Question 927: Thermodynamics & KTG

Question

For a gas molecule with 6 degrees of freedom, which one of the following relation between gas constant '$$\mathrm{R}$$' and molar specific heat '$$\mathrm{C}_{\mathrm{v}}$$' is correct?

Accepted Answer

To determine the correct relationship between the gas constant $$\mathrm{R}$$ and the molar specific heat at constant volume $$\mathrm{C}_{\mathrm{v}}$$ for a gas molecule with 6 degrees of freedom, we can use the concepts from the kinetic theory of gases. For a gas molecule, the degrees of freedom (f) directly influence the molar specific heat capacities.
Firstly, let's recall the general relationship between the degrees of freedom (f), the molar specific heat at constant volume ($$\mathrm{C}_{\mathrm{v}}$$), and the molar specific heat at constant pressure ($$\mathrm{C}_{\mathrm{p}}$$):
$$\mathrm{C}_{\mathrm{v}} = \frac{f}{2} \mathrm{R}$$
$$\mathrm{C}_{\mathrm{p}} = \mathrm{C}_{\mathrm{v}} + \mathrm{R}$$
For a gas molecule with 6 degrees of freedom:
$$\mathrm{C}_{\mathrm{v}} = \frac{6}{2} \mathrm{R} = 3\mathrm{R}$$
Now substituting $$\mathrm{C}_{\mathrm{v}}$$ in the first equation:
$$3 \mathrm{R} = \mathrm{C}_{\mathrm{v}}$$
Therefore, solving for $$\mathrm{R}$$ we get:
$$\mathrm{R} = \frac{\mathrm{C}_{\mathrm{v}}}{3}$$
Given this derivation, the correct relation is
Option A
:
$$\mathrm{R}=\frac{\mathrm{C}_{\mathrm{v}}}{3}$$

Answer

$$R=\frac{C_v}{3}$$

Answer

$$\mathrm{R}=\frac{5 \mathrm{C}_{\mathrm{v}}}{4}$$

Answer

$$\mathrm{R}=\frac{\mathrm{C}_{\mathrm{v}}}{2}$$

Answer

$$\mathrm{R}=\frac{3 \mathrm{C}_{\mathrm{v}}}{4}$$

Thermodynamics & KTG

For a gas molecule with 6 degrees of freedom, which one of the following relation between gas constant '$\mathrm{R}' and molar specific heat '\mathrm{C}_{\mathrm{v}}$' is correct?

Choose the correct answer: