1. Specific heat of gases

2. Specific heat of monatomic gas I
Add heat keeping the volume constant. Energy can only go into kinetic energy of atoms:
Conclusion:

3. Specific heat of a monatomic gas II
Add heat keeping the pressure constant. Energy can go into kinetic energy of atoms, or into work when the gas expands:
Some mathematical gymnastics:

4. Specific heat of a monatomic gas III
Substitute:
Conclusion:
Valid for all ideal gases

5. Equipartition of energy
According to classical mechanics, molecules in thermal equilibrium have an average energy of associated with each independent degree of freedom of their motion provided that the expression for energy is quadratic.
monatomic gas: x, y, z motion: since the terms are quadratic

6. Degrees of freedom for molecules
We take m to be the mass of the molecule, then as for monatomic gases
However, there is now internal energy due to rotation and vibration:

7. Diatomic molecule: rotation
Think of molecule as dumbbell:
Centre-of-mass (CM) moves with K.E.
Rotation about CM with K.E.
All quadratic: f = 6

8. Diatomic molecule: vibration
Atoms attract each other via Lennard-Jones potential
Atoms stay close to equilibrium
Can be approximated by a harmonic oscillator potential

9. Harmonic oscillator Average P.E. = average K.E.
P.E is quadratic so associated with it
K.E. is the same so also
Total: kT or f = 2.

10. Diatomic molecules: conclusion?
According to classical physics there should be 8 degrees of freedom for each diatomic molecule…
From Cp and CV measurements for H2:
below 70 K: f = 3
at room temperature: f = 5
above 5000 K: f = 7

11. Question
If e.g. vibration of H2 does not contribute to the specific heat, then that is because
a) The vibration is independent of temperature
b) The vibration doesn’t change in collisions
c) The vibration is like a very stiff spring

12. Quantum mechanical effects
Translation: 3 degrees
Rotation: 2 degrees
Vibration:
at room temperature often also “frozen out”
at high temperatures K.E. + P.E. = 2 degrees
Rotation about these
axes imparts energy
Rotation about this
axis is “frozen out”

13. Rotational states
Quantum mechanics: rotational and vibrational energies not continuous but quantised
Rotation: spacing ~1/I
small steps, near-continuous about two axes
big steps about axis through atoms
If step >> kT : not excited, no rotation
Even at T=104 K no rotation about third axis

14. Vibrational states Vibration: harmonic oscillator Spacing ~1/m
step >> kT for light molecules (O2, N2) but excitations possible for heavy molecules such as I2
Potential energy: on average equal to kinetic energy
vibration contributes kT when it does

15. Solids Model for solid:
Elemental (“monatomic”) solids have K.E. and P.E. to give C = 3R (Rule of Dulong and Petit)

16. Freezing out - again
At lower temperatures, some vibrational levels cannot be reached with kT energy. At very low temperatures C  T3.
For metals the electrons (which can move freely through the metal) contribute and C = aT + bT3 (Einstein-Debye model)

17. Specific molar heat
Dulong & Petit: elemental solids C  25 J mol-1 K-1

18. Question 2
A room measures 4.002.402.40 m3. Assume the “air molecules” all have a velocity of 360 ms-1 and a mass of 510-26 kg. The density of air is about 1 kg m-3. Calculate the pressure in this room.

19. Isothermal expansion of an ideal gas
Isothermal expansion: pressure drops as volume increases since pV = nRT = constant
The internal energy only depends on the temperature so it doesn’t change
Two equations hold at the same time:
pV = nRT and T2 = T1

20. Question
An identical volume of the same gas expands adiabatically to the same volume. The pressure drops
a) more than in the isothermal process
b) by the same amount
c) less than in the isothermal process

21. Adiabatic expansion of an ideal gas I
Q = 0 so U + W = 0
Remember: U = nCV T
Use ideal gas law:
Substitute:

22. Adiabatic expansion of an ideal gas II
Divide by nCV T:
Prepare for integration:
Define so that

23. Adiabatic expansion of an ideal gas III
Integrate:
Play around with it:
Use

24. Question
Consider isothermal and adiabatic expansion of a Van der Waals gas. Do T2 = T1 and pVg = C hold for this gas?
a) yes; yes
b) yes; no
c) no; yes
d) no; no

25. Question A gas is compressed adiabatically. The temperature
a) rises because work is done on the gas
b) rises depending on how much heat is added
c) drops because work is done by the gas
d) drops depending on how much heat is added

26. PS225 – Thermal Physics topics
The atomic hypothesis
Heat and heat transfer
Kinetic theory
The Boltzmann factor
The First Law of Thermodynamics
Specific Heat
Entropy
Heat engines
Phase transitions