Start free

Physics · Thermodynamics

JEE Main 2024 — 4 April, Shift 2 — Question 46

A sample of gas at temperature T is adiabatically expanded to double its volume. Adiabatic constant for the gas is γ=3/2\gamma=3 / 2. The work done by the gas in the process is : (μ=1(\mu=1 mole ))

  1. Option A:

    RT[2−2]\mathrm{RT}[\sqrt{2}-2]

  2. Option B:

    RT[1−22]\mathrm{RT}[1-2 \sqrt{2}]

  3. Option C:

    RT⁡[22−1]\operatorname{RT}[2 \sqrt{2}-1]

  4. Option D:

    RT[2−2]\mathrm{RT}[2-\sqrt{2}]

    Correct

Answer: D

Step-by-step solution

W=nRΔT1−γ\mathrm{W}=\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\gamma}

TVγ−1=\mathrm{TV}^{\gamma-1}= cons tan⁡t=Tf(2 V)γ−1\tan \mathrm{t}=\mathrm{T}_{\mathrm{f}}(2 \mathrm{~V})^{\gamma-1}

Tf=T(12)1/2=T2\mathrm{T}_{\mathrm{f}}=\mathrm{T}\left(\frac{1}{2}\right)^{1 / 2}=\frac{\mathrm{T}}{\sqrt{2}} W=R(T2−T)1−32=2RT(2−1)2\mathrm{W}=\frac{\mathrm{R}\left(\frac{\mathrm{T}}{\sqrt{2}}-\mathrm{T}\right)}{1-\frac{3}{2}}=2 R T \frac{(\sqrt{2}-1)}{\sqrt{2}} =RT(2−2)=\mathrm{RT}(2-\sqrt{2})

Answer key and solution verified before publishing.

Practise Thermodynamics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
4 April, Shift 2
Subject
Physics
Chapter
Thermodynamics
Topic
Different Thermodynamic Processes