JEE Mains · Physics · STD 11 - Thermodynamics
- JEE Mains 2024
- Hard
- MCQ
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 _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\mathrm{RT}[2-\sqrt{2}]\)
Step-by-step Solution
Detailed explanation
\(\mathrm{TV}^{\gamma-1}=\text { constant }\)
\(\Rightarrow \mathrm{T}(\mathrm{V})^{\frac{3}{2}-1}=\mathrm{T}_{\mathrm{f}}(2 \mathrm{~V})^{\frac{3}{2}-1}\)
\(\Rightarrow \mathrm{TV}^{\frac{1}{2}}=\mathrm{T}_{\mathrm{f}}(2)^{\frac{1}{2}}(\mathrm{~V})^{\frac{1}{2}}\)
\(\Rightarrow \mathrm{T}_{\mathrm{f}}=\left(\frac{\mathrm{T}}{\sqrt{2}}\right)\)
\(\text { Now, W.D. }=\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\gamma}=\frac{1 \cdot \mathrm{R}\left[\frac{\mathrm{T}}{\sqrt{2}}-\mathrm{T}\right]}{1-\frac{3}{2}}\)
\(\Rightarrow \text { W.D. }=2 \mathrm{RT}\left[1-\frac{1}{\sqrt{2}}\right]\)
\(\Rightarrow \text { W.D. }=\mathrm{RT}[2-\sqrt{2}]\)
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