Question:
According to law of equipartition of energy the molar specific heat of a non rigid diatomic gas at constant volume is
According to law of equipartition of energy the molar specific heat of a non rigid diatomic gas at constant volume is
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A non-rigid diatomic molecule has 7 degrees of freedom, so \(C_V=\frac72R\).
Updated On: Oct 1, 2026
- \((9/2)R\)
- \((5/2)R\)
- \((3/2)R\)
- \((7/2)R\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
By the law of equipartition, each quadratic energy term contributes \(\frac12RT\) per mole.
Step 2: Count the terms:
A non-rigid diatomic gas has 3 translational and 2 rotational degrees of freedom. It also vibrates, which adds 2 more terms (kinetic and potential), so the energy has \(3+2+2 = 7\) terms.
\[ U = \frac72RT,\quad C_V = \frac{dU}{dT} = \frac72R \]
Step 3: Other options:
\(\frac52R\) is for a rigid diatomic molecule. \(\frac32R\) is for a monatomic gas.
Final Answer:
\(C_V = \frac72R\), option (D). \[ \boxed{\frac{7}{2}R} \]
By the law of equipartition, each quadratic energy term contributes \(\frac12RT\) per mole.
Step 2: Count the terms:
A non-rigid diatomic gas has 3 translational and 2 rotational degrees of freedom. It also vibrates, which adds 2 more terms (kinetic and potential), so the energy has \(3+2+2 = 7\) terms.
\[ U = \frac72RT,\quad C_V = \frac{dU}{dT} = \frac72R \]
Step 3: Other options:
\(\frac52R\) is for a rigid diatomic molecule. \(\frac32R\) is for a monatomic gas.
Final Answer:
\(C_V = \frac72R\), option (D). \[ \boxed{\frac{7}{2}R} \]
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