# 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?

May 1, 2017

${p}_{\textrm{\to t}} = \text{400 torr"; p_text(O₂) = "130 torr"; p_text(N₂) = "270 torr}$

#### Explanation:

Total pressure

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

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {p}_{1} {V}_{1} = {p}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

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

${p}_{2} = \text{800 torr" × stackrelcolor(blue)(1)(color(red)(cancel(color(black)(V_1))))/(2color(red)(cancel(color(black)(V_1)))) = "400 torr}$

Partial pressures

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

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {p}_{i} = {\chi}_{1} {P}_{\textrm{\to t}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

The partial pressure of a gas $i$ in a mixture equals its mole fraction ${\chi}_{i}$ times the total pressure ${p}_{\textrm{\to t}}$.

Originally, ${p}_{\textrm{\to t}} = \text{800 torr}$

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"