Question

What pressure would a gas mixture in a 10.0 L tank exert if it were composed of 53.8 g He and 91.8 g CO2 at 398 K? Record your answer to 1 decimal place

Answer 1

Molar mass of He = 4.003 g/mol


Molar mass of CO2,
MM = 1*MM(C) + 2*MM(O)
= 1*12.01 + 2*16.0
= 44.01 g/mol

n(He) = mass of He/molar mass of He
= 53.8/4.003
= 13.4399

n(CO2) = mass of CO2/molar mass of CO2
= 91.8/44.01
= 2.0859

n(He),n1 = 13.4399 mol
n(CO2),n2 = 2.0859 mol

Total number of mol = n1+n2
= 13.4399 + 2.0859
= 15.5258 mol

Given:
V = 10.0 L
n = 15.5258 mol
T = 398.0 K

use:
P * V = n*R*T
P * 10 L = 15.5258 mol* 0.08206 atm.L/mol.K * 398 K
P = 50.7071 atm

Answer: 50.7 atm