Question

Six moles of an ideal gas performs a cycle shown in figure. If the temperature are $T_A = 600\, K, T_B = 800\, K, T_C = 2200 \,K$ and $T_D = 1200\, K$, the work done per cycle is approximately

• A. $20\, kJ$
• B. $30\, kJ$
• C. $40\, kJ$
• D. $60\, kJ$

Answer 1

Answer: (C)

Processes A to B and C to D are parts of straight line graphs passing through origin $P ∝T.$
So, volume remains constant for the graph AB and CD
So, no work is done during processes for A to B and C to D
i.e., $W_{AB} = W_{CD} = 0$
and $W_{BC} = P_2(V_C - V_B) = nR(T_C - T_B)$
$= 6R(2200 - 800) = 6R ×1400\, J$
Also, $W_{DA} = P_1(V_A - V_D) = nR(T_A - T_D)$
$= 6R(600 - 1200) = - 6R ×600\,J$
Hence, work done in complete cycle
$W = W_{AB} + W_{BC} + W_{CD} + W_{DA}$
$= 0+ 6R ×1400 + 0 - 6R ×600$
$= 6R ×800 = 6 ×8.3 × 800 = 40 \,kJ$