Question

An ideal monoatomic gas is taken round the cycle $ABCD$ as shown in the figure. The work done during the cycle is:

• A. $PV$
• B. $2PV$
• C. $\frac{1}{2}PV$
• D. Zero

Answer 1

The correct option is A $PV$

Here, $A⟶B$ and $C⟶D$ are isobaric processes.

$\therefore \Delta P=0$

$W_{AB}=-P\Delta V$

$=-P(2V-V)$

$=-PV$

$W_{CD}=-P\Delta V$

$=-2P(V-2V)$

$==+2PV$
$W_{BC}=W_{AD}=0$

$\therefore$ Total work done =$W_{AB}+W_{BC}+W_{CD}+W_{DA}$

$=|-PV+0+2PV+0|$

$=PV$
Option A is correct.