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← JEE Main 2022 · 28 Jul · Shift 2
PhysicsSingle correct+4 / -1Question 8 of 90

JEE Main 2022 · 28 Jul · Shift 2, Physics question 8

A vessel contains 14 g of nitrogen gas at a temperature of 27°C. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be :

Take R = 8.32 J mol−1^{-1} k−1^{-1}.

Choose one option
Answer and worked solution

Official answer: C

Double vrmsv_{rms} means 4 times T: 300 K to 1200 K.

Heat at constant volume (rigid vessel). nCVΔT=0.5×52×8.32×900=9360nC_V\Delta T=0.5\times\frac52\times8.32\times900=9360 J.

(C)