Question

A gas mixture of nonane and carbon dioxide with a total mass of 3.36 kg is placed inside a batch vessel at 37.2 atm and 129.2°C such that the mixture has a volume of 1073.6 L(STP).


Determine the moles of the mixture in the vessel, the average molecular weight, and the composition as mole percentage. Then choose the best answer below.

the average MW of mixture is 84.1 lbm /lb-mol

the total moles of mixture is 47.9

the mol% of nonane is 37.3

Two choices, B and C, are correct

None of the A-B-C choices are correct

Answer 1

first calculate volume at 37.2 atm and 129.2⁰C (your given volume at STP)

we know the equation

P₁V₁/T₁ = P₂V₂/T₂ then

V₁ = P₂V₂T₁/T₂P₁

where, P₁ = initial pressure = 37.2 atm

V₁ = initial volume = ?

T₁ = initial tempreture = 402.35 K(129.2 + 273.15)

P₂ = final pressure = 1 atm (at STP)

T₂ = final tempreture = 273.15 K (at STP)

V₂ = final volume = 1073.6 L

substitute value in above equation

V₁ = 1 1073.6 402.35 / 273.15 37.2 = 42.51 liter

Use ideal gas equation for calculation of mole of gas

Ideal gas equation

PV = nRT             where, P = atm pressure = 37.2 atm,

V = volume in Liter = 42.51 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 129.2⁰C = 273.15+ 129.2 = 402.35 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (37.2 42.51) / (0.08205 402.35) = 47.90 mole

the total mole of mixture is 47.9