Question

A mixture at 30.0 °C contains 0.850 moles of oxygen, 0.850 moles of argon and 12.5 moles of nitrogen. The mixture has a volume of 15.0 L. Calculate the mole fractions and the partial pressures of the gases in the mixture.

Answer 1

1)

n(O2),n1 = 0.85 mol

n(Ar),n2 = 0.85 mol

n(N2),n3 = 12.5 mol

Total number of mol = n1+n2+n3

= 0.85 + 0.85 + 12.5

= 14.2 mol

Mole fraction of each components are

X(O2) = n1/total mol

= 0.85/14.2

= 0.0599

X(Ar) = n2/total mol

= 0.85/14.2

= 0.0599

X(N2) = n3/total mol

= 12.5/14.2

= 0.8803

2)

Given:

V = 15.0 L

n = 14.2 mol

T = 30.0 oC

= (30.0+273) K

= 303 K

use:

P * V = n*R*T

P * 15 L = 14.2 mol* 0.08206 atm.L/mol.K * 303 K

P = 23.54 atm

This is total pressure of gases

p(O2) = X(O2)*P

= 0.0599* 23.54 atm

= 1.41 atm

p(Ar) = X(Ar)*P

= 0.0599* 23.54 atm

= 1.41 atm

p(N2) = X(N2)*P

= 0.8803* 23.54 atm

= 20.7 atm