Thermal Physics - Heat Capacity

For an ideal diatomic gas in thermal equilibrium, the ratio of the molar heat capacity at constant volume at very high temperatures to that at very low temperatures is equal to

A. 1
B. 5/3
C. 2
D. 7/3
E. 3

(GR9677 #79)

Solution:

C_V  for diatomic gas: C_V  = C_Vtrans + C_Vrot  + C_Vvib

For very low T, only translational component contributes
→ C_V(T≪) = ³/₂ Nk

For very high T, all 3 components (translational, vibrational, rotational) contribute
→ C_V(T≫) = (³/₂ + 1 + 1) R = ⁷/₂ Nk

Ratio C_V(T≪) / C_V(T≫)  = 7/3

Answer: D

Thermal Physics - Thermal Equilibrium

See Problem #46

In which region are the liquid and the vapor in equilibrium with each other?

A. A
B. B
C. C
D. D
E. E

(GR9277 #47)

Solution:

Region A: V small  → Liquid
Region E: P large, V small → Liquid
Region D and C: V large →  Gas
Phase equilibrium exists only if the temperature of isotherm lies below critical temperature.

Answer: B

Thermal Physics - Probability

Which of the following is true if the arrangement of an isolated thermodynamic system is of maximal probability?

A. Spontaneous change to a lower probability occurs
B. The entropy is minimum
C. Botzmann’s constant approaches zero
D. No spontaneous change occurs
E. The entropy is zero

(GR9277 #63)

Solution:

Maximal Probability → highest/maximum entropy → most stable state /equilibrium → No spontaneous change occurs. Spontaneous change occurs when the system is far from equilibrium.

Answer: D

Thermal Physics - Maxwell-Boltzmann Statistics

A large isolated of N weakly interacting particles is in thermal equilibrium. Each particle has only 3 possible non degenerate states of energies 0, ε, and 3ε. When the system is at an absolute temperature T ≫ ε/k, where k is Boltzmann’s constant, the average energy of each particle is

A. 0
B. ε
C. 4ε /3
D. 2ε
E. 3ε

(GR8677 #67)

Solution:

In thermal equilibrium:
→ Each particle has roughly equal probability of being in any of the three states.
→ The average energy of each particle is total energy/3 = (0+ε+3ε)/3 = 4ε/3

Answer: C

Note:

Complete Calculation:

T ≫ ε/k → 1 ≫ ε/kT 
ε/kT → 0
e^-ε/kT → e⁰  = 1

Partition function: Z = ∑_i e^-ε_i/kT
i = 3 → Z = 3

Probability: P_i = e^-ε_i/kT/Z
→ P₁ = P₂  = P₃  = 1/3

Energy: E = ∑_i E_iP_i
→ E = E₁P₁ + E₂P₂ + E₃P₃ = (0+ε+3ε)/3 = 4ε/3

Thermal Physics - Isothermal vs Adiabatic

An ideal monatomic gas expands quasi-statically to twice its volume. If the process is isothermal, the work done by the gas is W_i. If the process is adiabatic, the work done by the gas is W_a. Which is the following is true?

A. W_i = W_a
B. 0 = W_i  W_a 
C. 0  W_i  W_a 
D. 0 = W_a  W_i
E. 0  W_a  W_i

(GR0177 #06)

Solution:

Isothermal and Adiabatic P-V Diagram

• Adiabatic connects high-T isotherm and low-T isotherm.
• Isothermal line is always higher than the adiabatic line and they both end at the same volume
• The area under the isothermal line is bigger than the adiabatic → W_a  W_i

Answer: E

Calculation:

Isothermal:

PV = constant
P_i V_i  = P_f V_f  = constant
Given V_f  = 2V_i

P_i V_i  = 2P_f V_i

P_f (iso) = ½ P_i

Adiabatic:

PV^γ = c
P_i V_i^γ = P_f V_f ^γ = constant
V_f  = 2V_i
P_iV_i^γ  = P_f (2V_i)^γ = 2^γP_f V_i^γ 
P_f (adi) = (1/2^γ)P_i 
  
Therefore,

P_{f (iso)} /P_{f (adi)} = 2^γ / 2
P_f (iso) = 2^γ−1P_{f (adi)}

→ P_adi  P_iso

W = ∫ PdV → W_adi  W_iso

Thermal Physics - Maxwell-Boltzmann Statistics

An ensemble of system is in thermal equilibrium with a reservoir for which kT = 0.025 eV. State A has an energy that is 0.1 eV above that of state B. If it is assumed the systems obey Maxwell-Boltzmann statistics and that the degeneracies of the two states are the same, then the ratio of the number of the systems in state A to the number in state B is

A. e⁺⁴
B. e^+0.25
C. 1
D. e^-0.25
E. e^-4

(GR0177 #77)

Solution:

Maxwell-Boltzmann distribution:





Answer: E

Thermal Physics - Thermal Equilibrium

Suppose that a system in a quantum state i has energy E_i. In thermal equilibrium, the expression

represents which of the following?

A. The average energy of the system
B. The partition function
C. Unity
D. The probability to find system with energy E_i
E. The entropy of the system

 (GR0177 #98)

Solution:

The units of the given expression are the units of E_i which as given is energy.

The average energy is given by

<E> = ^{∑_i E_i e−E_i/kT} / _Z

where

Z = ∑_i e^−E_i/kT

is the partition function.

Answer: A