Question

If 3.68 g Ar are added to 4.79 atm He in a 2.00 L cylinder at 27.0°C, what is the total pressure of the resulting gaseous mixture? Pool

Answer 1

1)

1st calculate the pressure due to Ar

Molar mass of Ar = 39.95 g/mol

mass(Ar)= 3.68 g

use:

number of mol of Ar,

n = mass of Ar/molar mass of Ar

=(3.68 g)/(39.95 g/mol)

= 9.212*10^-2 mol

Given:

V = 2.0 L

n = 0.0921 mol

T = 27.0 oC

= (27.0+273) K

= 300 K

use:

P * V = n*R*T

P * 2 L = 0.0921 mol* 0.08206 atm.L/mol.K * 300 K

P = 1.13 atm

Use:

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

= 1.13 atm + 4.79 atm

= 5.92 atm

Answer: 5.92 atm

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