Question

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 A total of $1200J$ heat is withdrawn from the sample in the process.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

Answer 1

From first law of thermodynamics,

$\Delta U=q+W$

The change in internal energy during the cyclic process is zero. Therefore, heat supplied to the gas is equal to work done by it, i.e.,

$q=-W$

As the heat is withdrawn from the sample, thus

$q=-1200J$

Therefore,

$W_{Total}=-(-1200)=+1200J$

As the cyclic process is traced anticlockwise, the net work done by the system is negative.

$\therefore W=-1200$

Now from fig.,

$W=W_{A\to B}+W_{B\to C}+W_{C\to A}$

$\Rightarrow W_{A\to B}+W_{B\to C}+W_{C\to A}=-1200.....\left(1\right)(\because W=-1200J)$

Now, work done during process $A\to B$,

$W_{A\to B}=P\Delta V=nR\Delta T(\text{From ideal gas equation, i.e.,}PV=nRT)$

$\therefore W_{A\to B}=nR(T_B-T_A)$

$\Rightarrow W_{A\to B}=2\times 8.314\times (500-300)$

$\Rightarrow W_{A\to B}=3325.6J$

Now work done during process $B\to C$,

$W_{B\to C}=?$

Now work done during process $C\to A$,

$W_{C\to A}=0(\because \text{V is constant, i.e.,}\Delta V=0)$

Now from ${eq}^n\left(1\right)$, we have

$3325.6+W_{B\to C}+0=-1200$

$\Rightarrow W_{B\to C}=-1200+(-3325.6)=-4525.6J$

Hence the work done by the gas in process $B\to C$ is $-4525.6J$.