For A :
R=Cp−Cv=29−22=7
Cv=2fR
⇒22=2f×7
⇒f=744=6.3
⇒f≈6
(5 rotational + translational, 1 vibrational)
For B :
R=Cp−CV=30−21=9
Cv=2fR=21
⇒f=921×2=942=4.66≃5
∴f≃5
(5 rotational + transitional, 0 vibrational)
The specific heats, C_ and C_v of a gas of diatomic molecules, A, are given (in units of J mo ^-1 K^-1 ) by 29 and 22, respectively. Another gas of…
JEE Main 2019 — Physics Thermodynamics
2019mcqmedium
The specific heats, Cp and Cv of a gas of diatomic molecules, A, are given (in units of Jmol−1K−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:
Official previous-year question
Held on 9 Apr 2019 · Verified 6 Jul 2026.
Options
- A
A has one vibrational mode and B has two
- B
A has a vibrational mode but B has none.
- C
Both A and B have a vibrational mode each.
- D
A is rigid but B has a vibrational mode.
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