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

We have a 10 g mixture of hydrogen ($$H_2$$) and helium ($$He$$) in a vessel of capacity 0.0125 m$$^3$$ at 6 bar and 27 °C and need to find the mass of helium.

Let the mass of helium = $$x$$ g, so the mass of hydrogen = $$(10 - x)$$ g. Hence the moles of helium are $$\frac{x}{4}$$ (molar mass of He = 4 g/mol) and the moles of hydrogen are $$\frac{10 - x}{2}$$ (molar mass of $$H_2$$ = 2 g/mol).

Since we can find the total number of moles using the ideal gas equation, $$PV = nRT$$, we convert units: P = 6 bar = 6 × 10$$^5$$ Pa, V = 0.0125 m$$^3$$, T = 27 °C = 300 K, and R = 8.3 J K$$^{-1}$$ mol$$^{-1}$$ = 8.3 Pa m$$^3$$ K$$^{-1}$$ mol$$^{-1}$$.

Substituting these into $$n = \frac{PV}{RT} = \frac{6 \times 10^5 \times 0.0125}{8.3 \times 300}$$ gives a numerator of 6 × 10$$^5$$ × 0.0125 = 7500 and a denominator of 8.3 × 300 = 2490. This gives $$n = \frac{7500}{2490} = 3.012 \approx 3 \text{ mol}$$.

Therefore, the total moles satisfy $$\frac{x}{4} + \frac{10 - x}{2} = 3$$. Multiplying through by 4 yields $$x + 2(10 - x) = 12$$, which simplifies to $$x + 20 - 2x = 12$$ hence $$-x = -8$$ and $$x = 8$$.

Therefore, the mass of helium in the mixture is 8 g.