At what temperature,the rate of effusion of N | vedclass

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(C) According to Graham's Law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$ and directly pro

Vedclass · Accepted Answer

$373$

Vedclass · Answer

(C) According to Graham's Law of effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$ and directly proportional to the square root of the temperature $T$ for a given pressure.$\frac{r_{N_2}}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_{N_2}} 	imes \frac{T_{N_2}}{T_{SO_2}}}$Given: $r_{N_2} = 1.625 	imes r_{SO_2}$,$T_{SO_2} = 50 + 273 = 323\,K$,$M_{N_2} = 28\,g/mol$,$M_{SO_2} = 64\,g/mol$.Substituting the values:$1.625 = \sqrt{\frac{64}{28} 	imes \frac{T_{N_2}}{323}}$Squaring both sides:$(1.625)^2 = \frac{64}{28} 	imes \frac{T_{N_2}}{323}$$2.640625 = 2.2857 	imes \frac{T_{N_2}}{323}$$T_{N_2} = \frac{2.640625 	imes 323}{2.2857} \approx 373\,K$.

At what temperature,the rate of effusion of $N_2$ would be $1.625$ times that of $SO_2$ at $50\,^{\circ}C$ ............ $K$

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