Specific Heat With Vibration
Consider a diatomic gas heated to a temperature high enough that both rotational and vibrational modes are fully active; determine its molar specific heat at constant volume.
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Solution
29.1J/mol\cdotK
When a diatomic molecule's vibrational mode becomes fully active at high temperature, it contributes two additional degrees of freedom—one kinetic and one potential—bringing the total to seven. By equipartition, the molar specific heat at constant volume is Cv=2fR=27R. Numerically, Cv=3.5×8.314=29.1 J/mol\cdotK. Vibrational modes are quantised and stay frozen until the thermal energy kBT becomes comparable to the spacing between vibrational levels, which is why measured Cv rises in steps as temperature increases. The value 20.8 J/mol\cdotK equals 25R, the rigid diatomic case without vibration. The value 12.5 J/mol\cdotK equals 23R, valid only for a monatomic gas. The value 37.4 J/mol\cdotK corresponds to 29R, which over-counts the vibrational contribution. This applies the NCERT law of equipartition, noting that each vibrational mode counts twice because it stores both kinetic and potential energy in the oscillating bond. The corresponding ratio of specific heats falls toward γ=9/7≈1.29 once vibration is fully active, since extra storage modes always lower γ. As a plausibility check, real diatomic gases show Cv rising from about 25R toward 27R as temperature increases and vibrations switch on, so the value 29.1 J/mol\cdotK is physically reasonable for the fully excited diatomic case described.
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About This Question
- Subject
- physics
- Chapter
- kinetic theory of gases
- Topic
- specific heat with vibration
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
29.1J/mol\cdotK
When a diatomic molecule's vibrational mode becomes fully active at high temperature, it contributes two additional degrees of freedom—one kinetic and one potential—bringing the total to seven. By equipartition, the molar specific heat at constant volume is Cv=2fR=27R. Numerically, Cv=3.5×8.314=29.1 J/mol\cdotK. Vibrational modes are quantised and stay frozen until the thermal energy kBT becomes comparable to the spacing between vibrational levels, which is why measured Cv rises in steps as temperature increases. The value 20.8 J/mol\cdotK equals 25R, the rigid diatomic case without vibration. The value 12.5 J/mol\cdotK equals 23R, valid only for a monatomic gas. The value 37.4 J/mol\cdotK corresponds to 29R, which over-counts the vibrational contribution. This applies the NCERT law of equipartition, noting that each vibrational mode counts twice because it stores both kinetic and potential energy in the oscillating bond. The corresponding ratio of specific heats falls toward γ=9/7≈1.29 once vibration is fully active, since extra storage modes always lower γ. As a plausibility check, real diatomic gases show Cv rising from about 25R toward 27R as temperature increases and vibrations switch on, so the value 29.1 J/mol\cdotK is physically reasonable for the fully excited diatomic case described.
This medium difficulty physics question is from the chapter kinetic theory of gases, covering the topic of specific heat with vibration. It appeared in the 2025 exam.
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