500 gram of a diatomic gas is enclosed at a pressure of 10⁵ N m⁻². The density of the gas is 5 kg m⁻³. The energy of one mole of the gas due to its thermal motion is [consider the gas molecule as a rigid rotator] (a) 1.5 × 10⁴ J (b) 2.5 × 10⁴ J (c) 1.5 × 10⁷ J (d) 2.5 × 10⁷ J

Answer: (b). Volume V = m/ρ = 0.5/5 = 0.1 m³. Rigid diatomic: U = 5/2nRT = 5/2PV. U = 5/2 × 10⁵ × 0.1 = 2.5 × 10⁴ J So the answer is (b) 2.5 × 10⁴ J.

500 gram of a diatomic gas is enclosed at a pressure of 10⁵ N m⁻²… | MHT CET 2025 (5 May, Shift 2)

Q7MHT CET 2025 · 5 MAY · SHIFT 2Medium

Degrees of Freedom & Internal Energy

500 gram of a diatomic gas is enclosed at a pressure of $10^{5}\text{ N m}^{-2}$ . The density of the gas is $5\text{ kg m}^{-3}$ . The energy of one mole of the gas due to its thermal motion is [consider the gas molecule as a rigid rotator]

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Answer: (b)

Volume $V = \frac{m}{\rho} = \frac{0.5}{5} = 0.1\ \text{m}^3$ .
Rigid diatomic: $U = \frac{5}{2}nRT = \frac{5}{2}PV$ .
$U = \frac{5}{2} \times 10^5 \times 0.1 = 2.5 \times 10^4$ J
So the answer is (b) $2.5 \times 10^4$ J.

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