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The temperature at which the root mean square speed of ${SO}_{2}$ is half that of helium at 300 K is

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300 K

600 K

900 K

1200 K

answer is D.

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Detailed Solution

Since $u_{rms} = \sqrt{3 R T / M}$ ,

we have

$\frac{u_{1}}{u_{2}} = \sqrt{\frac{T_{1}}{T_{2}} \frac{M_{2}}{M_{1}}}$

This gives

$T_{2} = (\frac{M_{2}}{M_{1}}) {(\frac{u_{2}}{u_{1}})}^{2} T_{1} = (\frac{64}{4}) {(\frac{1}{2})}^{2} (300 K) = 1200 K$

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