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JEE Main 2021 (25 Feb, Shift 1) · Degrees of Freedom & Internal Energy

Degrees of Freedom & Internal Energy question from JEE Main 2021 (25 Feb, Shift 1)

Q27JEE Main 2021 · 25 FEB · SHIFT 1Medium

Degrees of Freedom & Internal Energy

A monoatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with velocity 30 m/s. If container is suddenly stopped then change in temperature of the gas (R = gas constant) is $\frac{x}{3R}$ . Value of x is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 3600

Per mole: $\frac{1}{2}Mv^2 = \frac{3}{2}R\,\Delta T \Rightarrow \Delta T = \frac{Mv^2}{3R}$ .
$Mv^2 = 4 \times 30^2 = 3600$ (taking $M = 4$ as in the official answer).
$\Delta T = \frac{3600}{3R}$ , so $x = 3600$ .

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