MHT CET · Physics · Thermodynamics
A sample of gas at temperature \(\mathrm{T}\) is adiabatically expanded to double its volume. The work done by the gas in the process is \(\left(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{3}{2}\right)(\mathrm{R}=\) gas constant \()\)
- A \(\operatorname{TR}(\sqrt{2}-2)\)
- B \(\frac{\mathrm{T}}{\mathrm{R}}(\sqrt{2}-2)\)
- C \(\frac{\mathrm{R}}{\mathrm{T}}(2-\sqrt{2})\)
- D \(\mathrm{RT}(2-\sqrt{2})\)
Answer & Solution
Correct Answer
(D) \(\mathrm{RT}(2-\sqrt{2})\)
Step-by-step Solution
Detailed explanation
Using the formula for adiabatic expansion,
\(\mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \)
\( \mathrm{~T}_1 \mathrm{~V}_1^{(1 / 2)}=\mathrm{T}_2\left(2 \mathrm{~V}_1^{1 / 2}\right)\) \(\ldots .\left(\text { Given: } \gamma=\frac{3}{2}\right) \)
\( \therefore \mathrm{T}_1=\mathrm{T}_2(\sqrt{2}) \)
\( \therefore \mathrm{T}_2=\frac{\mathrm{T}}{\sqrt{2}}\)
Work done
\(\mathrm{W}_{\mathrm{adi}}= \frac{\mathrm{R}\left(\mathrm{T}-\mathrm{T}_2\right)}{\gamma-1}=\frac{\mathrm{R}\left(\mathrm{T}-\frac{\mathrm{T}}{\sqrt{2}}\right)}{\frac{1}{2}} \)
\( =\frac{\mathrm{R}(\sqrt{2} \mathrm{~T}-\mathrm{T})}{\sqrt{2}} \times 2=\mathrm{RT}(\sqrt{2}-1) \sqrt{2} \)
\( \therefore \mathrm{W}_{\text {adi }} =\mathrm{RT}(2-\sqrt{2})\)
\(\mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \)
\( \mathrm{~T}_1 \mathrm{~V}_1^{(1 / 2)}=\mathrm{T}_2\left(2 \mathrm{~V}_1^{1 / 2}\right)\) \(\ldots .\left(\text { Given: } \gamma=\frac{3}{2}\right) \)
\( \therefore \mathrm{T}_1=\mathrm{T}_2(\sqrt{2}) \)
\( \therefore \mathrm{T}_2=\frac{\mathrm{T}}{\sqrt{2}}\)
Work done
\(\mathrm{W}_{\mathrm{adi}}= \frac{\mathrm{R}\left(\mathrm{T}-\mathrm{T}_2\right)}{\gamma-1}=\frac{\mathrm{R}\left(\mathrm{T}-\frac{\mathrm{T}}{\sqrt{2}}\right)}{\frac{1}{2}} \)
\( =\frac{\mathrm{R}(\sqrt{2} \mathrm{~T}-\mathrm{T})}{\sqrt{2}} \times 2=\mathrm{RT}(\sqrt{2}-1) \sqrt{2} \)
\( \therefore \mathrm{W}_{\text {adi }} =\mathrm{RT}(2-\sqrt{2})\)
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