Verified
(i)
(a) The amount in kg, of superpure grade N_{2}, per container is calculated below,
PV = nRT
n = PVT_{1}/(TP_{1}V_{1}/n_{1})) ... 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 N_{2}.
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 O_{2}, per container is calculated below,
PV = nRT
n = PVT_{1}/(TP_{1}V_{1}/n_{1})) ... 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 O_{2}.
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
x_{1} = mole fraction of acetylene = (30/26)/2.01 = 0.57
x_{2} = mole fraction of n-butane = 0.43
Redlich-Kwong parameters (Note that P is in kPa and T is in K)
acetylene:
a_{1} = 0.427R^{2}T_{c}^{2.5}/Pc = 0.427(8.314)^{2}(308.2)^{2.5}/6273 = 7846
b_{1} = 0.0866RT_{c}/P_{c} = 0.0866(8.314)(308.2)/6273 = 0.0354
n-butane:
a_{2} = 0.427R^{2}T_{c}^{2.5}/Pc = 0.427(8.314)^{2}(425)^{2.5}/3850 = 28547
b_{2} = 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}^{1/2}a_{j}^{1/2 }and a = ΣΣx_{i}x_{j}a_{ij} ; b = Σx_{i}b_{i } ......(1)
a_{12} = a_{21} = (1 – 0.092)(7846)1/2(28547)1/2 = 13589
a_{11} = a_{1}; and a_{22} = a_{2}.
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_{m}^{3} – 79465V_{m}^{2} – 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.