Question

1. A sample of methane gas that occupies a volume of 12.8 L at a temperature of 0 °C and a pressure of 1 atm contains ________ moles of gas.

2. A mixture of argon and helium gases, in a 9.49 L flask at 39 °C, contains 6.75 grams of argon and 1.61 grams of helium. The partial pressure of helium in the flask is _______ atm and the total pressure in the flask is _______ atm.

Answer 1

1.

V = 12.8L

T = 0 +273 = 273K

P = 1atm

PV = nRT

n = PV/RT

       = 1*12.8/0.0821*273   = 0.57 moles of gas

2.

no of moles of Ar (nAr) = W/G.A.Wt

                                        = 6.75/40   = 0.16875moles

no of moles of He ( nHe)   = W/G.A.Wt

                                           = 1.61/4   = 0.4025 moles

Total no of moles of Ar and He (n Ar + n He) = 0.16875 +0.4025   = 0.57125moles

n = 0.57125moles

V = 9.49L

T = 39+273   = 312K

PV = nRT

P   = nRT/V

      = 0.57125*0.0821*312/9.49

       = 1.54atm

total pressure of flask= 1.54atm

mole fraction of He ( X He)   = n He/n He + n Ar

                                             = 0.16875/0.57125

                                              = 0.2954

partial pressure of He ( P He) = X He * total pressure

                                                 = 0.2954*1.54   = 0.455atm