JEE Main 2019PhysicsKinetic Theory of GasesMediumMCQ

JEE Main 2019 — Kinetic Theory of Gases Question with Solution

JEE Main 2019 (09 Apr Shift 2)

Question

The specific heats, Cp and Cv of a gas of diatomic molecules, A, are given (in units of J mol-1 K-1 ) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then:

Choose an option

Show full solutionCorrect option: B
Correct answer
BA has a vibrational mode but B has none.

Step-by-step explanation

For A :
R=Cp-Cv=29-22=7
Cv=fR2
⇒22=f×72
⇒f=447=6.3
⇒f≈6

(5 rotational + translational, 1 vibrational)


For B :
R=Cp-CV=30-21=9
Cv=fR2=21
⇒f=21×29=429=4.66≃5
∴ f≃5

 (5 rotational + transitional, 0 vibrational)

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About this question

This is a previous-year question from JEE Main 2019, 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.