Question

65. A gas mixture contains 1.25 g N2 and 0.85 g 02 in a 1.55 L con tainer at 18 °C. Calculate the mole fraction and partial pressur of each component in the gas mixture. MISSED THIS? Read Section 6.6; Watch KCV 6.6, IWE 6.10

Answer 1

Step 1: find mole fraction of both components

Molar mass of N2 = 28.02 g/mol

Molar mass of O2 = 32 g/mol

n(N2) = mass of N2/molar mass of N2

= 1.25/28.02

= 0.0446

n(O2) = mass of O2/molar mass of O2

= 0.85/32

= 0.0266

n(N2),n1 = 0.0446 mol

n(O2),n2 = 0.0266 mol

Total number of mol = n1+n2

= 0.0446 + 0.0266

= 0.0712 mol

Mole fraction of each components are

X(N2) = n1/total mol

= 0.0446/0.0712

= 0.6268

X(O2) = n2/total mol

= 0.0266/0.0712

= 0.3732

Step 2: find total pressure

Given:

V = 1.55 L

n = 0.0712 mol

T = 18.0 oC

= (18.0+273) K

= 291 K

use:

P * V = n*R*T

P * 1.55 L = 0.0712 mol* 0.08206 atm.L/mol.K * 291 K

P = 1.0965 atm

Step 3: find partial pressure of both components

p(N2) = X(N2) * P

= 0.6268 * 1.0965

= 0.6873 atm

p(O2) = X(O2) * P

= 0.3732 * 1.0965

= 0.4092 atm

Answer:

mole fraction of N2 = 0.627

p(N2) = 0.687 atm

mole fraction of O2 = 0.373

p(O2) = 0.409 atm