At what temperature the RMS velocity of $S O_{2}$ be same as that of $O_{2}$ at $303 K$ ?

273 K

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606 K

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303 K

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403 K

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Solution

The correct option is D $606 K$
At constant temperature, $V_{r m s} = \sqrt{\frac{3 R T}{M}}$

According to question, $(V_{r m s})_{O_{2}} = (V_{r m s})_{S O_{2}}$

$∴$ $\sqrt{\frac{3 R T_{O_{2}}}{M_{O_{2}}}}$ = $\sqrt{\frac{3 R T_{S O_{2}}}{M_{S O_{2}}}}$

$⟹ T_{S O_{2}} = \frac{T_{O_{2}} \times M_{S O_{2}}}{M_{O_{2}}}$

$∴$ $T_{S O_{2}} = \frac{303 \times 64}{32}$

$∴$ $T_{S O_{2}} = 606 K$

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