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JEE Main 2019 (10 Jan, Shift 2) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2019 (10 Jan, Shift 2)

Q34JEE Main 2019 · 10 JAN · SHIFT 2Medium

Degrees of Freedom & Internal Energy

Two kg of a monoatomic gas is at a pressure of 4 $\times$ 104 N/m2. The density of the gas is 8 kg/m3. What is the order of energy of the gas due to its thermal motion?

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Answer: (a)

$V = \frac{m}{\rho} = \frac{2}{8} = 0.25\ \text{m}^3$
Monoatomic: $U = \frac{3}{2}nRT = \frac{3}{2}PV = \frac{3}{2} \times 4 \times 10^4 \times 0.25$
$U = 1.5 \times 10^4$ J
So the order is (a) $10^4$ J.

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