At what temperature T (in K) will the rms spe | vedclass

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(A) The $rms$ speed of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temp

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(A) The $rms$ speed of a gas is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$,where $R$ is the universal gas constant,$T$ is the absolute temperature in Kelvin,and $M$ is the molar mass.Given that the $rms$ speed of $H_2$ equals the $rms$ speed of $O_2$:$v_{rms(H_2)} = v_{rms(O_2)}$$\sqrt{\frac{3RT_{H_2}}{M_{H_2}}} = \sqrt{\frac{3RT_{O_2}}{M_{O_2}}}$Squaring both sides and canceling common terms $(3R)$:$\frac{T_{H_2}}{M_{H_2}} = \frac{T_{O_2}}{M_{O_2}}$Given values:$T_{O_2} = 47^{\circ}C = 47 + 273 = 320 \ K$$M_{O_2} = 32 \ g/mol$$M_{H_2} = 2 \ g/mol$Substituting the values:$\frac{T_{H_2}}{2} = \frac{320}{32}$$T_{H_2} = \frac{320}{32} 	imes 2$$T_{H_2} = 10 	imes 2 = 20 \ K$.

At what temperature $T$ (in $K$) will the $rms$ speed of $H_2$ be equal to the $rms$ speed of $O_2$ at $47^{\circ}C$?

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