500 gram of a diatomic gas is enclosed at a pressure of $10^5 \mathrm{Nm}^{-2}$. The density of the gas is…

500 gram of a diatomic gas is enclosed at a pressure of $10^5 \mathrm{Nm}^{-2}$. The density of the gas is $5 \mathrm{kgm}^{-3}$. The energy of one mole of the gas due to its thermal motion is [consider the gas molecule as a rigid rotator]
  1. $1.5 \times 10^4 \mathrm{~J}$
  2. $2.5 \times 10^4 \mathrm{~J}$
  3. $1.5 \times 10^7 \mathrm{~J}$
  4. $2.5 \times 10^7 \mathrm{~J}$

Solution

Energy of a diatomic gas due to thermal motion

The volume is obtained from the mass and density: $V = m / \rho = 0.5 \text{ kg} / 5 \text{ kg} \cdot \text{m}^{-3} = 0.1 \text{ m}^3$.

From the ideal gas law, $PV = nRT = 10^5 \text{ N} \cdot \text{m}^{-2} \times 0.1 \text{ m}^3 = 10^4 \text{ J}$.

A rigid diatomic molecule has five degrees of freedom: three translational and two rotational, so $f = 5$.

The internal energy becomes $U = \frac{5}{2} nRT = \frac{5}{2} \times 10^4 \text{ J} = 2.5 \times 10^4 \text{ J}$.

Since the problem refers to "the gas" and the mass is given, this result corresponds to the total internal energy of the sample, matching option B.

Asked in: MHT CET 2025 (05 May Shift 2)

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