2. Kinetic theory of gases The macroscopic behavior of an ideal gas can be explained by the kinetic theory of gases which involves the application of the laws of classical mechanics to the microscopic constituents of the gas (i.e. the molecules) statistically. This theory shows that the macroscopic relations linking the various properties of gases are consistent with the those relations derived microscopically.

3. An ideal gas – microscopic definition From the microscopic point of view, an ideal gas is defined with the following assumptions: 1.A gas consists of a large number of identical molecules and the average separation between molecules is great compared with their dimensions. 2.The molecules obey Newton’s laws of motion and they move randomly. 3.Collisions between molecules and between the molecules and the walls of the container are elastic and are of negligible duration. 4.No appreciable forces act on the molecules except during collisions.

4. Kinetic calculation of pressure Consider an ideal gas consisting of N molecules in a container of volume V, and the container is a cube with edges of length d. The mass of each molecule is m. When a molecule of velocity v (v x, v y, v z ) collides with the face of the container parallel to the y-z plane, v x is reversed while v y and v z remain unaltered. The change in momentum of the molecule is  p x = -2mv x

5. The subscript i indicates that this force on the wall is exerted by one molecule. For the molecule to collide with the same wall again, it must travel a distance 2d in the x direction. Assuming no collisions in between, this round trip takes a time 2d/ v x. The force exerted by the molecule to this wall is equal to the rate at which it transfers momentum to the wall: F i = 2mv x 2d/vx2d/vx = mv x 2 d m F = d (v x1 2 + v x2 2 + · · · · + v xN 2 ) The total force exerted by all N molecules on the wall is:

6. The average value of the square of the velocity in the x direction for N molecules is: v x 2 = (v x1 2 + v x2 2 + · · · · + v xN 2 ) N Thus the total force on the wall can be written as: Nm F = d vx2vx2 Since v 2 = v x 2 + v y 2 + v z 2 and because the motion is completely random v 2 = v x 2 + v y 2 + v z 2 = 3 v x 2 The total force can also be written as:

7. The total pressure exerted on the wall is simply: P = F A = F d 2 Hence This result indicates that the pressure is proportional to the number of molecules per unit volume and to the average translational kinetic energy of the molecules

8. The equation for pressure just derived can be written as: Comparing this equation with the one we have introduced earlier based on the experimental facts: PV = N k B T We can relate the temperature to the kinetic energy of the molecules:

9. Because it follows that In the similar manner, it follows that the motions in the y and z direction give: and The generalization of this result is known as the theorem of equipartition of energy, meaning that each degree of freedom contributes (1/2)k B T to the energy of the system.

10. The total translational kinetic energy of N molecules of gas is simply N times the average energy per molecule, which is given by: E trans depends ONLY on the temperature, NOT on the pressure, volume, or molecular species.

11. The square root of v 2 is called the root-mean-square (rms) speed of the molecules, which can be written as: where M = m N A is the molar mass in kilograms per mole. This equation reveals that, at a given temperature, lighter molecules move faster, on the average, than do heavier molecules.

12. Example : What is the average kinetic energy of a molecule of a gas at a temperature of 25 o C? The answer can be obtained using the equation: k B = 1.38 x 10 -23 J·K -1 T = 25 + 273.15 K = 298.15 K Average kinetic energy of a gas molecule = 1.5 x 1.38 x 298.15 x 10 -23 J = 6.17 x 10 -21 J

13. Example : What is the total kinetic energy of a the molecules in 1 mole of a gas at a temperature of 25 o C? R = 8.135 J·mol -1 ·K -1 T = 25 + 273.15 K = 298.15 K Total kinetic energy of the molecules in one mode of a gas = 1.5 x 8.135 x 298.15 J = 3638 J The answer can be obtained using the equation:

14. Molar specific heat of an ideal gas The heat energy required to raise the temperature of n moles of gas from from T to T +  T is path dependent. This can be easily seen from the 1 st law of thermodynamics: Q =  E int + W As shown earlier,  E int is path independent but W is path dependent. Hence, Q is path dependent. Thus, a gas does not have a unique molar specific heat.

15. Constant volume/pressure molar specific heats While there is no unique value of heat associated with a given change of temperature in general, there is one for each specific process. Two common processes the frequently occurred are: the isochoric process and the isobaric process. If we define: C V = molar specific heat at constant volume C P = molar specific heat at constant pressure then the heat required to raise the temperature of a gas by  T by these two processes are: Q V = n C V  T Q P = n C P  T

16. Value of C V for an ideal gas The heat energy transferred to a system at constant volume is simply: Q = n C V  T =  E int since there is no work done. Hence  E int = nR  T 3 2 and C V = = R = 12.47 J·mol -1 ·K -1 3 2  E int n  T E int = E trans = nRT 3 2 Let us assume that the internal energy is equal to the total translational energy of the molecules, then

17. Value of C P for an ideal gas The heat energy transferred to a system at constant pressure is: Q = n C P  T =  E int + W Since  E int = nR  T, 3 2 and W = P  V = nR  T Hence C P = R = 20.79 J·mol -1 ·K -1 5 2 Also C P - C V = R = 8.315 J·mol -1 ·K -1  = C P / C V = 1.667 &

18. Adiabatic process for an ideal gas For the adiabatic process ( Q = 0 ): dE int = n C V dT = -W = - P dV The total differential of the equation of state of an ideal gas PV = nRT: P dV + V dP = n R dT The above 2 equations give: P dV + V dP = - (RP/ C V ) dV Substituting R = C P – C V, and diving by PV, we obtain: dP dV P V +  = 0 Integrating both sides of the equation, we have: ln P +  ln V = constant or PV  = constant

19. Molar specific heats of some gases (at 20 o C and 1 atm) Gas C V C P C V –C P C V /C P ( J·mol -1 ·K -1 ) Monoatomic gases He 12.520.88.331.67 Ar 12.520.88.331.67 Ne12.720.88.121.64 Kr12.320.88.491.69 Diatomic gases H 2 20.428.88.331.41 N 2 20.829.18.331.40 O 2 21.129.48.331.40 CO21.029.38.331.40 Cl 2 25.734.78.961.35 Polyatomic gases CO 2 28.537.08.501.30 SO 2 32.440.49.001.29 H 2 O27.035.48.371.30

20. 1.They agree quite well for monatomic gases. 2.The model predictions do not agree at all with the experimental values, except for C P – C V, for other types of gases. 3.The number of atoms in the molecule is an important parameter. For gases having the same number of atoms in their molecules, their values are close although not the same. 4.The more number of atoms in the molecule of a gas the larger are its C V and C P. Comparing the experimental values of C V, C P, C P - C V and C P /C V, to those predicted by the ideal gas model, one notes that:

21. The discrepancies are mainly due to the wrong general assumption that the internal energy of the gas is equal to the total translational kinetic energy of the molecules. This assumption is only correct for monatomic gases. For diatomic and polyatomic gases, the molecule can rotate as well as vibrate about its centre of mass, hence giving rise to rotational energy and vibration energy. These additional energies should be included as parts of the internal energy. The model prediction is good for C P – C V (which is the same for all gases) because the difference between these two molar specific heats is the result of work done by the gas, which is independent of the molecular structure.

22. The branch of physics known as statistical mechanics has shown that, for a large number of particles obeying the laws of Newtonian mechanics, the internal energy of a gas is shared equally by each independent degree of freedom of motion. You may recall what we have shown earlier that if a molecule in a gas has only translational energy, it has 3 degrees of freedom of motion and each degree of freedom contributes ½k B T of energy. The equipartition of energy

23. Consider a diatomic gas whose molecules have the shape of a dumbbell as shown in the figure below: Internal energy of an ideal diatomic gas Such a molecule has 8 degrees of freedom: Translational motion of the centre of mass – 3 (x, y and z directions) Rotational motion – 3 (about x-, y- and z-axis) Vibrational motion – 2 (kinetic and potential energy associated with the vibrations along the length of the molecule) The theorem of equipartition of energy predicts that for the diatomic gases, C V = (8/2)R = 33.216 J·mol -1 ·K -1 and this value is much higher than those listed in the table shown earlier.

24. The experimental values of C V for diatomic gases suggest that the number of degrees of freedom at 20 o C at 1 atm is about (5/2)R. This implies that 3 degrees of freedom are negligible. The obvious one that can be neglected is the rotational motion about the y-axis because the moment of inertia about this axis is indeed small compared with those about the x- and z-axis. If we also neglect the 2 degrees of freedom for the vibrational motion, then we would have the “correct” values. The failure of the equipartition theorem to give the correct predictions for the molar specific heats of diatomic and polyatomic gases is due to the inadequacy of classical mechanics applied to molecular systems. A more adequate model to use is the quantum-mechanical model.

25. The equipartition theorem does not account for the observed temperature variation in molar specific heats. The diagram below shows the observed values of C V for H 2 as a function of temperature: The temperature variation observed suggests that the contribution by the vibrational motion only becomes significant at very high temperature and that the molecule has only translational energy at low temperature.

26. 1.The molar specific heats of solids are temperature dependent. 2.Generally, they decrease in a nonlinear manner with decreasing temperature and approach zero as the temperature approaches absolute zero. 3.At high temperature (>300 K), the molar specific heats approach the values of 3R  25 J·mol -1 ·K -1 (Dulong-Petit law). 4.The molar specific heat at high temperature can also be explained by the equipartition theorem. Each atom has 6 degrees of freedom, corresponding to the kinetic and potential energy associated with the vibrational motion in the x, y and z directions for small displacements from its equilibrium position. 5.The difference between the molar specific heat at constant volume and that at constant pressure is small. Molar specific heat of solids

27. Molar specific heat of Ge and Si

28. Example : Derive an expression for the atmospheric pressure at an altitude y, assuming that the atmosphere is at a constant temperature T. Let the atmospheric pressure at y be P and consider an atmospheric layer of thickness dy and cross-sectional area A, as shown in the figure on the right. Because the air is in static equilibrium, PA = m g n V A dy + (P+dP) A where = N/V = number of air molecules per unit volume. This equation can be reduced to: dP = -m g n V dy

29. Since PV = N k B T, P = n V k B T or n V = P/ k B T Thus dP/P = - (m g / k B T) dy Hence Integrating the equation, we have: or

30. Because P = n V k B T, the density of air molecules at the altitude y can also be written as: where n 0 is the density of air molecules at sea level.