Question

A mixture containing 2.91 g each of CH₄(g), C₂H₄(g) and C₄H₁₀(g) is contained in a 1.50 L flask at a temperature of 35°C.

(a) Calculate the partial pressure of each of the gases in the mixture.

PCH4	= ___ atm
PC2H4	= ___ atm
PC4H10	= ___ atm

(b) Calculate the total pressure of the mixture.
___ atm

Answer 1

Let's calculate total pressure first

according to ideal gas law PV = nRT

V = 1.5 L = 1.5 X 10^-3 m³

total no. of moles n = moles of CH4 + moles of C2H4 + moles of C4H10 =

moles = mass/molar mass

moles of CH4 = 2.91/16 = 0.182 moles

moles of C2H4 = 2.19/28 = 0.078 moles

moles of C4H10 = 2.19/58 = 0.0377 moles

therefore n = 0.182 + 0.078 + 0.0377 = 0.2977 moles

T = 35 + 273 = 308K

P = nRT/V

therefore P = (0.2977 moles X 8.314 JK^-1mol^-1 X  308K)/0.0015m³ = 508316 Pa

101325 Pa = 1 atm

therefore 508316 Pa = (1/101325) X 508316 = 5 atm

(b) Total pressure of the mixture = 5 atm.

(a) Partial pressure of one of gas of mixture = mole fraction of that gas X total pressure

Lets calculate mole fracton (X) of gases

X_CH4 = Moles of CH4/(total no. of moles of all gases) = 0.182/(0.182 + 0.078 + 0.0377) =`0.182/0.2977 = 0.611

X_C2H4 = 0.078/0.2977 = 0.262

X_C4H10 = 0.0377/0.2977 = 0.1266

P_CH4 = X_CH4 X P_Total = 0.611 X 5 = 3.055 atm

P_C2H4 = X_C2H4 X P_Total = 0.262 X 5 = 1.31 atm

P_C4H10 = X_C4H10 X P_Total = 0.1266 X 5 = 0.633 atm