Question

Two moles of Helium gas are taken over the cycle ABCDA as shown in the P-T diagram. The net work done on the gas in the cycle $ABCDA$ is :

• A. $0$
• B. $276$ $R$
• C. $1076$ $R$
• D. $1094$ $R$

Answer 1

The correct option is B $276$ $R$

$W_{ABCD}=W_{AB}+W_{BC}+W_{CD}+W_{DA}$.......(1)

$W_{AB}=nR\Delta T=2\times R\times (500-300)=400R$

$W_{BC}=nRT\ln \left(\frac{P_B}{P_C}\right)=2R\times 500\ln 2=690R$

$W_{CD}=2\times R\times (300-500)$
$=-400R$

$W_{DA}=nRT\ln \left(\frac{P_D}{P_A}\right)=2R\times 300\ln 0.5=414R$

From equation 1.

$W_{ABCD}=400R+690R+(-400R)-414R$
$=276$ $R$