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(B) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$ of the gas

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$2$

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(B) According to Graham's Law of Diffusion,the rate of diffusion $(r)$ is inversely proportional to the square root of the molar mass $(M)$ of the gas: $r_1 / r_2 = \sqrt{M_2 / M_1}$.Here,$r_1$ is the rate of helium $(He)$ and $r_2$ is the rate of methane $(CH_4)$.The molar mass of helium $(M_1)$ is $4 \ g/mol$.The molar mass of methane $(M_2)$ is $16 \ g/mol$.Substituting the values: $r_{He} / r_{CH_4} = \sqrt{16 / 4} = \sqrt{4} = 2$.Therefore,the rate of diffusion of helium is $2$ times that of methane.

List-$I$	List-$II$
$(A)$ Viscosity	$(I)$ Critical temperature
$(B)$ Ideal gas behaviour	$(II)$ Isobars
$(C)$ Liquefaction of gases	$(III)$ Compressibility factor
$(D)$ Charles' law	$(IV)$ $kg \ s^{-2}$
	$(V)$ $kg \ m^{-1} \ s^{-1}$

At the same temperature and pressure,the rate of diffusion of helium with respect to methane is ...... .

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