Thermodynamics

JEE Main level, single correct, medium

85 of 190

Question 85 of 190

A sample of 1 mole gas at temperature T\mathrm{T} is adiabatically expanded to double its volume. If adiabatic constant for the gas is γ=32\gamma=\frac{3}{2}, then the work done by the gas in the process is:

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Answer: (C) RT [2−2][2-\sqrt{2}]

TVγ−1= constant ⇒T(V)32−1=Tf(2 V)32−1⇒TV12=Tf(2)12( V)12⇒Tf=(T2) Now, W.D. =nRΔT1−γ=1⋅R[T2−T]1−32⇒ W.D. =2RT[1−12]⇒ W.D. =RT[2−2]\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}

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