It takes 38 , s for 1 , mol of N_2 gas to eff | vedclass

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(C) For gases at different pressures,the rate of effusion $r$ is given by Graham's Law modified for pressure: $r = \frac{P 	imes A}{\sqrt{2 \pi M R T

Vedclass · Accepted Answer

$252$

Vedclass · Answer

(C) For gases at different pressures,the rate of effusion $r$ is given by Graham's Law modified for pressure: $r = \frac{P 	imes A}{\sqrt{2 \pi M R T}} \propto \frac{P}{\sqrt{M}}$Thus,$\frac{r_1}{r_2} = \frac{P_1}{P_2} \sqrt{\frac{M_2}{M_1}}$Given: $P_{N_2} = 0.8 \, atm$,$t_{N_2} = 38 \, s$,$M_{N_2} = 28 \, g/mol$,$P_X = 1.6 \, atm$,$t_X = 57 \, s$.Since $r = \frac{n}{t}$,we have $\frac{r_{N_2}}{r_X} = \frac{t_X}{t_{N_2}} = \frac{57}{38} = 1.5$.Substituting into the formula: $1.5 = \frac{0.8}{1.6} \sqrt{\frac{M_X}{28}}$$1.5 = 0.5 	imes \sqrt{\frac{M_X}{28}}$$3 = \sqrt{\frac{M_X}{28}}$$9 = \frac{M_X}{28}$$M_X = 9 	imes 28 = 252 \, g/mol$.

It takes $38 \, s$ for $1 \, mol$ of $N_2$ gas to effuse through a pinhole at a pressure of $0.8 \, atm$. It takes $57 \, s$ for $1 \, mol$ of an unknown gas $X$ to effuse through the same pinhole at a pressure of $1.6 \, atm$. Calculate the molar mass of $X$.

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