Question

For a diatomic gas, if \( \gamma_1 = \frac{C_P}{C_V} \) for rigid molecules and \( \gamma_2 = \frac{C_P}{C_V} \) for another diatomic molecules, but also having vibrational modes. Then, which one of the following options is correct? (where \( C_P \) and \( C_V \) are specific heats of the gas at constant pressure and volume)

• A. \( \gamma_2 = \gamma_1 \)
• B. \( \gamma_2 > \gamma_1 \)
• C. \( 2 \gamma_2 = \gamma_1 \)
• D. \( \gamma_2 < \gamma_1 \)

Hint:

For diatomic gases with vibrational modes, the specific heat ratio \( \gamma \) will decrease compared to rigid molecules, as vibrational modes add additional degrees of freedom, which reduces the overall energy increase per unit temperature.

Answer 1

The Correct Option is D

For diatomic molecules: - For rigid molecules, the specific heat ratio \( \gamma_1 = \frac{C_P}{C_V} \) is typically 5/3 for a monoatomic gas. - For diatomic gases with vibrational modes included, the value of \( \gamma_2 \) will be lower, since vibrational modes contribute more degrees of freedom which lower the specific heat ratio. Thus, \( \gamma_2 \) is smaller than \( \gamma_1 \), as vibrational modes lead to higher internal energy without increasing the temperature as much. Therefore, the correct answer is \( \boxed{\gamma_2 < \gamma_1} \).

Answer 2

Step 1: Recall the formula for \( \gamma \). 
\[ \gamma = \frac{C_P}{C_V} = \frac{C_V + R}{C_V}. \] Hence, \[ \gamma = 1 + \frac{R}{C_V}. \]

Step 2: For a rigid diatomic molecule (no vibrational modes).
For a rigid diatomic gas, the degrees of freedom \( f = 5 \) (3 translational + 2 rotational). \[ C_V = \frac{f}{2}R = \frac{5}{2}R. \] Thus, \[ \gamma_1 = 1 + \frac{R}{C_V} = 1 + \frac{R}{(5/2)R} = 1 + \frac{2}{5} = \frac{7}{5} = 1.4. \]

Step 3: For a diatomic molecule with vibrational modes.
When vibrational modes are also active, additional degrees of freedom are present. Each vibrational mode contributes 2 degrees of freedom (one kinetic + one potential). So total \( f = 7 \). \[ C_V = \frac{f}{2}R = \frac{7}{2}R. \] Hence, \[ \gamma_2 = 1 + \frac{R}{C_V} = 1 + \frac{R}{(7/2)R} = 1 + \frac{2}{7} = \frac{9}{7} \approx 1.29. \]

Step 4: Compare \( \gamma_1 \) and \( \gamma_2 \).
\[ \gamma_1 = 1.4, \quad \gamma_2 = 1.29. \] Clearly, \[ \boxed{\gamma_2 < \gamma_1.} \]

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Final Answer:

\[ \boxed{\gamma_2 < \gamma_1} \]