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Graham's Law of Effusion

At what tempe...

Question

At what temperature will the r.m.s velocity of $S O_{2}$ be the same as that of $O_{2}$ at 303 K :

403 K

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303 K

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606 K

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273 K

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Solution

The correct option is C 606 K
$V_{r m s} = \sqrt{\frac{3 R T}{M_{w}}}$ $M_{w S O_{2}} = 32 + 32 = 64$
$M_{w O_{2}} = 32$
$\sqrt{\frac{3 R T_{S O_{2}}}{32 + 32}} = \sqrt{\frac{3 R T_{O_{2}}}{32}}$
$\sqrt{\frac{T_{S O_{2}}}{2}} = \sqrt{303}$
$T_{S O_{2}} = 606 k$

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