If 3.25 g Ar are added to 3.61 atm He in a 2.00 L cylinder at 27.0 °C, what is the total pressure of the resulting gaseous mixture?

Answer 1

Find pressure exerted by Ar and add it to the pressure exerted by He.

To find pressure of Ar, we'll use the ideal gas law after determining how many moles of Ar we have.

moles Ar = 3.25 g Ar x 1 mol Ar/39.9 g = 0.0815 moles Ar

PV = nRT

P = nRT/V = (0.0815 mol)(0.0821 Latm/Kmol)(300K) / 2.00 L = 2.00 atm

Total pressure = 2.00 atm + 3.61 atm = 5.61 atm