Question

Three moles of an ideal gas are taken through a cyclic process $ABCA$ as shown in $T-V$ indicator diagram. The gas loses $2510\text{J}$ of heat in the complete cycle. If $T_A=100\text{K}$ and $T_B=200\text{K}$, the work done by the gas during the process $BC$ is (line AB if extended passes through origin)

• A. $5000\text{J}$
• B. $-5000\text{J}$
• C. $4000\text{J}$
• D. $-2500\text{J}$

Answer 1

The correct option is B $-5000\text{J}$

From the diagram we can deduce that, cyclic process is in anti-clockwise direction. So, the work done by the gas during the cyclic process is negative.
In process $AB$, the volume $V$ increases linearly with temperature $T$. Hence process AB is isobaric.
$\therefore$ Work done during process $AB$ is given by
$W_{AB}=P\Delta V=nR\Delta T(\because PV=nRT$)
From the data given in the question,
$W_{AB}=3\times 8.3\times (200-100)=+2490\text{J}$
From the figure we can say that, process CA is isochoric. Hence, work done in this process $W_{CA}=0$.
Since, the whole process $ABCA$ is cyclic, the change in internal energy in complete cycle is zero, i.e. $\Delta U=0$.
Now, from the first law of thermodynamics,
$Q=\Delta U+W_{net}$
Since, gas loses heat energy , Heat energy lost by gas over the complete cycle is $Q=-2510\text{J}$
$\therefore Q=\Delta U+W_{AB}+W_{BC}+W_{CA}$
Substituting the values we get, $-2510=0+2490+W_{BC}+0$
$\Rightarrow W_{BC}=-2510-2490=-5000\text{J}$
Thus, the correct choice is option (b).