If V_{rms} of O_2 molecule is equal to V_{rms | vedclass

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(A) The root mean square velocity is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.Given that $(V_{rms})_{O_2} = (V_{rms})_{H_2}$,we have $\sqrt{\frac{3RT

Vedclass · Accepted Answer

$320 K$

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(A) The root mean square velocity is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.Given that $(V_{rms})_{O_2} = (V_{rms})_{H_2}$,we have $\sqrt{\frac{3RT_{O_2}}{M_{O_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}$.Squaring both sides and canceling common terms,we get $\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$.Here,$M_{O_2} = 32 \ g/mol$,$M_{H_2} = 2 \ g/mol$,and $T_{H_2} = 20 K$.Substituting the values: $\frac{T_{O_2}}{32} = \frac{20}{2}$.$T_{O_2} = 10 	imes 32 = 320 K$.

If $V_{rms}$ of $O_2$ molecule is equal to $V_{rms}$ of $H_2$ molecule at $20 K$,the temperature of $O_2$ is:

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