HARD PROBLEM

Question

A 4.36 liter container holds a mixture of gases. The gases in the mixture are helium gas and argon gas. There are 0.27 moles of helium gas and 0.12 moles of argon gas in the gas sample and the temperature of the gas sample is 12oC. What is the partial pressure of the helium gas in the sample?

A 4.36 liter container holds a mixture of gases.  The gases in the mixture are helium gas and argon gas.  There are 0.27 moles of helium gas and 0.12 moles of argon gas in the gas sample and the temperature of the gas sample is 12^oC.  What is the partial pressure of the helium gas in the sample?

Answer 1

moles He = 0.27 moles

moles Ar = 0.12 moles

Total moles = 0.27 + 0. 12 = 0.48

PV = nRT

P = nRT/V = (0.48 mol)(0.0821 Latm/Kmol)(285ºK)/4.36 L

P(total) = 2.58 atm

mole fraction Ar = 0.12/0.48 = 0.25

Partial pressure Ar is proportional to mole fraction of Ar.  Thus, 0.25 x 2.58 atm = 0.645 atm = Partial pressure of Ar