Thermodynamics

JEE Main level, single correct, medium

111 of 190

Question 111 of 190

A monoatomic gas at pressure PP and volume VV is suddenly compressed to one eighth of its original volume. The final pressure at constant entropy will be

Show answer and solution

Answer: (C) 32P32P

Change in entropy is given by dS=dQTdS=\dfrac{dQ}{T}. Since, the process is dS=0dS=0, dQ=0dQ=0. Therefore, constant entropy means the process is adiabatic.
PVγ=P{V}^{\gamma }=constant
⇒P1V1γ=P2V2γ\Rightarrow {P}_{1}{V}_{1}^{\gamma }={P}_{2}{V}_{2}^{\gamma }
For a monatomic gas γ=53\gamma =\dfrac{5}{3}
⇒P1V153=P2(V18)53\Rightarrow {P}_{1}{V}_{1}^{\dfrac{5}{3}}={P}_{2}{\left(\dfrac{{V}_{1}}{8}\right)}^{\dfrac{5}{3}}
⇒P1V153=P2V15332\Rightarrow {P}_{1}{V}_{1}^{\dfrac{5}{3}}=\dfrac{{P}_{2}{V}_{1}^{\dfrac{5}{3}}}{32}
⇒P2=32P1\Rightarrow {P}_{2}=32{P}_{1}

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.