Question

Consider the cyclic process $ABCA$ on a sample of $2.0\text{mol}$ of an ideal gas as shown in figure. The temperature of the gas at $\left(A\right)$ and $\left(B\right)$ are $300\text{K}$ and $500\text{K}$ respectively. A total of $1200\text{J}$ heat is withdrawn from the sample in the process. Find the work done by the gas in process $BC$. Take $R=8.3\text{J/mol –K}$

• A. $-3000\text{J}$
• B. $+3000\text{J}$
• C. $-4520\text{J}$
• D. $+4500\text{J}$

Answer 1

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

The change in internal energy during the cyclic process is zero. Therefore, heat supplied to the gas is equal to work done by
it.
$\therefore W_{AB}+W_{BC}+W_{CA}=–1200$ ...(i)
The work done during the process $AB$ is
$W_{AB}=P_A(V_B–V_A)=nR(T_B–T_A).(\because PV=nRT)$
$W_{AB}=2\times 8.3(500–300)=3320\text{J}$... (ii)
$R=$ universal gas constant
$n=$ No. of moles
Since in this process, the volume increases, the work done by the gas is positive.
Now, $W_{CA}=0$
(volume of gas remains constant)
$\therefore 3320+W_{BC}+0=–1200$
(using equation (i) and (ii),
$W_{BC}=–1200–3320$
$W_{BC}=–4520\text{J}$