A gas is taken from state-1 to state-2 along the path shown in figure. If 70 cal of heat is extracted from the gas in the process, calculate the change in internal energy of the system (in J).

We know work done by a gas is given by the area under PV-curve or the area between PV-curve and the volume axis. Generally, we take volume on x-axis while plotting  V-curve but in figure it is taken on y-axis thus the work done is given by the shaded area shown in figure. In this process volume of gas decreases thus work is done on the gas and it is given as

$\mathrm{W}=-\frac{1}{2}\times 1.5\times {10}^{-4}\times (2+5)\times {10}^5=-52.5\mathrm{J}$

It is given that heat extracted in the process is 70 cal, thus

    $\mathrm{Q}=-70 \mathrm{cal}=70\times 4.2 \mathrm{J}=-294 \mathrm{J}$

Now from first law of thermodynamics, we have

$\mathrm{Q}=\mathrm{W}+\mathrm{\Delta U}$

or $\mathrm{\Delta U}=\mathrm{Q}-\mathrm{W}=(-294)-(-52.5)=-241.5\mathrm{J}$

Thus in the process internal energy of gas decreases by 241.5 J