Question 215: Thermodynamics & KTG

Question

For a diatomic gas, if $\gamma_1=\left(\frac{C p}{C v}\right)$ for rigid molecules and $\gamma_2=\left(\frac{C p}{C v}\right)$ for another diatomic molecules, but also having vibrational modes. Then, which one of the following options is correct? (Cp and Cv are specific heats of the gas at constant pressure and volume)

Accepted Answer

Let's analyze the problem step by step:
For a rigid diatomic molecule (without vibrational modes):
Degrees of freedom:
Translation: 3
Rotation: 2
Total: 5
The molar specific heat at constant volume is given by:
$$C_v = \frac{5}{2}R$$
At constant pressure, it is:
$$C_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R$$
Therefore, the ratio is:
$$\gamma_1 = \frac{C_p}{C_v} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5} \approx 1.4$$
For the diatomic molecule that also has vibrational modes (assuming the vibrational mode is fully excited):
Additional vibrational mode contributions:
Each vibrational mode contributes 2 degrees of freedom (one kinetic and one potential).
Therefore, the degrees of freedom become:
$$3\ (	ext{translation}) + 2\ (	ext{rotation}) + 2\ (	ext{vibration}) = 7$$
The molar specific heat at constant volume now is:
$$C_v = \frac{7}{2}R$$
And at constant pressure:
$$C_p = C_v + R = \frac{7}{2}R + R = \frac{9}{2}R$$
Thus, the ratio is:
$$\gamma_2 = \frac{C_p}{C_v} = \frac{\frac{9}{2}R}{\frac{7}{2}R} = \frac{9}{7} \approx 1.2857$$
Comparison:
For the rigid molecule, $$\gamma_1 \approx 1.4$$.
For the molecule with vibrational modes, $$\gamma_2 \approx 1.2857$$.
Clearly, $$\gamma_2 < \gamma_1$$.
Therefore, the correct option is:
Option A: $$\gamma_2 < \gamma_1$$

Answer

$\gamma_2<\gamma_1$

Answer

$\gamma_2>\gamma_1$

Answer

$\gamma_2=\gamma_1$

Answer

$2 \gamma_2=\gamma_1$

Thermodynamics & KTG

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