The temperature of O_2 gas at which its r.m.s | vedclass

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(C) The $r.m.s.$ speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.The most probable speed of a gas is given by $v_{mp} = \sqrt{\frac{2RT}{M

Vedclass · Accepted Answer

$3200$

Vedclass · Answer

(C) The $r.m.s.$ speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.The most probable speed of a gas is given by $v_{mp} = \sqrt{\frac{2RT}{M}}$.For $H_2$ gas at $T_1 = 300 \ K$,$M_{H_2} = 2 \ g/mol$,so $v_{mp} = \sqrt{\frac{2R(300)}{2}} = \sqrt{300R}$.For $O_2$ gas at temperature $T_2$,$M_{O_2} = 32 \ g/mol$,so $v_{rms} = \sqrt{\frac{3RT_2}{32}}$.Equating the two speeds: $\sqrt{\frac{3RT_2}{32}} = \sqrt{300R}$.Squaring both sides: $\frac{3RT_2}{32} = 300R$.$T_2 = \frac{300 	imes 32}{3} = 100 	imes 32 = 3200 \ K$.

The temperature of $O_2$ gas at which its $r.m.s.$ speed is equal to the most probable speed of $H_2$ gas at $300 \ K$ is ............... $K$.

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