Problem on Redlich-Kwong equation

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.

E

Expert

##### Verified

(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
Pc 61.91 bar
Tc 35.1 oC

n-butane

MW 58.12
Pc   38 bar
T  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 = 0.427R2Tc2.5/Pc = 0.427(8.314)2(308.2)2.5/6273 = 7846
b1 = 0.0866RTc/Pc = 0.0866(8.314)(308.2)/6273 = 0.0354

n-butane:

a2 = 0.427R2Tc2.5/Pc = 0.427(8.314)2(425)2.5/3850 = 28547

b2 = 0.0866RTc/Pc = 0.0866(8.314)(425)/3850 = 0.0795

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

aij = (1 – kij)ai1/2aj1/2  and a = ΣΣxixjaij  ; b = Σxib  ......(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 Vm.
64483Vm3 – 79465Vm2 – 4479Vm – 782 = 0

We obtain the roots using MATLAB's roots function,

1.29
-0.0305 + 0.0919i
-0.0305 - 0.0919i

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

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