Question

Consider the cyclic process ABCA on a sample of 2.0 mol of an ideal gas as shown. The temperature of the gas at A and B are 300 K and 500 K respectively. A total of 1200 J heat is withdrawn from the sample in this process. Find the work done by the gas in part BC. Take $R=8.3J{K}^{-1}mo{l}^{-1}$

• A. $-4520J$
• B. $-4200J$
• C. $4520J$
• D. None of these

Answer 1

The correct option is A $-4520J$

$\Delta W_{AB}=P\Delta V=nR\Delta T=2\times 8.3\times (500-300)=3320J$
$\Delta W_{CA}=0$ as volume is constant
Now, for the cyclic process ABCA, $\Delta U=0$
Hence, $\Delta Q=\Delta W=\Delta W_{AB}+\Delta W_{BC}+\Delta W_{CA}\Rightarrow -1200=3320+\Delta W_{BC}\Rightarrow \Delta W_{BC}=-4520J$