Verified Solution • 2025–26 Session moderate

QJEEM23Phe47d: JEE Main 2023 Physics - Class 11 JEEM23Phe47d Solution: A sample of gas at temperature is adiabatically expanded to…

A sample of gas at temperature TT is adiabatically expanded to double its volume. The work done by the gas in the process is (given,γ=32)\left(\mathrm{given}, \gamma=\frac{3}{2}\right) :

Multiple Choice Options

  • A W=TR[2−2]W=T R[\sqrt{2}-2]
  • B W=TR[2−2]W=\frac{T}{R}[\sqrt{2}-2]
  • C W=RT[2−2]W=\frac{R}{T}[2-\sqrt{2}]
  • D W=RT[2−2]W=R T[2-\sqrt{2}]

Step-by-Step Verified Solution

γ=32\gamma=\frac{3}{2}

amp;W=nRΔT1−γ=nRTf−nRTi1−γamp;=(PV)f−(PVi)1−γ… (1) amp;PVγ= constant amp;PiViγ=Pf(2Vi)γ⇒Pf=Pi2γ=Pi22......(2)\begin{aligned}& W =\frac{n R \Delta T}{1-\gamma}=\frac{n R T_{f}-n R T_{i}}{1-\gamma} \\\\& =\frac{(P V)_{f}-\left(P V_{i}\right)}{1-\gamma} \quad \ldots \text { (1) } \\\\& P V^{\gamma}=\text { constant } \\\\& P_{i} V_{i}^{\gamma}=P_{f}\left(2 V_{i}\right)^{\gamma} \Rightarrow P_{f}=\frac{P_{i}}{2^{\gamma}}=\frac{P_{i}}{2 \sqrt{2}} ......(2)\end{aligned}

From (1) and (2)

amp;W=Pi222Vi−PiVi1−γ=PiVi−1/2(12−1)amp;=−nRT(2−2)amp;=nRT(2−2)\begin{aligned}& W=\frac{\frac{P_{i}}{2 \sqrt{2}} 2 V_{i}-P_{i} V_{i}}{1-\gamma}=\frac{P_{i} V_{i}}{-1 / 2}\left(\frac{1}{\sqrt{2}}-1\right) \\\\& =-n R T(\sqrt{2}-2) \\\\& =n R T(2-\sqrt{2})\end{aligned}

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