KVPY SB SX Exam 2011 Question Paper

Question : 95 of 120

Marks: +1, -0

A constant amount of an ideal gas undergoes the cyclic process ABCA in the PV diagram shown below.

The path BC is an isothermal. The work done by the gas during one complete cycle, beginning and ending at A is nearly

Solution:

W_{C→A} = 0

Process BC → T = Constant

P_CV_C = P_B V_B

500 × 2 = 200 × V_B

V_B = 5m^3

W_{A→ B} = 200 [V_B - V_A]

= 200 [5 – 2] = 600

W_{B→C} > W_{A→B}

∴ Net work done is –ve
∵

W_{B→C} < 1200 \KJ

∴ Total W < – 600 KJ

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