Question

Find work done by the gas in the process shown in figure :

• A. $\frac{5}{2}\pi atmL$
• B. $\frac{5}{2}atmL$
• C. $-\frac{3}{2}\pi atmL$
• D. $-\frac{5}{4}\pi atmL$

Answer 1

Answer: (D)

The work done by the gas in the process will be the area under the loop $ABCDA$ . We can divide the loop $ABCDA$ In two loops, which are $ABCA$ and $DEFD$ . Loops $ABCA$ and $DEFD$ Are semi-elliptical shapes.
The area of an ellipse is $\pi ab$ , where $a$ and $b$ are the lengths of semi major axis and the length of semi minor axis. Length of $a$ for semi ellipse $ABCA$ is $4Litre-3Litre=1Litre$ . Length of $b$ for semi ellipse $ABCA$ is $4atm-2atm=2atm$ . Length of $a$ for semi ellipse $DEFD$ is $3.5Litre-3Litre=0.5Litre$ . Length of $b$ for semi ellipse $ABCA$ is $2atm-1atm=1atm$ .
So the work done $W$ by the gas is,
$W=W_{ABCA}+W_{DEFD}$
$W=\frac{1}{2}\left(\pi \times 1Litre\times 2atm\right)+\frac{1}{2}\left(\pi \times 0.5Litre\times 1atm\right)$
$W=\frac{5}{4}\pi atmL$
As the curve $ABCDA$ is created by anti-clockwise movement, the work done will be negative. So, $W=-\frac{5}{4}\pi atmL$