Kinetic Theory

JEE Main level, numerical answer, medium

52 of 168

Question 52 of 168

The root-mean-square speed of molecules of a given mass of a gas at 27∘C27^{\circ}C and 11 atmosphere pressure is 200ms−1.200m{s}^{-1}. The root-mean-square speed of molecules of the gas at 127∘C127^{\circ}C and 22 atmosphere pressure is x3ms−1.\dfrac{x}{\sqrt{3}}m{s}^{-1}. The value of xx will be __________.

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Answer: 400

vrms=3RTM{v}_{rms}=\sqrt{\dfrac{3RT}{M}}⇒vrms∝T\Rightarrow {v}_{rms}\propto \sqrt{T}⇒(vrms)2(vrms)1=T2T1\Rightarrow \dfrac{{\left({v}_{rms}\right)}_{2}}{{\left({v}_{rms}\right)}_{1}}=\sqrt{\dfrac{{T}_{2}}{{T}_{1}}}=400300=\sqrt{\dfrac{400}{300}}=23=\dfrac{2}{\sqrt{3}}
(vrms)2=23(vrms)1{\left({v}_{rms}\right)}_{2}=\dfrac{2}{\sqrt{3}}{\left({v}_{rms}\right)}_{1}=23×200=\dfrac{2}{\sqrt{3}}\times 200⇒(vrms)2=4003ms−1\Rightarrow {\left({v}_{rms}\right)}_{2}=\dfrac{400}{\sqrt{3}}m{s}^{-1}.

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