2F1 :  An Application of Equations of State  10 pts 

Estimate
the pressure of ammonia at a temperature of 27^{o}C and a specific
volume of 0.626 m^{3}/kg. a.) Ideal Gas EOS b.) Virial EOS c.) van der Waal EOS d.) SoaveRedlichKwong EOS 

e.) Compressibility
Factor EOS f.) Steam Tables. 

Read :  Not much to say here.  
Given:  T  27  ^{o}C  Find:  P  ???  kPa  
V  0.626  m^{3}/kg  
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/molK  T_{c}  405.55  K  
MW  17.03  g NH_{3} / mol NH_{3}  P_{c}  1.128E+07  Pa  
w  0.250  
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  
Therefore : 

Eqn 5  
V  1.066E02  m^{3}/mol  
Now, plug values back into Eqn 2.  T  300.15  K  
P  2.341E+05  Pa  
Be careful with the units.  P  234.1  kPa  
Part b.)  Truncated Virial EOS : 

Eqn 6  
We can estimate B using : 

Eqn 7  

Eqn 8  

Eqn 9  
Where : 

Eqn 10  
We can solve Eqn 6 for P : 

Eqn 11  
Plugging numbers into Eqns 10, 8, 9, 7 and 11 (in that order) yields :  
T_{R}  0.740  B  2.14E04  m^{3}/mol  
B_{0}  0.6000  Z  9.80E01  
B_{1}  0.4698  P  229.4  kPa  
Part c.)  van der Waal EOS : 

Eqn 12  
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 13 

Eqn 14  
T_{c}  405.55  K  a  0.4252  Pamol^{2}/m^{6}  
P_{c}  1.128E+07  Pa  b  3.74E05  m^{3}/mol  
Now, we can plug the constants a and b into Eqn 12 to determine the pressure.  
P  231.2  kPa  
Part d.)  SoaveRedlichKwong EOS : 

Eqn 15  
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 16 

Eqn 17  

Eqn 18 

Eqn 19  

Eqn 20  
Where :  w  0.250  
Now, plug values into Eqns 15  20 :  
T_{R}  0.7401  a  0.43084  Pamol^{2}/m^{6}  
m  0.8633  b  2.590E05  m^{3}/mol  
a  1.25575  
P  229.9  kPa  
Part e.)  Compressibility EOS :  
Given T_{R} and the ideal reduced molar volume, use the compressibility charts to evaluate either P_{R} or the compressibility, Z  

Eqn 21  From part c :  T_{R}  0.7401  
Defiition of the ideal reduced molar volume : 

Eqn 22  
V_{R}^{ideal}  35.67  
Read the Generalized Compressibility Chart for P_{R} = 0 to 1 (not easy!): 
P_{R}  0.02  
Z  0.983  
We can use the definition of P_{R} to calculate P :  

Eqn 23 

Eqn 24  
P  225.6  kPa  
Or, we can use Z and its definition to determine P :  

Eqn 25  
P  230.1  kPa  
Part f.)  The Ammonia Tables provide the best available estimate of the pressure.  
We begin by determining the state of the system. In this case, it would be easiest to lookup the V_{sat vap} and V_{sat liq} at the given temperature.  
If :  V > V_{sat vap}  Then :  The system contains a superheated vapor.  
If :  V < V_{sat liq}  Then :  The system contains a subcooled liquid.  
If :  V_{sat vap} > V > V_{sat liq}  Then :  The system contains an equilibrium mixture of saturated liquid and saturated vapor.  
Data :  P*(kPa)  T (^{o}C)  V_{sat liq} (m^{3}/kg)  V_{sat vap} (m^{3}/kg)  H_{sat liq} (kJ/kg)  H_{sat vap} (kJ/kg)  
1066.56  27  1.67E03  0.12066  308.11  1465.42  
Because V > V_{sat vap}, the ammonia is superheated 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 ad) or we can scan the superheated vapor tables to determine which two pressures bracket our known value of the specific volume. The given specific volume of 0.626 m^{3}/kg lies between 200 kPa and 250 kPa and T = 27^{o}C lies between 25^{o}C and 50^{o}C. This is a tricky multiple interpolation problem !  
The Superheated Ammonia Table gives us :  P*(kPa)  T (^{o}C)  V (m^{3}/kg)  H (kJ/kg)  
200  20  0.6995  1510.1  
200  30  0.7255  1532.5  
250  25  0.5668  1518.2  
250  50  0.6190  1574.7  
We can now interpolate on this data to determine values of the specific volume at T = 27^{o}C at BOTH 200 kPa and 250 kPa. This will help us setup a second interpolation to determine the pressure that corresponds to T = 27^{o}C and V = 0.626 m^{3}/kg.  
At 200 kPa :  T (^{o}C)  V (m^{3}/kg)  
20  0.6995  
30  0.7255  

Eqn 26  

Eqn 27  
slope  2.600E03  (m^{3}/kg)/^{o}C  V  0.71766  m^{3}/kg  
At 250 kPa :  T (^{o}C)  V (m^{3}/kg)  
25  0.5668  
50  0.6190  

Eqn 28  

Eqn 29  
slope  2.085E03  (m^{3}/kg)/^{o}C  V  0.57102  m^{3}/kg  
Now, we must interpolate one more time to determine the pressure which, at 27^{o}C, yields a spoecific volume of 0.626 m^{3}/kg :  
At 27^{o}C :  P (kPa)  V (m^{3}/kg)  
200  0.7177  
250  0.5710  

Eqn 30  

Eqn 31  
slope  340.96  (m^{3}/kg)/kPa  P  231.3  kPa  
Verify:  No assumptions to verify.  
Answers :  a.)  P  234.1  kPa  d.)  P  229.9  kPa  
b.)  P  229.4  kPa  e.)  P  230.1  kPa  
c.)  P  231.2  kPa  f.)  P  231.3  kPa  