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(A) The formula for $rms$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given that the $rms$ speed of hydrogen $(H_2)$ is equal to the $rms$ speed of oxyg

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$20$

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(A) The formula for $rms$ speed is $v_{rms} = \sqrt{\frac{3RT}{M}}$.Given that the $rms$ speed of hydrogen $(H_2)$ is equal to the $rms$ speed of oxygen $(O_2)$:$\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$Squaring both sides and canceling common terms $(3R)$:$\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$Given $T_{O_2} = 47^{\circ}C = 47 + 273 = 320 \ K$.Molar mass of $H_2$ $(M_{H_2})$ = $2 \ g/mol$.Molar mass of $O_2$ $(M_{O_2})$ = $32 \ g/mol$.Substituting the values:$\frac{T_{H_2}}{2} = \frac{320}{32}$$\frac{T_{H_2}}{2} = 10$$T_{H_2} = 20 \ K$.

If the $rms$ speed of hydrogen gas is equal to the $rms$ speed of oxygen gas at $47^{\circ}C$,find the temperature of the hydrogen gas in $K$.

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