 12.3 Experimental T-V curves (solid lines) for CO2 at three pressures compared with Charles’ law (dotted lines). T is in units of 300 K and V in units of 0.13 litres. We next consider some examples which give us information about the volume occupied by the molecules and the volume of a single molecule. ⊳ Example 12.1 The density of water is 1000 kg m–3. The density of water vapour at 100 °C and 1 atm pressure is 0.6 kg m–3. The volume of a molecule multiplied by the total Fig.12.2 Experimental P-V curves (solid lines) for number gives ,what is called, molecular steam at three temperatures compared with volume. Estimate the ratio (or fraction) of Boyle’s law (dotted lines). P is in units of 22 the molecular volume to the total volume atm and V in units of 0.09 litres. occupied by the water vapour under the Finally, consider a mixture of non-interacting above conditions of temperature and pressure.ideal gases: µ moles of gas 1, µ moles of gas 2, 1 2 Reprint 2025-26 248 PHYSICS Answer For a given mass of water molecules, number of molecules and (ii) mass density the density is less if volume is large. So the of neon and oxygen in the vessel. Atomic volume of the vapour is 1000/0.6 = 1/(6 × 10 -4 ) mass of Ne = 20.2 u, molecular mass of O2times larger. If densities of bulk water and water = 32.0 u. molecules are same, then the fraction of molecular volume to the total volume in liquid Answer Partial pressure of a gas in a mixture is state is 1. As volume in vapour state has the pressure it would have for the same volume increased, the fractional volume is less by the and temperature if it alone occupied the vessel. same amount, i.e. 6×10-4. ⊳ (The total pressure of a mixture of non-reactive ⊳ gases is the sum of partial pressures due to its Example 12.2 Estimate the volume of a constituent gases.) Each gas (assumed ideal) water molecule using the data in Example obeys the gas law. Since V and T are common to 12.1. the two gases, we have P1V = µ 1 RT and P2V = Answer In the liquid (or solid) phase, the µ2 RT, i.e. (P1/P2) = (µ1 / µ2). Here 1 and 2 refer to neon and oxygen respectively. Since (P1/P2) =molecules of water are quite closely packed. The (3/2) (given), (µ1/ µ2) = 3/2.density of water molecule may therefore, be (i) By definition µ1 = (N1/NA ) and µ2 = (N2/NA)regarded as roughly equal to the density of bulk where N1 and N2 are the number of moleculeswater = 1000 kg m–3. To estimate the volume of of 1 and 2, and NA is the Avogadro’s number.a water molecule, we need to know the mass of Therefore, (N1/N2) = (µ1 / µ2) = 3/2.a single water molecule. We know that 1 mole (ii) We can also write µ1 = (m1/M1) and µ2 =of water has a mass approximately equal to (2 + 16)g = 18 g = 0.018 kg. (m2/M2) where m1 and m2 are the masses of 1 and 2; and M1 and M2 are their molecular Since 1 mole contains about 6 × 1023 masses. (Both m1 and M1; as well as m2 andmolecules (Avogadro’s number), the mass of a molecule of water is (0.018)/(6 × 1023) kg = M2 should be expressed in the same units). If ρ1 and ρ2 are the mass densities of 1 and3 × 10–26 kg. Therefore, a rough estimate of the 2 respectively, we havevolume of a water molecule is as follows : Volume of a water molecule ρ1 m 1 / V m 1 µ1  M 1  = (3 × 10–26 kg)/ (1000 kg m–3) = = = × ρ2 m 2 / V m 2 µ2  M 2  = 3 × 10–29 m3 = (4/3) π (Radius)3 3 20.2 Hence, Radius ≈ 2 ×10-10 m = 2 Å ⊳ = × = 0.947 2 32.0 ⊳ ⊳ Example 12.3 What is the average distance between atoms (interatomic distance) in water? Use the data given in 12.4 KINETIC THEORY OF AN IDEAL GAS Examples 12.1 and 12.2. Kinetic theory of gases is based on the molecular picture of matter. A given amount of gas is a Answer : A given mass of water in vapour state collection of a large number of molecules has 1.67×103 times the volume of the same mass (typically of the order of Avogadro’s number) that of water in liquid state (Ex. 12.1). This is also are in incessant random motion. At ordinary the increase in the amount of volume available pressure and temperature, the average distance for each molecule of water. When volume between molecules is a factor of 10 or more than increases by 103 times the radius increases by the typical size of a molecule (2 Å). Thus, V1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So the interaction between molecules is negligible and average distance is 2 × 20 = 40 Å. ⊳ we can assume that they move freely in straight lines according to Newton’s first law. However,⊳ Example 12.4 A vessel contains two non- occasionally, they come close to each other, reactive gases : neon (monatomic) and experience intermolecular forces and their oxygen (diatomic). The ratio of their partial velocities change. These interactions are called pressures is 3:2. Estimate the ratio of (i) collisions. The molecules collide incessantly against each other or with the walls and change Reprint 2025-26 KINETIC THEORY 249 their velocities. The collisions are considered to the wall. Thus, the number of molecules with be elastic. We can derive an expression for the velocity (vx, vy, vz ) hitting the wall in time ∆t is pressure of a gas based on the kinetic theory. ½A vx ∆t n, where n is the number of molecules We begin with the idea that molecules of a per unit volume. The total momentum gas are in incessant random motion, colliding transferred to the wall by these molecules in against one another and with the walls of the time ∆t is: container. All collisions between molecules Q = (2mvx) (½ n A vx ∆t ) (12.10) among themselves or between molecules and the The force on the wall is the rate of momentum walls are elastic. This implies that total kinetic transfer Q/∆t and pressure is force per unit energy is conserved. The total momentum is area : conserved as usual. P = Q /(A ∆t) = n m vx 2 (12.11) Actually, all molecules in a gas do not have 12.4.1 Pressure of an Ideal Gas the same velocity; there is a distribution in velocities. The above equation, therefore, standsConsider a gas enclosed in a cube of side l. Take for pressure due to the group of molecules withthe axes to be parallel to the sides of the cube, speed vx in the x-direction and n stands for theas shown in Fig. 12.4. A molecule with velocity number density of that group of molecules. The (vx, vy, vz ) hits the planar wall parallel to yz- total pressure is obtained by summing over theplane of area A (= l2). Since the collision is elastic, contribution due to all groups:the molecule rebounds with the same velocity; its y and z components of velocity do not change P = n m v x2 (12.12) in the collision but the x-component reverses where v 2x is the average of vx 2 . Now the gas sign. That is, the velocity after collision is is isotropic, i.e. there is no preferred direction (-vx, vy, vz ) . The change in momentum of the of velocity of the molecules in the vessel. molecule is: –mvx – (mvx) = – 2mvx . By the Therefore, by symmetry, principle of conservation of momentum, the momentum imparted to the wall in the collision v 2x = v y2 = v z2 = 2mvx . 2 2 2 2 = (1/3) [ v x + v y + v z ] = (1/3) v (12.13) where v is the speed and v 2 denotes the mean of the squared speed. Thus P = (1/3) n m v 2 (12.14) Some remarks on this derivation. First, though we choose the container to be a cube, the shape of the vessel really is immaterial. For a vessel of arbitrary shape, we can always choose a small infinitesimal (planar) area and carry through the steps above. Notice that both A and ∆t do not appear in the final result. By Pascal’s law, given in Ch. 9, pressure in one portion of Fig. 12.4 Elastic collision of a gas molecule with the the gas in equilibrium is the same as anywhere wall of the container. else. Second, we have ignored any collisions in To calculate the force (and pressure) on the the derivation. Though this assumption is wall, we need to calculate momentum imparted difficult to justify rigorously, we can qualitatively to the wall per unit time. In a small time interval see that it will not lead to erroneous results. The ∆t, a molecule with x-component of velocity vx number of molecules hitting the wall in time ∆t will hit the wall if it is within the distance vx ∆t was found to be ½ n Avx ∆t. Now the collisions from the wall. That is, all molecules within the are random and the gas is in a steady state. volume Avx ∆t only can hit the wall in time ∆t. Thus, if a molecule with velocity (vx, vy, vz ) But, on the average, half of these are moving acquires a different velocity due to collision with towards the wall and the other half away from some molecule, there will always be some other Reprint 2025-26 250 PHYSICS molecule with a different initial velocity which P = (1/3) [n1m1 v1 2 + n2 m2 v 22 +… ] (12.20) after a collision acquires the velocity (vx, vy, vz ). In equilibrium, the average kinetic energy of If this were not so, the distribution of velocities the molecules of different gases will be equal. would not remain steady. In any case we are That is, finding v x2 . Thus, on the whole, molecular ½ m1 v1 2 = ½ m2 v 22 = (3/2) kB Tcollisions (if they are not too frequent and the so thattime spent in a collision is negligible compared to time between collisions) will not affect the P = (n1 + n2 +… ) kB T (12.21) calculation above. which is Dalton’s law of partial pressures. From Eq. (12.19), we can get an idea of the12.4.2 Kinetic Interpretation of Temperature typical speed of molecules in a gas. At a Equation (13.14) can be written as temperature T = 300 K, the mean square speed PV = (1/3) nV m v 2 (12.15a) of a molecule in nitrogen gas is : PV = (2/3) N x ½ m v 2 (12.15b) M N 2 28 –26 where N (= nV) is the number of molecules in m = = 26 = 4.65 × 10 kg. N A 6.02 × 10the sample. The quantity in the bracket is the average v 2 = 3 kB T / m = (516)2 m2s-2 translational kinetic energy of the molecules in 2 The square root of v is known as root mean the gas. Since the internal energy E of an ideal square (rms) speed and is denoted by vrms,gas is purely kinetic*, 2 ( We can also write v 2 as < v2 >.) E = N × (1/2) m v (12.16) vrms = 516 m s-1 Equation (12.15) then gives : The speed is of the order of the speed of sound PV = (2/3) E (12.17) in air. It follows from Eq. (12.19) that at the same We are now ready for a kinetic interpretation temperature, lighter molecules have greater rms of temperature. Combining Eq. (12.17) with the speed. ⊳ideal gas Eq. (12.3), we get Example 12.5 A flask contains argon and E = (3/2) kB NT (12.18) chlorine in the ratio of 2:1 by mass. The or E/ N = ½ m v 2 = (3/2) kBT (12.19) temperature of the mixture is 27 °C. Obtain i.e., the average kinetic energy of a molecule is the ratio of (i) average kinetic energy per proportional to the absolute temperature of the molecule, and (ii) root mean square speed gas; it is independent of pressure, volume or vrms of the molecules of the two gases. the nature of the ideal gas. This is a fundamental Atomic mass of argon = 39.9 u; Molecular result relating temperature, a macroscopic mass of chlorine = 70.9 u. measurable parameter of a gas (a thermodynamic variable as it is called) to a Answer The important point to remember is thatmolecular quantity, namely the average kinetic the average kinetic energy (per molecule) of anyenergy of a molecule. The two domains are connected by the Boltzmann constant. We note (ideal) gas (be it monatomic like argon, diatomic in passing that Eq. (12.18) tells us that internal like chlorine or polyatomic) is always equal to energy of an ideal gas depends only on (3/2) kBT. It depends only on temperature, and temperature, not on pressure or volume. With is independent of the nature of the gas. this interpretation of temperature, kinetic theory (i) Since argon and chlorine both have the same of an ideal gas is completely consistent with the temperature in the flask, the ratio of average ideal gas equation and the various gas laws kinetic energy (per molecule) of the two gasesbased on it. is 1:1. For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gas (ii) Now ½ m vrms2 = average kinetic energy per in the mixture. Equation (12.14) becomes molecule = (3/2) ) kBT where m is the mass * E denotes the translational part of the internal energy U that may include energies due to other degrees of freedom also. See section 12.5. Reprint 2025-26 KINETIC THEORY 251 of a molecule of the gas. Therefore, v 2 ( rms ) Ar (m )Cl ( M )Cl 70.9 = = 2 v rms (m ) Ar ( M ) Ar = 39.9 =1.77 ( )Cl where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.) Taking square root of both sides, v ( rms ) Ar ( vrms )Cl = 1.33 You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains Fig. 12.5 Molecules going through a porous wall. unaltered. ⊳ ⊳ ⊳ Example 12.7 (a) When a molecule (or an Example 12.6 Uranium has two isotopes elastic ball) hits a ( massive) wall, it of masses 235 and 238 units. If both are rebounds with the same speed. When a ball present in Uranium hexafluoride gas which hits a massive bat held firmly, the same would have the larger average speed ? If thing happens. However, when the bat is atomic mass of fluorine is 19 units, moving towards the ball, the ball rebounds estimate the percentage difference in with a different speed. Does the ball move speeds at any temperature. faster or slower? (Ch.5 will refresh your Answer At a fixed temperature the average memory on elastic collisions.) energy = ½ m <v2 > is constant. So smaller the (b) When gas in a cylinder is compressed mass of the molecule, faster will be the speed. by pushing in a piston, its temperature The ratio of speeds is inversely proportional to rises. Guess at an explanation of this in the square root of the ratio of the masses. The terms of kinetic theory using (a) above. masses are 349 and 352 units. So (c) What happens when a compressed gas v349 / v352 = ( 352/ 349)1/2 = 1.0044 . pushes a piston out and expands. What ∆ V would you observe ? Hence difference = 0.44 %. (d) Sachin Tendulkar used a heavy cricket V bat while playing. Did it help him in [235U is the isotope needed for nuclear fission. anyway ?To separate it from the more abundant isotope 238U, the mixture is surrounded by a porous cylinder. The porous cylinder must be thick and Answer (a) Let the speed of the ball be u relative narrow, so that the molecule wanders through to the wicket behind the bat. If the bat is moving individually, colliding with the walls of the long towards the ball with a speed V relative to the pore. The faster molecule will leak out more than wicket, then the relative speed of the ball to bat the slower one and so there is more of the lighter is V + u towards the bat. When the ball rebounds molecule (enrichment) outside the porous (after hitting the massive bat) its speed, relative cylinder (Fig. 12.5). The method is not very to bat, is V + u moving away from the bat. So efficient and has to be repeated several times relative to the wicket the speed of the rebounding for sufficient enrichment.]. ⊳ ball is V + (V + u) = 2V + u, moving away from the When gases diffuse, their rate of diffusion is wicket. So the ball speeds up after the collision inversely proportional to square root of the with the bat. The rebound speed will be less than masses (see Exercise 12.12 ). Can you guess the u if the bat is not massive. For a molecule this explanation from the above answer? would imply an increase in temperature. Reprint 2025-26 252 PHYSICS You should be able to answer (b) (c) and (d) to the axis joining the two oxygen atoms about based on the answer to (a). which the molecule can rotate*. The molecule (Hint: Note the correspondence, pistonà bat, thus has two rotational degrees of freedom, each of which contributes a term to the total energycylinder à wicket, molecule à ball.) ⊳ consisting of translational energy tε and rotational energy εr.12.5 LAW OF EQUIPARTITION OF ENERGY The kinetic energy of a single molecule is εt + εr = 1 mv x2 + 1 mv y2 + 1 mv z2 + 1 I 1ω12 + 1 I 2ω22 (12.25) 2 2 2 2 2 1 2 1 2 1 2 εt = mv x + mv y + mv z (12.22) 2 2 2 For a gas in thermal equilibrium at temperature T the average value of energy denoted by < tε > is 1 2 1 2 1 2 3 εt = mv x + mv y + mv z = k B T (12.23) 2 2 2 2 Since there is no preferred direction, Eq. (12.23) implies 1 2 1 1 2 1 mv x = k B T , mv y = k B T , 2 2 2 2 Fig. 12.6 The two independent axes of rotation of a diatomic molecule 1 2 1 mv z = k B T (12.24) 2 2 where ω1 and ω2 are the angular speeds about A molecule free to move in space needs three the axes 1 and 2 and I1, I2 are the corresponding coordinates to specify its location. If it is moments of inertia. Note that each rotational constrained to move in a plane it needs two; and degree of freedom contributes a term to the if constrained to move along a line, it needs just energy that contains square of a rotational one coordinate to locate it. This can also be variable of motion. expressed in another way. We say that it has We have assumed above that the O2 molecule one degree of freedom for motion in a line, two is a ‘rigid rotator’, i.e., the molecule does not for motion in a plane and three for motion in vibrate. This assumption, though found to be space. Motion of a body as a whole from one true (at moderate temperatures) for O2, is notpoint to another is called translation. Thus, a always valid. Molecules, like CO, even at molecule free to move in space has three moderate temperatures have a mode of translational degrees of freedom. Each vibration, i.e., its atoms oscillate along the translational degree of freedom contributes a interatomic axis like a one-dimensional term that contains square of some variable of 2 oscillator, and contribute a vibrational energymotion, e.g., ½ mvx and similar terms in term εv to the total energy:vy and vz. In, Eq. (12.24) we see that in thermal equilibrium, the average of each such term is 1  d y  2 1 2 εv = m + ky½ kBT . 2  d t  2 Molecules of a monatomic gas like argon have only translational degrees of freedom. But what ε = εt + εr + ε v (12.26) about a diatomic gas such as O2 or N2? A where k is the force constant of the oscillator molecule of O2 has three translational degrees and y the vibrational co-ordinate. of freedom. But in addition it can also rotate Once again the vibrational energy terms in about its centre of mass. Figure 12.6 shows the Eq. (12.26) contain squared terms of vibrational two independent axes of rotation 1 and 2, normal variables of motion y and dy/dt . * Rotation along the line joining the atoms has very small moment of inertia and does not come into play for quantum mechanical reasons. See end of section 12.6. Reprint 2025-26 KINETIC THEORY 253 At this point, notice an important feature in where Cp is the molar specific heat at constant Eq.(12.26). While each translational and pressure. Thus, rotational degree of freedom has contributed only 5 one ‘squared term’ in Eq.(12.26), one vibrational Cp = R (12.30) mode contributes two ‘squared terms’ : kinetic 2 and potential energies. C p 5 The ratio of specific heats γ = = (12.31) Each quadratic term occurring in the C v 3 expression for energy is a mode of absorption of energy by the molecule. We have seen that in 12.6.2 Diatomic Gases thermal equilibrium at absolute temperature T, As explained earlier, a diatomic molecule treated for each translational mode of motion, the as a rigid rotator, like a dumbbell, has 5 degrees average energy is ½ kBT. The most elegant of freedom: 3 translational and 2 rotational. principle of classical statistical mechanics (first Using the law of equipartition of energy, the total proved by Maxwell) states that this is so for each internal energy of a mole of such a gas is mode of energy: translational, rotational and 5 5 vibrational. That is, in equilibrium, the total U = k B T × N A = RT (12.32) 2 2 energy is equally distributed in all possible The molar specific heats are then given by energy modes, with each mode having an average energy equal to ½ kBT. This is known as the law 5 7 Cv (rigid diatomic) = R, Cp = R (12.33)of equipartition of energy. Accordingly, each 2 2 translational and rotational degree of freedom 7 of a molecule contributes ½ kBT to the energy, γ (rigid diatomic) = (12.34) while each vibrational frequency contributes 5 If the diatomic molecule is not rigid but has 2 × ½ kBT = kBT , since a vibrational mode has in addition a vibrational mode both kinetic and potential energy modes.  5  7 The proof of the law of equipartition of energy U =  k B T + k B T  N A = RT is beyond the scope of this book. Here, we shall  2  2 apply the law to predict the specific heats of gases 7 9 9 theoretically. Later, we shall also discuss briefly, C v = R , C p = R , γ = R (12.35) 2 2 7 the application to specific heat of solids. 12.6.3 Polyatomic Gases 12.6 SPECIFIC HEAT CAPACITY In general a polyatomic molecule has 3 12.6.1 Monatomic Gases translational, 3 rotational degrees of freedom and a certain number ( f ) of vibrational modes.The molecule of a monatomic gas has only three According to the law of equipartition of energy,translational degrees of freedom. Thus, the it is easily seen that one mole of such a gas hasaverage energy of a molecule at temperature 3T is (3/2)kBT . The total internal energy of a mole  3 U = kBT + kBT + f kBT  NA of such a gas is  2 2 3 3 U = k B T × N A = RT (12.27) i.e.,Cv = (3 + f ) R, Cp = (4 + f ) R, 2 2 ( 4 + f ) γ = (12.36) The molar specific heat at constant volume, ( 3 + f ) Cv, is Note that Cp – Cv = R is true for any ideal d U 3 gas, whether mono, di or polyatomic. Cv (monatomic gas) = = RT (12.28) d T 2 Table 12.1 summarises the theoretical For an ideal gas, predictions for specific heats of gases ignoring Cp – Cv = R (12.29) any vibrational modes of motion. The values are Reprint 2025-26 254 PHYSICS in good agreement with experimental values of Answer Using the gas law PV = µRT, you can specific heats of several gases given in Table 12.2. easily show that 1 mol of any (ideal) gas at Of course, there are discrepancies between standard temperature (273 K) and pressure predicted and actual values of specific heats of (1 atm = 1.01 × 105 Pa) occupies a volume of 22.4 several other gases (not shown in the table), such litres. This universal volume is called molar volume. Thus the cylinder in this exampleas Cl2, C2H6 and many other polyatomic gases. contains 2 mol of helium. Further, since heliumUsually, the experimental values for specific is monatomic, its predicted (and observed) molar heats of these gases are greater than the specific heat at constant volume, Cv = (3/2) R,predicted values as given in Table12.1 suggesting and molar specific heat at constant pressure, that the agreement can be improved by including Cp = (3/2) R + R = (5/2) R. Since the volume of vibrational modes of motion in the calculation. the cylinder is fixed, the heat required is The law of equipartition of energy is, thus, well determined by Cv. Therefore, verified experimentally at ordinary temperatures. Heat required = no. of moles × molar specific heat × rise in temperature Table 12.1 Predicted values of specific heat = 2 × 1.5 R × 15.0 = 45 R capacities of gases (ignoring vibrational modes) = 45 × 8.31 = 374 J. ⊳ 12.6.4 Specific Heat Capacity of Solids Nature of Cv Cp Cp - Cv g Gas (J mol-1 K-1) (J mol-1 K-1) (J mol-1 K-1) We can use the law of equipartition of energy to determine specific heats of solids. Consider a Monatomic 12.5 20.8 8.31 1.67 solid of N atoms, each vibrating about its mean Diatomic 20.8 29.1 8.31 1.40 position. An oscillation in one dimension has average energy of 2 × ½ kBT = kBT . In three Triatomic 24.93 33.24 8.31 1.33 dimensions, the average energy is 3 kBT. For a mole of solid, N = NA, and the total energy is Table12.2 Measured values of specific heat U = 3 kBT × NA = 3 RT capacities of some gases Now at constant pressure ∆Q = ∆U + P∆V = ∆U, since for a solid ∆V is negligible. Hence, ∆Q ∆ U C = = = 3 R (12.37) ∆T ∆T Table 12.3 Specific Heat Capacity of some solids at room temperature and atmospheric pressure As Table 12.3 shows the prediction generally ⊳ Example 12.8 A cylinder of fixed capacity agrees with experimental values at ordinary 44.8 litres contains helium gas at standard temperature (Carbon is an exception). temperature and pressure. What is the amount of heat needed to raise the 12.7 MEAN FREE PATH temperature of the gas in the cylinder by Molecules in a gas have rather large speeds of 15.0 °C ? (R = 8.31 J mo1–1 K–1). the order of the speed of sound. Yet a gas leaking Reprint 2025-26 KINETIC THEORY 255 from a cylinder in a kitchen takes considerable are moving and the collision rate is determined time to diffuse to the other corners of the room. by the average relative velocity of the molecules. The top of a cloud of smoke holds together for Thus we need to replace <v> by <v r> in Eq. hours. This happens because molecules in a gas (12.38). A more exact treatment gives have a finite though small size, so they are bound 2 2 nπd (12.40)to undergo collisions. As a result, they cannot l = 1/ ( ) move straight unhindered; their paths keep Let us estimate l and τ for air molecules with getting incessantly deflected. average speeds <v> = ( 485m/s). At STP 0.02 × 1023 ( ) n = –3 22.4 × 10 ( ) = 2.7 × 10 25 m -3. Taking, d = 2 × 10–10 m, τ = 6.1 × 10–10 s t and l = 2.9 × 10–7 m ≈ 1500 d (12.41) v As expected, the mean free path given by d Eq. (12.40) depends inversely on the number density and the size of the molecules. In a highly evacuated tube n is rather small and the mean d free path can be as large as the length of the tube. ⊳ Example 12.9 Estimate the mean free path for a water molecule in water vapour at 373 K. Use information from Exercises 12.1 and Eq. Fig. 12.7 The volume swept by a molecule in time ∆t (12.41) above. in which any molecule will collide with it. Answer The d for water vapour is same as that Suppose the molecules of a gas are spheres of of air. The number density is inverselydiameter d. Focus on a single molecule with the proportional to absolute temperature.average speed <v>. It will suffer collision with any molecule that comes within a distance d 25 273 25 –3 So n = 2.7 × 10 × = 2 × 10 mbetween the centres. In time ∆t, it sweeps a 373 volume πd2 <v> ∆t wherein any other molecule –7 Hence, mean free path l = 4 × 10 m ⊳will collide with it (see Fig. 12.7). If n is the number of molecules per unit volume, the Note that the mean free path is 100 times the molecule suffers nπd2 <v> ∆t collisions in time interatomic distance ~ 40 Å = 4 × 10-9 m calculated ∆t. Thus the rate of collisions is nπd2 <v> or the earlier. It is this large value of mean free path that time between two successive collisions is on the leads to the typical gaseous behaviour. Gases can average, not be confined without a container. τ = 1/(nπ <v> d2 ) (12.38) Using, the kinetic theory of gases, the bulk The average distance between two successive measurable properties like viscosity, heat collisions, called the mean free path l, is : conductivity and diffusion can be related to the l = <v> τ = 1/(nπd2) (12.39) microscopic parameters like molecular size. It is In this derivation, we imagined the other through such relations that the molecular sizes molecules to be at rest. But actually all molecules were first estimated. Reprint 2025-26 256 PHYSICS SUMMARY 1. The ideal gas equation connecting pressure (P), volume (V) and absolute temperature (T ) is PV = µ RT = kB NT where µ is the number of moles and N is the number of molecules. R and kB are universal constants. R R = 8.314 J mol–1 K–1, kB = N A = 1.38 × 10–23 J K–1 Real gases satisfy the ideal gas equation only approximately, more so at low pressures and high temperatures. 2. Kinetic theory of an ideal gas gives the relation 1 2 P = n m v 3 where n is number density of molecules, m the mass of the molecule and v 2 is the mean of squared speed. Combined with the ideal gas equation it yields a kinetic interpretation of temperature. 1 2 3 2 1/2 3k B T m v = k B T , v rms = v = 2 2 ( ) m This tells us that the temperature of a gas is a measure of the average kinetic energy of a molecule, independent of the nature of the gas or molecule. In a mixture of gases at a fixed temperature the heavier molecule has the lower average speed. 3. The translational kinetic energy 3 E = kB NT. 2 This leads to a relation 2 PV = E 3 4. The law of equipartition of energy states that if a system is in equilibrium at absolute temperature T, the total energy is distributed equally in different energy modes of absorption, the energy in each mode being equal to ½ kB T. Each translational and rotational degree of freedom corresponds to one energy mode of absorption and has energy ½ kB T. Each vibrational frequency has two modes of energy (kinetic and potential) with corresponding energy equal to 2 × ½ kB T = kB T. 5. Using the law of equipartition of energy, the molar specific heats of gases can be determined and the values are in agreement with the experimental values of specific heats of several gases. The agreement can be improved by including vibrational modes of motion. 6. The mean free path l is the average distance covered by a molecule between two successive collisions : 1 l = 2 2 n πd where n is the number density and d the diameter of the molecule. Reprint 2025-26 KINETIC THEORY 257 POINTS TO PONDER 1. Pressure of a fluid is not only exerted on the wall. Pressure exists everywhere in a fluid. Any layer of gas inside the volume of a container is in equilibrium because the pressure is the same on both sides of the layer. 2. We should not have an exaggerated idea of the intermolecular distance in a gas. At ordinary pressures and temperatures, this is only 10 times or so the interatomic distance in solids and liquids. What is different is the mean free path which in a gas is 100 times the interatomic distance and 1000 times the size of the molecule. 3. The law of equipartition of energy is stated thus: the energy for each degree of freedom in thermal equilibrium is ½ k T. Each quadratic term in the total energy expression of a molecule is to be counted asB a degree of freedom. Thus, each vibrational mode gives 2 (not 1) degrees of freedom (kinetic and potential energy modes), corresponding to the energy 2 × ½ k T = k T. B B 4. Molecules of air in a room do not all fall and settle on the ground (due to gravity) because of their high speeds and incessant collisions. In equilibrium, there is a very slight increase in density at lower heights (like in the atmosphere). The effect is small since the potential energy (mgh) for ordinary heights is much less than the average kinetic energy ½ mv2 of the molecules. 5. < v2 > is not always equal to ( < v >)2. The average of a squared quantity is not necessarily the square of the average. Can you find examples for this statement. EXERCISESEXERCISESEXERCISESEXERCISESEXERCISES 12.112.112.112.112.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å. 12.212.212.212.212.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, 0 °C). Show that it is 22.4 litres. 12.312.312.312.312.3 Figure 12.8 shows plot of PV/T versus P for 1.00×10–3 kg of oxygen gas at two different temperatures. y T1 PV –1 T2 (J K ) T x P Fig.Fig.Fig.Fig.Fig. 12.812.812.812.812.8 (a) What does the dotted plot signify? (b) Which is true: T1 > T2 or T1 < T2? (c) What is the value of PV/T where the curves meet on the y-axis? Reprint 2025-26 258 PHYSICS (d) If we obtained similar plots for 1.00×10–3 kg of hydrogen, would we get the same value of PV/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot) ? (Molecular mass of H2 = 2.02 u, of O2 = 32.0 u, R = 8.31 J mo1–1 K–1.) 12.412.412.412.412.4 An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32 u). 12.512.512.512.512.5 An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ? 12.612.612.612.612.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m3 at a temperature of 27 °C and 1 atm pressure. 12.712.712.712.712.7 Estimate the average thermal energy of a helium atom at (i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star). 12.812.812.812.812.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root mean square speed of molecules the same in the three cases? If not, in which case is vrms the largest ? 12.912.912.912.912.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u). 12.1012.1012.1012.1012.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 0C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u). Reprint 2025-26 CHAPTER THIRTEEN OSCILLATIONS 13.1 INTRODUCTION In our daily life we come across various kinds of motions. You have already learnt about some of them, e.g., rectilinear 13.1 Introduction motion and motion of a projectile. Both these motions are non-repetitive. We have also learnt about uniform circular13.2 Periodic and oscillatory motions motion and orbital motion of planets in the solar system. In 13.3 Simple harmonic motion these cases, the motion is repeated after a certain interval of 13.4 Simple harmonic motion time, that is, it is periodic. In your childhood, you must have and uniform circular enjoyed rocking in a cradle or swinging on a swing. Both motion these motions are repetitive in nature but different from the 13.5 Velocity and acceleration periodic motion of a planet. Here, the object moves to and fro in simple harmonic motion about a mean position. The pendulum of a wall clock executes 13.6 Force law for simple a similar motion. Examples of such periodic to and fro harmonic motion motion abound: a boat tossing up and down in a river, the

8.3 HOOKE’S LAW Stress and strain take different forms in the situations depicted in the Fig. (8.1). For small deformations the stress and strain are proportional to each other. This is known as Fig. 8.2 A typical stress-strain curve for a metal. Hooke’s law. Thus, In the region from A to B, stress and strain stress ∝ strain are not proportional. Nevertheless, the body still stress = k × strain (8.6) returns to its original dimension when the load where k is the proportionality constant and is is removed. The point B in the curve is known known as modulus of elasticity. as yield point (also known as elastic limit) and Hooke’s law is an empirical law and is found the corresponding stress is known as yield to be valid for most materials. However, there strength (σy) of the material. are some materials which do not exhibit this If the load is increased further, the stress linear relationship. developed exceeds the yield strength and strain increases rapidly even for a small change in the 8.4 STRESS-STRAIN CURVE stress. The portion of the curve between B and D shows this. When the load is removed, say at The relation between the stress and the strain some point C between B and D, the body does for a given material under tensile stress can be not regain its original dimension. In this case, found experimentally. In a standard test of even when the stress is zero, the strain is not tensile properties, a test cylinder or a wire is zero. The material is said to have a permanent stretched by an applied force. The fractional set. The deformation is said to be plastic change in length (the strain) and the applied deformation. The point D on the graph is the force needed to cause the strain are recorded. ultimate tensile strength (σu) of the material. The applied force is gradually increased in steps Beyond this point, additional strain is produced and the change in length is noted. A graph is even by a reduced applied force and fracture plotted between the stress (which is equal in occurs at point E. If the ultimate strength and magnitude to the applied force per unit area) and fracture points D and E are close, the material the strain produced. A typical graph for a metal is said to be brittle. If they are far apart, the is shown in Fig. 8.2. Analogous graphs for material is said to be ductile. Reprint 2025-26 170 PHYSICS 8.5 ELASTIC MODULI The proportional region within the elastic limit of the stress-strain curve (region OA in Fig. 8.2) is of great importance for structural and manufacturing engineering designs. The ratio of stress and strain, called modulus of elasticity, is found to be a characteristic of the material. 8.5.1 Young’s Modulus Experimental observation show that for a given material, the magnitude of the strain produced is same whether the stress is tensile or compressive. The ratio of tensile (or compressive) stress (σ) to the longitudinal strain (ε) is defined as Fig. 8.3 Stress-strain curve for the elastic tissue of Young’s modulus and is denoted by the symbol Y. Aorta, the large tube (vessel) carrying blood from the heart. σ Y = (8.7) As stated earlier, the stress-strain behaviour ε varies from material to material. For example, From Eqs. (8.1) and (8.2), we have rubber can be pulled to several times its original length and still returns to its original shape. Fig. Y = (F/A)/(∆L/L) 8.3 shows stress-strain curve for the elastic = (F × L) /(A × ∆L) (8.8) tissue of aorta, present in the heart. Note that Since strain is a dimensionless quantity, the although elastic region is very large, the material unit of Young’s modulus is the same as that of does not obey Hooke’s law over most of the region. stress i.e., N m–2 or Pascal (Pa). Table 8.1 gives Secondly, there is no well defined plastic region. the values of Young’s moduli and yield strengths Substances like tissue of aorta, rubber etc. of some material. which can be stretched to cause large strains From the data given in Table 8.1, it is noticed are called elastomers. that for metals Young’s moduli are large. Table 8.1 Young’s moduli and yield strenghs of some material # Substance tested under compression Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 171 Therefore, these materials require a large force Answer The copper and steel wires are under to produce small change in length. To increase a tensile stress because they have the same the length of a thin steel wire of 0.1 cm2 cross- tension (equal to the load W) and the same area sectional area by 0.1%, a force of 2000 N is of cross-section A. From Eq. (8.7) we have stress required. The force required to produce the same = strain × Young’s modulus. Therefore strain in aluminium, brass and copper wires W/A = Yc × (∆Lc/Lc) = Ys × (∆Ls/Ls) having the same cross-sectional area are 690 N, where the subscripts c and s refer to copper 900 N and 1100 N respectively. It means that and stainless steel respectively. Or, steel is more elastic than copper, brass and ∆Lc/∆Ls = (Ys/Yc) × (Lc/Ls) aluminium. It is for this reason that steel is Given Lc = 2.2 m, Ls = 1.6 m, preferred in heavy-duty machines and in From Table 9.1 Yc = 1.1 × 1011 N.m–2, and structural designs. Wood, bone, concrete and Ys = 2.0 × 1011 N.m–2. glass have rather small Young’s moduli. ∆Lc/∆Ls = (2.0 × 1011/1.1 × 1011) × (2.2/1.6) = 2.5. The total elongation is given to be u Example 8.1 A structural steel rod has a ∆Lc + ∆Ls = 7.0 × 10-4 m radius of 10 mm and a length of 1.0 m. A Solving the above equations, 100 kN force stretches it along its length. ∆Lc = 5.0 × 10-4 m, and ∆Ls = 2.0 × 10-4 m. Calculate (a) stress, (b) elongation, and (c) Therefore strain on the rod. Young’s modulus, of W = (A × Yc × ∆Lc)/Lc structural steel is 2.0 × 1011 N m-2. = π (1.5 × 10-3)2 × [(5.0 × 10-4 × 1.1 × 1011)/2.2] = 1.8 × 102 N ⊳ Answer We assume that the rod is held by a clamp at one end, and the force F is applied at uExample 8.3 In a human pyramid in a the other end, parallel to the length of the rod. circus, the entire weight of the balanced Then the stress on the rod is given by group is supported by the legs of a performer F F who is lying on his back (as shown in Fig. Stress = = 2 8.4). The combined mass of all the persons A πr 3 performing the act, and the tables, plaques 100 × 10 N = etc. involved is 280 kg. The mass of the −2 2 3.14 × 10 m performer lying on his back at the bottom of ( ) = 3.18 × 108 N m–2 the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm andThe elongation, an effective radius of 2.0 cm. Determine the ( F/A ) L ∆L = amount by which each thighbone gets Y compressed under the extra load. 8 –2 3.18 × 10 N m 1m ) ( )( = 11 –2 2 × 10 N m = 1.59 × 10–3 m = 1.59 mm The strain is given by Strain = ∆L/L = (1.59 × 10–3 m)/(1m) = 1.59 × 10–3 = 0.16 % ⊳ u Example 8.2 A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied. Fig. 8.4 Human pyramid in a circus. Reprint 2025-26 172 PHYSICS Answer Total mass of all the performers, tables, Table 8.2 Shear moduli (G) of some common materialsplaques etc. = 280 kg Mass of the performer = 60 kg Material G (109 Nm–2 Mass supported by the legs of the performer or GPa) at the bottom of the pyramid Aluminium 25 = 280 – 60 = 220 kg Brass 36 Weight of this supported mass Copper 42 = 220 kg wt. = 220 × 9.8 N = 2156 N. Glass 23 Weight supported by each thighbone of the Iron 70 performer = ½ (2156) N = 1078 N. Lead 5.6 From Table 9.1, the Young’s modulus for bone Nickel 77 is given by Steel 84 Y = 9.4 × 109 N m–2. Tungsten 150 Wood 10 Length of each thighbone L = 0.5 m the radius of thighbone = 2.0 cm u Example 8.4 A square lead slab of side 50 Thus the cross-sectional area of the thighbone A = π × (2 × 10-2)2 m2 = 1.26 × 10-3 m2. cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × Using Eq. (9.8), the compression in each 104 N. The lower edge is riveted to the floor. thighbone (∆L) can be computed as How much will the upper edge be displaced? ∆L = [(F × L)/(Y × A)] = [(1078 × 0.5)/(9.4 × 109 × 1.26 × 10-3)] Answer The lead slab is fixed and the force is = 4.55 × 10-5 m or 4.55 × 10-3 cm. applied parallel to the narrow face as shown in This is a very small change! The fractional Fig. 8.6. The area of the face parallel to which decrease in the thighbone is ∆L/L = 0.000091 or this force is applied is 0.0091%. ⊳ A = 50 cm × 10 cm = 0.5 m × 0.1 m8.5.2 Shear Modulus = 0.05 m2 The ratio of shearing stress to the corresponding Therefore, the stress applied is shearing strain is called the shear modulus of = (9.4 × 104 N/0.05 m2) the material and is represented by G. It is also = 1.80 × 106 N.m–2 called the modulus of rigidity. G = shearing stress (σs)/shearing strain G = (F/A)/(∆x/L) = (F × L)/(A × ∆x) (8.10) Similarly, from Eq. (9.4) G = (F/A)/θ = F/(A × θ) (8.11) The shearing stress σs can also be expressed as σs = G × θ (8.12) aaaaaaaaaaaaaaaaaaaaa aaaaaaaaaaaaaaaaaaaaa SI unit of shear modulus is N m–2 or Pa. The Fig. 8.5 shear moduli of a few common materials are given in Table 9.2. It can be seen that shear We know that shearing strain = (∆x/L)= Stress /G. modulus (or modulus of rigidity) is generally less Therefore the displacement ∆x = (Stress × L)/G than Young’s modulus (from Table 9.1). For most = (1.8 × 106 N m–2 × 0.5m)/(5.6 × 109 N m–2) materials G ≈ Y/3. = 1.6 × 10–4 m = 0.16 mm ⊳ Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 173 8.5.3 Bulk Modulus Table 8.3 Bulk moduli (B) of some common Materials In Section (8.3), we have seen that when a body is submerged in a fluid, it undergoes a hydraulic Material B (109 N m–2 or GPa) stress (equal in magnitude to the hydraulic Solids pressure). This leads to the decrease in the Aluminium 72volume of the body thus producing a strain called volume strain [Eq. (8.5)]. The ratio of hydraulic Brass 61 stress to the corresponding hydraulic strain is Copper 140called bulk modulus. It is denoted by symbol B. B = – p/(∆V/V) (8.12) Glass 37 The negative sign indicates the fact that with Iron 100 an increase in pressure, a decrease in volume occurs. That is, if p is positive, ∆V is negative. Nickel 260 Thus for a system in equilibrium, the value of Steel 160 bulk modulus B is always positive. SI unit of bulk modulus is the same as that of pressure Liquids i.e., N m–2 or Pa. The bulk moduli of a few common Water 2.2 materials are given in Table 8.3. The reciprocal of the bulk modulus is called Ethanol 0.9 compressibility and is denoted by k. It is defined Carbon disulphide 1.56 as the fractional change in volume per unit increase in pressure. Glycerine 4.76 k = (1/B) = – (1/∆p) × (∆V/V) (8.13) Mercury 25 It can be seen from the data given in Table Gases8.3 that the bulk moduli for solids are much larger than for liquids, which are again much Air (at STP) 1.0 × 10–4 larger than the bulk modulus for gases (air). Table 8.4 Stress, strain and various elastic moduli Type of Stress Strain Change in Elastic Name of State of stress shape volume Modulus Modulus Matter Tensile Two equal and Elongation or Yes No Y = (F×L)/ Young’s Solid or opposite forces compression (A×∆L) modulus compressive perpendicular to parallel to force (σ = F/A) opposite faces direction (∆L/L) (longitudinal strain) Shearing Two equal and Pure shear, θ Yes No G = F/(A×θ) Shear Solid (σs = F/A) opposite forces modulus parallel to oppoiste or modulus surfaces forces of rigidity in each case such that total force and total torque on the body vanishes Hydraulic Forces perpendicular Volume change No Yes B = –p/(∆V/V) Bulk Solid, liquid everywhere to the (compression or modulus and gas surface, force per elongation) unit area (pressure) (∆V/V) same everywhere. Reprint 2025-26 174 PHYSICS Thus, solids are the least compressible, whereas, 8.5.5 Elastic Potential Energy gases are the most compressible. Gases are about in a Stretched Wire a million times more compressible than solids! When a wire is put under a tensile stress, work Gases have large compressibilities, which vary is done against the inter-atomic forces. This with pressure and temperature. The work is stored in the wire in the form of elastic incompressibility of the solids is primarily due potential energy. When a wire of original length to the tight coupling between the neighbouring L and area of cross-section A is subjected to a atoms. The molecules in liquids are also bound deforming force F along the length of the wire, with their neighbours but not as strong as in let the length of the wire be elongated by l. Then solids. Molecules in gases are very poorly from Eq. (8.8), we have F = YA × (l/L). Here Y is coupled to their neighbours. the Young’s modulus of the material of the wire. Table 8.4 shows the various types of stress, Now for a further elongation of infinitesimal strain, elastic moduli, and the applicable state small length dl, work done dW is F × dl or YAldl/ of matter at a glance. L. Therefore, the amount of work done (W) in increasing the length of the wire from L to L + l, u Example 8.5 The average depth of Indian that is from l = 0 to l = l is Ocean is about 3000 m. Calculate the l YAl YA l 2 dl = × fractional compression, ∆V/V, of water at W = ∫0 2 L L the bottom of the ocean, given that the bulk 2 modulus of water is 2.2 × 109 N m–2. (Take 1  l  W = × Y ×   × AL g = 10 m s–2) 2  L  1 Answer The pressure exerted by a 3000 m = × Young’s modulus × strain2 × column of water on the bottom layer 2 p = hρ g = 3000 m × 1000 kg m–3 × 10 m s–2 volume of the wire = 3 × 107 kg m–1 s-2 1 × stress × strain × volume of the = 3 × 107 N m–2 = 2 Fractional compression ∆V/V, is wire ∆V/V = stress/B = (3 × 107 N m-2)/(2.2 × 109 N m–2) This work is stored in the wire in the form of = 1.36 × 10-2 or 1.36 % ⊳ elastic potential energy (U). Therefore the elastic potential energy per unit volume of the wire (u) is 1 8.5.4 POISSON’S RATIO u = ×σε (8.14) 2 The strain perpendicular to the applied force is called lateral strain. Simon Poisson pointed out 8.6 APPLICATIONS OF ELASTIC that within the elastic limit, lateral strain is BEHAVIOUR OF MATERIALS directly proportional to the longitudinal strain. The elastic behaviour of materials plays an The ratio of the lateral strain to the longitudinal important role in everyday life. All engineering strain in a stretched wire is called Poisson’s designs require precise knowledge of the elastic ratio. If the original diameter of the wire is d behaviour of materials. For example while and the contraction of the diameter under stress designing a building, the structural design of is ∆d, the lateral strain is ∆d/d. If the original the columns, beams and supports require length of the wire is L and the elongation under knowledge of strength of materials used. Have stress is ∆L, the longitudinal strain is ∆L/L. you ever thought why the beams used in Poisson’s ratio is then (∆d/d)/(∆L/L) or (∆d/∆L) construction of bridges, as supports etc. have × (L/d). Poisson’s ratio is a ratio of two strains; a cross-section of the type I? Why does a heap it is a pure number and has no dimensions or of sand or a hill have a pyramidal shape? units. Its value depends only on the nature of Answers to these questions can be obtained material. For steels the value is between 0.28 and from the study of structural engineering which 0.30, and for aluminium alloys it is about 0.33. is based on concepts developed here. Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 175 Cranes used for lifting and moving heavy loads This relation can be derived using what you from one place to another have a thick metal have already learnt and a little calculus. From rope to which the load is attached. The rope is Eq. (8.16), we see that to reduce the bending pulled up using pulleys and motors. Suppose we for a given load, one should use a material with want to make a crane, which has a lifting a large Young’s modulus Y. For a given material, capacity of 10 tonnes or metric tons (1 metric increasing the depth d rather than the breadth ton = 1000 kg). How thick should the steel rope b is more effective in reducing the bending, since be? We obviously want that the load does not δ is proportional to d -3 and only to b-1(of course deform the rope permanently. Therefore, the the length l of the span should be as small as extension should not exceed the elastic limit. possible). But on increasing the depth, unlessFrom Table 8.1, we find that mild steel has a yield strength (σy) of about 300 × 106 N m–2. Thus, the load is exactly at the right place (difficult to the area of cross-section (A) of the rope should arrange in a bridge with moving traffic), the at least be deep bar may bend as shown in Fig. 8.7(b). This A ≥ W/σy = Mg/σy (8.15) is called buckling. To avoid this, a common = (104 kg × 9.8 m s-2)/(300 × 106 N m-2) compromise is the cross-sectional shape shown = 3.3 × 10-4 m2 in Fig. 8.7(c). This section provides a large load- corresponding to a radius of about 1 cm for a bearing surface and enough depth to prevent rope of circular cross-section. Generally a bending. This shape reduces the weight of the large margin of safety (of about a factor of ten beam without sacrificing the strength and in the load) is provided. Thus a thicker rope of hence reduces the cost. radius about 3 cm is recommended. A single wire of this radius would practically be a rigid rod. So the ropes are always made of a number of thin wires braided together, like in pigtails, for ease in manufacture, flexibility and strength. A bridge has to be designed such that it can withstand the load of the flowing traffic, the force of winds and its own weight. Similarly, in the design of buildings the use of beams and columns is very common. In both the cases, the overcoming of the problem of bending of beam under a load is of prime importance. The beam should not bend too much or break. Let us consider the case of a beam loaded at the centre and supported near its ends as shown in Fig. 8.6. A bar of length l, breadth b, and depth d when loaded at the centre by a load W sags by (a) (b) (c) an amount given by Fig. 8.7 Different cross-sectional shapes of a δ = W l 3/(4bd 3Y) (8.16) beam. (a) Rectangular section of a bar; (b) A thin bar and how it can buckle; (c) Commonly used section for a load bearing bar. The use of pillars or columns is also very common in buildings and bridges. A pillar with rounded ends as shown in Fig. 8.9(a) supports less load than that with a distributed shape at the ends [Fig. 8.9(b)]. The precise design of a bridge or a building has to take into account the conditions under which it will function, the Fig. 8.6 A beam supported at the ends and loaded cost and long period, reliability of usable at the centre. material, etc. Reprint 2025-26 176 PHYSICS shearing stress to the rocks under which they can flow. The stress due to all the material on the top should be less than the critical shearing stress at which the rocks flow. At the bottom of a mountain of height h, the force per unit area due to the weight of the mountain is hρg where ρ is the density of the material of the mountain and g is the acceleration due to gravity. The material at the bottom experiences this force in the vertical direction, and the sides of the mountain are free. Therefore, this is not a case of pressure or bulk compression. (a) (b) There is a shear component, approximately hρg Fig. 8.8 Pillars or columns: (a) a pillar with rounded itself. Now the elastic limit for a typical rock is ends, (b) Pillar with distributed ends. 30 × 107 N m-2. Equating this to hρg, with The answer to the question why the maximum ρ = 3 × 103 kg m-3 gives height of a mountain on earth is ~10 km can hρg = 30 × 107 N m-2 . also be provided by considering the elastic h = 30 × 107 N m-2/(3 × 103 kg m-3 × 10 m s-2) properties of rocks. A mountain base is not under = 10 km uniform compression and this provides some which is more than the height of Mt. Everest! SUMMARY 1. Stress is the restoring force per unit area and strain is the fractional change in dimension. In general there are three types of stresses (a) tensile stress — longitudinal stress (associated with stretching) or compressive stress (associated with compression), (b) shearing stress, and (c) hydraulic stress. 2. For small deformations, stress is directly proportional to the strain for many materials. This is known as Hooke’s law. The constant of proportionality is called modulus of elasticity. Three elastic moduli viz., Young’s modulus, shear modulus and bulk modulus are used to describe the elastic behaviour of objects as they respond to deforming forces that act on them. A class of solids called elastomers does not obey Hooke’s law. 3. When an object is under tension or compression, the Hooke’s law takes the form F/A = Y∆L/L where ∆L/L is the tensile or compressive strain of the object, F is the magnitude of the applied force causing the strain, A is the cross-sectional area over which F is applied (perpendicular to A) and Y is the Young’s modulus for the object. The stress is F/A. 4. A pair of forces when applied parallel to the upper and lower faces, the solid deforms so that the upper face moves sideways with respect to the lower. The horizontal displacement ∆L of the upper face is perpendicular to the vertical height L. This type of deformation is called shear and the corresponding stress is the shearing stress. This type of stress is possible only in solids. In this kind of deformation the Hooke’s law takes the form F/A = G × ∆L/L where ∆L is the displacement of one end of object in the direction of the applied force F, and G is the shear modulus. 5. When an object undergoes hydraulic compression due to a stress exerted by a surrounding fluid, the Hooke’s law takes the form p = B (∆V/V), where p is the pressure (hydraulic stress) on the object due to the fluid, ∆V/V (the volume strain) is the absolute fractional change in the object’s volume due to that pressure and B is the bulk modulus of the object. Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 177 POINTS TO PONDER 1. In the case of a wire, suspended from celing and stretched under the action of a weight (F) suspended from its other end, the force exerted by the ceiling on it is equal and opposite to the weight. However, the tension at any cross-section A of the wire is just F and not 2F. Hence, tensile stress which is equal to the tension per unit area is equal to F/A. 2. Hooke’s law is valid only in the linear part of stress-strain curve. 3. The Young’s modulus and shear modulus are relevant only for solids since only solids have lengths and shapes. 4. Bulk modulus is relevant for solids, liquid and gases. It refers to the change in volume when every part of the body is under the uniform stress so that the shape of the body remains unchanged. 5. Metals have larger values of Young’s modulus than alloys and elastomers. A material with large value of Young’s modulus requires a large force to produce small changes in its length. 6. In daily life, we feel that a material which stretches more is more elastic, but it a is misnomer. In fact material which stretches to a lesser extent for a given load is considered to be more elastic. 7. In general, a deforming force in one direction can produce strains in other directions also. The proportionality between stress and strain in such situations cannot be described by just one elastic constant. For example, for a wire under longitudinal strain, the lateral dimensions (radius of cross section) will undergo a small change, which is described by another elastic constant of the material (called Poisson ratio). 8. Stress is not a vector quantity since, unlike a force, the stress cannot be assigned a specific direction. Force acting on the portion of a body on a specified side of a section has a definite direction. EXERCISES 8.1 A steel wire of length 4.7 m and cross-sectional area 3.0 × 10-5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper? 8.2 Figure 8.9 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material? Fig. 8.9 Reprint 2025-26 178 PHYSICS 8.3 The stress-strain graphs for materials A and B are shown in Fig. 8.10. Fig. 8.10 The graphs are drawn to the same scale. (a) Which of the materials has the greater Young’s modulus? (b) Which of the two is the stronger material? 8.4 Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young’s modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus. 8.5 Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 8.11. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires. Fig. 8.11 8.6 The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face? 8.7 Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. 8.8 A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain? 8.9 A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N m–2, what is the maximum load the cable can support ? 8.10 A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratios of their diameters if each is to have the same tension. 8.11 A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path. Reprint 2025-26 MECHANICAL PROPERTIES OF SOLIDS 179 8.12 Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large. 8.13 What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3? 8.14 Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm. 8.15 Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 106 Pa. 8.16 How much should the pressure on a litre of water be changed to compress it by 0.10%? carry one quarter of the load. Reprint 2025-26 CHAPTER NINE MECHANICAL PROPERTIES OF FLUIDS 9.1 INTRODUCTION In this chapter, we shall study some common physical properties of liquids and gases. Liquids and gases can flow 9.1 Introduction and are therefore, called fluids. It is this property that 9.2 Pressure distinguishes liquids and gases from solids in a basic way. 9.3 Streamline flow Fluids are everywhere around us. Earth has an envelop of 9.4 Bernoulli’s principle air and two-thirds of its surface is covered with water. Water is not only necessary for our existence; every mammalian9.5 Viscosity body constitute mostly of water. All the processes occurring9.6 Surface tension in living beings including plants are mediated by fluids. Thus Summary understanding the behaviour and properties of fluids is Points to ponder important. Exercises How are fluids different from solids? What is common in Additional exercises liquids and gases? Unlike a solid, a fluid has no definite Appendix shape of its own. Solids and liquids have a fixed volume, whereas a gas fills the entire volume of its container. We have learnt in the previous chapter that the volume of solids can be changed by stress. The volume of solid, liquid or gas depends on the stress or pressure acting on it. When we talk about fixed volume of solid or liquid, we mean its volume under atmospheric pressure. The difference between gases and solids or liquids is that for solids or liquids the change in volume due to change of external pressure is rather small. In other words solids and liquids have much lower compressibility as compared to gases. Shear stress can change the shape of a solid keeping its volume fixed. The key property of fluids is that they offer very little resistance to shear stress; their shape changes by application of very small shear stress. The shearing stress of fluids is about million times smaller than that of solids. 9.2 PRESSURE A sharp needle when pressed against our skin pierces it. Our skin, however, remains intact when a blunt object with a wider contact area (say the back of a spoon) is pressed against it with the same force. If an elephant were to step on a man’s chest, his ribs would crack. A circus performer across whose Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 181 chest a large, light but strong wooden plank is In principle, the piston area can be made placed first, is saved from this accident. Such arbitrarily small. The pressure is then defined everyday experiences convince us that both the in a limiting sense as force and its coverage area are important. Smaller lim ∆F the area on which the force acts, greater is the P = ∆A → 0 (9.2) ∆Aimpact. This impact is known as pressure. Pressure is a scalar quantity. We remind the When an object is submerged in a fluid at reader that it is the component of the force rest, the fluid exerts a force on its surface. This normal to the area under consideration and not force is always normal to the object’s surface. the (vector) force that appears in the numerator This is so because if there were a component of in Eqs. (9.1) and (9.2). Its dimensions are force parallel to the surface, the object will also [ML–1T–2]. The SI unit of pressure is N m–2. It has exert a force on the fluid parallel to it; as a been named as pascal (Pa) in honour of the consequence of Newton’s third law. This force French scientist Blaise Pascal (1623-1662) who will cause the fluid to flow parallel to the surface. carried out pioneering studies on fluid pressure. Since the fluid is at rest, this cannot happen. A common unit of pressure is the atmosphere Hence, the force exerted by the fluid at rest has (atm), i.e. the pressure exerted by the to be perpendicular to the surface in contact atmosphere at sea level (1 atm = 1.013 × 105 Pa). with it. This is shown in Fig.9.1(a). Another quantity, that is indispensable in The normal force exerted by the fluid at a point describing fluids, is the density ρ. For a fluid of may be measured. An idealised form of one such mass m occupying volume V, pressure-measuring device is shown in Fig. m ρ = (9.3) 9.1(b). It consists of an evacuated chamber with V a spring that is calibrated to measure the force The dimensions of density are [ML–3]. Its SI acting on the piston. This device is placed at a unit is kg m–3. It is a positive scalar quantity. A point inside the fluid. The inward force exerted liquid is largely incompressible and its density by the fluid on the piston is balanced by the is therefore, nearly constant at all pressures. outward spring force and is thereby measured. Gases, on the other hand exhibit a large variation in densities with pressure. The density of water at 4oC (277 K) is 1.0 × 103 kg m–3. The relative density of a substance is the ratio of its density to the density of water at 4oC. It is a dimensionless positive scalar quantity. For example the relative density of aluminium is 2.7. Its density is 2.7 × 103 kg m–3. The densities of some common fluids are displayed in Table 9.1. Table 9.1 Densities of some common fluids (a) (b) at STP* Fig. 9.1 (a) The force exerted by the liquid in the beaker on the submerged object or on the walls is normal (perpendicular) to the surface at all points. (b) An idealised device for measuring pressure. If F is the magnitude of this normal force on the piston of area A then the average pressure Pav is defined as the normal force acting per unit area. F Pav = (9.1) A * STP means standard temperature (00C) and 1 atm pressure. Reprint 2025-26 182 PHYSICS ⊳ this element of area corresponding to the normal Example 9.1 The two thigh bones (femurs), forces Fa, Fb and Fc as shown in Fig. 9.2 on the each of cross-sectional area10 cm2 support faces BEFC, ADFC and ADEB denoted by Aa, Ab the upper part of a human body of mass 40 and Ac respectively. Then kg. Estimate the average pressure Fb sinθ = Fc, Fb cosθ = Fa (by equilibrium) sustained by the femurs. Ab sinθ = Ac, Ab cosθ = Aa (by geometry) Thus, Answer Total cross-sectional area of the Fb Fc Fafemurs is A = 2 × 10 cm2 = 20 × 10–4 m2. The = = ; Pb = Pc = Pa (9.4) force acting on them is F = 40 kg wt = 400 N Ab A c A a (taking g = 10 m s–2). This force is acting Hence, pressure exerted is same in all vertically down and hence, normally on the directions in a fluid at rest. It again reminds us femurs. Thus, the average pressure is that like other types of stress, pressure is not a F 5 −2 vector quantity. No direction can be assigned Pav = = 2 × 10 N m ⊳ A to it. The force against any area within (or bounding) a fluid at rest and under pressure is 9.2.1 Pascal’s Law normal to the area, regardless of the orientation of the area. The French scientist Blaise Pascal observed that Now consider a fluid element in the form of a the pressure in a fluid at rest is the same at all horizontal bar of uniform cross-section. The bar points if they are at the same height. This fact is in equilibrium. The horizontal forces exerted may be demonstrated in a simple way. at its two ends must be balanced or the pressure at the two ends should be equal. This proves that for a liquid in equilibrium the pressure is same at all points in a horizontal plane. Suppose the pressure were not equal in different parts of the fluid, then there would be a flow as the fluid will have some net force acting on it. Hence in the absence of flow the pressure in the fluid must be same everywhere in a horizontal plane. 9.2.2 Variation of Pressure with Depth Fig. 9.2 Proof of Pascal’s law. ABC-DEF is an Consider a fluid at rest in a container. In element of the interior of a fluid at rest. Fig. 9.3 point 1 is at height h above a point 2. This element is in the form of a right- The pressures at points 1 and 2 are P1 and P2 angled prism. The element is small so that respectively. Consider a cylindrical element of the effect of gravity can be ignored, but it fluid having area of base A and height h. As the has been enlarged for the sake of clarity. fluid is at rest the resultant horizontal forces Fig. 9.2 shows an element in the interior of a should be zero and the resultant vertical forces fluid at rest. This element ABC-DEF is in the should balance the weight of the element. The form of a right-angled prism. In principle, this forces acting in the vertical direction are due to prismatic element is very small so that every the fluid pressure at the top (P1A) acting part of it can be considered at the same depth downward, at the bottom (P2A) acting upward. from the liquid surface and therefore, the effect If mg is weight of the fluid in the cylinder we of the gravity is the same at all these points. have But for clarity we have enlarged this element. (P2 − P1) A = mg (9.5) The forces on this element are those exerted by Now, if ρ is the mass density of the fluid, we the rest of the fluid and they must be normal to have the mass of fluid to be m = ρV= ρhA so the surfaces of the element as discussed above. that Thus, the fluid exerts pressures Pa, Pb and Pc on P2 − P1= ρgh (9.6) Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 183 Fig 9.4 Illustration of hydrostatic paradox. The three vessels A, B and C contain different amounts of liquids, all upto the same height. ⊳ Example 9.2 What is the pressure on a swimmer 10 m below the surface of a lake? Answer HereFig.9.3 Fluid under gravity. The effect of gravity is illustrated through pressure on a vertical h = 10 m and ρ = 1000 kg m-3. Take g = 10 m s–2 cylindrical column. From Eq. (9.7) P = Pa + ρgh Pressure difference depends on the vertical = 1.01 × 105 Pa + 1000 kg m–3 × 10 m s–2 × 10 m = 2.01 × 105 Padistance h between the points (1 and 2), mass ≈ 2 atmdensity of the fluid ρ and acceleration due to This is a 100% increase in pressure from gravity g. If the point 1 under discussion is surface level. At a depth of 1 km, the increase shifted to the top of the fluid (say, water), which in pressure is 100 atm! Submarines are designed is open to the atmosphere, P1 may be replaced to withstand such enormous pressures. ⊳ by atmospheric pressure (Pa) and we replace P2 by P. Then Eq. (9.6) gives 9.2.3 Atmospheric Pressure and Gauge Pressure P = Pa + ρgh (9.7) Thus, the pressure P, at depth below the The pressure of the atmosphere at any point is equal to the weight of a column of air of unitsurface of a liquid open to the atmosphere is cross-sectional area extending from that pointgreater than atmospheric pressure by an to the top of the atmosphere. At sea level, it is amount ρgh. The excess of pressure, P − Pa, at 1.013 × 105 Pa (1 atm). Italian scientist depth h is called a gauge pressure at that point. Evangelista Torricelli (1608–1647) devised for The area of the cylinder is not appearing in the first time a method for measuring the expression of absolute pressure in Eq. (9.7). atmospheric pressure. A long glass tube closed Thus, the height of the fluid column is important at one end and filled with mercury is inverted and not cross-sectional or base area or the shape into a trough of mercury as shown in Fig.9.5 (a). of the container. The liquid pressure is the same This device is known as ‘mercury barometer’. at all points at the same horizontal level (same The space above the mercury column in the tube depth). The result is appreciated through the contains only mercury vapour whose pressure example of hydrostatic paradox. Consider three P is so small that it may be neglected. Thus, vessels A, B and C [Fig.9.4] of different shapes. the pressure at Point A=0. The pressure inside They are connected at the bottom by a horizontal the coloumn at Point B must be the same as the pipe. On filling with water, the level in the three pressure at Point C, which is atmospheric vessels is the same, though they hold different pressure, Pa. amounts of water. This is so because water at Pa = ρgh (9.8) where ρ is the density of mercury and h is thethe bottom has the same pressure below each height of the mercury column in the tube.section of the vessel. Reprint 2025-26 184 PHYSICS In the experiment it is found that the mercury column in the barometer has a height of about 76 cm at sea level equivalent to one atmosphere (1 atm). This can also be obtained using the value of ρ in Eq. (9.8). A common way of stating pressure is in terms of cm or mm of mercury (Hg). A pressure equivalent of 1 mm is called a torr (after Torricelli). 1 torr = 133 Pa. The mm of Hg and torr are used in medicine and physiology. In meteorology, a common unit is the bar and millibar. 1 bar = 105 Pa An open tube manometer is a useful (b) The open tube manometer instrument for measuring pressure differences. Fig 9.5 Two pressure measuring devices. It consists of a U-tube containing a suitable Pressure is same at the same level on both liquid i.e., a low density liquid (such as oil) for sides of the U-tube containing a fluid. For measuring small pressure differences and a liquids, the density varies very little over wide high density liquid (such as mercury) for large ranges in pressure and temperature and we can pressure differences. One end of the tube is open treat it safely as a constant for our present to the atmosphere and the other end is purposes. Gases on the other hand, exhibits connected to the system whose pressure we want large variations of densities with changes in to measure [see Fig. 9.5 (b)]. The pressure P at A pressure and temperature. Unlike gases, liquids is equal to pressure at point B. What we are, therefore, largely treated as incompressible. normally measure is the gauge pressure, which ⊳ Example 9.3 The density of the is P − Pa, given by Eq. (9.8) and is proportional to atmosphere at sea level is 1.29 kg/m3. manometer height h. Assume that it does not change with altitude. Then how high would the atmosphere extend? Answer We use Eq. (9.7) ρgh = 1.29 kg m–3 × 9.8 m s2 × h m = 1.01 × 105 Pa ∴ h = 7989 m ≈ 8 km In reality the density of air decreases with height. So does the value of g. The atmospheric cover extends with decreasing pressure over 100 km. We should also note that the sea level atmospheric pressure is not always 760 mm of Hg. A drop in the Hg level by 10 mm or more is a sign of an approaching storm. ⊳ ⊳ Example 9.4 At a depth of 1000 m in an ocean (a) what is the absolute pressure? (b) What is the gauge pressure? (c) Find the force acting on the window of area 20 cm × 20 cm of a submarine at this depth, the interior of which is maintained at sea- level atmospheric pressure. (The density of sea water is 1.03 × 103 kg m-3, g = 10 m s–2.) Fig 9.5 (a) The mercury barometer. Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 185 Answer Here h = 1000 m and ρ = 1.03 × 103 kg m-3. law. In these devices, fluids are used for (a) From Eq. (9.6), absolute pressure transmitting pressure. In a hydraulic lift, as P = Pa + ρgh shown in Fig. 9.6 (b), two pistons are separated = 1.01 × 105 Pa by the space filled with a liquid. A piston of small + 1.03 × 103 kg m–3 × 10 m s–2 × 1000 m cross-section A1 is used to exert a force F1 directly F1 = 104.01 × 105 Pa ≈ 104 atm on the liquid. The pressure P = A1 is (b) Gauge pressure is P − Pa = ρgh = Pg transmitted throughout the liquid to the larger Pg = 1.03 × 103 kg m–3 × 10 ms2 × 1000 m cylinder attached with a larger piston of area A2, = 103 × 105 Pa which results in an upward force of P × A2. ≈ 103 atm Therefore, the piston is capable of supporting a (c) The pressure outside the submarine is large force (large weight of, say a car, or a truck, P = Pa + ρgh and the pressure inside it is Pa. F1 A 2 Hence, the net pressure acting on the placed on the platform) F2 = PA2 = A1 . By window is gauge pressure, Pg = ρgh. Since the area of the window is A = 0.04 m2, the changing the force at A1, the platform can be force acting on it is moved up or down. Thus, the applied force has F = Pg A = 103 × 105 Pa × 0.04 m2 = 4.12 × 105 N A 2 ⊳ been increased by a factor of A1 and this factor is the mechanical advantage of the device. The9.2.4 Hydraulic Machines example below clarifies it. Let us now consider what happens when we change the pressure on a fluid contained in a vessel. Consider a horizontal cylinder with a piston and three vertical tubes at different points [Fig. 9.6 (a)]. The pressure in the horizontal cylinder is indicated by the height of liquid column in the vertical tubes. It is necessarily the same in all. If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them. Fig 9.6 (b) Schematic diagram illustrating the principle behind the hydraulic lift, a device used to lift heavy loads. Fig 9.6 (a) Whenever external pressure is applied ⊳ Example 9.5 Two syringes of different on any part of a fluid in a vessel, it is cross-sections (without needles) filled with equally transmitted in all directions. water are connected with a tightly fitted rubber tube filled with water. Diameters of This indicates that when the pressure on the the smaller piston and larger piston are cylinder was increased, it was distributed 1.0 cm and 3.0 cm respectively. (a) Find uniformly throughout. We can say whenever the force exerted on the larger piston when external pressure is applied on any part of a a force of 10 N is applied to the smaller fluid contained in a vessel, it is transmitted piston. (b) If the smaller piston is pushed undiminished and equally in all directions. in through 6.0 cm, how much does the This is another form of the Pascal’s law and it larger piston move out? has many applications in daily life. A number of devices, such as hydraulic lift Answer (a) Since pressure is transmitted and hydraulic brakes, are based on the Pascal’s undiminished throughout the fluid, Reprint 2025-26 186 PHYSICS –2 2 important advantage of the system is that the 3/2 10 m × π A ( ) set up by pressing pedal is transmitted F2 = 2 F1 = 2 × 10 N pressure A1 equally to all cylinders attached to the four 1/2 10 –2 m × π ( ) wheels so that the braking effort is equal on = 90 N all wheels.(b) Water is considered to be perfectly incompressible. Volume covered by the 9.3 STREAMLINE FLOWmovement of smaller piston inwards is equal to volume moved outwards due to the larger piston. So far we have studied fluids at rest. The study L1 A1 = L 2 A2 of the fluids in motion is known as fluid dynamics. When a water tap is turned on slowly, the water flow is smooth initially, but loses its smoothness when the speed of the outflow is increased. In studying the motion of fluids, we j 0.67 × 10-2 m = 0.67 cm focus our attention on what is happening to Note, atmospheric pressure is common to both various fluid particles at a particular point in pistons and has been ignored. ⊳ space at a particular time. The flow of the fluid ⊳ is said to be steady if at any given point, the Example 9.6 In a car lift compressed air velocity of each passing fluid particle remains exerts a force F1 on a small piston having constant in time. This does not mean that the a radius of 5.0 cm. This pressure is velocity at different points in space is same. The transmitted to a second piston of radius velocity of a particular particle may change as it 15 cm (Fig 9.7). If the mass of the car to be moves from one point to another. That is, at some lifted is 1350 kg, calculate F1. What is the other point the particle may have a different pressure necessary to accomplish this velocity, but every other particle which passes task? (g = 9.8 ms-2). the second point behaves exactly as the previous particle that has just passed that point. Each Answer Since pressure is transmitted particle follows a smooth path, and the paths of undiminished throughout the fluid, the particles do not cross each other. = 1470 N ≈ 1.5 × 103 N The air pressure that will produce this force is This is almost double the atmospheric pressure. ⊳ Fig. 9.7 The meaning of streamlines. (a) A typical Hydraulic brakes in automobiles also work on trajectory of a fluid particle. the same principle. When we apply a little force (b) A region of streamline flow. on the pedal with our foot the master piston moves inside the master cylinder, and the The path taken by a fluid particle under a pressure caused is transmitted through the steady flow is a streamline. It is defined as a brake oil to act on a piston of larger area. A large curve whose tangent at any point is in the force acts on the piston and is pushed down direction of the fluid velocity at that point. expanding the brake shoes against brake lining. Consider the path of a particle as shown in In this way, a small force on the pedal produces Fig.9.7 (a), the curve describes how a fluid a large retarding force on the wheel. An particle moves with time. The curve PQ is like a Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 187 permanent map of fluid flow, indicating how the but their directions are parallel. Figure 9.8 (b) fluid streams. No two streamlines can cross, for gives a sketch of turbulent flow. if they do, an oncoming fluid particle can go either one way or the other and the flow would not be steady. Hence, in steady flow, the map of flow is stationary in time. How do we draw closely spaced streamlines ? If we intend to show streamline of every flowing particle, we would end up with a continuum of lines. Consider planes perpendicular to the direction of fluid flow e.g., at three points P, R and Q in Fig.9.7 (b). The plane Fig. 9.8 (a) Some streamlines for fluid flow. pieces are so chosen that their boundaries be (b) A jet of air striking a flat plate placed determined by the same set of streamlines. This perpendicular to it. This is an example means that number of fluid particles crossing of turbulent flow. the surfaces as indicated at P, R and Q is the same. If area of cross-sections at these points 9.4 BERNOULLI’S PRINCIPLE are AP,AR and AQ and speeds of fluid particles are vP, vR and vQ, then mass of fluid ∆mP crossing Fluid flow is a complex phenomenon. But we at AP in a small interval of time ∆t is ρPAPvP ∆t. can obtain some useful properties for steady Similarly mass of fluid ∆mR flowing or crossing or streamline flows using the conservation at AR in a small interval of time ∆t is ρRARvR ∆t of energy. and mass of fluid ∆mQ is ρQAQvQ ∆t crossing at Consider a fluid moving in a pipe of varying AQ. The mass of liquid flowing out equals the cross-sectional area. Let the pipe be at varying mass flowing in, holds in all cases. Therefore, heights as shown in Fig. 9.9. We now suppose ρPAPvP∆t = ρRARvR∆t = ρQAQvQ∆t (9.9) that an incompressible fluid is flowing through For flow of incompressible fluids the pipe in a steady flow. Its velocity must ρP = ρR = ρQ change as a consequence of equation of Equation (9.9) reduces to continuity. A force is required to produce this APvP = ARvR = AQvQ (9.10) acceleration, which is caused by the fluid which is called the equation of continuity and surrounding it, the pressure must be different it is a statement of conservation of mass in flow in different regions. Bernoulli’s equation is a of incompressible fluids. In general general expression that relates the pressure Av = constant (9.11) difference between two points in a pipe to both Av gives the volume flux or flow rate and velocity changes (kinetic energy change) and remains constant throughout the pipe of flow. elevation (height) changes (potential energy Thus, at narrower portions where the change). The Swiss Physicist Daniel Bernoulli streamlines are closely spaced, velocity developed this relationship in 1738. increases and its vice versa. From (Fig 9.7b) it Consider the flow at two regions 1 (i.e., BC) is clear that AR > AQ or vR < vQ, the fluid is and 2 (i.e., DE). Consider the fluid initially lying accelerated while passing from R to Q. This is between B and D. In an infinitesimal time associated with a change in pressure in fluid interval ∆t, this fluid would have moved. Suppose flow in horizontal pipes. v1 is the speed at B and v2 at D, then fluid initially Steady flow is achieved at low flow speeds. at B has moved a distance v1∆t to C (v1∆t is small Beyond a limiting value, called critical speed, enough to assume constant cross-section along this flow loses steadiness and becomes BC). In the same interval ∆t the fluid initially at turbulent. One sees this when a fast flowing D moves to E, a distance equal to v2∆t. Pressures stream encounters rocks, small foamy P1 and P2 act as shown on the plane faces of whirlpool-like regions called ‘white water areas A1 and A2 binding the two regions. The rapids are formed. work done on the fluid at left end (BC) is W1 = Figure 9.8 displays streamlines for some P1A1(v1∆t) = P1∆V. Since the same volume ∆V typical flows. For example, Fig. 9.8(a) describes passes through both the regions (from the a laminar flow where the velocities at different equation of continuity) the work done by the fluid points in the fluid may have different magnitudes at the other end (DE) is W2 = P2A2(v2∆t) = P2∆V or, Reprint 2025-26 188 PHYSICS the work done on the fluid is –P2∆V. So the total In words, the Bernoulli’s relation may be work done on the fluid is stated as follows: As we move along a streamline W1 – W2 = (P1− P2) ∆V the sum of the pressure (P), the kinetic energy Part of this work goes into changing the kinetic  ρv2 energy of the fluid, and part goes into changing per unit volume and the potential energythe gravitational potential energy. If the density  2  of the fluid is ρ and ∆m = ρA1v1∆t = ρ∆V is the per unit volume (ρgh) remains a constant. mass passing through the pipe in time ∆t, then Note that in applying the energy conservation change in gravitational potential energy is principle, there is an assumption that no energy ∆U = ρg∆V (h2 − h1) is lost due to friction. But in fact, when fluids The change in its kinetic energy is flow, some energy does get lost due to internal 1 friction. This arises due to the fact that in a fluid ∆K = ρ ∆V (v2 2 − v1 2) flow, the different layers of the fluid flow with 2 different velocities. These layers exert frictional We can employ the work – energy theorem forces on each other resulting in a loss of energy. (Chapter 6) to this volume of the fluid and This property of the fluid is called viscosity and this yields is discussed in more detail in a later section. The 1 lost kinetic energy of the fluid gets converted into (P1− P2) ∆V = ρ ∆V (v2 2 − v1 2) + ρg∆V (h2 − h1) heat energy. Thus, Bernoulli’s equation ideally 2 applies to fluids with zero viscosity or non- We now divide each term by ∆V to obtain viscous fluids. Another restriction on application of Bernoulli theorem is that the fluids must be 1 incompressible, as the elastic energy of the fluid (P1− P2) = ρ (v2 2 − v1 2) + ρg (h2 − h1) is also not taken into consideration. In practice, 2 it has a large number of useful applications and We can rearrange the above terms to obtain can help explain a wide variety of phenomena 1 1 2 2 for low viscosity incompressible fluids. P1 + ρv1 + ρgh1 = P2+ ρv2 + ρgh2 2 2 Bernoulli’s equation also does not hold for non- (9.12) steady or turbulent flows, because in that This is Bernoulli’s equation. Since 1 and 2 situation velocity and pressure are constantly refer to any two locations along the pipeline, we fluctuating in time. may write the expression in general as When a fluid is at rest i.e., its velocity is zero everywhere, Bernoulli’s equation becomes 1 P + ρv2 + ρgh = constant (9.13) P1 + ρgh1 = P2 + ρgh2 2 (P1− P2) = ρg (h2 − h1) which is same as Eq. (9.6). 9.4.1 Speed of Efflux: Torricelli’s Law The word efflux means fluid outflow. Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body. Consider a tank containing a liquid of density ρ with a small hole in its side at a height y1 from the bottom (see Fig. 9.10). The air above the liquid, whose surface is at height y2, is at pressure P. From the equation of continuity [Eq. (9.10)] we have Fig. 9.9 The flow of an ideal fluid in a pipe of varying v1 A1 = v2 A2 cross section. The fluid in a section of length v1∆t moves to the section of length v2∆t in A1 v 2 = v1 time ∆t. A 2 Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 189 from its parabolic trajectory as it moves through air. This deviation can be partly explained on the basis of Bernoulli’s principle. (i) Ball moving without spin: Fig. 9.11(a) shows the streamlines around a non-spinning ball moving relative to a fluid. From the symmetry of streamlines it is clear that the velocity of fluid (air) above and below the ball at corresponding points is the same resulting in zero pressure difference. The air therefore, exerts no upward or downward force on the ball. Fig. 9.10 Torricelli’s law. The speed of efflux, v1, (ii) Ball moving with spin: A ball which is from the side of the container is given by the application of Bernoulli’s equation. spinning drags air along with it. If the If the container is open at the top to the surface is rough more air will be dragged. atmosphere then v1 = 2 g h . Fig 9.11(b) shows the streamlines of air for a ball which is moving and spinning at the same time. The ball is moving If the cross-sectional area of the tank A2 is forward and relative to it the air is moving much larger than that of the hole (A2 >>A1), then backwards. Therefore, the velocity of air we may take the fluid to be approximately at rest above the ball relative to the ball is larger at the top, i.e., v2 = 0. Now, applying the Bernoulli and below it is smaller (see Section 9.3).equation at points 1 and 2 and noting that at the hole P1 = Pa, the atmospheric pressure, we The stream lines, thus, get crowded above have from Eq. (9.12) and rarified below. This difference in the velocities of air results 1 2 Pa + ρ v1 + ρ g y1 = P + ρ g y 2 in the pressure difference between the lower and 2 upper faces and there is a net upward force on Taking y2 – y1 = h we have the ball. This dynamic lift due to spining is called 2 ( P − Pa ) Magnus effect. v1 = 2 g h + (9.14) ρ Aerofoil or lift on aircraft wing: Figure 9.11 When P >>Pa and 2 g h may be ignored, the (c) shows an aerofoil, which is a solid piece speed of efflux is determined by the container shaped to provide an upward dynamic lift pressure. Such a situation occurs in rocket when it moves horizontally through air. The propulsion. On the other hand, if the tank is cross-section of the wings of an aeroplane open to the atmosphere, then P = Pa and looks somewhat like the aerofoil shown in Fig. v1 = 2g h (9.15) 9.11 (c) with streamlines around it. When the This is also the speed of a freely falling body. aerofoil moves against the wind, the Equation (9.15) represents Torricelli’s law. orientation of the wing relative to flow direction causes the streamlines to crowd together 9.4.2 Dynamic Lift above the wing more than those below it. The Dynamic lift is the force that acts on a body, flow speed on top is higher than that below it. such as airplane wing, a hydrofoil or a spinning There is an upward force resulting in a ball, by virtue of its motion through a fluid. In dynamic lift of the wings and this balances many games such as cricket, tennis, baseball, the weight of the plane. The following example or golf, we notice that a spinning ball deviates illustrates this. Reprint 2025-26 190 PHYSICS (a) (b) (c) Fig 9.11 (a) Fluid streaming past a static sphere. (b) Streamlines for a fluid around a sphere spinning clockwise. (c) Air flowing past an aerofoil. ⊳ Example 9.7 A fully loaded Boeing aircraft vav = (v2 + v1)/2 = 960 km/h = 267 m s-1, has a mass of 3.3 × 105 kg. Its total wing we have area is 500 m2. It is in level flight with a ∆ P speed of 960 km/h. (a) Estimate the (v 2 – v1 ) / v av = 2 ≈ 0.08 ρ v av pressure difference between the lower and The speed above the wing needs to be only 8 upper surfaces of the wings (b) Estimate % higher than that below. ⊳ the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface. [The density of air is ρ 9.5 VISCOSITY = 1.2 kg m-3] Most of the fluids are not ideal ones and offer some resistance to motion. This resistance to fluid motion is like an internal friction analogous to friction when Answer (a) The weight of the Boeing aircraft is a solid moves on a surface. It is called viscosity. balanced by the upward force due to the This force exists when there is relative motion pressure difference between layers of the liquid. Suppose we consider ∆P × A = 3.3 × 105 kg × 9.8 a fluid like oil enclosed between two glass plates ∆P = (3.3 × 105 kg × 9.8 m s–2) / 500 m2 as shown in Fig. 9.12 (a). The bottom plate is fixed while the top plate is moved with a constant = 6.5 ×103 Nm-2 velocity v relative to the fixed plate. If oil is (b) We ignore the small height difference replaced by honey, a greater force is required to between the top and bottom sides in Eq. (9.12). move the plate with the same velocity. Hence The pressure difference between them is we say that honey is more viscous than oil. The then fluid in contact with a surface has the same ρ 2 2 velocity as that of the surfaces. Hence, the layer ∆P = ( v 2 – v1 ) of the liquid in contact with top surface moves 2 where v2 is the speed of air over the upper with a velocity v and the layer of the liquid in surface and v1 is the speed under the bottom contact with the fixed surface is stationary. The surface. velocities of layers increase uniformly from bottom (zero velocity) to the top layer (velocity 2 ∆ P (v 2 – v1 )= v). For any layer of liquid, its upper layer pulls ρ ( v 2 + v1 ) it forward while lower layer pulls it backward. Taking the average speed This results in force between the layers. This Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 191 type of flow is known as laminar. The layers of change of strain’ or ‘strain rate’ i.e. ∆x/(l ∆t) or liquid slide over one another as the pages of a v/l instead of strain itself. The coefficient of book do when it is placed flat on a table and a viscosity (pronounced ‘eta’) for a fluid is defined horizontal force is applied to the top cover. When as the ratio of shearing stress to the strain rate. a fluid is flowing in a pipe or a tube, then velocity (9.16)of the liquid layer along the axis of the tube is maximum and decreases gradually as we move The SI unit of viscosity is poiseiulle (Pl). Its towards the walls where it becomes zero, other units are N s m-2 or Pa s. The dimensions Fig. 9.12 (b). The velocity on a cylindrical surface of viscosity are [ML-1T-1]. Generally, thin liquids, in a tube is constant. like water, alcohol, etc., are less viscous than thick liquids, like coal tar, blood, glycerine, etc. The coefficients of viscosity for some common fluids are listed in Table 9.2. We point out two facts about blood and water that you may find interesting. As Table 9.2 indicates, blood is ‘thicker’ (more viscous) than water. Further, the relative viscosity (η/ηwater) of blood remains constant between 0 oC and 37 oC. (a) Fig. 9.13 Measurement of the coefficient of viscosity of a liquid. (b) Fig 9.12 (a) A layer of liquid sandwiched between The viscosity of liquids decreases with two parallel glass plates, in which the temperature, while it increases in the case of gases. lower plate is fixed and the upper one is moving to the right with velocity v ⊳ (b) velocity distribution for viscous flow in a pipe. Example 9.8 A metal block of area 0.10 m2 is connected to a 0.010 kg mass via a string On account of this motion, a portion of liquid, that passes over an ideal pulley (considered which at some instant has the shape ABCD, massless and frictionless), as in Fig. 9.13. take the shape of AEFD after short interval of A liquid with a film thickness of 0.30 mm time (∆t). During this time interval the liquid has is placed between the block and the table. undergone a shear strain of ∆x/l. Since, the When released the block moves to the right strain in a flowing fluid increases with time with a constant speed of 0.085 m s-1. Find continuously. Unlike a solid, here the stress is the coefficient of viscosity of the liquid. found experimentally to depend on ‘rate of Reprint 2025-26 192 PHYSICS Answer The metal block moves to the right This is known as Stokes’ law. We shall not because of the tension in the string. The tension derive Stokes’ law. T is equal in magnitude to the weight of the This law is an interesting example of retarding suspended mass m. Thus, the shear force F is force, which is proportional to velocity. We can study its consequences on an object falling F = T = mg = 0.010 kg × 9.8 m s–2 = 9.8 × 10-2 N through a viscous medium. We consider a Shear stress on the fluid = F/A = N/m2 raindrop in air. It accelerates initially due to gravity. As the velocity increases, the retarding Strain rate = force also increases. Finally, when viscous force plus buoyant force becomes equal to the force due to gravity, the net force becomes zero and so does the acceleration. The sphere (raindrop) then descends with a constant velocity. Thus, in = equilibrium, this terminal velocity vt is given by 6πηavt = (4π/3) a3 (ρ-σ)g = 3.46 ×10-3 Pa s where ρ and σ are mass densities of sphere and ⊳ the fluid, respectively. We obtain Table 9.2 The viscosities of some fluids vt = 2a2 (ρ-σ)g / (9η) (9.18) Fluid T(oC) Viscosity (mPl) So the terminal velocity vt depends on the Water 20 1.0 square of the radius of the sphere and inversely 100 0.3 on the viscosity of the medium. Blood 37 2.7 You may like to refer back to Example 6.2 in Machine Oil 16 113 this context. 38 34 ⊳ Glycerine 20 830 Example 9.9 The terminal velocity of a Honey – 200 copper ball of radius 2.0 mm falling through Air 0 0.017 a tank of oil at 20oC is 6.5 cm s-1. Compute the viscosity of the oil at 20oC. Density of 40 0.019 oil is 1.5 ×103 kg m-3, density of copper is 8.9 × 103 kg m-3. 9.5.1 Stokes’ Law When a body falls through a fluid it drags the Answer We have vt = 6.5 × 10-2 ms-1, a = 2 × 10-3 m,layer of the fluid in contact with it. A relative g = 9.8 ms-2, ρ = 8.9 × 103 kg m-3,motion between the different layers of the fluid is set and, as a result, the body experiences a σ =1.5 ×103 kg m-3. From Eq. (9.18) retarding force. Falling of a raindrop and swinging of a pendulum bob are some common examples of such motion. It is seen that the viscous force is proportional to the velocity of = 9.9 × 10-1 kg m–1 s–1 ⊳ the object and is opposite to the direction of motion. The other quantities on which the force F depends are viscosity η of the fluid and radius 9.6 SURFACE TENSION a of the sphere. Sir George G. Stokes (1819– You must have noticed that, oil and water do 1903), an English scientist enunciated clearly not mix; water wets you and me but not ducks; the viscous drag force F as mercury does not wet glass but water sticks to F = 6 π ηav (9.17) it, oil rises up a cotton wick, inspite of gravity, Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 193 Sap and water rise up to the top of the leaves of Let us consider a molecule near the surface the tree, hair of a paint brush do not cling Fig. 9.14(b). Only lower half side of it is together when dry and even when dipped in surrounded by liquid molecules. There is some water but form a fine tip when taken out of it. negative potential energy due to these, but All these and many more such experiences are obviously it is less than that of a molecule in related with the free surfaces of liquids. As bulk, i.e., the one fully inside. Approximately liquids have no definite shape but have a it is half of the latter. Thus, molecules on a definite volume, they acquire a free surface when liquid surface have some extra energy in poured in a container. These surfaces possess comparison to molecules in the interior. A some additional energy. This phenomenon is liquid, thus, tends to have the least surface known as surface tension and it is concerned area which external conditions permit. with only liquid as gases do not have free Increasing surface area requires energy. Most surfaces. Let us now understand this surface phenomenon can be understood in phenomena. terms of this fact. What is the energy required for having a molecule at the surface? As9.6.1 Surface Energy mentioned above, roughly it is half the energy A liquid stays together because of attraction required to remove it entirely from the liquid between molecules. Consider a molecule well i.e., half the heat of evaporation. inside a liquid. The intermolecular distances are Finally, what is a surface? Since a liquid such that it is attracted to all the surrounding consists of molecules moving about, there cannot molecules [Fig. 9.14(a)]. This attraction results be a perfectly sharp surface. The density of the in a negative potential energy for the molecule, liquid molecules drops rapidly to zero around which depends on the number and distribution z = 0 as we move along the direction indicated of molecules around the chosen one. But the Fig 9.14 (c) in a distance of the order of a few average potential energy of all the molecules is molecular sizes. the same. This is supported by the fact that to take a collection of such molecules (the liquid) 9.6.2 Surface Energy and Surface Tension and to disperse them far away from each other As we have discussed that an extra energy is in order to evaporate or vaporise, the heat of associated with surface of liquids, the creation evaporation required is quite large. For water it of more surface (spreading of surface) keeping is of the order of 40 kJ/mol. other things like volume fixed requires a Fig. 9.14 Schematic picture of molecules in a liquid, at the surface and balance of forces. (a) Molecule inside a liquid. Forces on a molecule due to others are shown. Direction of arrows indicates attraction of repulsion. (b) Same, for a molecule at a surface. (c) Balance of attractive (AI and repulsive (R) forces. Reprint 2025-26 194 PHYSICS horizontal liquid film ending in bar free to slide We make the following observations from over parallel guides Fig (9.15). above: (i) Surface tension is a force per unit length (or surface energy per unit area) acting in the plane of the interface between the plane of the liquid and any other substance; it also is the extra energy that the molecules at the interface have as compared to molecules in the interior. (ii) At any point on the interface besides the Fig. 9.15 Stretching a film. (a) A film in equilibrium; boundary, we can draw a line and imagine (b) The film stretched an extra distance. equal and opposite surface tension forces Suppose that we move the bar by a small S per unit length of the line acting distance d as shown. Since the area of the perpendicular to the line, in the plane of surface increases, the system now has more the interface. The line is in equilibrium. To energy, this means that some work has been be more specific, imagine a line of atoms or done against an internal force. Let this internal molecules at the surface. The atoms to the force be F, the work done by the applied force is left pull the line towards them; those to the F.d = Fd. From conservation of energy, this is right pull it towards them! This line of stored as additional energy in the film. If the atoms is in equilibrium under tension. If surface energy of the film is S per unit area, the the line really marks the end of the extra area is 2dl. A film has two sides and the interface, as in Figure 9.14 (a) and (b) there liquid in between, so there are two surfaces and is only the force S per unit length the extra energy is acting inwards. Table 9.3 gives the surface tension of various S (2dl) = Fd (9.19) liquids. The value of surface tension depends Or, S=Fd/2dl = F/2l (9.20) on temperature. Like viscosity, the surface This quantity S is the magnitude of surface tension of a liquid usually falls with tension. It is equal to the surface energy per unit temperature. area of the liquid interface and is also equal to Table 9.3 Surface tension of some liquids at thethe force per unit length exerted by the fluid on temperatures indicated with the the movable bar. heats of the vaporisation So far we have talked about the surface of one liquid. More generally, we need to consider Liquid Temp (oC) Surface Heat of fluid surface in contact with other fluids or solid Tension vaporisation (N/m) (kJ/mol)surfaces. The surface energy in that case depends on the materials on both sides of the surface. For example, if the molecules of the Helium –270 0.000239 0.115 materials attract each other, surface energy is Oxygen –183 0.0132 7.1 reduced while if they repel each other the surface energy is increased. Thus, more Ethanol 20 0.0227 40.6 appropriately, the surface energy is the energy Water 20 0.0727 44.16 of the interface between two materials and Mercury 20 0.4355 63.2 depends on both of them. Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 195 A fluid will stick to a solid surface if the by θ. It is different at interfaces of different pairs surface energy between fluid and the solid is of liquids and solids. The value of θ determines smaller than the sum of surface energies whether a liquid will spread on the surface of a between solid-air, and fluid-air. Now there is solid or it will form droplets on it. For example, attraction between the solid surface and the water forms droplets on lotus leaf as shown in liquid. It can be directly measured Fig. 9.17 (a) while spreads over a clean plastic experimentaly as schematically shown in Fig. plate as shown in Fig. 9.17(b). 9.16. A flat vertical glass plate, below which a vessel of some liquid is kept, forms one arm of the balance. The plate is balanced by weights on the other side, with its horizontal edge just over water. The vessel is raised slightly till the liquid just touches the glass plate and pulls it down a little because of surface tension. Weights (a)are added till the plate just clears water. (b) Fig. 9.17 Different shapes of water drops with interfacial tensions (a) on a lotus leaf (b) on a clean plastic plate. Fig. 9.16 Measuring Surface Tension. We consider the three interfacial tensions at all the three interfaces, liquid-air, solid-air and Suppose the additional weight required is W. solid-liquid denoted by Sla, Ssa and Ssl , respectivelyThen from Eq. 9.20 and the discussion given as given in Fig. 9.17 (a) and (b). At the line of there, the surface tension of the liquid-air contact, the surface forces between the three media interface is must be in equilibrium. From the Fig. 9.17(b) the Sla = (W/2l) = (mg/2l ) (9.21) following relation is easily derived. where m is the extra mass and l is the length of Sla cos θ + Ssl = Ssa (9.22) the plate edge. The subscript (la) emphasises The angle of contact is an obtuse angle if the fact that the liquid-air interface tension is involved. Ssl > Sla as in the case of water-leaf interface while it is an acute angle if Ssl < Sla as in the case of water-plastic interface. When θ is an 9.6.3 Angle of Contact obtuse angle then molecules of liquids are The surface of liquid near the plane of contact, attracted strongly to themselves and weakly to with another medium is in general curved. The those of solid, it costs a lot of energy to create a angle between tangent to the liquid surface at liquid-solid surface, and liquid then does not the point of contact and solid surface inside the wet the solid. This is what happens with water liquid is termed as angle of contact. It is denoted on a waxy or oily surface, and with mercury on Reprint 2025-26 196 PHYSICS any surface. On the other hand, if the molecules so that of the liquid are strongly attracted to those of (Pi – Po) = (2 Sla/ r) (9.25) the solid, this will reduce Ssl and therefore, In general, for a liquid-gas interface, the cos θ may increase or θ may decrease. In this convex side has a higher pressure than the case θ is an acute angle. This is what happens concave side. For example, an air bubble in a for water on glass or on plastic and for kerosene liquid, would have higher pressure inside it. oil on virtually anything (it just spreads). Soaps, See Fig 9.18 (b). detergents and dying substances are wetting agents. When they are added the angle of contact becomes small so that these may penetrate well and become effective. Water proofing agents on the other hand are added to create a large angle of contact between the water and fibres. 9.6.4 Drops and Bubbles Fig. 9.18 Drop, cavity and bubble of radius r. One consequence of surface tension is that free A bubble Fig 9.18 (c) differs from a drop liquid drops and bubbles are spherical if effects and a cavity; in this it has two interfaces. Applying of gravity can be neglected. You must have seen the above argument we have for a bubble this especially clearly in small drops just formed (Pi – Po) = (4 Sla/ r) (9.26) in a high-speed spray or jet, and in soap bubbles blown by most of us in childhood. Why are drops This is probably why you have to blow hard, and bubbles spherical? What keeps soap but not too hard, to form a soap bubble. A little bubbles stable? extra air pressure is needed inside! As we have been saying repeatedly, a liquid- air interface has energy, so for a given volume 9.6.5 Capillary Rise the surface with minimum energy is the one with One consequence of the pressure difference the least area. The sphere has this property. across a curved liquid-air interface is the well- Though it is out of the scope of this book, but known effect that water rises up in a narrow you can check that a sphere is better than at tube in spite of gravity. The word capilla means least a cube in this respect! So, if gravity and hair in Latin; if the tube were hair thin, the rise other forces (e.g. air resistance) were ineffective, would be very large. To see this, consider a liquid drops would be spherical. vertical capillary tube of circular cross section Another interesting consequence of surface (radius a) inserted into an open vessel of water tension is that the pressure inside a spherical (Fig. 9.19). The contact angle between water and drop Fig. 9.18(a) is more than the pressure outside. Suppose a spherical drop of radius r is in equilibrium. If its radius increase by ∆r. The extra surface energy is [4π(r + ∆r) 2- 4πr2] Sla = 8πr ∆r Sla (9.23) If the drop is in equilibrium this energy cost is balanced by the energy gain due to expansion under the pressure difference (Pi – Po) between the inside of the bubble and the outside. The work done is Fig. 9.19 Capillary rise, (a) Schematic picture of a narrow tube immersed water. W = (Pi – Po) 4πr2∆r (9.24) (b) Enlarged picture near interface. Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 197 glass is acute. Thus the surface of water in the ⊳ Example 9.10 The lower end of a capillary capillary is concave. This means that there is tube of diameter 2.00 mm is dipped 8.00 a pressure difference between the two sides cm below the surface of water in a beaker. of the top surface. This is given by What is the pressure required in the tube in order to blow a hemispherical bubble at (Pi – Po) =(2S/r) = 2S/(a sec θ ) its end in water? The surface tension of = (2S/a) cos θ (9.27) water at temperature of the experiments is Thus the pressure of the water inside the 7.30 × 10-2 Nm-1. 1 atmospheric pressure = 1.01 × 105 Pa, density of water = 1000 kg/m3,tube, just at the meniscus (air-water interface) g = 9.80 m s-2. Also calculate the excess is less than the atmospheric pressure. Consider pressure. the two points A and B in Fig. 9.19(a). They must be at the same pressure, namely Answer The excess pressure in a bubble of gas P0 + h ρ g = Pi = PA (9.28) in a liquid is given by 2S/r, where S is the where ρ is the density of water and h is called surface tension of the liquid-gas interface. You the capillary rise [Fig. 9.19(a)]. Using should note there is only one liquid surface in Eq. (9.27) and (9.28) we have this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for h ρ g = (Pi – P0) = (2S cos θ )/a (9.29) excess pressure in that case is 4S/r.) The The discussion here, and the Eqs. (9.24) and radius of the bubble is r. Now the pressure (9.25) make it clear that the capillary rise is due outside the bubble Po equals atmospheric pressure plus the pressure due to 8.00 cm ofto surface tension. It is larger, for a smaller a. water column. That is Typically it is of the order of a few cm for fine Po = (1.01 × 105 Pa + 0.08 m × 1000 kg m–3 capillaries. For example, if a = 0.05 cm, using × 9.80 m s–2) the value of surface tension for water (Table 9.3), = 1.01784 × 105 Pa we find that Therefore, the pressure inside the bubble is h = 2S/(ρ g a) Pi = Po + 2S/r -1 = 1.01784 × 105 Pa + (2 × 7.3 × 10-2 Pa m/10-3 m) 2×(0.073 N m ) = (1.01784 + 0.00146) × 105 Pa = 3 -3 -2 -4 (10 kg m ) (9.8 m s )(5 × 10 m) = 1.02 × 105 Pa where the radius of the bubble is taken = 2.98 × 10–2 m = 2.98 cm to be equal to the radius of the capillary tube, Notice that if the liquid meniscus is convex, since the bubble is hemispherical ! (The answer as for mercury, i.e., if cos θ is negative then from has been rounded off to three significant figures.) Eq. (9.28) for example, it is clear that the liquid The excess pressure in the bubble is 146 Pa. will be lower in the capillary ! ⊳ SUMMARY 1. The basic property of a fluid is that it can flow. The fluid does not have any resistance to change of its shape. Thus, the shape of a fluid is governed by the shape of its container. 2. A liquid is incompressible and has a free surface of its own. A gas is compressible and it expands to occupy all the space available to it. 3. If F is the normal force exerted by a fluid on an area A then the average pressure Pav is defined as the ratio of the force to area F Pav = A Reprint 2025-26 198 PHYSICS 4. The unit of the pressure is the pascal (Pa). It is the same as N m-2. Other common units of pressure are 1 atm = 1.01×105 Pa 1 bar = 105 Pa 1 torr = 133 Pa = 0.133 kPa 1 mm of Hg = 1 torr = 133 Pa 5. Pascal’s law states that: Pressure in a fluid at rest is same at all points which are at the same height. A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel. 6. The pressure in a fluid varies with depth h according to the expression P = Pa + ρgh where ρ is the density of the fluid, assumed uniform. 7. The volume of an incompressible fluid passing any point every second in a pipe of non uniform crossection is the same in the steady flow. v A = constant ( v is the velocity and A is the area of crossection) The equation is due to mass conservation in incompressible fluid flow. 8. Bernoulli’s principle states that as we move along a streamline, the sum of the pressure (P), the kinetic energy per unit volume (ρv2/2) and the potential energy per unit volume (ρgy) remains a constant. P + ρv2/2 + ρgy = constant The equation is basically the conservation of energy applied to non viscuss fluid motion in steady state. There is no fluid which have zero viscosity, so the above statement is true only approximately. The viscosity is like friction and converts the kinetic energy to heat energy. 9. Though shear strain in a fluid does not require shear stress, when a shear stress is applied to a fluid, the motion is generated which causes a shear strain growing with time. The ratio of the shear stress to the time rate of shearing strain is known as coefficient of viscosity, η. where symbols have their usual meaning and are defined in the text. 10. Stokes’ law states that the viscous drag force F on a sphere of radius a moving with velocity v through a fluid of viscosity is, F = 6πηav. 11. Surface tension is a force per unit length (or surface energy per unit area) acting in the plane of interface between the liquid and the bounding surface. It is the extra energy that the molecules at the interface have as compared to the interior. POINTS TO PONDER 1. Pressure is a scalar quantity. The definition of the pressure as “force per unit area” may give one false impression that pressure is a vector. The “force” in the numerator of the definition is the component of the force normal to the area upon which it is impressed. While describing fluids as a concept, shift from particle and rigid body mechanics is required. We are concerned with properties that vary from point to point in the fluid. 2. One should not think of pressure of a fluid as being exerted only on a solid like the walls of a container or a piece of solid matter immersed in the fluid. Pressure exists at all points in a fluid. An element of a fluid (such as the one shown in Fig. 9.4) is in equilibrium because the pressures exerted on the various faces are equal. Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 199 3. The expression for pressure P = Pa + ρgh holds true if fluid is incompressible. Practically speaking it holds for liquids, which are largely incompressible and hence is a constant with height. 4. The gauge pressure is the difference of the actual pressure and the atmospheric pressure. P – Pa = Pg Many pressure-measuring devices measure the gauge pressure. These include the tyre pressure gauge and the blood pressure gauge (sphygmomanometer). 5. A streamline is a map of fluid flow. In a steady flow two streamlines do not intersect as it means that the fluid particle will have two possible velocities at the point. 6. Bernoulli’s principle does not hold in presence of viscous drag on the fluid. The work done by this dissipative viscous force must be taken into account in this case, and P2 [Fig. 9.9] will be lower than the value given by Eq. (9.12). 7. As the temperature rises the atoms of the liquid become more mobile and the coefficient of viscosity, η falls. In a gas the temperature rise increases the random motion of atoms and η increases. 8. Surface tension arises due to excess potential energy of the molecules on the surface in comparison to their potential energy in the interior. Such a surface energy is present at the interface separating two substances at least one of which is a fluid. It is not the property of a single fluid alone. EXERCISES 9.1 Explain why (a) The blood pressure in humans is greater at the feet than at the brain (b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km (c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area. 9.2 Explain why (a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute. (b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.) Reprint 2025-26 200 PHYSICS (c) Surface tension of a liquid is independent of the area of the surface (d) Water with detergent disolved in it should have small angles of contact. (e) A drop of liquid under no external forces is always spherical in shape 9.3 Fill in the blanks using the word(s) from the list appended with each statement: (a) Surface tension of liquids generally ... with temperatures (increases / decreases) (b) Viscosity of gases ... with temperature, whereas viscosity of liquids ... with temperature (increases / decreases) (c) For solids with elastic modulus of rigidity, the shearing force is proportional to ... , while for fluids it is proportional to ... (shear strain / rate of shear strain) (d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle) (e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for turbulence for an actual plane (greater / smaller) 9.4 Explain why (a) To keep a piece of paper horizontal, you should blow over, not under, it (b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers (c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection (d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel (e) A spinning cricket ball in air does not follow a parabolic trajectory 9.5 A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor ? 9.6 Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m–3. Determine the height of the wine column for normal atmospheric pressure. 9.7 A vertical off-shore structure is built to withstand a maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean ? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents. 9.8 A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear ? 9.9 A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit ? 9.10 In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6) 9.11 Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river ? Explain. 9.12 Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation ? Explain. 9.13 Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube ? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct]. 9.14 In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s–1and 63 m s-1 respectively. What is the lift on the wing if its area is 2.5 m2 ? Take the density of air to be 1.3 kg m–3. 9.15 Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ? Reprint 2025-26 MECHANICAL PROPERTIES OF FLUIDS 201 Fig. 9.20 9.16 The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes ? 9.17 A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film ? 9.18 Figure 9.21 (a) shows a thin liquid film supporting a small weight = 4.5 × 10–2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c) ? Explain your answer physically. Fig. 9.21 9.19 What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature ? Surface tension of mercury at that temperature (20 °C) is 4.65 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop. 9.20 What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10–2 N m–1 ? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble ? (1 atmospheric pressure is 1.01 × 105 Pa). Reprint 2025-26 CHAPTER TEN THERMAL PROPERTIES OF MATTER 10.1 INTRODUCTION We all have common sense notions of heat and temperature. Temperature is a measure of ‘hotness’ of a body. A kettle 10.1 Introduction with boiling water is hotter than a box containing ice. In 10.2 Temperature and heat physics, we need to define the notion of heat, temperature, 10.3 Measurement of etc., more carefully. In this chapter, you will learn what heat temperature is and how it is measured, and study the various proceses by 10.4 Ideal-gas equation and which heat flows from one body to another. Along the way, absolute temperature you will find out why blacksmiths heat the iron ring before 10.5 Thermal expansion fitting on the rim of a wooden wheel of a horse cart and why 10.6 Specific heat capacity the wind at the beach often reverses direction after the sun 10.7 Calorimetry goes down. You will also learn what happens when water boils or freezes, and its temperature does not change during these10.8 Change of state processes even though a great deal of heat is flowing into or10.9 Heat transfer out of it.10.10 Newton’s law of cooling Summary 10.2 TEMPERATURE AND HEAT Points to ponder We can begin studying thermal properties of matter with Exercises definitions of temperature and heat. Temperature is a relative Additional Exercises measure, or indication of hotness or coldness. A hot utensil is said to have a high temperature, and ice cube to have a low temperature. An object that has a higher temperature than another object is said to be hotter. Note that hot and cold are relative terms, like tall and short. We can perceive temperature by touch. However, this temperature sense is somewhat unreliable and its range is too limited to be useful for scientific purposes. We know from experience that a glass of ice-cold water left on a table on a hot summer day eventually warms up whereas a cup of hot tea on the same table cools down. It means that when the temperature of body, ice-cold water or hot tea in this case, and its surrounding medium are different, heat transfer takes place between the system and the surrounding medium, until the body and the surrounding medium are at the same temperature. We also know that in the case of glass tumbler of ice-cold water, heat flows from the environment to Reprint 2025-26 THERMAL PROPERTIES OF MATTER 203 the glass tumbler, whereas in the case of hot tea, it flows from the cup of hot tea to the environment. So, we can say that heat is the form of energy transferred between two (or more) systems or a system and its surroundings by virtue of temperature difference. The SI unit of heat energy transferred is expressed in joule (J) while SI unit of temperature is Kelvin (K), and degree Celsius (oC) is a commonly used unit of temperature. When an object is heated, many changes may take place. Its temperature may rise, it may expand or change state. We will study the effect of heat on different bodies in later sections. Fig. 10.1 A plot of Fahrenheit temperature (tF) versus10.3 MEASUREMENT OF TEMPERATURE Celsius temperature (tc). A measure of temperature is obtained using a thermometer. Many physical properties of A relationship for converting between the two materials change sufficiently with temperature. scales may be obtained from a graph of Some such properties are used as the basis for Fahrenheit temperature (tF) versus celsius constructing thermometers. The commonly used temperature (tC) in a straight line (Fig. 10.1), property is variation of the volume of a liquid whose equation is with temperature. For example, in common t F – 32 t C = (10.1)liquid–in–glass thermometers, mercury, alcohol 180 100 etc., are used whose volume varies linearly with temperature over a wide range. 10.4 IDEAL-GAS EQUATION AND Thermometers are calibrated so that a ABSOLUTE TEMPERATURE numerical value may be assigned to a given temperature in an appropriate scale. For the Liquid-in-glass thermometers show different definition of any standard scale, two fixed readings for temperatures other than the fixed reference points are needed. Since all points because of differing expansion properties. substances change dimensions with A thermometer that uses a gas, however, gives temperature, an absolute reference for the same readings regardless of which gas is expansion is not available. However, the used. Experiments show that all gases at low necessary fixed points may be correlated to the densities exhibit same expansion behaviour. The physical phenomena that always occur at the variables that describe the behaviour of a given same temperature. The ice point and the steam quantity (mass) of gas are pressure, volume, and point of water are two convenient fixed points temperature (P, V, and T )(where T = t + 273.15; and are known as the freezing and boiling t is the temperature in °C). When temperature points, respectively. These two points are the is held constant, the pressure and volume of a temperatures at which pure water freezes and quantity of gas are related as PV = constant. boils under standard pressure. The two familiar This relationship is known as Boyle’s law, after temperature scales are the Fahrenheit Robert Boyle (1627–1691), the English Chemist temperature scale and the Celsius temperature who discovered it. When the pressure is held scale. The ice and steam point have values constant, the volume of a quantity of the gas is 32 °F and 212 °F, respectively, on the Fahrenheit related to the temperature as V/T = constant. scale and 0 °C and 100 °C on the Celsius scale. This relationship is known as Charles’ law, On the Fahrenheit scale, there are 180 equal after French scientist Jacques Charles (1747– intervals between two reference points, and on 1823). Low-density gases obey these the Celsius scale, there are 100. laws, which may be combined into a single Reprint 2025-26 204 PHYSICS Fig. 10.3 A plot of pressure versus temperature and Fig. 10.2 Pressure versus temperature of a low extrapolation of lines for low density gases density gas kept at constant volume. indicates the same absolute zero temperature. relationship. Notice that since PV = constant named after the British scientist Lord Kelvin. On and V/T = constant for a given quantity of gas, this scale, – 273.15 °C is taken as the zero point, then PV/T should also be a constant. This that is 0 K (Fig. 10.4). relationship is known as ideal gas law. It can be written in a more general form that applies not just to a given quantity of a single gas but to any quantity of any low-density gas and is known as ideal-gas equation: or PV = µRT (10.2) where, µ is the number of moles in the sample of gas and R is called universal gas constant: R = 8.31 J mol–1 K–1 In Eq. 10.2, we have learnt that the pressure and volume are directly proportional to temperature : PV ∝ T. This relationship allows a gas to be used to measure temperature in a constant volume gas thermometer. Holding the volume of a gas constant, it gives P ∝T. Thus, with a constant-volume gas thermometer, Fig. 10.4 Comparision of the Kelvin, Celsius and temperature is read in terms of pressure. A plot Fahrenheit temperature scales. The size of unit in Kelvin and Celsius of pressure versus temperature gives a straight temperature scales is the same. So, temperature line in this case, as shown in Fig. 10.2. on these scales are related by However, measurements on real gases deviate from the values predicted by the ideal gas law T = tC + 273.15 (10.3) at low temperature. But the relationship is linear over a large temperature range, and it looks as 10.5 THERMAL EXPANSION though the pressure might reach zero with You may have observed that sometimes sealed decreasing temperature if the gas continued to bottles with metallic lids are so tightly screwed be a gas. The absolute minimum temperature that one has to put the lid in hot water for some for an ideal gas, therefore, inferred by time to open it. This would allow the metallic lid extrapolating the straight line to the axis, as in to expand, thereby loosening it to unscrew Fig. 10.3. This temperature is found to be easily. In case of liquids, you may have observed – 273.15 °C and is designated as absolute zero. that mercury in a thermometer rises, when the Absolute zero is the foundation of the Kelvin thermometer is put in slightly warm water. If temperature scale or absolute scale temperature we take out the thermometer from the warm Reprint 2025-26 THERMAL PROPERTIES OF MATTER 205 water the level of mercury falls again. Similarly, Table 10.1 Values of coefficient of linear in case of gases, a balloon partially inflated in a expansion for some material cool room may expand to full size when placed in warm water. On the other hand, a fully Material αl (10–5 K–1) inflated balloon when immersed in cold water Aluminium 2.5 would start shrinking due to contraction of the Brass 1.8 air inside. Iron 1.2 It is our common experience that most Copper 1.7 substances expand on heating and contract on Silver 1.9 cooling. A change in the temperature of a body Gold 1.4 Glass (pyrex) 0.32causes change in its dimensions. The increase Lead 0.29in the dimensions of a body due to the increase in its temperature is called thermal expansion. The expansion in length is called linear Similarly, we consider the fractional change expansion. The expansion in area is called area expansion. The expansion in volume is called ∆V in volume, , of a substance for temperature volume expansion (Fig. 10.5). V change ∆T and define the coefficient of volume expansion (or volume expansivity), as (10.5) Here αV is also a characteristic of the substance but is not strictly a constant. It depends in general on temperature (Fig 10.6). It ∆l ∆A ∆V = a l ∆ T = 2a l ∆T = 3a l ∆ T is seen that αV becomes constant only at a high l A V temperature. (a) Linear expansion (b) Area expansion (c) Volume expansion Fig. 10.5 Thermal Expansion. If the substance is in the form of a long rod, then for small change in temperature, ∆T, the fractional change in length, ∆l/l, is directly proportional to ∆T. (10.4) where α1 is known as the coefficient of linear expansion (or linear expansivity) and is characteristic of the material of the rod. In Table Fig. 10.6 Coefficient of volume expansion of copper 10.1, typical average values of the coefficient of as a function of temperature. linear expansion for some material in the temperature range 0 °C to 100 °C are given. From Table 10.2 gives the values of coefficient of this Table, compare the value of αl for glass and volume expansion of some common substances copper. We find that copper expands about five in the temperature range 0–100 °C. You can see times more than glass for the same rise in that thermal expansion of these substances temperature. Normally, metals expand more and (solids and liquids) is rather small, with material, have relatively high values of αl. Reprint 2025-26 206 PHYSICS like pyrex glass and invar (a special iron-nickel lakes and ponds, freeze at the top first. As a lake alloy) having particularly low values of αV. From cools toward 4 °C, water near the surface loses this Table we find that the value of αv for energy to the atmosphere, becomes denser, and alcohol (ethanol) is more than mercury and sinks; the warmer, less dense water near the expands more than mercury for the same rise bottom rises. However, once the colder water on in temperature. top reaches temperature below 4 °C, it becomes less dense and remains at the surface, where it Table 10.2 Values of coefficient of volume freezes. If water did not have this property, lakes expansion for some substances and ponds would freeze from the bottom up, which would destroy much of their animal and Material αv ( K–1) plant life. Aluminium 7 × 10–5 Gases, at ordinary temperature, expand more Brass 6 × 10–5 than solids and liquids. For liquids, the Iron 3.55 × 10–5 coefficient of volume expansion is relatively Paraffin 58.8 × 10–5 independent of the temperature. However, for Glass (ordinary) 2.5 × 10–5 gases it is dependent on temperature. For an Glass (pyrex) 1 × 10–5 ideal gas, the coefficient of volume expansion at Hard rubber 2.4 × 10–4 constant pressure can be found from the ideal Invar 2 × 10–6 gas equation: Mercury 18.2 × 10–5 PV = µRT Water 20.7 × 10–5 At constant pressure Alcohol (ethanol) 110 × 10–5 P∆V = µR ∆T ∆V ∆ T Water exhibits an anomalous behaviour; it = V T contracts on heating between 0 °C and 4 °C. The volume of a given amount of water decreases as 1 i.e., αv = for ideal gas (10.6) it is cooled from room temperature, until its T temperature reaches 4 °C, [Fig. 10.7(a)]. Below At 0 °C, αv = 3.7 × 10–3 K–1, which is much 4 °C, the volume increases, and therefore, the larger than that for solids and liquids. density decreases [Fig. 10.7(b)]. Equation (10.6) shows the temperature This means that water has the maximum dependence of αv; it decreases with increasing density at 4 °C. This property has an important temperature. For a gas at room temperature and environmental effect: bodies of water, such as constant pressure, αv is about 3300 × 10–6 K–1, as Temperature (°C) Temperature (°C) (a) (b) Fig. 10.7 Thermal expansion of water. Reprint 2025-26 THERMAL PROPERTIES OF MATTER 207 much as order(s) of magnitude larger than the Answer coefficient of volume expansion of typical liquids. ∆A3 = (∆a) (∆b) a (∆b) There is a simple relation between the ∆Al =coefficient of volume expansion (αv) and coefficient of linear expansion (αl). Imagine a ∆b cube of length, l, that expands equally in all directions, when its temperature increases by b ∆T. We have ∆a a ∆l = αl l ∆T so, ∆V = (l+∆l)3 – l3 ≃ 3l2 ∆l (10.7) In Equation (10.7), terms in (∆l)2 and (∆l)3 have been neglected since ∆l is small compared to l. So ∆A2 = b (∆a) 3V ∆ l Fig. 10.8 ∆ V = = 3V α l ∆T (10.8) l Consider a rectangular sheet of the solid which gives material of length a and breadth b (Fig. 10.8 ). αv = 3αl (10.9) When the temperature increases by ∆T, a increases by ∆a = αl a∆T and b increases by ∆b What happens by preventing the thermal = αlb ∆T. From Fig. 10.8, the increase in area expansion of a rod by fixing its ends rigidly? ∆A = ∆A1 +∆A2 + ∆A3 Clearly, the rod acquires a compressive strain ∆A = a ∆b + b ∆a + (∆a) (∆b) due to the external forces provided by the rigid = a αlb ∆T + b αl a ∆T + (αl)2 ab (∆T)2 support at the ends. The corresponding stress = αl ab ∆T (2 + αl ∆T) = αl A ∆T (2 + αl ∆T) set up in the rod is called thermal stress. For Since αl ≃ 10–5 K–1, from Table 10.1, theexample, consider a steel rail of length 5 m and product αl ∆T for fractional temperature is smallarea of cross-section 40 cm2 that is prevented in comparision with 2 and may be neglected. from expanding while the temperature rises by Hence, 10 °C. The coefficient of linear expansion of steel is αl(steel) = 1.2 × 10–5 K–1. Thus, the compressive ⊳ ∆l strain is = αl(steel) ∆T = 1.2 × 10–5 × 10=1.2 × 10–4. l ⊳ Youngs modulus of steel is Y (steel) = 2 × 1011 N m–2. Example 10.2 A blacksmith fixes iron ring on the rim of the wooden wheel of a horseTherefore, the thermal stress developed is cart. The diameter of the rim and the iron ∆F  ∆l  = Y steel 2.4 × 107 N m –2, which ring are 5.243 m and 5.231 m, respectively A  l = at 27 °C. To what temperature should the corresponds to an external force of ring be heated so as to fit the rim of the wheel?  ∆l  ∆F = AYsteel  l  = 2.4 × 107 × 40 × 10–4 j 105N. If Answer two such steel rails, fixed at their outer ends, Given, T1 = 27 °Care in contact at their inner ends, a force of this magnitude can easily bend the rails. LT1 = 5.231 m ⊳ LT2 = 5.243 m Example 10.1 Show that the coefficient of So, area expansion, (∆A/A)/∆T, of a LT2 =LT1 [1+αl (T2–T1)] rectangular sheet of the solid is twice its linear expansivity, αl. 5.243 m = 5.231 m [1 + 1.20×10–5 K–1 (T2–27 °C)] or T2 = 218 °C. ⊳ Reprint 2025-26 208 PHYSICS

14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.