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 Physic - Critical Isotherm

Questions 46-47.

Isotherms and coexistence curves are shown in the PV diagram above for a liquid-gas system. The dashed lines are the boundaries of the labeled regions. Which numbered curve is the critical isotherm? 

A. 1
B. 2
C. 3
D. 4
E. 5

(GR9277 #46)

Solution:

Critical isotherm (critical temperature) → dP/dV = 0

• the derivative of the curve is zero
• the tangent to the curve results in a horizontal line
• the point where the vertical and horizontal dashed lines cross

Answer: B

Thermal Physics - Isothermal

A mole of ideal gas initially at temperature T₀ and volume V₀ undergoes a reversible isothermal expansion to volume V₁. If the ratio of specific heats is c_p/c_v = γ and if R is the gas constant, the work done by the gas is

A. Zero
B. RT₀ (V₁/V₀ )^γ
C. RT₀ (V₁/V₀ − 1)
D. c_v T₀ [1 − (V₁/V₀ )^(γ−1)]
E. RT₀ ln (V₁/V₀ )

(GR9277 #62)

Solution:

W = ∫ P dV

Ideal Gas: PV = nRT  → P = nRT / V

W = _V0∫^V₁ nRT (1/V) dV

1 mole → n = 1
Isothermal, constant T = T₀

→ W = RT_0 V0∫^V₁ (1/V) dV = RT₀ ln (V₁/V₀ )

Answer: E  

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 - Adiabatic Expansion

Consider the quasi-static adiabatic expansion of an ideal gas from an initial state i to a final state f. Which is the following statements is NOT true?

A. No heat flows into or out of the gas.
B. The entropy of state i equals the entropy of state f.
C. The change of internal energy of the gas is −∫ PdV.
D. The mechanical work done by the gas is ∫ PdV.
E. The temperature of the gas remains constant.

(GR0177 #36)

Solution:

A (not) dia (through) batic (passable) = no heat flow, container is well insulated
(A) and (B) are TRUE.

Q = 0
∆U = −W = −∫ PdV
(C) TRUE.

Definition of mechanical work: W = ∫ PdV
(D) TRUE.

T constant is Isothermal not adiabatic
(E) FALSE

Answer: E

Thermal Physics - Ideal Gas

A constant amount of an ideal gas undergoes the cyclic process ABCA in the PV diagram shown above. The path BC is isothermal. The work done by the gas during one complete cycle, beginning and ending at A, is most nearly

A. 600 kJ
B. 300 kJ
C. 0
D. −300 kJ
E. −600 kJ

(GR0177 #37)

Solution:

B-C Isotherm → T Constant
PV = constant
P_BV_B = P_CV_C

V_B = P_CV_C / P_B = 500 × 2/200 = 5

The work done ≈ area of ∆CAB = ½ (CA × AB) = ½ [(500-200) × (5-2)] = 450.

Since BC is curved inside the ∆CAB, the work done is less than 450.
And, since the arrow is counterclockwise, the work done is negative.
Thus, the work done is less than −450

Answer: D


Complete calculation:

A-B Isobaric → P constant
W = PdV = 200(V_B −  2)

C-A Isovolume → V constant
W = PdV = 0

B-C Isotherm → T constant
PV = constant
P_BV_B = P_CV_C
V_B = P_CV_C/P_B  = (500)(2)/(200) = 5

W_isobaric = 200(5 − 2) = 600

W_isotherm =  nRT ln(V_f /V_i )

For isotherm, P_BV_B = P_CV_C = nRT
W_isotherm = P_BV_B  ln(V_f /V_i )
= 200 × 5 × ln (2/5)
= 1000 (−0.916)
= −916

W_total = 600 − 916 = −316 kJ

Thermal Physics - Entropy

A sealed and thermally insulated container of total volume V is divided into two equal volumes by an impermeable wall. The left half of the container is initially occupied by n moles of a ideal gas at temperature T. Which of the following gives the change in entropy of the system when the wall is suddenly removed and the gas expands to fill the entire volume?

A. 2nR ln 2
B. nR ln 2
C. ½nR ln 2
D. -nR ln 2
E. -2nR ln 2

(GR0177 #47)

Solution:

The container is thermally insulated, dT = 0 or T constant → Isothermal expansion

Entropy for Isothermal expansion: S = Nk ln (V_f / V_i )

Since V_f = 2V_i → S = Nk ln 2 = nR ln 2

Answer: B