1. Molar Specific Heat of Ideal Gases
Since Q depends on process, C  dQ/dT also depends on process.
Define a) molar specific heat at constant volume:
CV  (1/n) dQ/dT for constant V process.
b) molar specific heat at constant at constant pressure:
CP = (1/n) dQ/dT for constant P process.
Consider constant V process: W = 0 and Q = Eint .
Q = n CV dT (= nCV T if CV = constant)
Therefore Eint = nCV dT
Since Eint/n only depends on temperature end points (“Joule effect”),
Eint = nCV dT for any process.
Consider constant P process:.
Q = nCP dT and W = -PdV Since P = constant, dV = nR dT/P,
 dW = -  P (nR dT/P) = -nRdT
Q = Eint – W
nCP dT = nCV dT + nRdT  CP = CV + R (for any ideal gas)

2. aValues at T = 300 K and P = 1 atm.
Ideal Gas: CP = CV + R
a) CP > CV: true for everything (not just ideal gas)
(For liquids and solids, the difference is usually small.)
b) If CV = constant, then CP = constant.
i) Constant V process, Q = nCV dT = nCV T
ii) Constant P process: Q = nCP dT = nCP T
iii) Any process: Eint = = nCV dT = nCV T
c) monoatomic ideal gas: no internal excitation of molecule so
Eint = N<Ktrans> = 3/2 NkBT = 3/2 nRT
 CV = 3/2 R = 12.5 J/molK, CP = 5/2 R = 20.8 J/molK, CP/CV = 5/3 = 1.67
aValues at T = 300 K and P = 1 atm.
Small deviations of CV due to non-ideal properties – interactions between molecules

3. Equipartition of Energy Theorem:
The average (i.e. thermal) value of the energy for each “classical”, “quadratic” degree of freedom = ½ kBT.
Consider a diatomic molecule (e.g. H2, N2, O2, HCl, … [since air is 98% nitrogen and oxygen, it is mostly diatomic]).
In addition to CM translation (3 quadratic degrees of freedom), it has rotational and vibrational energy.
Krot = ½ Iz2 + ½ Ix2 [no energy for rotation about inter-atomic (y) axis since Iy = 0]: 2 quadratic degrees of freedom.
Evib = Kvib + Uvib = ½  (ds/dt)2 + ½ k (s-s0)2
effective (reduced) mass
effective spring constant
Therefore, total of 7 quadratic degrees of freedom, expect Eint = 7/2 NkBT = 7/2 nRT  CV = 7/2 R = 29.1 J/molK
CP = 9/2 R = 37.4 J/molK s

4. Molar Specific Heat of Hydrogen
CV steps from 3/2 R (T < ~ 100 K) to 5/2 R (T < 1000 K) to 7/2 R.
only translations translations + rotations translations, rotations, and vibrations

5. Equipartition of Energy Theorem:
Hydrogen
Equipartition of Energy Theorem:
The average (i.e. thermal) value of the energy for each “classical”, “quadratic” degree of freedom = ½ kBT.
classical: kBT >>  = the quantum of energy
Vibrations: vib  h(k/)1/2
Rotations: rot  h2/I,
where Planck’s constant h = 1.05 x Js.
For kBT < vib, vibrations “freeze out” – cannot be thermally excited, and CV  5/2 R.
For kBT < rot, rotations “freeze out” – cannot be thermally excited, and CV  3/2 R.
For nitrogen and oxygen, which are much heavier than hydrogen, rotations theoretically freeze out a few Kelvin, i.e. below boiling point, so not observable in gas, but vibrations freeze out near 1000 K (like H2).
Near room temperature, most diatomic molecules are on the CV ~ 5/2 R shelf.

6. Expect CP - CV = R = 8.314 J/molKF
[Polyatomic may have 3 rotational and several vibrational degrees of freedom, some of which are frozen out at room temperature and some not.]
[Deviations from expected values due to interactions between molecules (i.e. non-ideal behavior) and because CV = 5/2 “shelf” not perfectly flat.]

7. Problem: J of energy are transferred by heat into 10 moles of an ideal diatomic gas near room temperature. Find the work and changes in internal energy and temperature for the following cases:
Constant V process
Constant P process (isobar)
Constant T process (isotherm)

8. Problem: J of energy are transferred by heat into 10 moles of an ideal diatomic gas near room temperature. Find the work and changes in internal energy and temperature for the following cases:
Constant V process
Constant P process (isobar)
Constant T process (isotherm)
Diatomic near room temp.: CV = 5/2 R, CP = 7/2 R
c) T = 0  Eint = 0  W = - Q = J
a) V = 0  W = 0  Eint = Q = 3000 J
Eint = n CV T = 5/2 nR T
 T = (2/5) (3000 J) / [(10 moles)(8.314 J/molK)] = 14.4 K

9. Problem: J of energy are transferred by heat into 10 moles of an ideal diatomic gas near room temperature. Find the work and changes in internal energy and temperature for the following cases:
Constant V process
Constant P process (isobar)
Constant T process (isotherm)
Diatomic near room temp.: CV = 5/2 R, CP = 7/2 R
c) T = 0  Eint = 0  W = - Q = J
a) V = 0  W = 0  Eint = Q = 3000 J
But Eint = n CV T = 5/2 nR T
 T = (2/5) (3000 J) / [(10 moles)(8.314 J/molK)] = 14.4 K
b) Q = n CP T = 7/2 nR T
 T = (2/7) (3000 J) / [[(10 moles)(8.314 J/molK)] = 10.3 K
Eint = nCV T = (5/2) nR T = 2143 J
W = Eint – Q = (2143 – 3000) J = J

10.  CP / CV = (CV + R)/CV = 1 + R/CV, ( = constant (> 1) if CV = constant).
CV = 3/2 R   = 5/3 = 1.67 (its maximum value for ideal gas in 3D)
CV = 5/2 R   = 7/5 = 1.40
CV = 7/2 R   = 9/7 = 1.29

11. Significance of   CP/CV
Consider an adiabatic (Q = 0: well insulated and/or too rapid for heat flow, but still “quasi-static”) process in an ideal gas with constant CV (e.g. on a “shelf”).
dEint = dW
nCV dT = -P dV
dT = - (1/nCV) PdV
But PV = nRT  nR dT = P dV + V dP
(P dV + V dP)/nR = - (1 / nCV) P dV
(P dV + V dP) = - (R / CV) P dV
(1 + R/CV) P dV = - V dP
 P dV = - V dP
 dV / V = - dP / P
 ln V = const – ln P
exp( ln V) = exp (const – ln P)
V = const/P
PV = constant = PiVi
(for an adiabatic process in an ideal gas with constant CV.)

12. adiabats PV = constant
since  > 1, adiabats are steeper than isotherms

13. Problem: Consider a diatomic ideal gas at high temperature, so CV = 7/2 R and  = 9/7. It is adiabatically compressed so that Pf = 3 Pi. What are the ratios of Vf/Vi and Tf/Ti?
PfVf = PiVi  Vf/Vi = (Pi/Pf)1/ = (1/3)7/9 = 0.426
Tf = PfVf/nR = (3Pi) (0.426 Vi) / nR = 1.28 Pi Vi/nR = 1.28 Ti
Tf/Ti = 1.28

14. Problem: Consider the quasi-static cycle on the right, for n moles of an ideal gas with molar specific heat ratio . AB is an adiabatic process. Find the work, heat, and Eint for each step. [Give all results in terms of n, , P0, V0.]
Preliminaries
Find CP and CV in terms of :
CP = CV + R
Therefore  = (CV+R)/CV = 1 +R/CV
 CV = R/(-1)
CP = R/(-1)
2) Can find T’s in terms of PV’s, so first must find PB:
PAVA = PBVB
PB = PA(VA/VB) = P0/3
TA = P0V0/nR
TB = (P0/3)(3V0)/nR = P0V0/(3-1nR)
TC = (P0/3)(V0)/nR = P0V0/(3nR)

15. Problem: Consider the quasi-static cycle on the right, for n moles of an ideal gas with molar specific heat ratio . AB is an adiabatic process. Find the work, heat, and Eint for each step. [Give all results in terms of n, , P0, V0.]
CV = R/(-1)
CP = R/(-1)
PB = P0/3
TA = P0V0/nR, TB = P0V0/(3-1nR)
TC = P0V0/(3nR)
CA is constant volume, so WCA = 0 and QCA = nCV(TA-TC) = [P0V0/(-1)](1-1/3) (heat in)
ECA = WCA + QCA = P0V0/(-1)](1-1/3)
BC is constant pressure, so WBC = -PB(VC-VB) = (2P0V0/3)
QBC = nCP(TC-TB) = - [P0V0/(-1))](1/3-1 – 1/3) = -2P0V0/[3(-1)] (heat out)
EBC = WBC + QBC = (2P0V0/3) [1-/(-1)] = -2P0V0/[3 (-1)]
AB is adiabatic, so QAB=0. WAB = - PdV = - P0 (V0/V) dV
WAB = -P0V0  dV/V = [P0V0/(-1)][1/(3V0)-1 – 1/V0-1] = -[P0V0/(-1)] [1 - 1/3-1]
EAB = WAB + QAB = -[P0V0/(-1)] [1 - 1/3-1]

16. WCA = 0, QCA = (P0V0 /(-1)](1-1/3), ECA = [P0V0 /(-1)](1-1/3)
WBC = (2P0V0/3), QBC = -2P0V0/[3(-1)], EBC = -2P0V0/[3 (-1)]
WAB = -[P0V0/(-1)] [1 - 1/3-1], QAB = 0, EAB = -[P0V0/(-1)] [1 - 1/3-1]
Note that Ecycle = EAB + EBC + ECA = [P0V0 /(-1)] [1- 1/3 -2/3 /3-1]
= [P0V0 /(-1)] [-3/3 + 1/3-1] = 0 , as expected!

17. WCA = 0, QCA = (P0V0 /(-1)](1-1/3), ECA = [P0V0 /(-1)](1-1/3)
WBC = (2P0V0/3), QBC = -2P0V0/[3(-1)], EBC = -2P0V0/[3 (-1)]
WAB = -[P0V0/(-1)] [1 - 1/3-1], QAB = 0, EAB = -[P0V0/(-1)] [1 - 1/3-1]
e.g. if diatomic near room temperature,  = 7/5:
WCA = 0, QCA = ECA = 1.96 P0V0
WBC = 0.43 P0V0, QBC = P0V0, EBC = P0V0
WAB = P0V0, QAB = 0, EAB = P0V0