Thermodynamics

JEE Main level, single correct, medium

101 of 190

Question 101 of 190

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, (given γ=32\gamma =\dfrac{3}{2}) :

Show answer and solution

Answer: (D) W=RT[2−2]W=RT\left[2-\sqrt{2}\right]

Work done by gas in adiabatic process is:
W=−(P2V2−P1V1)γ−1W=-\dfrac{\left({P}_{2}{V}_{2}-{P}_{1}{V}_{1}\right)}{\gamma -1}
=−nR(T2−T1)γ−1=-\dfrac{nR\left({T}_{2}-{T}_{1}\right)}{\gamma -1}
For an adiabatic process, T1V1γ−1=T2V2γ−1{T}_{1}{{V}_{1}}^{\gamma -1}={T}_{2}{{V}_{2}}^{\gamma -1}
⇒TV32−1=T2(2V)32−1⇒TV12=T2(2V)12⇒T2=T2\Rightarrow T{V}^{\dfrac{3}{2}-1}={T}_{2}{\left(2V\right)}^{\dfrac{3}{2}-1} \Rightarrow T{V}^{\dfrac{1}{2}}={T}_{2}{\left(2V\right)}^{\dfrac{1}{2}} \Rightarrow {T}_{2}=\dfrac{T}{\sqrt{2}}
Now, work done is W=R(T1−T2)γ−1=R(T−T2)12=RT(2−2)W=\dfrac{R\left({T}_{1}-{T}_{2}\right)}{\gamma -1}=\dfrac{R\left(T-\dfrac{T}{\sqrt{2}}\right)}{\dfrac{1}{2}}=RT\left(2-\sqrt{2}\right)

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Thermodynamics (Thermodynamic Processes), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.