Question

An ideal gas is taken around the cycle $ABCA$ shown in $P-V$ diagram. The network done by the gas during the cycle is equal to:

• A. $12P_1{V}_1$
• B. $6P_1{V}_1$
• C. $3P_1{V}_1$
• D. $P_1{V}_1$

Answer 1

The correct option is C $3P_1{V}_1$

$A⟶B$ is an isobaric process.

$\therefore \Delta V=0\therefore W_{AB}=P\left(\Delta V\right)=0$

$B⟶C$ is an isothermal process.

$\therefore W_{BC}=-nRTln\left(\frac{3V_1}{V_1}\right)$
$=-P_1{V}_1$

$C⟶A$ is an isobaric process.

$\therefore W_{AC}=-P_1\Delta V$

$=-P_1(3V_1-V_1)$

$=-2P_1{V}_1$

$\therefore$ Work done on the gas=$W_{AB}+W_{BC}+W_{AC}$

$=0-P_1{V}_1-2P_1{V}_1$

$=-3P_1{V}_1$

But the work done by the gas$=-$Work done on the gas

=$-(-3P_1{V}_1)$

$=3P_1{V}_1$