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, the specific heat ratio \( \gamma \) is defined as \( \frac{C_P}{C_V} \). For rigid molecules (monoatomic gases), this ratio is typically \( \frac{5}{3} \), denoted as \( \gamma_1 \). When vibrational modes are considered in diatomic gases, the specific heat ratio, denoted as \( \gamma_2 \), decreases. This is because vibrational modes add degrees of freedom, leading to increased internal energy without a proportional temperature rise, thus lowering the specific heat ratio. Consequently, \( \gamma_2 \) is less than \( \gamma_1 \). The conclusion is that \( \gamma_2<\gamma_1 \), which is the correct answer. Therefore, the correct answer is \( \boxed{\gamma_2 < \gamma_1} \).