Question

5. Natural gas is a mixture of hydrocarbons, primarily methane (CH4) with a small amount of propane (C3H8). A sample of natural gas in a 2.00 L container exerts a pressure of 0.950 atm at 35°C. If there are 0.0020 moles of C3H8 present:

a. What is the partial pressure of C3H8 and CH4?

b. How many moles of CH4 are present?

Answer 1

PV = nRT

P = 0.95atm

V = 2L

T = 35+273   = 308K

PV = nRT

n   = PV/RT

      = 0.95*2/0.0821*308

n    = 0.075 moles

total no of moles of ( CH4, C3H8) of natural gas = 0.075 moles

b.

nCH4 + nC3H8   = 0.075

nCH4 + 0.002    = 0.075

n CH4             = 0.075-0.002   = 0.073 moles

a. mole fraction of CH4 XCH4   = nCH4/nCH4+ nC3H8

                                                 = 0.073/0.075   = 0.973

mole fraction of C3H8   XC3H8   = nC3H8/nCH4+ nC3H8

                                                      = 0.002/0.075   = 0.027

partial pressure of CH4 (PCH4)   = XCH4* total pressure

                                                       = 0.973*0.95    = 0.924 atm

partial pressure of C3H8 (PC3H8)   = XC3H8* total pressure

                                                       = 0.027*0.95      = 0.026atm