Kinetic Theory

JEE Main level, single correct, medium

110 of 168

Question 110 of 168

A 15g15g mass of nitrogen gas is enclosed in a vessel at a temperature, 27oC{27}^{o}C. The amount of heat transferred to the gas, so that R.M.S.R.M.S. velocity of molecules is doubled, is about.
[R=8.3J(Kmole)−1]\left[R=8.3J{\left(Kmole\right)}^{-1}\right]

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Answer: (B) 10kJ10kJ

The formula of root-mean-square speed, vrms=3RTM{v}_{rms}=\sqrt{\dfrac{3RT}{M}}. For doubling the vrms,{v}_{rms}, temperature, should be made 44-time is, Tf=(273+27)×4K=1200K{T}_{f}=\left(273+27\right)\times 4K=1200K, the formula of, ∆Q=∆U+W(W=0)∆Q=∆U+W\left(W=0\right),
⇒∆Q=∆U\Rightarrow ∆Q=∆U=nCV∆T=n{C}_{V}∆T=1528×52R∆T=\dfrac{15}{28}\times \dfrac{5}{2}R∆T
⇒∆Q=1528×52×8.3×(1200−300)J\Rightarrow ∆Q=\dfrac{15}{28}\times \dfrac{5}{2}\times 8.3\times \left(1200-300\right)J≈10KJ.\approx 10KJ.

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