What is the presser exerted by a mixture of `3.5 g` of nitrogen and 2 g of helium when confined in a vessel of volume `40` lires at `0^(@)C`?

Answer

`0.354 "bar"`

Explanation

Moles of `N_(2) (.^(n)N_(2)) = ("Mass of" N_(2))/("Molar mass") = ((3.5 g))/((28 g mol^(-1))) = 0.125` mol
Moles of `He (.^(n)He) = ("Mass of" He)/("Molar mass") = ((2.0 g))/((4 g mol^(-1))) = 0.5` mol.
Partial pressure of `N_(2) (.^(n)N_(2)) = (.^(n)N_(32)RT)/(V) = ((0.125 mol) xx (0.083 L "bar" K^(-1) mol^(-1)) xx (273 K))/((40 L)) = 0.071 "bar"`.
Partial pressure of `He (.^(p)He) = (.^(n)HeRT)/(V) = ((0.5 mol) xx (0.083 L "bar"K^(-1) mol^(-1)) xx (273 K))/((40 L)) = 0.283 "bar"`.
Total pressure of gaseous mixture ` = 0.071 + 0.283 = 0.354 "bar"`