Question 729: Thermodynamics & KTG

Question

According to the law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is

Accepted Answer

The law of equipartition of energy states that each quadratic degree of freedom contributes $$\frac{1}{2}R$$ per mole to the internal energy, where $$R$$ is the gas constant.
For a diatomic molecule with one additional vibrational mode, we count the degrees of freedom as follows:
Translational Motion:
3 degrees of freedom
Contribution: $$3 	imes \frac{1}{2}R = \frac{3}{2}R$$
Rotational Motion:
Diatomic molecules have 2 rotational degrees of freedom (rotation about the two axes perpendicular to the bond)
Contribution: $$2 	imes \frac{1}{2}R = R$$
Vibrational Mode:
A vibrational mode contributes 2 degrees of freedom (one for kinetic energy and one for potential energy)
Contribution: $$2 	imes \frac{1}{2}R = R$$
Now, we simply add these contributions:
$$ C_v = \frac{3}{2}R + R + R = \frac{7}{2}R. $$
Thus, the molar specific heat of the diatomic gas at constant volume, with one additional vibrational mode, is:
Option D: $$\frac{7}{2}R.$$

Answer

$\frac{9}{2} \mathrm{R}$

Answer

$\frac{5}{2} R$

Answer

$\frac{3}{2} R$

Answer

$\frac{7}{2} R$

Thermodynamics & KTG

According to the law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is

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