JEE Physics 2025 Question with Solution

Given: Helium and argon are at the same temperature, $T = 300 \, \text{K}$.

Find: The ratio of average kinetic energies per molecule of helium and argon.

For an ideal gas, the average kinetic energy per molecule is

$$K.E. = \frac{3}{2} kT$$

This expression depends only on temperature and not on the mass or identity of the gas.

Since both helium and argon are at the same temperature,

$$K_{\text{avg, He}} = \frac{3}{2}k(300)$$

and

$$K_{\text{avg, Ar}} = \frac{3}{2}k(300)$$

Therefore,

$$\frac{K_{\text{avg, He}}}{K_{\text{avg, Ar}}} = \frac{\frac{3}{2}k(300)}{\frac{3}{2}k(300)} = 1$$

So, the ratio is $1 : 1$.

Therefore, the correct option is D.

Given: Helium and argon are monatomic gases at the same temperature, $300 \, \text{K}$.

Find: The ratio of their average kinetic energies per molecule.

By equipartition theorem,

$$K_{avg} = \frac{f}{2} k_B T$$

For monatomic gases, the degrees of freedom are

$$f = 3$$

Hence,

$$K_{avg} = \frac{3}{2} k_B T$$

For helium,

$$K_{avg,He} = \frac{3}{2} k_B (300)$$

For argon,

$$K_{avg,Ar} = \frac{3}{2} k_B (300)$$

Taking the ratio,

$$\frac{K_{avg,He}}{K_{avg,Ar}} = 1$$

Thus, the average kinetic energies per molecule are equal. The molar masses given do not affect this result.

Therefore, the ratio is $1 : 1$, so the correct option is D.