Kinetic Theory

JEE Main level, single correct, medium

32 of 168

Question 32 of 168

If the r.m.s speed of chlorine molecule is 490ms−1490m{s}^{-1} at 27oC{27}^{o}C, the r.m.s speed of argon molecules at the same temperature will be (Atomic mass of argon =39.9u=39.9u, molecular mass of chlorine =70.9u=70.9u)

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Answer: (B) 651.7ms−1651.7m{s}^{-1}

The formula to calculate the rms speed of the argon molecules is given by
vAr=3RTMAr...(1){v}_{Ar}=\sqrt{\dfrac{3RT}{{M}_{Ar}}}...\left(1\right)
The formula to calculate the rms speed of chlorine molecule is given by
vCl=3RTMCl...(2){v}_{Cl}=\sqrt{\dfrac{3RT}{{M}_{Cl}}}...\left(2\right)
Divide equation (1) by equation (2) and simplify to obtain the required ratio.
vArvCl=MClMAr...(3)\dfrac{{v}_{Ar}}{{v}_{Cl}}=\sqrt{\dfrac{{M}_{Cl}}{{M}_{Ar}}}...\left(3\right)
Substitute the values of the known parameters into equation (3) and solve to calculate the required value.
vAr490ms−1=7140⇒vAr=7140×490ms−1≈651.7ms−1\begin{matrix}\dfrac{{v}_{Ar}}{490m{s}^{-1}} & = & \sqrt{\dfrac{71}{40}} \\ & \Rightarrow & {v}_{Ar}=\sqrt{\dfrac{71}{40}}\times 490m{s}^{-1} \\ & \approx & 651.7m{s}^{-1}\end{matrix}

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