The temperature at which $R M S$ velocity of $S O_{2}$ molecules is half that of helium molecules at $300 K$ is:

150 K

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600 K

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900 K

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1200 K

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Solution

The correct option is D $1200 K$
$\frac{C_{1}}{C_{2}} = \sqrt{\frac{T_{1} M_{2}}{T_{2} M_{1}}}$
$\frac{C_{S O_{2}}}{C_{H e}} = \frac{1}{2} = \sqrt{\frac{T_{1} \times 4}{300 \times 64}}$

= $1200 K$

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The temperature at which RMS velocity of $S O_{2}$ molecules is half that of He molecules at 300 K is

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