Medium MCQ +4 / -1 PYQ · JEE Mains 2023

A sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is $\left(\mathrm{given}, \gamma=\frac{3}{2}\right)$ :

  1. A $W=T R[\sqrt{2}-2]$
  2. B $W=\frac{T}{R}[\sqrt{2}-2]$
  3. C $W=\frac{R}{T}[2-\sqrt{2}]$
  4. D $W=R T[2-\sqrt{2}]$ Correct answer

Solution

$\gamma=\frac{3}{2}$ <br/><br/>$$ \begin{aligned} & W =\frac{n R \Delta T}{1-\gamma}=\frac{n R T_{f}-n R T_{i}}{1-\gamma} \\\\ & =\frac{(P V)_{f}-\left(P V_{i}\right)}{1-\gamma} \quad \ldots \text { (1) } \\\\ & P V^{\gamma}=\text { constant } \\\\ & P_{i} V_{i}^{\gamma}=P_{f}\left(2 V_{i}\right)^{\gamma} \Rightarrow P_{f}=\frac{P_{i}}{2^{\gamma}}=\frac{P_{i}}{2 \sqrt{2}} ......(2) \end{aligned} $$ <br/><br/>From (1) and (2) <br/><br/>$$ \begin{aligned} & W=\frac{\frac{P_{i}}{2 \sqrt{2}} 2 V_{i}-P_{i} V_{i}}{1-\gamma}=\frac{P_{i} V_{i}}{-1 / 2}\left(\frac{1}{\sqrt{2}}-1\right) \\\\ & =-n R T(\sqrt{2}-2) \\\\ & =n R T(2-\sqrt{2}) \end{aligned} $$

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law

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