Question

An ideal gas undergoes cyclic process $\text{ABCDA}$ as shown in given $P-V$ diagram (Fig.). The amount of work done by the gas is

• A. $+2P_0{V}_0$
• B. $6P_0{V}_0$
• C. $+4P_0{V}_0$
• D. $-2P_0{V}_0$

Answer 1

The correct option is D $-2P_0{V}_0$

Let $P$ be the pressure of the gas in the cylinder, then
The force $F$ exerted by the gas on the piston of the cylinder,
$F=PA$
In a small displacement of piston through $dx$, work done by the gas,
$dw=F.dx=PAdx=PdV$
As we know,
$\text{Volume}=\text{Area}\times \text{distance}=A.dx$
Therefore,
Total amount of work done is given by,

$\Delta W=\int dW={\int }_{V_i}^{V_f}PdV=P(V_f-V_i)$

As per the $P-V$ diagram or indicator diagram, the area under $P-V$ curve represents work done $\left(W\right)$.
$\text{Work done(W)= Area under}$ $(P-V)$ diagram


According to the $P-V$ diagram given in the question,
Work done in the process $\text{ABCDA}=\text{Area of rectangle ABCDA}$

$l\times b=AB\times BC$
$=(3V_0-V_0))\times (2P_0-P_0)$

$W=2V_0\times P_0$

$W=2P_0{V}_0$

Since, the cyclic process is anti-clockwise, work done by the gas is negative that is, $-2P_0{V}_0$. Hence there is a net compression in the gas.

Correct Answer: (b).