A sample of 1 mole gas at temperature $\mathrm{T}$ is adiabatically expanded to double its volume. If…

A sample of 1 mole gas at temperature $\mathrm{T}$ is adiabatically expanded to double its volume. If adiabatic constant for the gas is $\gamma=\frac{3}{2}$, then the work done by the gas in the process is:
  1. $\frac{R}{T}[2-\sqrt{2}]$
  2. $\frac{T}{R}[2+\sqrt{2}]$
  3. RT $[2-\sqrt{2}]$
  4. $\mathrm{RT}[2+\sqrt{2}]$

Solution

$\begin{aligned} & \mathrm{TV}^{\gamma-1}=\text { constant } \\ & \Rightarrow \mathrm{T}(\mathrm{V})^{\frac{3}{2}-1}=\mathrm{T}_{\mathrm{f}}(2 \mathrm{~V})^{\frac{3}{2}-1} \\ & \Rightarrow \mathrm{TV}^{\frac{1}{2}}=\mathrm{T}_{\mathrm{f}}(2)^{\frac{1}{2}}(\mathrm{~V})^{\frac{1}{2}} \\ & \Rightarrow \mathrm{T}_{\mathrm{f}}=\left(\frac{\mathrm{T}}{\sqrt{2}}\right) \\ & \text { Now, W.D. }=\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\gamma}=\frac{1 \cdot \mathrm{R}\left[\frac{\mathrm{T}}{\sqrt{2}}-\mathrm{T}\right]}{1-\frac{3}{2}} \\ & \Rightarrow \text { W.D. }=2 \mathrm{RT}\left[1-\frac{1}{\sqrt{2}}\right] \\ & \Rightarrow \text { W.D. }=\mathrm{RT}[2-\sqrt{2}]\end{aligned}$

Asked in: JEE Main 2024 (09 Apr Shift 1)

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