JEE Main 2019ChemistryThermodynamics (C)MediumMCQ

JEE Main 2019 — Thermodynamics (C) Question with Solution

JEE Main 2019 (08 Apr Shift 2)

Question

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV=28 J K-1, calculate ∆U and ∆pV for the process. (R=8.0 J K-1 mol-1)

Choose an option

Show full solutionCorrect option: C
Correct answer
C∆U=14 kJ; ∆pV=4 kJ

Step-by-step explanation

ΔU=nCVΔT
=5×28×100=14000 J
=14 kJ
Δ(PV)=nRΔT=5×8×100=4000 J=4 kJ

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Thermodynamics (C) chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2019, covering the Thermodynamics (C) chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.