Question

A mixture of 2.00 g of H₂ and 1.00 g of He is placed in a 1.00 L container at 310 K. Calculate the partial pressure of each gas and the total pressure.

Answer 1

Molar mass of H2 = 2.016 g/mol

Molar mass of He = 4.003 g/mol

n(H2) = mass of H2/molar mass of H2

= 2.0/2.016

= 0.9921

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

= 1.0/4.003

= 0.2498

n(H2),n1 = 0.9921 mol

n(He),n2 = 0.2498 mol

Total number of mol = n1+n2

= 0.9921 + 0.2498

= 1.2419 mol

we have:

V = 1.0 L

n = 1.2419 mol

T = 310.0 K

we have below equation to be used:

P * V = n*R*T

P * 1 L = 1.2419 mol* 0.08206 atm.L/mol.K * 310 K

P = 31.5922 atm

Partial pressure of each components are

p(H2),p1 = (n1*Ptotal)/total mol

= (0.9921 * 31.5922)/1.2419

= 25.2 atm

p(He),p2 = (n2*Ptotal)/total mol

= (0.2498 * 31.5922)/1.2419

= 6.36 atm

partial pressure of H2 = 25.2 atm

partial pressure of He = 6.36 atm