0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127… | JEE Main 2022 (24 Jun, Shift 1)

Q57JEE Main 2022 · 24 JUN · SHIFT 1Medium

Molecular Speeds & Maxwell Distribution

0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127 $^\circ$ C. The amount of heat required to double the speed of its molecules is ____________ k cal.

(Take R = 2 cal mole ${}^{-1}$ K ${}^{-1}$ )

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

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Answer: 12

$n = \dfrac{56}{28} = 2$ mol, $T_1 = 127 + 273 = 400$ K.
$v \propto \sqrt{T}$ , so doubling the speed needs $T_2 = 4T_1 = 1600$ K, $\Delta T = 1200$ K.
Gas is in a closed vessel (constant volume), N $_2$ is diatomic: $C_V = \dfrac{5}{2}R = 5$ cal mol $^{-1}$ K $^{-1}$ .
$Q = nC_V\Delta T = 2 \times 5 \times 1200 = 12000$ cal
So the heat required is $12$ kcal.

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