The temperature at which the r.m.s velocity of oxygen molecules is equal to that of nitrogen molecules at $100^{o} C$ is

426.3 K

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456.3 K

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436.3 K

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446.3 K

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Solution

The correct option is A 426.3 K
Given that $V_{r m s} (O_{2}) = V_{r m s} (N_{2})$

$\Rightarrow \sqrt{\frac{3 R T_{1}}{M_{1}}} = \sqrt{\frac{3 R T_{2}}{M_{2}}}$
Where $M_{1}$ is Molar mass of $O_{2}$ = 32
and $M_{2}$ is Molar mass of $N_{2}$ = 28
thus $O_{2}$ is at $T_{1}$ temperature
$\Rightarrow N_{2}$ is at $T_{2} = 373 K$ temperature.
$T_{1} = \frac{M_{1}}{M_{2}} \times T_{2}$

$T_{1} = \frac{32}{28} \times 373$
$T_{1} = 426.3 K$

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