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