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

11.4 HEAT, INTERNAL ENERGY AND WORK associated with various random motions of its The Zeroth Law of Thermodynamics led us to molecules. We will see in the next chapter that the concept of temperature that agrees with our in a gas this motion is not only translational commonsense notion. Temperature is a marker (i.e. motion from one point to another in the of the ‘hotness’ of a body. It determines the volume of the container); it also includes rotational and vibrational motion of thedirection of flow of heat when two bodies are molecules (Fig. 11.3).placed in thermal contact. Heat flows from the body at a higher temperature to the one at lower temperature. The flow stops when the temperatures equalise; the two bodies are then in thermal equilibrium. We saw in some detail how to construct temperature scales to assign temperatures to different bodies. We now describe the concepts of heat and other relevant quantities like internal energy and work. Fig. 11.3 (a) Internal energy U of a gas is the sum of the kinetic and potential energies of its The concept of internal energy of a system is molecules when the box is at rest. Kinetic not difficult to understand. We know that every energy due to various types of motion bulk system consists of a large number of (translational, rotational, vibrational) is to molecules. Internal energy is simply the sum of be included in U. (b) If the same box is the kinetic energies and potential energies of moving as a whole with some velocity, these molecules. We remarked earlier that in the kinetic energy of the box is not to be thermodynamics, the kinetic energy of the included in U. system, as a whole, is not relevant. Internal energy is thus, the sum of molecular kinetic and potential energies in the frame of reference relative to which the centre of mass of the system is at rest. Thus, it includes only the (disordered) energy associated with the random motion of molecules of the system. We denote the internal energy of a system by U. Though we have invoked the molecular picture to understand the meaning of internal energy, as far as thermodynamics is concerned, U is simply a macroscopic variable of the system. The important thing about internal energy is that it depends only on the state of the system, not on how that state was achieved. Internal energy U of a system is an example of a thermodynamic ‘state variable’ – its value depends only on the given state of the system, not on history i.e. not on the ‘path’ taken to arrive at that state. Thus, the internal energy of a given mass of gas depends on its state described by Fig. 11.4 Heat and work are two distinct modes of specific values of pressure, volume and energy transfer to a system that results in temperature. It does not depend on how this change in its internal energy. (a) Heat is energy transfer due to temperaturestate of the gas came about. Pressure, volume, difference between the system and the temperature, and internal energy are surroundings. (b) Work is energy transfer thermodynamic state variables of the system brought about by means (e.g. moving the (gas) (see section 11.7). If we neglect the small piston by raising or lowering some weight intermolecular forces in a gas, the internal connected to it) that do not involve such a energy of a gas is just the sum of kinetic energies temperature difference. Reprint 2025-26 230 PHYSICS What are the ways of changing internal 11.5 FIRST LAW OF THERMODYNAMICS energy of a system ? Consider again, for We have seen that the internal energy U of a simplicity, the system to be a certain mass of system can change through two modes of energy gas contained in a cylinder with a movable transfer : heat and work. Let piston as shown in Fig. 11.4. Experience shows ∆Q = Heat supplied to the system by thethere are two ways of changing the state of the surroundingsgas (and hence its internal energy). One way is to put the cylinder in contact with a body at a ∆W = Work done by the system on the higher temperature than that of the gas. The surroundings temperature difference will cause a flow of ∆U = Change in internal energy of the system energy (heat) from the hotter body to the gas, The general principle of conservation of thus increasing the internal energy of the gas. energy then implies that The other way is to push the piston down i.e. to ∆Q = ∆U + ∆W (11.1) do work on the system, which again results in i.e. the energy (∆Q) supplied to the system goesincreasing the internal energy of the gas. Of in partly to increase the internal energy of thecourse, both these things could happen in the system (∆U) and the rest in work on thereverse direction. With surroundings at a lower environment (∆W). Equation (11.1) is known astemperature, heat would flow from the gas to the First Law of Thermodynamics. It is simplythe surroundings. Likewise, the gas could push the general law of conservation of energy applied the piston up and do work on the surroundings. to any system in which the energy transfer from In short, heat and work are two different modes or to the surroundings is taken into account. of altering the state of a thermodynamic system Let us put Eq. (11.1) in the alternative form and changing its internal energy. The notion of heat should be carefully ∆Q – ∆W = ∆U (11.2) distinguished from the notion of internal energy. Heat is certainly energy, but it is the energy in Now, the system may go from an initial state transit. This is not just a play of words. The to the final state in a number of ways. For distinction is of basic significance. The state of example, to change the state of a gas from a thermodynamic system is characterised by its (P1, V1) to (P2, V2), we can first change the volume of the gas from V1 to V2, keeping itsinternal energy, not heat. A statement like ‘a pressure constant i.e. we can first go the stategas in a given state has a certain amount of heat’ is as meaningless as the statement that (P1, V2) and then change the pressure of the gas from P1 to P2, keeping volume constant, to‘a gas in a given state has a certain amount take the gas to (P2, V2). Alternatively, we canof work’. In contrast, ‘a gas in a given state first keep the volume constant and then keep has a certain amount of internal energy’ is a the pressure constant. Since U is a state perfectly meaningful statement. Similarly, the variable, ∆U depends only on the initial and statements ‘a certain amount of heat is final states and not on the path taken by the supplied to the system’ or ‘a certain amount gas to go from one to the other. However, ∆Q of work was done by the system’ are perfectly and ∆W will, in general, depend on the path meaningful. taken to go from the initial to final states. From To summarise, heat and work in the First Law of Thermodynamics, Eq. (11.2), thermodynamics are not state variables. They it is clear that the combination ∆Q – ∆W, is are modes of energy transfer to a system however, path independent. This shows that resulting in change in its internal energy, if a system is taken through a process in which ∆U = 0 (for example, isothermal expansion ofwhich, as already mentioned, is a state variable. an ideal gas, see section 11.8), In ordinary language, we often confuse heat with internal energy. The distinction between ∆Q = ∆W them is sometimes ignored in elementary physics books. For proper understanding of i.e., heat supplied to the system is used up thermodynamics, however, the distinction is entirely by the system in doing work on the crucial. environment. Reprint 2025-26 THERMODYNAMICS 231 If the system is a gas in a cylinder with a If the amount of substance is specified in movable piston, the gas in moving the piston does terms of moles µ (instead of mass m in kg ), we work. Since force is pressure times area, and can define heat capacity per mole of the area times displacement is volume, work done substance by by the system against a constant pressure P is S 1 ∆Q C = = (11.6) ∆W = P ∆V µ µ ∆T C is known as molar specific heat capacity of where ∆V is the change in volume of the gas. the substance. Like s, C is independent of the Thus, for this case, Eq. (11.1) gives amount of substance. C depends on the nature ∆Q = ∆U + P ∆V (11.3) of the substance, its temperature and the conditions under which heat is supplied. The As an application of Eq. (11.3), consider the unit of C is J mo1–1 K–1. As we shall see later (in change in internal energy for 1 g of water when connection with specific heat capacity of gases), we go from its liquid to vapour phase. The additional conditions may be needed to define measured latent heat of water is 2256 J/g. i.e., C or s. The idea in defining C is that simple for 1 g of water ∆Q = 2256 J. At atmospheric predictions can be made in regard to molar pressure, 1 g of water has a volume 1 cm3 in specific heat capacities. liquid phase and 1671 cm3 in vapour phase. Table 11.1 lists measured specific and molar heat capacities of solids at atmospheric pressure Therefore, and ordinary room temperature. ∆W =P (Vg –Vl ) = 1.013 ×105 × (1671 × 10–6) =169.2 J We will see in Chapter 12 that predictions of specific heats of gases generally agree with Equation (11.3) then gives experiment. We can use the same law of equipartition of energy that we use there to ∆U = 2256 – 169.2 = 2086.8 J predict molar specific heat capacities of solids We see that most of the heat goes to increase (See Section 12.5 and 12.6). Consider a solid of the internal energy of water in transition from N atoms, each vibrating about its mean the liquid to the vapour phase. position. An oscillator in one dimension has

11.6 SPECIFIC HEAT CAPACITY average energy of 2 × ½ kBT = kBT. In three dimensions, the average energy is 3 kBT.Suppose an amount of heat ∆Q supplied to a For a mole of a solid, the total energy is substance changes its temperature from T to T + ∆T. We define heat capacity of a substance U = 3 kBT × NA = 3 RT (∵kBT × NA = R) (see Chapter 10) to be Now, at constant pressure, ∆Q = ∆U + P ∆V ≅ ∆ Q ∆U, since for a solid ∆V is negligible. Therefore, S = (11.4) ∆ T ∆ Q ∆U C = = = 3 R (11.7) We expect ∆Q and, therefore, heat capacity S ∆ T ∆ T to be proportional to the mass of the substance. Table 11.1 Specific and molar heat capacities Further, it could also depend on the of some solids at room temperature, i.e., a different amount of heat may temperature and atmospheric be needed for a unit rise in temperature at pressure different temperatures. To define a constant –v Speci"c heat Molar speci"c characteristic of the substance and Substance –1 –1 –1 –1 (J kg K ) heat (J mol K ) independent of its amount, we divide S by the mass of the substance m in kg : S  1  ∆Q s = (11.5) m =  m  ∆T s is known as the specific heat capacity of the substance. It depends on the nature of the As Table 11.1 shows, the experimentally substance and its temperature. The unit of measured values which generally agrees withspecific heat capacity is J kg–1 K–1. Reprint 2025-26 232 PHYSICS predicted value 3R at ordinary temperatures. ideal gas, we have a simple relation. (Carbon is an exception.) The agreement is known to break down at low temperatures. Cp – Cv = R (11.8) Specific heat capacity of water where Cp and Cv are molar specific heat capacities of an ideal gas at constant pressure The old unit of heat was calorie. One calorie and volume respectively and R is the universal was earlier defined to be the amount of heat gas constant. To prove the relation, we begin required to raise the temperature of 1g of water with Eq. (11.3) for 1 mole of the gas : by 1°C. With more precise measurements, it was found that the specific heat of water varies ∆Q = ∆U + P ∆V slightly with temperature. Figure 11.5 shows If ∆Q is absorbed at constant volume, ∆V = 0this variation in the temperature range 0 to 100 °C.  ∆Q   ∆ U   ∆ U  (11.9) C v =  ∆T  v =  ∆T  v =  ∆T  where the subscript v is dropped in the last step, since U of an ideal gas depends only on temperature. (The subscript denotes the quantity kept fixed.) If, on the other hand, ∆Q is absorbed at constant pressure,  ∆ Q   ∆ U   ∆V  (11.10) C p =  ∆T  p =  ∆T  p + P  ∆ T  p The subscript p can be dropped from the Fig. 11.5 Variation of specific heat capacity of first term since U of an ideal gas depends only water with temperature. on T. Now, for a mole of an ideal gas For a precise definition of calorie, it was, PV = RTtherefore, necessary to specify the unit temperature interval. One calorie is defined which gives to be the amount of heat required to raise the temperature of 1g of water from 14.5 °C to  ∆ V  P = R (11.11)15.5 °C. Since heat is just a form of energy,  ∆ T  p it is preferable to use the unit joule, J. In SI units, the specific heat capacity of water Equations (11.9) to (11.11) give the desired is 4186 J kg–1 K–1 i.e. 4.186 J g–1 K–1. The so relation, Eq. (11.8). called mechanical equivalent of heat defined 11.7THERMODYNAMIC STATE VARIABLES as the amount of work needed to produce AND EQUATION OF STATE 1 cal of heat is in fact just a conversion factor between two different units of energy : calorie Every equilibrium state of a thermodynamic to joule. Since in SI units, we use the unit joule system is completely described by specific for heat, work or any other form of energy, the values of some macroscopic variables, also term mechanical equivalent is now called state variables. For example, an superfluous and need not be used. equilibrium state of a gas is completely As already remarked, the specific heat specified by the values of pressure, volume, capacity depends on the process or the temperature, and mass (and composition if conditions under which heat capacity transfer there is a mixture of gases). A thermodynamic takes place. For gases, for example, we can system is not always in equilibrium. For example, define two specific heats : specific heat a gas allowed to expand freely against vacuum capacity at constant volume and specific is not an equilibrium state [Fig. 11.6(a)]. During heat capacity at constant pressure. For an the rapid expansion, pressure of the gas may Reprint 2025-26 THERMODYNAMICS 233 not be uniform throughout. Similarly, a mixture temperature do not. To decide which variable is of gases undergoing an explosive chemical extensive and which intensive, think of a reaction (e.g. a mixture of petrol vapour and relevant system in equilibrium, and imagine that air when ignited by a spark) is not an it is divided into two equal parts. The variables that remain unchanged for each part areequilibrium state; again its temperature and intensive. The variables whose values get halvedpressure are not uniform [Fig. 11.6(b)]. in each part are extensive. It is easily seen, forEventually, the gas attains a uniform example, that internal energy U, volume V, total temperature and pressure and comes to mass M are extensive variables. Pressure P, thermal and mechanical equilibrium with its temperature T, and density ρ are intensive surroundings. variables. It is a good practice to check the consistency of thermodynamic equations using this classification of variables. For example, in the equation ∆Q = ∆U + P ∆V quantities on both sides are extensive*. (The product of an intensive variable like P and an extensive quantity ∆V is extensive.) 11.8 THERMODYNAMIC PROCESSES 11.8.1 Quasi-static process Consider a gas in thermal and mechanical equilibrium with its surroundings. The pressure of the gas in that case equals the external Fig. 11.6 (a) The partition in the box is suddenly pressure and its temperature is the same as removed leading to free expansion of the that of its surroundings. Suppose that the gas. (b) A mixture of gases undergoing an external pressure is suddenly reduced (say by explosive chemical reaction. In both lifting the weight on the movable piston in the situations, the gas is not in equilibrium and container). The piston will accelerate outward. cannot be described by state variables. During the process, the gas passes through In short, thermodynamic state variables states that are not equilibrium states. The non- describe equilibrium states of systems. The equilibrium states do not have well-defined various state variables are not necessarily pressure and temperature. In the same way, if independent. The connection between the state a finite temperature difference exists between variables is called the equation of state. For the gas and its surroundings, there will be a example, for an ideal gas, the equation of state rapid exchange of heat during which the gas is the ideal gas relation will pass through non-equilibrium states. In due course, the gas will settle to an equilibrium P V = µ R T state with well-defined temperature and For a fixed amount of the gas i.e. given µ, there pressure equal to those of the surroundings. The are thus, only two independent variables, say P free expansion of a gas in vacuum and a mixture and V or T and V. The pressure-volume curve of gases undergoing an explosive chemical for a fixed temperature is called an isotherm. reaction, mentioned in section 11.7 are also Real gases may have more complicated examples where the system goes through non- equations of state. equilibrium states. The thermodynamic state variables are of two Non-equilibrium states of a system are difficult kinds: extensive and intensive. Extensive to deal with. It is, therefore, convenient to variables indicate the ‘size’ of the system. imagine an idealised process in which at every Intensive variables such as pressure and stage the system is an equilibrium state. Such a * As emphasised earlier, Q is not a state variable. However, ∆Q is clearly proportional to the total mass of system and hence is extensive. Reprint 2025-26 234 PHYSICS process is, in principle, infinitely slow, hence the A process in which the temperature of the name quasi-static (meaning nearly static). The system is kept fixed throughout is called an system changes its variables (P, T, V ) so slowly isothermal process. The expansion of a gas in that it remains in thermal and mechanical a metallic cylinder placed in a large reservoir of equilibrium with its surroundings throughout. fixed temperature is an example of an isothermal In a quasi-static process, at every stage, the process. (Heat transferred from the reservoir to difference in the pressure of the system and the the system does not materially affect the external pressure is infinitesimally small. The temperature of the reservoir, because of its very large heat capacity.) In isobaric processes thesame is true of the temperature difference pressure is constant while in isochoricbetween the system and its surroundings processes the volume is constant. Finally, if the(Fig.11.7). To take a gas from the state (P, T ) to another state (P ′, T ′ ) via a quasi-static process, system is insulated from the surroundings and no heat flows between the system and thewe change the external pressure by a very small surroundings, the process is adiabatic. The amount, allow the system to equalise its pressure definitions of these special processes are with that of the surroundings and continue the summarised in Table. 11.2 process infinitely slowly until the system achieves the pressure P ′. Similarly, to change Table 11.2 Some special thermodynamic the temperature, we introduce an infinitesimal processes temperature difference between the system and the surrounding reservoirs and by choosing reservoirs of progressively different temperatures T to T ′, the system achieves the temperature T ′. We now consider these processes in some detail : 11.8.2 Isothermal process For an isothermal process (T fixed), the ideal gas equation gives PV = constant i.e., pressure of a given mass of gas varies inversely as its volume. This is nothing but Boyle’s Law. Suppose an ideal gas goes isothermally (at temperature T ) from its initial state (P1, V1) to Fig. 11.7 In a quasi-static process, the temperature the final state (P2, V 2). At any intermediate stage of the surrounding reservoir and the with pressure P and volume change from V to external pressure differ only infinitesimally V + ∆V (∆V small) from the temperature and pressure of the system. ∆W = P ∆ V A quasi-static process is obviously a Taking (∆V → 0) and summing the quantity hypothetical construct. In practice, processes ∆W over the entire process, that are sufficiently slow and do not involve V2 accelerated motion of the piston, large W = ∫ P d V temperature gradient, etc., are reasonably V1 approximation to an ideal quasi-static process. V2 d V V2We shall from now on deal with quasi-static = µ RT = µRT In ∫ (11.12) Vprocesses only, except when stated otherwise. V1 V 1 Reprint 2025-26 THERMODYNAMICS 235 where in the second step we have made use of We can calculate, as before, the work done in the ideal gas equation PV = µ RT and taken the an adiabatic change of an ideal gas from the constants out of the integral. For an ideal gas, state (P1, V1, T1) to the state (P2, V2, T2). internal energy depends only on temperature. V 2Thus, there is no change in the internal energy W = ∫ P d Vof an ideal gas in an isothermal process. The V1First Law of Thermodynamics then implies that heat supplied to the gas equals the work done by the gas : Q = W. Note from Eq. (11.12) that for V2 > V1, W > 0; and for V2 < V1, W < 0. That is, in an isothermal expansion, the gas absorbs heat and does work while in an isothermal (11.15) compression, work is done on the gas by the environment and heat is released. From Eq. (11.14), the constant is P1V1γ or P2V2γ 11.8.3 Adiabatic process γ  γ  P2 V 2 P1V 1In an adiabatic process, the system is insulated W = 1 − 1   −1 γ γ − from the surroundings and heat absorbed or 1 − γ  V2 V1  released is zero. From Eq. (11.1), we see that work done by the gas results in decrease in its 1 µR(T1 − T2 ) = [ P2 V2 − P1V1 ] = (11.16)internal energy (and hence its temperature for 1 − γ γ − 1 an ideal gas). We quote without proof (the result that you will learn in higher courses) that for As expected, if work is done by the gas in an an adiabatic process of an ideal gas. adiabatic process (W > 0), from Eq. (11.16), P V γ = const (11.13) T2 < T1. On the other hand, if work is done on the gas (W < 0), we get T2 > T1 i.e., thewhere γ is the ratio of specific heats (ordinary temperature of the gas rises. or molar) at constant pressure and at constant volume. 11.8.4 Isochoric process Cp In an isochoric process, V is constant. No work γ = Cv is done on or by the gas. From Eq. (11.1), the heat absorbed by the gas goes entirely to change Thus if an ideal gas undergoes a change in its internal energy and its temperature. The its state adiabatically from (P1, V1) to (P2, V2) : change in temperature for a given amount of γ = P2 V2γ (11.14) heat is determined by the specific heat of the P1 V1 gas at constant volume. Figure11.8 shows the P-V curves of an ideal gas for two adiabatic processes connecting two 11.8.5 Isobaric process isotherms. In an isobaric process, P is fixed. Work done by the gas is W = P (V2 – V1) = µ R (T2 – T1) (11.17) Since temperature changes, so does internal energy. The heat absorbed goes partly to increase internal energy and partly to do work. The change in temperature for a given amount of heat is determined by the specific heat of the gas at constant pressure. 11.8.6 Cyclic process In a cyclic process, the system returns to its initial state. Since internal energy is a stateFig. 11.8 P-V curves for isothermal and adiabatic variable, ∆U = 0 for a cyclic process. From processes of an ideal gas. Reprint 2025-26 236 PHYSICS Eq. (11.1), the total heat absorbed equals the 11.10 REVERSIBLE AND IRREVERSIBLE work done by the system. PROCESSES Imagine some process in which a thermodynamic11.9 SECOND LAW OF THERMODYNAMICS system goes from an initial state i to a final state The First Law of Thermodynamics is the f. During the process the system absorbs heat Q principle of conservation of energy. Common from the surroundings and performs work W on experience shows that there are many it. Can we reverse this process and bring both conceivable processes that are perfectly the system and surroundings to their initial allowed by the First Law and yet are never states with no other effect anywhere ? Experience observed. For example, nobody has ever seen suggests that for most processes in nature this a book lying on a table jumping to a height by is not possible. The spontaneous processes ofitself. But such a thing would be possible if nature are irreversible. Several examples can bethe principle of conservation of energy were cited. The base of a vessel on an oven is hotterthe only restriction. The table could cool than its other parts. When the vessel is removed,spontaneously, converting some of its internal heat is transferred from the base to the otherenergy into an equal amount of mechanical parts, bringing the vessel to a uniformenergy of the book, which would then hop to a height with potential energy equal to the temperature (which in due course cools to the mechanical energy it acquired. But this never temperature of the surroundings). The process happens. Clearly, some additional basic cannot be reversed; a part of the vessel will not principle of nature forbids the above, even get cooler spontaneously and warm up the base. though it satisfies the energy conservation It will violate the Second Law of Thermodynamics, principle. This principle, which disallows if it did. The free expansion of a gas is irreversible. many phenomena consistent with the First The combustion reaction of a mixture of petrol Law of Thermodynamics is known as the and air ignited by a spark cannot be reversed. Second Law of Thermodynamics. Cooking gas leaking from a gas cylinder in the The Second Law of Thermodynamics gives kitchen diffuses to the entire room. The diffusion a fundamental limitation to the efficiency of a process will not spontaneously reverse and bring heat engine and the co-efficient of the gas back to the cylinder. The stirring of a performance of a refrigerator. In simple terms, liquid in thermal contact with a reservoir will it says that efficiency of a heat engine can convert the work done into heat, increasing the never be unity. For a refrigerator, the Second internal energy of the reservoir. The process Law says that the co-efficient of performance cannot be reversed exactly; otherwise it would can never be infinite. The following two amount to conversion of heat entirely into work, statements, one due to Kelvin and Planck violating the Second Law of Thermodynamics. denying the possibility of a perfect heat engine, Irreversibility is a rule rather an exception and another due to Clausius denying the in nature. possibility of a perfect refrigerator or heat Irreversibility arises mainly from two causes: pump, are a concise summary of these one, many processes (like a free expansion, or observations. an explosive chemical reaction) take the system Kelvin-Planck statement to non-equilibrium states; two, most processes No process is possible whose sole result is the involve friction, viscosity and other dissipative absorption of heat from a reservoir and the effects (e.g., a moving body coming to a stop and complete conversion of the heat into work. losing its mechanical energy as heat to the floor and the body; a rotating blade in a liquid coming Clausius statement to a stop due to viscosity and losing its No process is possible whose sole result is the mechanical energy with corresponding gain in transfer of heat from a colder object to a the internal energy of the liquid). Since hotter object. dissipative effects are present everywhere and It can be proved that the two statements can be minimised but not fully eliminated, most above are completely equivalent. processes that we deal with are irreversible. Reprint 2025-26 THERMODYNAMICS 237 A thermodynamic process (state i → state f ) in a reversible heat engine operating between is reversible if the process can be turned back two temperatures, heat should be absorbed such that both the system and the surroundings (from the hot reservoir) isothermally and return to their original states, with no other released (to the cold reservoir) isothermally. We change anywhere else in the universe. From the thus have identified two steps of the reversible preceding discussion, a reversible process is an heat engine : isothermal process at temperature idealised notion. A process is reversible only if T1 absorbing heat Q1 from the hot reservoir, and another isothermal process at temperature T2it is quasi-static (system in equilibrium with the releasing heat Q2 to the cold reservoir. Tosurroundings at every stage) and there are no complete a cycle, we need to take the systemdissipative effects. For example, a quasi-static from temperature T1 to T2 and then back fromisothermal expansion of an ideal gas in a temperature T2 to T1. Which processes shouldcylinder fitted with a frictionless movable piston we employ for this purpose that are reversible?is a reversible process. A little reflection shows that we can only adopt Why is reversibility such a basic concept in reversible adiabatic processes for these thermodynamics ? As we have seen, one of the purposes, which involve no heat flow from any concerns of thermodynamics is the efficiency reservoir. If we employ any other process that is with which heat can be converted into work. not adiabatic, say an isochoric process, to take The Second Law of Thermodynamics rules out the system from one temperature to another, we the possibility of a perfect heat engine with 100% shall need a series of reservoirs in the efficiency. But what is the highest efficiency temperature range T2 to T1 to ensure that at each possible for a heat engine working between two stage the process is quasi-static. (Remember reservoirs at temperatures T1 and T2? It turns again that for a process to be quasi-static and out that a heat engine based on idealised reversible, there should be no finite temperature reversible processes achieves the highest difference between the system and the reservoir.) efficiency possible. All other engines involving But we are considering a reversible engine that irreversibility in any way (as would be the case operates between only two temperatures. Thus for practical engines) have lower than this adiabatic processes must bring about the limiting efficiency. temperature change in the system from T1 to T2 and T2 to T1 in this engine.11.11 CARNOT ENGINE Suppose we have a hot reservoir at temperature T1 and a cold reservoir at temperature T2. What is the maximum efficiency possible for a heat engine operating between the two reservoirs and what cycle of processes should be adopted to achieve the maximum efficiency ? Sadi Carnot, a French engineer, first considered this question in 1824. Interestingly, Carnot arrived at the correct answer, even though the basic concepts of heat and thermodynamics had yet to be firmly established. We expect the ideal engine operating between two temperatures to be a reversible engine. Irreversibility is associated with dissipative Fig. 11.9 Carnot cycle for a heat engine with an effects, as remarked in the preceding section, ideal gas as the working substance. and lowers efficiency. A process is reversible if it is quasi-static and non-dissipative. We have A reversible heat engine operating between seen that a process is not quasi-static if it two temperatures is called a Carnot engine. We involves finite temperature difference between have just argued that such an engine must have the system and the reservoir. This implies that the following sequence of steps constituting one Reprint 2025-26 238 PHYSICS cycle, called the Carnot cycle, shown in Fig. 11.9. We have taken the working substance In  V 3  of the Carnot engine to be an ideal gas. T2   V4  = 1 −  T1   V2  (11.23)(a) Step 1 → 2 Isothermal expansion of the gas In  V1  taking its state from (P1, V1, T1) to (P2, V2, T1). Now since step 2 → 3 is an adiabatic process, The heat absorbed by the gas (Q1) from the γ −1 γ −1 reservoir at temperature T1 is given by T1 V 2 = T2 V3 Eq. (11.12). This is also the work done (W1 → 2) by the gas on the environment. 1 /( γ −1 ) V2  T2   V2  i.e. = (11.24) 3  T1  W1 → 2 = Q1 = µ R T1 ln  V1  (11.18) V (b) Step 2 → 3 Adiabatic expansion of the gas Similarly, since step 4 → 1 is an adiabatic from (P2, V2, T1) to (P3, V3, T2) process Work done by the gas, using γ −1 γ −1 Eq. (11.16), is T2 V 4 = T1 V1 µR ( T1 − T2 ) 1 /γ −1 W2 →=3 (11.19) V1  T2  γ − 1 i.e. = (11.25) V4  T1  (c) Step 3 → 4 Isothermal compression of the gas from (P3, V3, T2) to (P4, V4, T2). From Eqs. (11.24) and (11.25), 2 Heat released (Q2) by the gas to the reservoir V3 = V (11.26) at temperature T2 is given by Eq. (11.12). This V4 V1 is also the work done (W3 → 4) on the gas by the environment. Using Eq. (11.26) in Eq. (11.23), we get  V 3  2 W 3 → 4 = Q2 = µRT2 ln η = 1 −T (Carnot engine) (11.27)  V 4  (11.20) T1 (d) Step 4 → 1 Adiabatic compression of the gas from (P4, V4, T2) to (P1,V1, T1). We have already seen that a Carnot engine is a reversible engine. Indeed it is the only Work done on the gas, [using Eq.(11.16), is reversible engine possible that works between two reservoirs at different temperatures. Each  T1 − T2  W4 → 1 = µR (11.21) step of the Carnot cycle given in Fig. 11.9 can  γ -1  be reversed. This will amount to taking heat Q2 from the cold reservoir at T2, doing work W on From Eqs. (11.18) to (11.21) total work done by the gas in one complete cycle is the system, and transferring heat Q1 to the hot reservoir. This will be a reversible refrigerator. W = W1 → 2 + W2 → 3 – W3 → 4 – W4 → 1 We next establish the important result  V2   V3  (sometimes called Carnot’s theorem) that = µ RT1 ln  V1  – µ RT2 ln  V4  (11.22) (a) working between two given temperatures T1 and T2 of the hot and cold reservoirs respectively, The efficiency η of the Carnot engine is no engine can have efficiency more than that of the Carnot engine and (b) the efficiency of the W Q 2 η = = 1 − Carnot engine is independent of the nature of Q1 Q1 the working substance. Reprint 2025-26 THERMODYNAMICS 239 To prove the result (a), imagine a reversible reservoir and delivers the same amount of work (Carnot) engine R and an irreversible engine I in one cycle, without any change in the source working between the same source (hot reservoir) or anywhere else. This is clearly against the and sink (cold reservoir). Let us couple the Kelvin-Planck statement of the Second Law of engines, I and R, in such a way so that I acts Thermodynamics. Hence the assertion ηI > ηR like a heat engine and R acts as a refrigerator. is wrong. No engine can have efficiency greater Let I absorb heat Q1 from the source, deliver than that of the Carnot engine. A similar work W ′ and release the heat Q1- W′ to the sink. argument can be constructed to show that a We arrange so that R returns the same heat Q1 reversible engine with one particular substance to the source, taking heat Q2 from the sink and cannot be more efficient than the one using requiring work W = Q1 – Q2 to be done on it. another substance. The maximum efficiency of Now suppose ηR < ηI i.e. if R were to act a Carnot engine given by Eq. (11.27) is as an engine it would give less work output independent of the nature of the system performing the Carnot cycle of operations. Thus we are justified in using an ideal gas as a system in the calculation of efficiency η of a Carnot engine. The ideal gas has a simple equation of I state, which allows us to readily calculate η, but the final result for η, [Eq. (11.27)], is true for any Carnot engine. R This final remark shows that in a Carnot cycle, Q1 T1 = W (11.28) T2 Q 2 is a universal relation independent of the natureFig. 11.10 An irreversible engine (I) coupled to a reversible refrigerator (R). If W ′ > W, this of the system. Here Q1 and Q2 are respectively, would amount to extraction of heat the heat absorbed and released isothermally W′ – W from the sink and its full (from the hot and to the cold reservoirs) in a conversion to work, in contradiction with Carnot engine. Equation (11.28), can, therefore, the Second Law of Thermodynamics. be used as a relation to define a truly universal thermodynamic temperature scale that is than that of I i.e. W < W ′ for a given Q1. With R independent of any particular properties of the acting like a refrigerator, this would mean system used in the Carnot cycle. Of course, for Q2 = Q1 – W > Q1 – W ′. Thus, on the whole, an ideal gas as a working substance, this the coupled I-R system extracts heat universal temperature is the same as the ideal (Q1 – W) – (Q1 – W ′) = (W ′ – W ) from the cold gas temperature introduced in section 11.9. SUMMARY 1. The zeroth law of thermodynamics states that ‘two systems in thermal equilibrium with a third system separately are in thermal equilibrium with each other’. The Zeroth Law leads to the concept of temperature. 2. Internal energy of a system is the sum of kinetic energies and potential energies of the molecular constituents of the system. It does not include the over-all kinetic energy of the system. Heat and work are two modes of energy transfer to the system. Heat is the energy transfer arising due to temperature difference between the system and the surroundings. Work is energy transfer brought about by other means, such as moving the piston of a cylinder containing the gas, by raising or lowering some weight connected to it. Reprint 2025-26 240 PHYSICS 3. The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) is taken into account. It states that ∆Q = ∆U + ∆W where ∆Q is the heat supplied to the system, ∆W is the work done by the system and ∆U is the change in internal energy of the system. 4. The specific heat capacity of a substance is defined by 1 ∆Q s = m ∆T where m is the mass of the substance and ∆Q is the heat required to change its temperature by ∆T. The molar specific heat capacity of a substance is defined by 1 ∆ Q C = µ ∆T where µ is the number of moles of the substance. For a solid, the law of equipartition of energy gives C = 3 R which generally agrees with experiment at ordinary temperatures. Calorie is the old unit of heat. 1 calorie is the amount of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C. 1 cal = 4.186 J. 5. For an ideal gas, the molar specific heat capacities at constant pressure and volume satisfy the relation Cp – Cv = R where R is the universal gas constant. 6. Equilibrium states of a thermodynamic system are described by state variables. The value of a state variable depends only on the particular state, not on the path used to arrive at that state. Examples of state variables are pressure (P ), volume (V ), temperature (T ), and mass (m ). Heat and work are not state variables. An Equation of State (like the ideal gas equation PV = µ RT ) is a relation connecting different state variables. 7. A quasi-static process is an infinitely slow process such that the system remains in thermal and mechanical equilibrium with the surroundings throughout. In a quasi-static process, the pressure and temperature of the environment can differ from those of the system only infinitesimally. 8. In an isothermal expansion of an ideal gas from volume V1 to V2 at temperature T the heat absorbed (Q) equals the work done (W ) by the gas, each given by  V 2   Q = W = µ R T ln   V1  9. In an adiabatic process of an ideal gas γ PV = constant C p where γ = C v Work done by an ideal gas in an adiabatic change of state from (P1, V1, T1) to (P2, V2, T2) is µR ( T1 − T2 ) W = γ – 1 Reprint 2025-26 THERMODYNAMICS 241 10. The second law of thermodynamics disallows some processes consistent with the First Law of Thermodynamics. It states Kelvin-Planck statement No process is possible whose sole result is the absorption of heat from a reservoir and complete conversion of the heat into work. Clausius statement No process is possible whose sole result is the transfer of heat from a colder object to a hotter object. Put simply, the Second Law implies that no heat engine can have efficiency η equal to 1 or no refrigerator can have co-efficient of performance α equal to infinity. 11. A process is reversible if it can be reversed such that both the system and the surroundings return to their original states, with no other change anywhere else in the universe. Spontaneous processes of nature are irreversible. The idealised reversible process is a quasi-static process with no dissipative factors such as friction, viscosity, etc. 12. Carnot engine is a reversible engine operating between two temperatures T1 (source) and T2 (sink). The Carnot cycle consists of two isothermal processes connected by two adiabatic processes. The efficiency of a Carnot engine is given by T 2 η= 1 − (Carnot engine) T1 No engine operating between two temperatures can have efficiency greater than that of the Carnot engine. 13. If Q > 0, heat is added to the system If Q < 0, heat is removed to the system If W > 0, Work is done by the system If W < 0, Work is done on the system Quantity Symbol Dimensions Unit Remark Co-efficienty of volume αv [K–1] K–1 αv = 3 α1 expansion Heat supplied to a system ∆Q [ML2 T–2] J Q is not a state variable Specific heat capacity s [L2 T–2 K–1] J kg–1 K–1 dt Thermal Conductivity K [MLT–3 K–1] J s–1 K–1 H = – KA dx POINTS TO PONDER 1. Temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass. A bullet fired from a gun is not at a higher temperature because of its high speed. 2. Equilibrium in thermodynamics refers to the situation when macroscopic variables describing the thermodynamic state of a system do not depend on time. Equilibrium of a system in mechanics means the net external force and torque on the system are zero. Reprint 2025-26 242 PHYSICS 3. In a state of thermodynamic equilibrium, the microscopic constituents of a system are not in equilibrium (in the sense of mechanics). 4. Heat capacity, in general, depends on the process the system goes through when heat is supplied. 5. In isothermal quasi-static processes, heat is absorbed or given out by the system even though at every stage the gas has the same temperature as that of the surrounding reservoir. This is possible because of the infinitesimal difference in temperature between the system and the reservoir. EXERCISES 11.1 A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g ? 11.2 What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure ? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.) 11.3 Explain why (a) Two bodies at different temperatures T1 and T2 if brought in thermal contact do not necessarily settle to the mean temperature (T1 + T2 )/2. (b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat. (c) Air pressure in a car tyre increases during driving. (d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude. 11.4 A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume ? 11.5 In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case ? (Take 1 cal = 4.19 J) 11.6 Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (a) What is the final pressure of the gas in A and B ? (b) What is the change in internal energy of the gas ? (c) What is the change in the temperature of the gas ? (d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface ? 11.7 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing? Reprint 2025-26 THERMODYNAMICS 243 11.8 A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (11.13) Fig. 11.11 Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F Reprint 2025-26 CHAPTER TWELVE KINETIC THEORY 12.1 INTRODUCTION Boyle discovered the law named after him in 1661. Boyle, Newton and several others tried to explain the behaviour of 12.1 Introduction gases by considering that gases are made up of tiny atomic 12.2 Molecular nature of matter particles. The actual atomic theory got established more than 12.3 Behaviour of gases 150 years later. Kinetic theory explains the behaviour of gases 12.4 Kinetic theory of an ideal gas based on the idea that the gas consists of rapidly moving 12.5 Law of equipartition of energy atoms or molecules. This is possible as the inter-atomic forces, 12.6 Specific heat capacity which are short range forces that are important for solids and liquids, can be neglected for gases. The kinetic theory12.7 Mean free path was developed in the nineteenth century by Maxwell, Summary Boltzmann and others. It has been remarkably successful. It Points to ponder gives a molecular interpretation of pressure and temperature Exercises of a gas, and is consistent with gas laws and Avogadro’s hypothesis. It correctly explains specific heat capacities of many gases. It also relates measurable properties of gases such as viscosity, conduction and diffusion with molecular parameters, yielding estimates of molecular sizes and masses. This chapter gives an introduction to kinetic theory. 12.2 MOLECULAR NATURE OF MATTER Richard Feynman, one of the great physicists of 20th century considers the discovery that “Matter is made up of atoms” to be a very significant one. Humanity may suffer annihilation (due to nuclear catastrophe) or extinction (due to environmental disasters) if we do not act wisely. If that happens, and all of scientific knowledge were to be destroyed then Feynman would like the ‘Atomic Hypothesis’ to be communicated to the next generation of creatures in the universe. Atomic Hypothesis: All things are made of atoms - little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another. Speculation that matter may not be continuous, existed in many places and cultures. Kanada in India and Democritus Reprint 2025-26 KINETIC THEORY 245 Atomic Hypothesis in Ancient India and Greece Though John Dalton is credited with the introduction of atomic viewpoint in modern science, scholars in ancient India and Greece conjectured long before the existence of atoms and molecules. In the Vaiseshika school of thought in India founded by Kanada (Sixth century B.C.) the atomic picture was developed in considerable detail. Atoms were thought to be eternal, indivisible, infinitesimal and ultimate parts of matter. It was argued that if matter could be subdivided without an end, there would be no difference between a mustard seed and the Meru mountain. The four kinds of atoms (Paramanu — Sanskrit word for the smallest particle) postulated were Bhoomi (Earth), Ap (water), Tejas (fire) and Vayu (air) that have characteristic mass and other attributes, were propounded. Akasa (space) was thought to have no atomic structure and was continuous and inert. Atoms combine to form different molecules (e.g. two atoms combine to form a diatomic molecule dvyanuka, three atoms form a tryanuka or a triatomic molecule), their properties depending upon the nature and ratio of the constituent atoms. The size of the atoms was also estimated, by conjecture or by methods that are not known to us. The estimates vary. In Lalitavistara, a famous biography of the Buddha written mainly in the second century B.C., the estimate is close to the modern estimate of atomic size, of the order of 10 –10 m. In ancient Greece, Democritus (Fourth century B.C.) is best known for his atomic hypothesis. The word ‘atom’ means ‘indivisible’ in Greek. According to him, atoms differ from each other physically, in shape, size and other properties and this resulted in the different properties of the substances formed by their combination. The atoms of water were smooth and round and unable to ‘hook’ on to each other, which is why liquid /water flows easily. The atoms of earth were rough and jagged, so they held together to form hard substances. The atoms of fire were thorny which is why it caused painful burns. These fascinating ideas, despite their ingenuity, could not evolve much further, perhaps because they were intuitive conjectures and speculations not tested and modified by quantitative experiments - the hallmark of modern science. in Greece had suggested that matter may consist of matter. The theory is now well accepted by of indivisible constituents. The scientific ‘Atomic scientists. However even at the end of the Theory’ is usually credited to John Dalton. He nineteenth century there were famous scientists proposed the atomic theory to explain the laws who did not believe in atomic theory ! of definite and multiple proportions obeyed by From many observations, in recent times we elements when they combine into compounds. now know that molecules (made up of one or The first law says that any given compound has, more atoms) constitute matter. Electron a fixed proportion by mass of its constituents. microscopes and scanning tunnelling The second law says that when two elements microscopes enable us to even see them. The form more than one compound, for a fixed mass size of an atom is about an angstrom (10 -10 m). of one element, the masses of the other elements In solids, which are tightly packed, atoms are are in ratio of small integers. spaced about a few angstroms (2 Å) apart. In To explain the laws Dalton suggested, about liquids the separation between atoms is also 200 years ago, that the smallest constituents about the same. In liquids the atoms are not of an element are atoms. Atoms of one element as rigidly fixed as in solids, and can move are identical but differ from those of other around. This enables a liquid to flow. In gases elements. A small number of atoms of each the interatomic distances are in tens of element combine to form a molecule of the angstroms. The average distance a molecule compound. Gay Lussac’s law, also given in early can travel without colliding is called the mean 19th century, states: When gases combine free path. The mean free path, in gases, is of chemically to yield another gas, their volumes the order of thousands of angstroms. The atoms are in the ratios of small integers. Avogadro’s are much freer in gases and can travel long law (or hypothesis) says: Equal volumes of all distances without colliding. If they are not gases at equal temperature and pressure have enclosed, gases disperse away. In solids and the same number of molecules. Avogadro’s law, liquids the closeness makes the interatomic force when combined with Dalton’s theory explains important. The force has a long range attraction Gay Lussac’s law. Since the elements are often and a short range repulsion. The atoms attract in the form of molecules, Dalton’s atomic theory when they are at a few angstroms but repel when can also be referred to as the molecular theory they come closer. The static appearance of a gas Reprint 2025-26 246 PHYSICS is misleading. The gas is full of activity and the is 6.02 × 1023. This is known as Avogadro number equilibrium is a dynamic one. In dynamic and is denoted by NA. The mass of 22.4 litres of equilibrium, molecules collide and change their any gas is equal to its molecular weight in grams speeds during the collision. Only the average at S.T.P (standard temperature 273 K and properties are constant. pressure 1 atm). This amount of substance is Atomic theory is not the end of our quest, but called a mole (see Chapter 1 for a more precise the beginning. We now know that atoms are not definition). Avogadro had guessed the equality of indivisible or elementary. They consist of a numbers in equal volumes of gas at a fixed nucleus and electrons. The nucleus itself is made temperature and pressure from chemical up of protons and neutrons. The protons and reactions. Kinetic theory justifies this hypothesis. neutrons are again made up of quarks. Even The perfect gas equation can be written as quarks may not be the end of the story. There PV = µ RT (12.3)may be string like elementary entities. Nature always has surprises for us, but the search for where µ is the number of moles and R = NA truth is often enjoyable and the discoveries kB is a universal constant. The temperature T is absolute temperature. Choosing kelvin scale forbeautiful. In this chapter, we shall limit ourselves absolute temperature, R = 8.314 J mol–1K–1.to understanding the behaviour of gases (and a Herelittle bit of solids), as a collection of moving molecules in incessant motion. M N µ = = (12.4) M 0 N A