JEE MAIN - Physics (2024 - 8th April Morning Shift - No. 12)

A mixture of one mole of monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature $$(27^{\circ} \mathrm{C})$$. The ratio of specific heat of gases at constant volume respectively is:
$$\frac{3}{2}$$
$$\frac{3}{5}$$
$$\frac{7}{5}$$
$$\frac{5}{3}$$

Explanation

To find the ratio of specific heats at constant volume ($C_V$) of the gases, we need to understand the degrees of freedom each type of gas molecule has, as this determines their specific heat capacity at constant volume. Degrees of freedom refer to the number of independent ways in which a molecule can store energy.

A monoatomic gas molecule has 3 translational degrees of freedom. This is because it can move in three dimensions: x, y, and z. A diatomic gas, if we assume it to be rigid for this context, has 5 degrees of freedom: 3 translational like the monoatomic gas and 2 rotational since it can rotate around two axes perpendicular to the bond axis connecting the two atoms. Vibrational modes are not considered at room temperature for a rigid diatomic gas, as these require higher energy to become accessible.

The specific heat capacity at constant volume for a monoatomic gas is given by:

$$C_{V, mono} = \frac{3}{2} R$$

And for a diatomic gas, it's:

$$C_{V, diatomic} = \frac{5}{2} R$$

Where $R$ is the ideal gas constant.

To find the ratio of their specific heats at constant volume, we divide the specific heat of the monoatomic gas by that of the diatomic gas:

$$\frac{C_{V, mono}}{C_{V, diatomic}} = \frac{\frac{3}{2} R}{\frac{5}{2} R} = \frac{3}{2} \times \frac{2}{5} = \frac{3}{5}$$

Thus, the correct ratio of the specific heats at constant volume of the monoatomic to diatomic (rigid) gases is given by Option B: $\frac{3}{5}$.

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