Question

A 1-L flask is filled with 1.30 g of argon at 25 ∘C. A sample of ethane vapour is added to the same flask until the total pressure is 1.06 bar .

a) What is the partial pressure of argon, PAr, in the flask?

b)What is the partial pressure of ethane, Pethane, in the flask?

Answer 1

Given :

Mass of Ar = 1.30 g

T = 25 deg C= 25 deg C + 273.15 = 298.15 K

Total pressure = 1.06 bar

Calculation of moles of Ar

Number of moles of Ar = 1.30 g Ar / molar mass of Ar.

= 1.30 g / 39.95 g per mol

=0.0325 mol

Partial pressure of Ar = mol fraction x total pressure

We have total pressure we only need to get mol fraction.

Mole fraction = mole of Ar/ Total moles.

We get total mole by using pV = nRT

n = pV/ RT

R is gas constant = 0.08206 L atm per K per mol

T = 298.15 K

P = 1.06 bar = 1.06 bar x 1 atm/ 1.01325 bar

=1.0461 atm

V = 1.0 L

Lets put all these values

n = 1.0461 atm x 1.0 L / ( 0.08206 Latm(Kmol)^-1 x 298.15 K )

= 0.04276 mol

Mole fraction of Ar = moles of Ar / Total moles of Ar

= 0.0325 moles / 0.04276 mol

= 0.761

Partial pressure of Ar = Mole fraction of Ar x Total pressure

= 0.761 x 1.0461 atm

= 0.796 atm

Calculation of mole fraction of ethane.

Mole fraction of ethane = 1.0 – 0.761 = 0.239

Partial pressure of ethane = 0.239 x 1.0461 atm

=0.25 atm