Question

A sample of an ideal gas in taken through the cyclic process $abca$ as shown in figure. The change in the internal energy of the gas along the path $ca$ is $-180\text{J}$. The gas absorbs $250\text{J}$ of heat along the path $ab$ and $60\text{J}$ along the path $bc$. The work done by the gas along the $abc$ is

• A. $120\text{J}$
• B. $130\text{J}$
• C. $100\text{J}$
• D. $140\text{J}$

Answer 1

The correct option is B $130\text{J}$

$\begin{array}{cccc}\text{Path}& \Delta U& W& Q\\ \text{ab}& & & 250\text{J}\\ \text{bc}& & 0& 60\text{J}\\ \text{ca}& -180\text{J}& & \end{array}$

From the graph, we can see $bc\to$ isochoric process.

$\Rightarrow W_{bc}=0$

As, $Q_{bc}=\Delta U_{bc}+W_{bc}$

$\Rightarrow 60=\Delta U_{bc}$

For a cyclic process, $\Delta U=0$

$\Rightarrow \Delta U_{ab}+\Delta U_{bc}+\Delta U_{ca}=0$

$\Rightarrow \Delta U_{ab}=-60-(-180)=120J$

Applying 1st law of thermodynamics,

$\Delta Q_{ab}=\Delta U_{ab}+\Delta W_{ab}$

$\Rightarrow 250\text{J}=120\text{J}+\Delta W_{ab}$

$\Rightarrow \Delta W_{ab}=250-120=130\text{J}$

So work done along the path $abc$,

$W_{abc}=W_{ab}+W_{bc}$

but $W_{bc}=0$

$\Rightarrow W_{abc}=W_{ab}=130\text{J}$

Hence, the correct option is $\left(B\right)$