Problem on Redlich-Kwong equation

Question

i) Welcome to Beaver Gas Co.! Your first task is to calculate the annual gross sales of our superpure-grade nitrogen and oxygen gases.

a) The total gross sales of N₂ is 30,000 units. Take the volume of the cylinder to be 43 L, the pressure to be 12,400 kPa, and the cost to be $6.I/kg. Compare your result to that you would obtain using the ideal gas model.

b) Repeat for 30,000 units of O₂ at 15,000 kPa and $9/kg.

ii) Use the Redlich-Kwong equation to calculate the size of vessel you would need to contain 30 kg of acetylene mixed with 50 kg of n-butane at 30 bar and 450 K. The binary interaction coefficient is given by k₁₂ = 0.092.

Answer 1

(i)

(a) The amount in kg, of superpure grade N₂, per container is calculated below,

PV = nRT

n = PVT₁/(TP₁V₁/n₁)) ... where suffix 1 indicates conditions at STP.

n = (12400)(43x10^-3)(273)/((298)(101)(22.4)) = 0.22 kmol

m = Mn = 28 x 0.22 = 6.16 kg.

Hence according to Ideal gas law, there'll be 6.16 kg per unit of superpure-grade N₂.

And the annual gross sales will be $ 6.1 x 6.16 x 30000 = $1127280 = $1.13 million

(b) The amount in kg, of superpure grade O₂, per container is calculated below,

PV = nRT

n = PVT₁/(TP₁V₁/n₁)) ... where suffix 1 indicates conditions at STP.

n = (15000)(43x10^-3)(273)/((298)(101)(22.4)) = 0.27 kmol

m = Mn = 32 x 0.27 = 8.64 kg.

Hence according to Ideal gas law, there'll be 8.64 kg per unit of superpure-grade O₂.

And the annual gross sales will be $ 9 x 8.64 x 30000 = $ 2332800 = $2.33 million

(ii)

The following data is obtained from Internet.

Acetylene

MW 26 g/mol
P_c 61.91 bar
T_c 35.1 oC

n-butane

MW 58.12
P_c   38 bar
T_c   425 K

The total amount of mixture in kmol = 30/26 + 50/58.12 = 2.01

x₁ = mole fraction of acetylene = (30/26)/2.01 = 0.57

x₂ = mole fraction of n-butane = 0.43

Redlich-Kwong parameters (Note that P is in kPa and T is in K)

acetylene:

a₁ = 0.427R²T_c^2.5/Pc = 0.427(8.314)²(308.2)^2.5/6273 = 7846
b₁ = 0.0866RT_c/P_c = 0.0866(8.314)(308.2)/6273 = 0.0354

n-butane:

a₂ = 0.427R²T_c^2.5/Pc = 0.427(8.314)²(425)^2.5/3850 = 28547

b₂ = 0.0866RT_c/P_c = 0.0866(8.314)(425)/3850 = 0.0795

Using the following mixing rules, we'll find a and b for the binary mixture.

a_ij = (1 – k_ij)a_i^1/2a_j^1/2  and a = ΣΣx_ix_ja_ij  ; b = Σx_ib_i   ......(1)

a₁₂ = a₂₁ = (1 – 0.092)(7846)1/2(28547)1/2 = 13589

a₁₁ = a₁; and a₂₂ = a₂.

Now using equation (1)

a = (0.57)(0.57)(7846) + (0.57)(0.43)(13589) + (0.43)(0.43) (28547) + (0.43)(0.57)(13589) = 14489

b = 0.57x0.0354 + 0.43x0.0795 = 0.054

The Redlich Kwong equation,

P = {RT/(Vm – b)} - {a/(T1/2Vm(Vm+b))}

Use the given values,

P = 30 bar = 3030.75 kPa

T = 450 K

After rearraning the Redlich-Kwong equation we get a cubic polynomial in V_m.
64483V_m³ – 79465V_m² – 4479V_m – 782 = 0

We obtain the roots using MATLAB's roots function,

1.29
-0.0305 + 0.0919_i
-0.0305 - 0.0919_i

Hence the volume of the vessel is V_m x No of moles,
= 1.29 x 2.01 = 2.6 m3 = 2600 lit.