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 - 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