Question

Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 ∘C, the total pressure in the container is 4.40 atm . Calculate the partial pressure of each gas in the container. Express the pressure values numerically in atmospheres, separated by commas.

Enter the partial pressure of methane first, then ethane, then propane.

Answer 1

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Temperature

R = ideal gas constant

substitute and solve for "n"

n = PV/(RT)

n = (4.4*10)/(0.082*(23+273))

n =1.81278 total mol

mol of CH4 = mass/MW = 8/16 = 0.5

mol of C2H6 = mass/MW = 18/30.0690 = 0.59862

now..

total mol = CH4 +C2H6 + C3H8

1.81278 = 0.5+0.59862+C3H8

C3H8 = 1.81278 - ( 0.5+0.59862) = 0.71416 moles of C3H8

P-gas = x-gas * Ptotal

P-CH4 = (0.5/1.81278) * 4.4 = 1.213 atm

P-C2H6  = (0.59862/1.81278) * 4.4 = 1.452 atm

P-C3H8= (0.71416 /1.81278) * 4.4 = 1.7334 atm