JEE Main 2022 · 24 Jun · Shift 1 · Physics Q51

← JEE Main 2022 · 24 Jun · Shift 1

PhysicsInteger+4 / -1Question 51 of 90

0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127°C. The amount of heat required to double the speed of its molecules is ______ k cal.

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

Your answer

Answer and worked solution

Official answer: 12

Speed and temperature. $v_{\rm rms}\propto\sqrt T$ , so doubling the speed needs $T$ to go from 400 K to 1600 K: $\Delta T=1200$ K.

Moles. $n=\dfrac{56}{28}=2$ .

Heat at constant volume (closed vessel, diatomic): $C_v=\frac52R=5$ cal mol $^{-1}$ K $^{-1}$ , so $Q=2\times5\times1200=12000$ cal $=12$ kcal.

Answer: 12

Topics:Specific heats of gases Molecular speeds

← Previous question Next question →

JEE Main 2022 · 24 Jun · Shift 1, Physics question 51