Question

Consider the cyclic process $ABCA$ as shown in the figure, performed on a sample of $2.0$ mole of an ideal gas. A total of $1200\text{J}$ of heat is withdrawn from the sample in the process. Find the magnitude of the work done (in joules) by the gas during the part $BC$. $(R=8.3{\text{JK}}^{-1}{\text{mol}}^{-1})$

• A. $2580\text{J}$
• B. $3625\text{J}$
• C. $4520\text{J}$
• D. $1550\text{J}$

Answer 1

The correct option is C $4520\text{J}$

The net change in internal energy for a cyclic process is zero.
$\Delta Q=-1200J$
Using First law of thermodynamics, $\Delta Q_{net}=\Delta U_{net}+W_{net}$
$\therefore W_{net}=W_{AB}+W_{BC}+W_{CA}=-1200J$...(i)
Lets study the graph now, graph $AB$ is a straight line in $T$ v/s $V$ graph $\Rightarrow P$ is constant.
Hence it is an isobaric process.
$W_{AB}=nR\Delta T$
In graph from $C$ to $A$ volume is constant.
Hence it is an isochoric process.
$W_{CA}=0$
Substituting these values in equation (i), we get
$n\times 8.3\times (500-300)+W_{BC}+0=-1200$
$W_{BC}=-4520J$
Hence the magnitude of workdone is $4520J$