Question:
In non-rigid diatomic molecule with an additional vibrational mode
In non-rigid diatomic molecule with an additional vibrational mode
Show Hint
For a diatomic gas with vibrational mode active:
\[
f=7,\qquad C_V=\frac{7R}{2},\qquad C_P=\frac{9R}{2}
\]
Always use \(C_P=C_V+R\).
Updated On: Jun 22, 2026
- \(81C_V^2=49C_P^2\)
- \(49C_V^2=25C_P^2\)
- \(49C_V^2=81C_P^2\)
- \(25C_V^2=49C_P^2\)
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The Correct Option is A
Solution and Explanation
Step 1: Determine degrees of freedom.
For a non-rigid diatomic molecule with one vibrational mode active:
- Translational degrees of freedom \(=3\)
- Rotational degrees of freedom \(=2\)
- One vibrational mode contributes \(2\) degrees of freedom
Therefore, total degrees of freedom are \[ f=3+2+2=7 \]
Step 2: Write the molar specific heat at constant volume.
For an ideal gas, \[ C_V=\frac{f}{2}R \] Thus, \[ C_V=\frac{7}{2}R \]
Step 3: Find \(C_P\).
Using Mayer's relation, \[ C_P=C_V+R \] \[ C_P=\frac{7}{2}R+R \] \[ C_P=\frac{9}{2}R \]
Step 4: Find the relation between \(C_V\) and \(C_P\).
From \[ C_V=\frac{7}{2}R \] and \[ C_P=\frac{9}{2}R \] we get \[ \frac{C_V}{C_P}=\frac{7}{9} \] Squaring both sides, \[ \frac{C_V^2}{C_P^2}=\frac{49}{81} \] Cross multiplying, \[ 81C_V^2=49C_P^2 \]
Step 5: Final conclusion.
Hence, the correct relation is \[ \boxed{81C_V^2=49C_P^2} \]
For a non-rigid diatomic molecule with one vibrational mode active:
- Translational degrees of freedom \(=3\)
- Rotational degrees of freedom \(=2\)
- One vibrational mode contributes \(2\) degrees of freedom
Therefore, total degrees of freedom are \[ f=3+2+2=7 \]
Step 2: Write the molar specific heat at constant volume.
For an ideal gas, \[ C_V=\frac{f}{2}R \] Thus, \[ C_V=\frac{7}{2}R \]
Step 3: Find \(C_P\).
Using Mayer's relation, \[ C_P=C_V+R \] \[ C_P=\frac{7}{2}R+R \] \[ C_P=\frac{9}{2}R \]
Step 4: Find the relation between \(C_V\) and \(C_P\).
From \[ C_V=\frac{7}{2}R \] and \[ C_P=\frac{9}{2}R \] we get \[ \frac{C_V}{C_P}=\frac{7}{9} \] Squaring both sides, \[ \frac{C_V^2}{C_P^2}=\frac{49}{81} \] Cross multiplying, \[ 81C_V^2=49C_P^2 \]
Step 5: Final conclusion.
Hence, the correct relation is \[ \boxed{81C_V^2=49C_P^2} \]
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