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

$\frac{U_{S O_{2}}}{U_{H} e} = \frac{1}{2} = \sqrt{\frac{M_{H e}}{M_{S O_{2}}} \frac{T_{S O_{2}}}{T_{H e}}} = \sqrt{\frac{4}{64} \frac{T_{S O_{2}}}{300}}$

$= \frac{4}{64} \frac{T_{S O_{2}}}{300} = \frac{1}{4}; T_{S O_{2}} = 1200^{\circ} K$

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