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(A) Given: $P = 6 \ bar = 6 	imes 10^{5} \ Pa$,$V = 0.0125 \ m^{3}$,$T = 27 + 273 = 300 \ K$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$.Using the ideal gas eq

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(A) Given: $P = 6 \ bar = 6 	imes 10^{5} \ Pa$,$V = 0.0125 \ m^{3}$,$T = 27 + 273 = 300 \ K$,$R = 8.3 \ J \ K^{-1} \ mol^{-1}$.Using the ideal gas equation $PV = n_{mix}RT$:$n_{mix} = \frac{PV}{RT} = \frac{6 	imes 10^{5} 	imes 0.0125}{8.3 	imes 300} = \frac{7500}{2490} \approx 3.012 \ mol \approx 3 \ mol$.Let the moles of $He$ be $x$ and moles of $H_{2}$ be $(3-x)$.The total mass is $10 \ g$,so: $4x + 2(3-x) = 10$.$4x + 6 - 2x = 10 \implies 2x = 4 \implies x = 2 \ mol$.Mass of $He = n 	imes M = 2 \ mol 	imes 4 \ g \ mol^{-1} = 8 \ g$.

$A$ $10 \ g$ mixture of hydrogen and helium is contained in a vessel of capacity $0.0125 \ m^{3}$ at $6 \ bar$ and $27^{\circ}C$. The mass of helium in the mixture is $g$. (nearest integer) Given : $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ (Atomic masses of $H$ and $He$ are $1 \ u$ and $4 \ u$,respectively)

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