Question

If 1.51 g of Ar are added to 2.12 atm of He in a 2.00 L cylinder at 27.0 °C, what is the total pressure of the resulting gaseous mixture?

?total= _______________atm

Answer 1

Molar mass of Ar = 39.95 g/mol

mass(Ar)= 1.51 g

use:

number of mol of Ar,

n = mass of Ar/molar mass of Ar

=(1.51 g)/(39.95 g/mol)

= 3.78*10^-2 mol

Given:

V = 2.0 L

n = 0.0378 mol

T = 27.0 oC

= (27.0+273) K

= 300 K

use:

P * V = n*R*T

P * 2 L = 0.0378 mol* 0.08206 atm.L/mol.K * 300 K

P = 0.4653 atm

This is partial pressure of Ar

Total pressure = P(Ar) + P(He)

= 0.4653 atm + 2.12 atm

= 2.59 atm

Answer: 2.59 atm