Question

Consider a cyclic process ABCA performed by a sample of 2 moles of an ideal gas as shown in figure. If the net heat transfer during the cyclic process is -1200 J, the work done during the process BC
(Take R=8.314 J\mol-K)

• A. -3324.4 J
• B. -4526 J
• C. -6520 J
• D. +4526 J

Answer 1

The correct option is B -4526 J

Given:$Q_{net}=-1200J,T_A=300K,T_B=500K,n=2moles$

Since, it is a cyclic process,

$Q_{net}=W_{net}$

$\therefore W_{AB}+W_{BC}+W_{CA}=-1200$

$W_{BC}=-1200-W_{AB}-W_{CA}$

$\text{Now,}{W}_{CA}=0$,

Since it is a constant volume process.

$\therefore W_{BC}=-1200-W_{AB}$

Now, line AB passes through origin,

$T=KV$

$500=K{V}_2$

$300=K{V}_1$

$\therefore K(V_2-V_1)=200$

$\text{Now,}{W}_{AB}={\int }_{V_2}^{V_1}pdv$

Also, $PV=n\overline{R}T$

$P=\frac{n\overline{R}KV}{V}=n\overline{R}K$

$W_{AB}={\int }_{V_2}^{V_1}n\overline{R}KdV=n\overline{R}K(V_2-V_1)=2\times 8.314\times 200=3325.6J$

$W_{BC}=-1200-3325.6=-4525.6J$