Question

Consider the cyclic process $ABCA$, shown in Fig. performed on a sample of $2.0$ mole of an ideal gas. $A$ total of $1200J$ of heat is withdrawn from the sample in the process. Find the work done (in joule) by the gas during the part $BC$.

Answer 1

Answer: 4524

In the cyclic process $\Delta U=0$
From first law of thermodynamics for the cyclic process $Q=\Delta U+W$
$\therefore W=Q-A\Delta U=-1200-0$
$=-1200J$

From $C$ to $A,\Delta V=0\therefore W_{CA}=0$
For the whole cycle $W_{AB}+W_{BC}+W_{CA}=W=-1200$
As $W_{CA}=0,\therefore W_{AB}+W_{BC}=-1200J$....(i)
Work done from $A$ to $B$:
In the process $V\propto T$, so pressure remains constant. We know that $PV=nRT$
or $P\Delta V=nR\Delta T$
$\therefore W_{AB}=P\Delta V=nR\Delta T=2\times 8.31\times \left(500-300\right)$
$=3324J$
Substituting this value in equation (i), we get
$3324+W_{BC}=-1200$
$\therefore W_{BC}=-4524J$