Question

One mole of an ideal monoatomic gas is carried through the reversible cyclic process as shown in figure. Then which statements are correct?

• A. $\Delta U$ for complete cycle is zero
• B. $\Delta U$ along path $BC$ is $\frac{-3}{2}{P}_0{V}_0$
• C. Work done along $BC$ is $–2P_0{V}_0$
• D. Net heat absorbed by the gas in the path $BC$ is $\frac{1}{2}{P}_0{V}_0$

Answer 1

Answer: (A), (B), (C), (D)

Net heat absorbed by the gas in the path $BC$ is $\frac{1}{2}{P}_0{V}_0$

$\Delta =q+w$

$w_{BC}=\frac{1}{2}(2V^0-V^0)(P^0-3P^0)+(2V^0-V^0)(0-P^0)=-2P^0{V}^0$
$=1\times \frac{3}{2}R[\frac{P^0\left(2V^0\right)}{R}-\frac{3P^0{V}^0}{R}]$
$=\frac{-3}{2}{P}^0{V}^0$
$qBC=\Delta U-w$
$=\frac{-3}{2}{P}^0{V}^0+2P^0{V}^0$
$=\frac{1}{2}{P}^0{V}^0$