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

1.36 100 g of liquid A (molar mass 140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr. 29 Solutions Reprint 2025-26

1.34 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.