Nitrogen gas is at 300^{circ} C temperature. | vedclass

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(A) The root mean square $(rms)$ speed of a gas molecule is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.Given that the $rms$ speed of $H_{2

Vedclass · Accepted Answer

$41$

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(A) The root mean square $(rms)$ speed of a gas molecule is given by the formula: $V_{rms} = \sqrt{\frac{3RT}{M}}$.Given that the $rms$ speed of $H_{2}$ is equal to the $rms$ speed of $N_{2}$:$V_{rms(H_{2})} = V_{rms(N_{2})}$$\sqrt{\frac{3RT_{H_{2}}}{M_{H_{2}}}} = \sqrt{\frac{3RT_{N_{2}}}{M_{N_{2}}}}$Squaring both sides and canceling common terms $(3R)$:$\frac{T_{H_{2}}}{M_{H_{2}}} = \frac{T_{N_{2}}}{M_{N_{2}}}$Given $T_{N_{2}} = 300^{\circ}C = 300 + 273 = 573 \ K$,$M_{N_{2}} = 28 \ g/mol$,and $M_{H_{2}} = 2 \ g/mol$:$\frac{T_{H_{2}}}{2} = \frac{573}{28}$$T_{H_{2}} = \frac{573 	imes 2}{28} = \frac{573}{14} \approx 40.928 \ K$.Rounding to the nearest integer,we get $41 \ K$.

Nitrogen gas is at $300^{\circ} C$ temperature. The temperature (in $K$) at which the $rms$ speed of a $H_{2}$ molecule would be equal to the $rms$ speed of a nitrogen molecule is........ (Molar mass of $N_{2}$ gas = $28 \ g/mol$,Molar mass of $H_{2}$ gas = $2 \ g/mol$)

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