Question

A mixture of 10.0 grams of Ne and 10.00 grams of Ar are stored in a rigid
container at constant temperature with a total pressure of 11.8 atm. Determine the partial
pressure of each gas in the mixture in atm.

Answer 1

n(Ne)=10/20= 0.5

n(Ar)= 10/40 = 0.25

11.8 x 0.5/0.75=7.86666667 atm of Ne.

11.8 x 0.25 / 0.75 = 3.93333333 atm of Ar.