Degrees Of Freedom And Specific Heat
Treating a rigid diatomic gas molecule with only translational and rotational motion active, what value does the ratio of specific heats γ take for this gas?
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About This Question
- Subject
- physics
- Chapter
- thermodynamics
- Topic
- degrees of freedom and specific heat
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
7/5
By the equipartition theorem, each active degree of freedom contributes 21R to the molar specific heat at constant volume. A rigid diatomic molecule has three translational and two rotational degrees of freedom, giving f=5, so CV=25R. Using Mayer's relation, CP=CV+R=27R. The ratio is therefore γ=CVCP=5/27/2=57=1.4. The value 5/3 corresponds to a monatomic gas with only three degrees of freedom. The value 4/3 matches a nonlinear polyatomic gas with six degrees of freedom. The value 9/7 would apply if a vibrational mode were excited, raising f to 7, but the molecule here is specified as rigid. The general formula γ=1+f2 confirms the result: with f=5, γ=1+2/5=7/5, which matches the measured value for gases like oxygen and nitrogen at room temperature. The reason two rotational modes count for a diatomic molecule is that rotation about the bond axis carries negligible moment of inertia and stays unexcited. At high temperatures the vibrational mode can switch on, adding two more degrees of freedom and lowering γ toward 9/7, which is why measured specific heats are mildly temperature dependent.
This medium difficulty physics question is from the chapter thermodynamics, covering the topic of degrees of freedom and specific heat. It appeared in the 2025 exam.
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