Question

An ideal gas undergoes cyclic process $ABCDA$ as shown in given $PV$ diagram. The amount of work done by the gas is

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

Answer 1

The correct option is D $+4P_0{V}_0$

Let $P$ be the pressure of the gas in the cylinder, then the force exerted by the gas on the piston of the cylinder,

$F=PA$

In a small displacement of piston through $dc$, work done by the gas,

$dW=F.dx=PAdx=PdV$

$\therefore$ Total amount of work done $\Delta W=\int dW={\int }_{V_i}^{V_f}PdV=P(V_f-V_i)$

In $P-V$ diagram or indicator diagram, the area under $P-V$ curve represents work done.

$W=$ Area under $P-V$ diagram

According to the $P-V$ diagram given in the question,

Work done in the process $ABCD=$ Area of rectangle $ABCDA$

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

$=2V_0\times P_0=2P_0{V}_0$

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.