A 20ft^{3} tank contains natural gas with the composition given below. After removing 200 SCF from the tank, the pressure and temperature drop to 1800 psia and 80^{o}F. Determine the how many moles of gas were in the tank originally.
Component 
Mole % 

C_{1} 
89.50 

C_{2} 
4.50 

C_{3} 
3.50 

iC_{4} 
0.70 

nC_{4} 
0.60 

iC_{5} 
0.40 

nC_{5} 
0.20 

C_{6} 
0.10 

C_{7+} 
0.50 

M_{C7+}= 110 
ϒ _{C7+}= 0.75 
We know that Standard Cubic Foot = one cubic foot of gas at 21°C or 70°F and 101.35 kilopascals pressure or 14.7 psai = 760 mmHg
Moles of gas = 1.2 moles at SCF OR 0.002 lbs moles
Let initial moles = x
so decrease in moles = x0.002
So new gas equation will be
PV = nRT
R = gas constant = 998.9 ft3 mmHg K−1 lbmol−1
P = 1800psai = 93086.8 mmHg
V = 20 ft^3
n = x0.002
T = 80^{0}F = 299.8 K
x0.002 = 93086.8 X 20 / 998.9 X 299.8 = 6.216
x = 6.218 lb moles = 2819 moles
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