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?

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"#