The rate of diffusion of methane at 1.0 atm | vedclass

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(A) According to Graham's law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the density $(\rho)$ of the gas

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$44$

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(A) According to Graham's law of diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the density $(ho)$ of the gas: $r \propto \frac{1}{\sqrt{ho}}$.From the ideal gas equation,$PV = nRT = \frac{m}{MW} RT$,where $m$ is the mass and $MW$ is the molecular weight.Rearranging gives $\frac{m}{V} = ho = \frac{P 	imes MW}{RT}$.Since $T$ is constant,$ho \propto P 	imes MW$.Therefore,the rate of diffusion is given by $r \propto \frac{1}{\sqrt{P 	imes MW}}$.For two gases,$\frac{r_{CH_4}}{r_X} = \sqrt{\frac{P_X 	imes (MW)_X}{P_{CH_4} 	imes (MW)_{CH_4}}}$.Given $r_{CH_4} = 2r_X$,$P_{CH_4} = 1.0 \ atm$,$P_X = 1.45 \ atm$,and $(MW)_{CH_4} = 16 \ g/mol$:$2 = \sqrt{\frac{1.45 	imes (MW)_X}{1.0 	imes 16}}$.Squaring both sides: $4 = \frac{1.45 	imes (MW)_X}{16}$.$(MW)_X = \frac{4 	imes 16}{1.45} = \frac{64}{1.45} \approx 44.13 \approx 44$.

The rate of diffusion of methane at $1.0 \ atm$ pressure is twice that of another gas '$X$' kept at $1.45 \ atm$. The molecular weight of the gas '$X$' is
[$T$ kept constant]

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The rate of diffusion of methane at $1.0 \ atm$ pressure is twice that of another gas '$X$' kept at $1.45 \ atm$. The molecular weight of the gas '$X$' is[$T$ kept constant]