# Example Problem with Complete Solution

2F-2 : An Application of Equations of State 6 pts
Steam is contained in a 203 L tank at 500oC.  The mass of steam in the tank is 12.4 kg. Determine the pressure in the tank using:
a.) Ideal Gas EOS
b.) Virial EOS
c.) van der Waal EOS
d.) Redlich-Kwong EOS
e.) Compressibility Factor EOS
f.) Steam Tables.

Read : Not much to say here.
Given: m 12.4 kg V 203 L
T 500 oC 0.203 m3
Find: P ??? kPa
Assumptions: None.
Equations / Data / Solve:
Begin by collecting all of the constants needed for all the Equations of State in this problem.
R 8.314 J/mol-K Tc 647.4 K
MW 18.016 g NH3 / mol NH3 Pc 2.21E+07 Pa
Part a.) Ideal Gas EOS : Eqn 1 Solve for pressure : Eqn 2
We must determine the molar volume before we can use Eqn 2 to answer the question.
Use the definition of molar volume: Eqn 3
Where : Eqn 4
MW 18.016 g H2O / mol H2O n 688.28 mol H2O
V 2.95E-04 m3/mol
Now, plug values back into Eqn 2. R 8.314 J/mol-K
Be careful with the units. T 773.15 K
P 2.18E+07 Pa
P 21.8 MPa
Part b.) van der Waal EOS : Eqn 5
We can determine the values of a and b, which are constants that depend only on the chemical species in the system, from the following equations. Eqn 6 Tc 647.4 K
Pc 2.21E+07 Pa
a 0.5530 Pa-mol2/m6 Eqn 7 b 3.04E-05 m3/mol
Now, we can plug the constants a and b into Eqn 5 to determine the pressure.
P 17.9 MPa
Part c.) Redlich-Kwong EOS : Eqn 8
We can determine the values of a, b and a, which are constants that depend only on the chemical species in the system, from the following equations. Eqn 9 Eqn 10
Now, plug values into Eqns 8 -10 :
a 14.25855 Pa-m6-K1/2/mol2
b 2.110E-05 m3/mol P 18.0 MPa
Part d.) Compressibility EOS : Given TR and the ideal reduced molar volume, use the compressibility charts to evaluate either PR or the compressibility, Z Eqn 11 Eqn 12
TR 1.1942
Defiition of the ideal reduced molar volume : Eqn 13
VRideal 1.2110
Read the Generalized Compressibility Chart for PR = 0 to 1 : PR 0.78
Z 0.83
We can use the definition of PR to calculate P : Eqn 14 Eqn 15
P 17.2 MPa
Or, we can use Z and its definition to determine P : Eqn 16
P 18.1 MPa
Part e.) The Steam Tables provide the best available estimate of the pressure in the tank.
Because T > Tc, the properties of the water in the tank must be obtained from the superheated vapor table, even though the water is actually a supercritical fluid in this system.
At this point we can make use of the fact that we have a pretty good idea of what the actual pressure is in the tank (from parts a-d) or we can scan the spuerheated vapor tables to determine which two pressures bracket our known value of the specific volume.
In either case, we begin by converting the molar volume into a specific volume : Eqn 17
Using the MW of water from part (a) yields : v 1.637E-05 m3/g
v 0.016371 m3/kg
The Superheated Steam Table gives us :
At P = 20 MPa and At P = 40 MPa
v = 0.014793 m3/kg v = 0.005623 m3/kg
We can determine the pressure in our tank by interpolation : P 16.56 MPa
P 16.6 MPa
Verify: No assumptions to verify.
Answers : a.) P 21.8 kPa d.) P 18.1 kPa
b.) P 17.9 kPa e.) P 16.6 kPa
c.) P 18.0 kPa
None of these Equations of State did very well because steam at high pressure behaves in a very non-ideal manner due to the high polarity of the molecules and the resulting stron electrostatic interactions. 