Question

A mixture of 0.2 mole O2 and 0.4 mole N2 gas is enclosed in a vessel of certain volume at 27C and 800 torr. If volume of mixture is doubled at same temperature then determine pressure of mixture and individual gases?

Answer 1

`p_text(tot) = "400 torr"; p_text(O₂) = "130 torr"; p_text(N₂) = "270 torr"`

Explanation:

Total pressure

We can use Boyle's Law to solve this problem:

`color(blue)(bar(ul(|color(white)(a/a)p_1V_1= p_2V_2 color(white)(a/a)|)))" "`

This gives

`p_2 = p_1 ×V_1/V_2`

In this problem,

`p_1 = "800 torr"; color(white)(m)p_2 = ?`
`V_1 = V_1; color(white)(mmmll)V_2 = 2V_1`

`p_2 = "800 torr" × stackrelcolor(blue)(1)(color(red)(cancel(color(black)(V_1))))/(2color(red)(cancel(color(black)(V_1)))) = "400 torr"`

Partial pressures

Now, we can use a form of Dalton's Law of Partial Pressures:

`color(blue)(bar(ul(|color(white)(a/a)p_i = chi_1P_text(tot)color(white)(a/a)|)))" "`

The partial pressure of a gas `i` in a mixture equals its mole fraction `chi_i` times the total pressure `p_text(tot)`.

Originally, `p_text(tot) = "800 torr"`

`χ_text(O₂) = (0.2 color(red)(cancel(color(black)("mol"))))/((0.2 + 0.4) color(red)(cancel(color(black)("mol")))) = 0.2/0.6 = 0.33`

`χ_text(N₂) = (0.4 color(red)(cancel(color(black)("mol"))))/((0.2 + 0.4) color(red)(cancel(color(black)("mol")))) = 0.4/0.6 = 0.67`

The mole fractions are still the same in the new volume.

∴ `p_text(O₂) = "0.33 × 400 torr" = "130 torr"`

`p_text(N₂) = "0.67 × 400 torr" = "270 torr"`