Q.50. A sample of gas at temperature T is adiabatically expanded to double its volume. The work done by the gas in the process is ( \frac{C_{p}}{C_{v}}=\gamma=\frac{3}{2} ) (R= gas constant )
A. TR(\sqrt{2}-2)
B. \frac{T}{R}(\sqrt{2}-2)
C. \frac{R}{T}(2-\sqrt{2})
D. RT(2-\sqrt{2})
Answer :- D. RT(2-\sqrt{2})
Explanation :-
\text{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) \quad \ldots (\text{Given : } \gamma = \frac{3}{2}) \therefore \quad \mathrm{T}_1 = \mathrm{T}_2 (\sqrt{2}) \therefore \quad \mathrm{T}_2 = \frac{\mathrm{T}}{\sqrt{2}} \text{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 \quad \mathrm{~W}_{\mathrm{adi}} = \mathrm{RT}(2 - \sqrt{2})