An ideal gas goes through a process $A \rightarrow B \rightarrow C \rightarrow A$ cycle. The process, $A \rightarrow B$ is adiabatic. Calculate the work done in the process $A \rightarrow B$.

• A

  $p_0 V_0$
• Correct answer
  B

  $\frac{p_0 V_0\left(2^{1 / \gamma}-2\right)}{(1-\gamma)}$
• C

  $p_0 V_0 \ln (2)$
• D

  $\frac{p_0 V_0\left(2^{1 / \gamma}-1\right)}{(\gamma-1)}$

Solution

We know that, adiabatic relation, $$ p V^\gamma=\text { constant } $$ For adiabatic process $A \rightarrow B$ $$ \begin{aligned} 2 p_0 V_0^\gamma & =p_0 V_1^\gamma \\ \Rightarrow \quad V_1^\gamma & =2 V_0^\gamma \Rightarrow V_1=2^{1 / \gamma} V_0 \end{aligned} $$
$\therefore$ Work done in the process $A \rightarrow B$ is $$ \begin{aligned} W & =\frac{p_0 V_1-2 p_0 V_0}{1-\gamma} \\ & =\frac{p_0 V_0 2^{1 / \gamma}-2 p_0 V_0}{1-\gamma} \\ W & =\frac{p_0 V_0\left(2^{1 / \gamma}-2\right)}{1-\gamma} \end{aligned} $$

Hence, the correct answer is option: 'B'.