--%>

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.

   Related Questions in Chemistry

  • Q : Finding Active mass of urea Can someone

    Can someone please help me in getting through this problem. 10 litre solution of urea comprises of 240 gm urea. The active mass of urea is: (i) 0.04 (ii) 0.02 (iii) 0.4 (iv) 0.2

  • Q : Colligative property associated question

    Give me answer of this question. Which of the following is not a colligative property : (a)Optical activity (b)Elevation in boiling point (c)Osmotic pressure (d)Lowering of vapour pressure

  • Q : Calculating weight of acid Give me

    Give me answer of this question. The formula weight of H2SO4 is 98. The weight of the acid in 400mi of solution is: (a)2.45g (b) 3.92g (c) 4.90g (d) 9.8g

  • Q : Calculating amount of Sodium hydroxide

    Choose the right answer from following. The amount of NaOH in gms in 250cm3 of a0.100M NaOH solution would be : (a) 4 gm (b) 2 gm (c) 1 gm (d) 2.5 gm

  • Q : Problem on Osmotic Pressure of solution

    The osmotic pressure of a 5% solution of cane sugar at 150oC  is (mol. wt. of cane sugar = 342)(a) 4 atm (b) 3.4 atm (c) 5.07 atm (d) 2.45 atmAnswer: (c) Π = (5 x 0.0821 x 1000 x 423)/(342 x 100) = 5.07 atm

  • Q : Explain preparation and properties of

    It may be prepared by the action of phosphorus on thionyl chloride.P4 + 8SOCl2    4

  • Q : Molecular weight of substance The

    The boiling point of a solution of 0.11 gm of a substance in 15 gm of ether was found to be 0.1oC higher than that of the pure ether. The molecular weight of the substance will be (Kb = 2.16)       (a) 148 &nbs

  • Q : Question related to molarity Help me to

    Help me to go through this problem. Molarity of a solution containing 1g NaOH in 250ml of solution: (a) 0.1M (b) 1M (c) 0.01M (d) 0.001M

  • Q : Haloalkane how haloalkane can be

    how haloalkane can be prepared by refluxing alcohol with hydrohalic acids

  • Q : What do you mean by the term medicine

    What do you mean by the term medicine dropper? Explain briefly?