Thermal Physics - Isothermal

Suppose one mole of an ideal gas undergoes the reversible cycle ABCA shown in the P-V diagram above, where AB is an isotherm. The molar heat capacities are C_p at constant pressure and C_v at constant volume. The net heat added to the gas during the cycle is equal to

A. RT_h (V₂/V₁)
B. −C_p(T_h − T_c)
C. C_p(T_h − T_c)
D. RT_h ln (V₂/V₁) − C_p(T_h − T_c)
E. RT_h ln (V₂/V₁) − R(T_h − T_c)

(GR9677 #15)

Solution:

AB Isotherm → T Constant = T_h

Ideal Gas: PV = nRT 
P = nRT / V

n = 1 mole,

W_AB = _V1∫^V₂ P dV = RT_h V1∫^V₂ (1/V) dV  = RT_h ln (V₂/V₁)

BC Isobaric → P constant = P₂

W_BC = _V2∫^V₁ P dV = P₂ (V₁− V₂) =  P₂V₁ −  P₂V₂

From the diagram:

P₂V₁ = nRT_c 

P₂V₂ = nRT_h

W_BC = nR(T_c − T_h)  = R(T_c − T_h)

W_CA = 0  since V constant

Total W = W_AB + W_BC = RT_h ln (V₂/V₁) + R(T_c − T_h)

or

W = RT_h ln (V₂/V₁) − R(T_h − T_c)

Answer: E

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 - Heat Capacity

A Classical model of a diatomic molecule is a springy dumbbell, as shown above, where the dumbbell is free to rotate about axes perpendicular to the spring. In the limit of high temperature, what is the specific heat per mole at constant volume?

A. 3/2 R
B. 5/2 R
C. 7/2 R
D. 9/2 R
E. 11/2 R

(GR9277 #15) 

Solution:

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

At high T, all components contribute and each degree of freedom (DoF) contributes ½R to C_V .

DoF trans = 3 (since it's 3 dimension)
DoF rot = 2 (diatomic atom)
DoF vib = 2 (1 for kinetic energy + 1 potential energy)
→ 3 + 2 + 2 = 7 → 7 DoF contributes ⁷/₂R

Answer: C

Thermal Physics - Heat Capacity

Questions 71-73

A system in thermal equilibrium at temperature T consists of a large number N₀ of subsystems, each of which can exist only in two states of energy E₁ and E₂, where . In the expressions that follow, k is the Boltzmann constant.

The internal energy of this system at any temperature T is given by . The heat capacity of the system is given by which of the following expressions?

A.

B.

C.

D.

E.

(GR9277 #72)

Solution:

Heat Capacity (at constant  volume), 



Answer: A

Thermal Physics - Heat Capacity

Einstein’s formula for the molar heat capacity C of solids is given above. At high temperatures, C approaches which of the following?

A. 0
B. 3kN_A(hv/kT)
C. 3kN_Ahv
D. 3kN_A
E. N_Ahv

(GR0177 #65)

Solution:

At high temperature → Classical approach, there’s no hv involved.

Answer: D