Answer to Question #112146 in Physical Chemistry for Elhussein Aly

Question

We have a gas mixture and the total pressure of this mixture is 2 atm. Find the mole fractions and partial pressures of each component according to the data below.
[ ] N2 > % 75,52
O2 > % 22,95
Ar > % 1,28
CO2 > % 0,25
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A clean, dried and vacuumed glass tube weighs 40.1305g. When this tube is filled with water at 240C, it comes to 138.2410 g. When filled with propylene gas, 40,2959 g comes at 240C and under 740.3 mmHg pressure conditions. What is the mole mass of propylene gas?

Answer 1

1)

p_N2 = 2 atm × 0.7552 = 1.5104 atm

p_O2 = 2 atm × 0.2295 = 0.459 atm

p_Ar​ = 2 atm × 0.0128 = 0.0256 atm

p_CO2​ = 2 atm × 0.0025 = 0.005 atm

2)

m(H₂​O) = 138.2410 − 40.1305 = 98.1105 g

V(H_2​O) = 98.1105 ml = V_tube​ = V_gas

m(gas) = 40.2959 - 40.1305 = 0.1620 g

​p_1​ × V₁ / T₁​ = p₂ × V₂ / T₂

V₂​ = (760 × 22.4 × 297.15) / (740.3 × 273.15) ​= 25 L

d(gas) = m / V = 0.1620 g / 98.1105 mL = 0.00165 g / mL

M(gas) = d × V = 0.00165 g/mL × 25000 mL / mol = 41.28 g / mol