JEE Main 2020PhysicsKinetic Theory of GasesMediumMCQ

JEE Main 2020 — Kinetic Theory of Gases Question with Solution

JEE Main 2020 (03 Sep Shift 2)

Question

To raise the temperature of a certain mass of gas by 50 °C at a constant pressure,160 calories of heat is required. When the same mass of gas is cooled by 100°C at constant volume,240 calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?

Choose an option

Show full solutionCorrect option: B
Correct answer
B6

Step-by-step explanation

At constant pressure :

∆Q=nCP∆T ⇒   160=nCP.50  .....(1)

At constant volume

∆Q=nCv∆T ⇒  240=nCv.100......(2)

Equation (1) divided by (2)

160240=CPCV·50100

CPCV=43=1+2f

f=6

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Kinetic Theory of Gases 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 2020, covering the Kinetic Theory of Gases chapter of Physics. 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.