Question

A 1.00 L flask is filled with 1.35 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.300 atm .

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

Answer 1

Start with the ideal gas law:

PV = nRT

The following values were provided:

V = 1.00 L

T = 25 C = 298 K

R = 0.0821 liter·atm/mol·K

First we can solve for n for just the argon by just looking up the molecular weight on the periodic table

1.35 g argon x 1 mol / 39.948 g = 0.0337 mol argon

Now that we know all the variables for the ideal gas law we can solve for pressure which is what we need to know

Pargon = 0.0337 mol x 0.0821 liter·atm/mol·K x 298 K / 1.00 L

Pargon = 0.826 atm

Partial pressure is just the sum of all pressures exerted by all gases present. In this case:

Ptotal = Pargon + Pethane

Therefore:

Pethane = Ptotal - Pargon = 1.300 atm - 0.826 atm = 0.474 atm

The final answers are

Pargon = 0.826 atm

Pethane = 0.474 atm