Question

In non-rigid diatomic molecule with an additional vibrational mode

• A. \(81C_V^2=49C_P^2\)
• B. \(49C_V^2=25C_P^2\)
• C. \(49C_V^2=81C_P^2\)
• D. \(25C_V^2=49C_P^2\)

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 \(C_P=C_V+R\).

Answer 1

The Correct Option is A

Step 1: Count the degrees of freedom.
A non-rigid diatomic molecule with an active vibrational mode has translational $= 3$, rotational $= 2$, and vibrational $= 2$ degrees of freedom. So the total is \[ f = 3 + 2 + 2 = 7 \]
Step 2: Write the molar heat at constant volume.
By the equipartition theorem, \[ C_V = \frac{f}{2}R = \frac{7}{2}R \]
Step 3: Find the molar heat at constant pressure.
Using Mayer's relation $C_P = C_V + R$, \[ C_P = \frac{7}{2}R + R = \frac{9}{2}R \]
Step 4: Form the ratio of the two heats.
\[ \frac{C_V}{C_P} = \frac{\tfrac{7}{2}R}{\tfrac{9}{2}R} = \frac{7}{9} \]
Step 5: Cross-multiply to get a clean relation.
\[ 9 C_V = 7 C_P \] Squaring both sides, \[ 81 C_V^2 = 49 C_P^2 \]
Step 6: Match with the options.
This is exactly option (1). \[ \boxed{81 C_V^2 = 49 C_P^2} \]