Kinetic Theory

JEE Main level, single correct, medium

35 of 168

Question 35 of 168

The temperature of an ideal gas is increased from 200K200K to 800K800K. If r.m.s. speed of gas at 200K200K is v0{v}_{0}. Then, r.m.s. speed of the gas at 800K800K will be:

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Answer: (D) 2v02{v}_{0}

At 200K200K, the rms speed is given by
v0=3R×200M...(1){v}_{0}=\sqrt{\dfrac{3R\times 200}{M}}...(1)
At 800K800K, the rms speed is given by
v′=3R×800M...(2)v'=\sqrt{\dfrac{3R\times 800}{M}}...\left(2\right)
Divide equation (2) by equation (1) and solve to calculate the required speed.
v′v0=3R×800M3R×200M=800200=2⇒v′=2v0\begin{matrix}\dfrac{v'}{{v}_{0}} & = & \dfrac{\sqrt{\dfrac{3R\times 800}{M}}}{\sqrt{\dfrac{3R\times 200}{M}}} \\ & = & \sqrt{\dfrac{800}{200}} \\ & = & 2 \\ & \Rightarrow & v'=2{v}_{0}\end{matrix}

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