Question

An ideal monoatomic gas follows the path $ABCD$. the work done on the gas is :

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

Answer 1

The correct option is A $-PV$

Since the process $C$ to $A$ and $B$ to $D$ are isochoric(constant volume), the work done in each process is zero.

Now, since the process $A$ to $B$ and $D$ to $C$ is isobaric, the work done will be,

$W=P_i(V_f-V_i)$

Where $P_i$ is the isobaric pressure and $V_i$ and $V_f$ are the initial and final volume of the gas respectively

Therefore, $W_{AB}=2P(2V-V)$

$=2PV$

And, $W_{DC}=P(V-2V)$

$=-PV$

Hence the net-work done by the gas in the cycle is

$W_{AB}+W_{DC}=PV$

And the net-work done on the gas in the cycle is $-PV$