Search any question & find its solution
Question:
Answered & Verified by Expert
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 $A \rightarrow B$.
Options:
Solution:
1536 Upvotes
Verified Answer
The correct answer is:
$\frac{p_0 V_0\left(2^{1 / \gamma}-2\right)}{(1-\gamma)}$
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
$$
\begin{aligned}
W & =\frac{p_0 V_1-2 p_0 V_0}{1-\gamma} \\
& =\frac{p_0 V_0 2^{1 / \gamma}-2 p_0 V_0}{1-\gamma} \\
W & =\frac{p_0 V_0\left(2^{1 / \gamma}-2\right)}{1-\gamma}
\end{aligned}
$$
$$
p V^\gamma=\text { constant }
$$
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
$$
\begin{aligned}
W & =\frac{p_0 V_1-2 p_0 V_0}{1-\gamma} \\
& =\frac{p_0 V_0 2^{1 / \gamma}-2 p_0 V_0}{1-\gamma} \\
W & =\frac{p_0 V_0\left(2^{1 / \gamma}-2\right)}{1-\gamma}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.