EOS Example, How To Read Tables, and Basic Numerical Analysis
Thermodynamics  

Introduction  What is this thing called Thermodynamics???  Definitions  Thermal Equilibrium and Zeroth Law  Limitations 
First Law  Work, Heat, Energy, and the First Law  Work, Heat, Energy, and the First Law (simplied)  Derivatives  Derivatives Exercise  Reversibility, Enthalpy, and Heat Capacity 
Second Law  Things to Think About  Observations and Second Law of Thermodynamics  Alternative Approach  the Clausis Inequality  Consequences of the Second Law  Consequences of the Second Law (simplified)  Carnot Principle  motivation and examples  Equivalence of Second Law Statements* 
Third Law  Third Law of Thermodynamics  Consequences of Third Law* 
Development of Thermodynamics  The Thermodynamic Network  Network Exercise  Equations of State (EOS)  EOS Example, Reading Tables, and Numerical Analysis  EOS Exercises  Thermochemistry 
* Optional Section 

Example
Calculate the compressibility factor and the specific volume of formaldehyde at 100 °C and 15.0 bar using (a) the ideal gas equation, (b) the three parameter correlation, and (c) the Soave equation.
Solution
(a) The ideal gas equation is [math]P\hat{V}=RT[/math]. By inspection we can see that Z = 1. To calculate the specific volume we will use R = 83.14 cm^{3}/mol K. These units are the most useful in EOS calculations.
Then
[math]\hat{V}=\frac{RT}{P}=\frac{83.14\frac{cm^{3}bar}{mol\,K}373\,K}{15\,bar}=2067\frac{cm^{3}}{mol}[/math]
(b) From the data we get T_{C} = 408 K, P_{C} = 65.90 bar, ω = 0.282
With these values we get [math]T_{R}=\frac{373K}{408K}=0.914, P_{R}=\frac{15bar}{65.90bar}=0.228[/math]
We now need to read the values of Z^{0} and Z^{1} from the Z tables. But there is no listed values for T_{R} = 0.914 or P_{R} = 0.228.
Well, to determine the value from the table we use interpolation. This is based on the assumption that the function varies linearly between the values given in the table.
Say we have the following table:
(table1)
If we now wish to know the value of y' at x' we can use the following interpolation formula:
[math]\frac{y'y_{1}}{x'x_{1}}=\frac{y_{2}y_{1}}{x_{2}x_{2}}[/math]
In the case of our example we need to interpolate twice  once in T and once in P. The relevant part of the Z^{0} table is:
(table2)
Interpolating over T using P_{R} = 0.2000 gives for T_{R} = 0.914,
[math]\frac{Z^{0}0.9015}{0.9140.90}=\frac{0.91150.9015}{0.930.90}[/math]
or Z^{0 = .9062 }
Interpolating over T using P_{R} = 0.4000 gives for T_{R} = 0.914,
[math]\frac{Z^{0}0.7800}{0.9140.90}=\frac{0.80590.7800}{0.930.90}[/math]
or Z^{0} = .7921
Modifying our table:
(table3)
Interpolating over P using T_{R} = 0.914 gives for P_{R} = 0.228,
[math]\frac{Z^{0}0.9062}{0.2280.2000}=\frac{0.79210.9062}{0.40000.2000}[/math]
or Z^{0} = 0.8902
Doing the same process gives for Z^{1} = 0.0467
Then from Pitzer's relation,
[math]Z=Z^{0}+\omega Z^{1}=0.8902+0.282(0.0467)=0.8770[/math]
finally
[math]\hat{V}=\frac{ZRT}{P}=1813\frac{cm^{3}}{mol}[/math]
(c) Using the data we get f = 0.48 + 1.574(0.282)  0.176(0.282)^2 = 0.9099
[math]a=\frac{0.42748\left(83.14\frac{cm^{3}bar}{molK}\right)^{2}\left(408K\right)^{2}}{65.9bar}\left[1+0.9099\left(10.914^{0.5}\right)\right]^{2}=8.073x10^{6}cm^{6}bar/mol^{2}[/math]
[math]b=\frac{0.08664\left(83.14\frac{cm^{3}bar}{molK}\right)\left(408K\right)}{15bar}=44.60cm^{3}/mol[/math]
Note the odd units on a.
• The Soave equation is:
[math]P=\frac{RT}{\hat{V}b}\frac{a}{\hat{V}(\hat{V}+b)}[/math]
There are two ways to solve this:
1. Use a computer or calculator to solve the equation.
2. Solve the equation iteratively.
Let us use the iterative method to solve our example^{[1]}
The only unknown is [math]\hat{V}[/math]. Let us now rearrange the Soave equation by multiplying both sides by [math]\frac{\hat{V}b}{P}[/math]:
[math]\hat{V}b=\frac{RT}{P}\frac{a\left(\hat{V}b\right)}{P\hat{V}\left(\hat{V}+b\right)}[/math]
[math]\hat{V}=\frac{RT}{P}\frac{a\left(\hat{V}b\right)}{P\hat{V}\left(\hat{V}+b\right)}+b[/math]
Slightly rewrite it as:
[math]\hat{V}_{i+1}=\frac{RT}{P}\frac{a\left(\hat{V}_{i}b\right)}{P\hat{V}_{i}\left(\hat{V}_{i}+b\right)}+b[/math]
Where i is an iteration index.
The basic concept is that we guess a value for [math]\hat{V}_{i}[/math], put it into the right side of the equation, and then calculate [math]\hat{V}_{i+1}[/math]. If [math]\hat{V}_{i+1}[/math] and [math]\hat{V}_{i}[/math] are approximately equal then we are finished. If not, then we use [math]\hat{V}_{i+1}[/math] as the new guess. This is continued until we get convergence. The iteration index i is 1 for the first guess, etc.
As an initial guess it is usually best to use the ideal gas value. From above that gives [math]\hat{V}_{1}=2067[/math]. Now the iterative equation gives [math]\hat{V}_{2}=1863[/math]. We can then use the following table:
(table4)
Since [math]\hat{V}_{i+1}[/math] and [math]\hat{V}_{i}[/math] are now the same we will say the solution is [math]\hat{V}=1832cm^{3}/mol[/math]. Then [math]Z=\frac{P\hat{V}}{RT}=0.8861[/math].
Some observations on the above exercise. The difference between using the tables and using the equation is only about 1 percent, which is quite good. However, the difference between these and the ideal gas equation is around 15%, which can be significant. For someone designing equipment that can be very important, too big of a tank may cost a lot of extra money^{[2]}
Notes
 ↑ Yes, we could just plug the numbers in a computer, but it is instructive for students to know how the calculation is made (a computer also uses the iterative method)
 ↑ Note that you may still want a bigger tank for safety considerations. However, that safety factor added is after the calculation.