1. Course No: ME 5243- Advanced Thermodynamics
Estimating Different Thermodynamic Relations using
Redlich- Kwong-Soave Equation of State.
Final Project report
Abu Saleh Ahsan,Md. Saimon Islam, Syed Hasib Akhter Faruqui
5-5-2016

2. 1 | P a g e
Table of Contents
Nomenclature:..............................................................................................................................................2
Abstract:........................................................................................................................................................4
Introduction:.................................................................................................................................................5
Derivation......................................................................................................................................................6
(a) Evaluation of the Constants.................................................................................................................6
(b) Equation of State in Reduced Form.....................................................................................................9
c) Critical Compressibility Factor ............................................................................................................10
d) Express Z in terms TR, vR’:..................................................................................................................12
e) Accuracy of EOS from Equation (d).....................................................................................................13
f) Equation for Departure .......................................................................................................................15
𝒉 ∗ −𝒉𝑹𝑻𝒄 .........................................................................................................................................15
(u*-u)/RTc...........................................................................................................................................15
𝒔 ∗ −𝒔𝑹...............................................................................................................................................15
g) Accuracy of EOS for equation (C)........................................................................................................16
h) Derivation of Expressions: ..................................................................................................................17
𝒂 ∗ − 𝒂𝑹𝑻𝒄:.......................................................................................................................................17
𝒈 ∗ − 𝒈𝑹𝑻𝒄: ......................................................................................................................................17
i) Speed of sound ....................................................................................................................................18
(j) Derive the Properties..........................................................................................................................19
Cp ........................................................................................................................................................19
Cv ........................................................................................................................................................19
1/v vp ..................................................................................................................................19
k = v/p pv ....................................................................................................................................19
kT .........................................................................................................................................................20
J .........................................................................................................................................................20
Summary:....................................................................................................................................................21
Appendix .....................................................................................................................................................22
MATLAB Code .........................................................................................................................................22
Reference....................................................................................................................................................23

3. 2 | P a g e
Nomenclature:
P = Pressure
Pr = Reduced pressure
Pc = Critical pressure
v = Specific volume
vr = Reduced specific volume
vc = Critical volume
𝑣𝑟
∗
= Specific volume of ideal gas
vrf = Reduced volume at liquid state
vrg = Reduced volume at gaseous state
T = Temperature
Tr = Reduced temperature
Tc= Critical temperature
R = Molar gas constant
Z = Compressibility factor
Zr = Reduced compressibility factor
Zc = Critical compressibility factor
𝑔 = Gibbs free energy
𝑔0 = Gibbs free energy for ideal gas
ℎ = Specific enthalpy for real gas
ℎ0 = Specific enthalpy for ideal gas
𝑢 = Specific internal energy for real gas
𝑢0 = Specific internal energy for ideal gas

4. 3 | P a g e
𝑠 = Specific entropy for real gas
𝑠0 = Specific entropy for ideal gas
𝑘 𝑇 = Isothermal expansion exponent
𝑐 = Speed of sound
𝛽 =Volumetric co-efficient of thermal expansion
𝐶 𝑝 = Constant pressure specific heat
𝐶 𝑣 = Constant volume specific heat

5. 4 | P a g e
Abstract:
For the project we will be using “Redlich- Kwong-Soave” Equation of State (EOS) to derive to Estimating
Different Thermodynamic Relations. Starting from “Redlich- Kwong-Soave” equation we have calculated
the constants “a(T)” & “b”. The EOS is again represented in its reduced form. Compressibility factor for
the selected EOS is calculated and expressed in terms of TR & VR. By using the EOS different thermodynamic
relations such as departure enthalpy, entropy, and change in internal energy has been evaluated. Again,
we have expressed different important parameters such as speed of sound, isothermal expansion
exponent, CP & CV in reduced form. As for the substance in question we are using “Nitrogen”.

6. 5 | P a g e
Introduction:
Real gases are different from that of ideal gases. Thus the evaluated properties of ideal gas cannot be
used as the same for real gases. To understand the characteristics of real gases we have to consider the
following-
 compressibility effects;
 variable specific heat capacity;
 van der Waals forces;
 non-equilibrium thermodynamic effects;
 issues with molecular dissociation and elementary reactions with variable composition.
In our project, we have taken “Redlich- Kwong-Soave” equation of state (EOS) into consideration to
derive the various fundamental relations of thermodynamics. The compressibility, enthalpy departure
and entropy departure, can all be calculated if an equation of state for a fluid is known which is
“Nitrogen” in our case.
Now, the “Redlich- Kwong-Soave” equation of state (EOS) is almost similar to Van Der Walls equation of
state. The equation is-
𝑝 =
𝑅𝑇
𝑣−𝑏
−
𝑎(𝑇)
𝑣(𝑣+𝑏)
… …. … … … … … … … … (i)
In thermodynamics, a departure function is defined for any thermodynamic property as the difference
between the property as computed for an ideal gas and the property of the species as it exists in the real
world, for a specified temperature T and pressure P. Common departure functions include those for
enthalpy, entropy, and internal energy.
Departure functions are used to calculate real fluid extensive properties (i.e properties which are
computed as a difference between two states). A departure function gives the difference between the
real state, at a finite volume or non-zero pressure and temperature, and the ideal state, usually at zero
pressure or infinite volume and temperature.

7. 6 | P a g e
Derivation
(a) Evaluation of the Constants
Taking the first and second derivative of pressure WRT to volume be –
(i) =>
𝛿𝑝
𝛿𝑣
)
𝑇
= −
𝑅𝑇
(𝑣−𝑏)2 +
𝑎(𝑇)
𝑣2(𝑣+𝑏)
+
𝑎(𝑇)
𝑣(𝑣+𝑏)2 … … … … … … … … (ii)
And,
𝛿2 𝑝
𝛿𝑣2)
𝑇
=
2𝑅𝑇
(𝑣−𝑏)3 + 𝑎(𝑇) [−
1
𝑣2(𝑣+𝑏)2 −
2
𝑣3(𝑣+𝑏)
−
1
𝑣2(𝑣+𝑏)2 −
2
𝑣(𝑣+𝑏)3]
𝛿2 𝑝
𝛿𝑣2)
𝑇
=
2𝑅𝑇
(𝑣−𝑏)3 + 𝑎(𝑇) [−
2
𝑣2(𝑣+𝑏)2 −
2
𝑣3(𝑣+𝑏)
−
2
𝑣(𝑣+𝑏)3]
𝛿2 𝑝
𝛿𝑣2)
𝑇
=
2𝑅𝑇
(𝑣−𝑏)3 − 𝑎(𝑇) [
2
𝑣2(𝑣+𝑏)2 +
2
(𝑣+𝑏)
+
2
𝑣(𝑣+𝑏)3]
𝛿2 𝑝
𝛿𝑣2)
𝑇
=
2𝑅𝑇
(𝑣−𝑏)3 − 𝑎(𝑇) [
3𝑣2+3𝑣𝑏+𝑏2
𝑣3(𝑣+𝑏)3 ] … … …. …. …. …. …. …. (iii)
Now at critical point first and second derivative of pressure WRT to volume be zero. So from equation
(ii) and (iii) we can write,
𝛿𝑝
𝛿𝑣
)
𝑇
= −
𝑅𝑇𝑐
(𝑣 𝑐−𝑏)2 +
𝑎(𝑇)
𝑣 𝑐
2(𝑣 𝑐+𝑏)
+
𝑎(𝑇)
𝑣 𝑐(𝑣 𝑐+𝑏)2 = 0
𝑜𝑟, 0 = −
𝑅𝑇𝑐
(𝑣 𝑐−𝑏)2 +
𝑎(𝑇)
𝑣 𝑐
2(𝑣 𝑐+𝑏)
+
𝑎(𝑇)
𝑣 𝑐(𝑣 𝑐+𝑏)2
𝑜𝑟,
𝑅𝑇𝑐
(𝑣 𝑐−𝑏)2 =
𝑎(𝑇)
𝑣 𝑐
2(𝑣 𝑐+𝑏)
+
𝑎(𝑇)
𝑣 𝑐(𝑣 𝑐+𝑏)2
𝑜𝑟,
𝑅𝑇𝑐
(𝑣 𝑐−𝑏)2 =
𝑎(𝑇) (2𝑣 𝑐+𝑏)
𝑣 𝑐
2(𝑣 𝑐+𝑏)2 …. …. .… …. ….. ….. …. … (iv)
And,
𝛿2 𝑝
𝛿𝑣2)
𝑇
=
2𝑅𝑇𝑐
(𝑣 𝑐−𝑏)3 − 𝑎(𝑇) [
3𝑣 𝑐
2+3𝑣 𝑐 𝑏+𝑏2
𝑣 𝑐
3(𝑣 𝑐+𝑏)3 ] = 0
𝑜𝑟,
2𝑅𝑇𝑐
(𝑣 𝑐−𝑏)3 = 𝑎(𝑇) [
3𝑣 𝑐
2+3𝑣 𝑐 𝑏+𝑏2
𝑣 𝑐
3(𝑣 𝑐+𝑏)3 ] …. …. …. …. …. …. …. (v)
Now, Dividing Equation (iv) with equation (v) we get,
(𝑣𝑐 − 𝑏) =
(𝑣 𝑐+𝑏).𝑣 𝑐.(𝑣 𝑐+𝑏)
3𝑣 𝑐
2+3𝑣 𝑐 𝑏+𝑏2
𝑜𝑟, (𝑣𝑐 − 𝑏)(3𝑣𝑐
2
+ 3𝑣𝑐 𝑏 + 𝑏2
) = (𝑣𝑐 + 𝑏). 𝑣𝑐. (𝑣𝑐 + 𝑏)

8. 7 | P a g e
𝑜𝑟, 3𝑣𝑐
3
+ 3𝑣𝑐
2
+ 𝑏2
𝑣𝑐 − 3𝑏𝑣𝑐
2
− 3𝑣𝑐 𝑏2
− 𝑏3
= 2𝑣𝑐
3
+ 2𝑣𝑐
2
+ 𝑣𝑐
2
𝑏 + 𝑣𝑐 𝑏2
𝑜𝑟, 𝑏3
+ 3𝑏2
𝑣𝑐 + 3𝑏𝑣𝑐
2
− 𝑣𝑐
3
= 0
𝑜𝑟, (𝑏 + 𝑣𝑐)3
= 2𝑣𝑐
3
= (√2
3
𝑣𝑐)
3
𝑜𝑟, 𝑏 = (√2
3
− 1)𝑣𝑐 = (√2
3
− 1)𝑧 𝑐
𝑅𝑇𝑐
𝑝 𝑐
𝑜𝑟, 𝑏 = (√2
3
− 1).
1
3
.
𝑅𝑇𝑐
𝑝 𝑐
= 0.08664
𝑅𝑇𝑐
𝑝 𝑐
𝑜𝑟, 𝑏 = 0.08664
𝑅𝑇𝑐
𝑝 𝑐
…. …. …. …. …. …. …. (vi)
Note: Here, 𝑧 𝑐 =
1
3
. As the original equation is from Redlich-Kwong equation, we will use the value of 𝑧 𝑐
from the Redlich-Kwong equation.
Now putting the value of ‘b’ and 𝑣𝑐 = 𝑧 𝑐
𝑅𝑇𝑐
𝑝 𝑐
=
1
3
.
𝑅𝑇𝑐
𝑝 𝑐
in equation (iv) we get,
𝑅𝑇𝑐
(𝑣 𝑐−𝑏)2 =
𝑎(𝑇) (2𝑣 𝑐+𝑏)
𝑣 𝑐
2(𝑣 𝑐+𝑏)2
𝑜𝑟, 𝑎(𝑇) =
𝑅𝑇𝑐 𝑣 𝑐
2(𝑣 𝑐+𝑏)2
(2𝑣 𝑐+𝑏) (𝑣 𝑐−𝑏)2
𝑜𝑟, 𝑎(𝑇) =
𝑅𝑇𝑐 [𝑧 𝑐
𝑅𝑇 𝑐
𝑝 𝑐
]
2
(𝑧 𝑐
𝑅𝑇 𝑐
𝑝 𝑐
+0.08664
𝑅𝑇 𝑐
𝑝 𝑐
)
2
(2𝑧 𝑐
𝑅𝑇 𝑐
𝑝 𝑐
+0.08664
𝑅𝑇 𝑐
𝑝 𝑐
) (𝑧 𝑐
𝑅𝑇 𝑐
𝑝 𝑐
−0.08664
𝑅𝑇 𝑐
𝑝 𝑐
)
2
𝑜𝑟, 𝑎(𝑇) =
𝑅𝑇𝑐 [𝑧 𝑐
𝑅𝑇 𝑐
𝑝 𝑐
]
2
(
𝑅𝑇 𝑐
𝑝 𝑐
)
2
(
1
3
+0.08664 )
2
(
𝑅𝑇 𝑐
𝑝 𝑐
) (
2
3
+0.08664) (
𝑅𝑇 𝑐
𝑝 𝑐
)
2
(
1
3
−0.08664 )
2
𝑜𝑟, 𝑎(𝑇) =
[
1
3
.
𝑅𝑇 𝑐
𝑝 𝑐
]
2
∗ (0.176377600711111)
(
1
𝑝 𝑐
) (
2
3
+0.08664) (
1
3
−0.08664 )
2
𝑜𝑟, 𝑎(𝑇) = 0.42747
𝑅2 𝑇𝑐
2
𝑝 𝑐
… …. …. …. …. …. …. …. …. (vii)
Now, this is for the critical point only. As for other points for the assigned substance Nitrogen we get
a(T)= 0.42747
𝑅2 𝑇𝑐
2
𝑝 𝑐
(1 + [{0.0007T4
- 0.3012T3
+ 45.74036T2
- 3068.87T + 76836.9287}]*T)
at critical point T=TR=1 thus we get again,
a(T)= 0.42747
𝑅2 𝑇𝑐
2
𝑝 𝑐

9. 8 | P a g e
Figure-1: a(T) function determination
-50
0
50
100
150
200
250
300
0 20 40 60 80 100 120 140
a(T)
T

10. 9 | P a g e
(b) Equation of State in Reduced Form
𝑝 =
𝑅𝑇
𝑣−𝑏
−
𝑎(𝑇)
𝑣(𝑣+𝑏)
𝑜𝑟, 𝑝𝑟 𝑝𝑐 =
𝑅𝑇𝑟 𝑇𝑐
𝑣 𝑟 𝑣 𝑐−0.08664
𝑅𝑇 𝑐
𝑝 𝑐
−
0.42747
𝑅2 𝑇 𝑐
2
𝑝 𝑐
(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇)
𝑣 𝑟 𝑣 𝑐(𝑣 𝑟 𝑣 𝑐+0.08664
𝑅𝑇 𝑐
𝑝 𝑐
)
𝑜𝑟, 𝑝𝑟 =
𝑅𝑇𝑟 𝑇𝑐
(𝑣 𝑟 𝑣 𝑐−0.08664
𝑅𝑇 𝑐
𝑝 𝑐
)𝑝 𝑐
−
0.42747
𝑅2 𝑇 𝑐
2
𝑝 𝑐
(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇)
𝑝 𝑐 𝑣 𝑟 𝑣 𝑐(𝑣 𝑟 𝑣 𝑐+0.08664
𝑅𝑇 𝑐
𝑝 𝑐
)
𝑜𝑟, 𝑝𝑟 =
𝑅𝑇 𝑟 𝑇 𝑐
𝑝 𝑐
(𝑣 𝑟 𝑣 𝑐−0.08664
𝑅𝑇 𝑐
𝑝 𝑐
)
−
0.42747
𝑅2 𝑇 𝑐
2
𝑝 𝑐
2 (1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇)
𝑣 𝑟 𝑣 𝑐(𝑣 𝑟 𝑣 𝑐+0.08664
𝑣 𝑐
𝑧 𝑐
)
𝑜𝑟, 𝑝𝑟 =
𝑇𝑟 (
𝑣 𝑐
𝑧 𝑐
)
[𝑣 𝑟 𝑣 𝑐−0.08664 (
𝑣 𝑐
𝑧 𝑐
)]
−
0.42747 (
𝑣 𝑐
𝑧 𝑐
)
2
(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇)
𝑣 𝑟 𝑣 𝑐(𝑣 𝑟 𝑣 𝑐+0.08664
𝑣 𝑐
𝑧 𝑐
)
𝑜𝑟, 𝑝𝑟 =
𝑇𝑟 (
1
𝑧 𝑐
)
[𝑣 𝑟−0.08664 (
1
𝑧 𝑐
)]
−
0.42747 (
1
𝑧 𝑐
)
2
(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇)
𝑣 𝑟(𝑣 𝑟+0.08664
1
𝑧 𝑐
)
𝑜𝑟, 𝑝𝑟 =
3 𝑇𝑟
[𝑣 𝑟−0.25992]
−
0.0474967(1+[{0.0007(𝑇𝑟 𝑇𝑐)4 − 0.3012(𝑇𝑟 𝑇𝑐)3 + 45.74036(𝑇𝑟 𝑇𝑐)2 − 3068.87(𝑇𝑟 𝑇𝑐) + 76836.9287}]∗(𝑇𝑟 𝑇𝑐)
𝑣 𝑟(𝑣 𝑟+0.25992)
At critical point it becomes,
𝑜𝑟, 𝑝𝑟 =
𝑇𝑟
0.3333 𝑣 𝑟−0.08664
−
1
21.0541 𝑣 𝑟
2+5.472384 𝑣 𝑟

11. 10 | P a g e
c) Critical Compressibility Factor
𝑝𝑐 =
𝑅𝑇𝑐
𝑣𝑐 − 0.08664
𝑅𝑇𝑐
𝑝𝑐
−
0.42747
𝑅2
𝑇𝑐
2
𝑝𝑐
𝑣𝑐 (𝑣𝑐 + 0.08664
𝑅𝑇𝑐
𝑝𝑐
)
𝑜𝑟, 𝑝𝑐 =
𝑅𝑇𝑐
{𝑣𝑐 − 0.08664 (
𝑣𝑐
𝑧 𝑐
)}
−
0.42747
𝑅2
𝑇𝑐
2
𝑝𝑐
𝑣𝑐 (𝑣𝑐 + 0.08664 (
𝑣𝑐
𝑧 𝑐
))
𝑜𝑟, 1 =
𝑅𝑇𝑐
𝑝𝑐
𝑣𝑐 {1 − 0.08664 (
1
𝑧 𝑐
)}
−
0.42747
𝑅2
𝑇𝑐
2
𝑝𝑐
2
𝑣𝑐
2
(1 + 0.08664 (
1
𝑧 𝑐
))
𝑜𝑟, 1 =
(
𝑣𝑐
𝑧 𝑐
)
𝑣𝑐 {1 − 0.08664 (
1
𝑧 𝑐
)}
−
0.42747 (
𝑣𝑐
𝑧 𝑐
)
2
𝑣𝑐
2
(1 + 0.08664 (
1
𝑧 𝑐
))
𝑜𝑟, 1 =
(
1
𝑧 𝑐
)
{1−0.08664 (
1
𝑧 𝑐
)}
−
0.42747(
1
𝑧 𝑐
)
2
(1+0.08664 (
1
𝑧 𝑐
))
𝑜𝑟, 1 =
(
1
𝑧 𝑐
)(1+0.08664 (
1
𝑧 𝑐
))− {1−0.08664 (
1
𝑧 𝑐
)} {0.42747(
1
𝑧 𝑐
)
2
}
{{1}2−{0.08664 (
1
𝑧 𝑐
)}
2
}
𝑜𝑟, 1 − {0.08664 (
1
𝑧 𝑐
)}
2
= (
1
𝑧 𝑐
) + 0.08664 (
1
𝑧 𝑐
)
2
− 0.42747 (
1
𝑧 𝑐
)
2
+ (0.42747 ∗ .08664) (
1
𝑧 𝑐
)
3
𝑜𝑟, 0 = (
1
𝑧 𝑐
) + 0.08664 (
1
𝑧 𝑐
)
2
− 0.42747 (
1
𝑧 𝑐
)
2
+ (0.42747 ∗ .08664) (
1
𝑧 𝑐
)
3
− 1 + {0.08664 (
1
𝑧 𝑐
)}
2
𝑜𝑟,
𝑧 𝑐
2
+ 0.08664 𝑧 𝑐 − 0.42747 𝑧 𝑐 + (0.42747 ∗ .08664) − 𝑧 𝑐
3
+ (0.08664)2
𝑧 𝑐
𝑧 𝑐
3 = 0
𝑜𝑟, 𝑧 𝑐
2
+ 0.08664 𝑧 𝑐 − 0.42747 𝑧 𝑐 + (0.42747 ∗ .08664) − 𝑧 𝑐
3
+ (0.08664)2
𝑧 𝑐 = 0
𝑜𝑟, 𝑧 𝑐
3
− 𝑧 𝑐
2
+ (0.42747 − 0.08664 − 0.086642)𝑧 𝑐 − (0.42747 ∗ .08664) = 0
Solving this equation numerically, we get,
𝑧 𝑐 = 0.3471 (MATLAB Code at Appendix)

12. 11 | P a g e
Figure-2: Solution for Critical Compressibility Factor

13. 12 | P a g e
d) Express Z in terms TR, vR’:
𝑍 =
𝑝𝑣
𝑅𝑇
From equation (i) substituting the value of p we get,
𝑍 =
𝑣
𝑅𝑇
[
𝑅𝑇
𝑣−𝑏
−
𝑎(𝑇)
𝑣(𝑣+𝑏)
]
𝑜𝑟, 𝑍 =
𝑣
𝑅𝑇
[
𝑅𝑇
𝑣−0.08664
𝑅𝑇 𝑐
𝑝 𝑐
−
0.42747
𝑅2 𝑇 𝑐
2
𝑝 𝑐
(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇
𝑣(𝑣+0.08664
𝑅𝑇 𝑐
𝑝 𝑐
)
]
𝑜𝑟, 𝑍 = [
𝑣
𝑣−0.08664
𝑅𝑇 𝑐
𝑝 𝑐
−
0.42747
𝑅𝑇 𝑐
2
𝑇 𝑝 𝑐
(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇
𝑣(𝑣+0.08664
𝑅𝑇 𝑐
𝑝 𝑐
)
]
𝑜𝑟, 𝑍 =
[
𝑣
𝑣(1−0.08664
𝑅𝑇 𝑐
𝑣 𝑝 𝑐
)
−
0.42747
𝑇 𝑐
𝑇
(1+[{0.0007𝑇4 − 0.3012𝑇3 + 45.74036𝑇2 − 3068.87𝑇 + 76836.9287}]∗𝑇
(
𝑣
𝑅𝑇 𝑐
𝑝 𝑐
+0.08664 )
]
𝑜𝑟, 𝑍 = [
1
(1−0.08664
1
𝑣 𝑅
′ )
−
0.42747
1
𝑇 𝑅
(1+[{0.0007(𝑇𝑟 𝑇𝑐)4 − 0.3012(𝑇𝑟 𝑇𝑐)3 + 45.74036(𝑇𝑟 𝑇𝑐)2 − 3068.87(𝑇𝑟 𝑇𝑐) + 76836.9287}]∗(𝑇𝑟 𝑇𝑐)
(𝑣 𝑅
′ +0.08664 )
]
𝑜𝑟, 𝑍 = [
1
(1−0.08664
1
𝑣 𝑅
′ )
−
0.42747
1
𝑇 𝑅
(1+[{176433.1632(𝑇𝑟)4 − 604113.552(𝑇𝑟)3 + 726168.24(𝑇𝑟)2 − 386568(𝑇𝑟) + 76836.9287}]∗(𝑇𝑟∗126)
(𝑣 𝑅
′ +0.08664 )
]
At critical point,
𝑜𝑟, 𝑍 = [
1
(1−0.08664
1
𝑣 𝑅
′ )
−
0.42747
1
𝑇 𝑅
(𝑣 𝑅
′ +0.08664 )
]

14. 13 | P a g e
e) Accuracy of EOS from Equation (d)
TR = 1
TR VR
'
From
Table
From
Equation % Error
1 0.7 0.58 1.086951409 87.40542
1 0.8 0.635 0.919693232 44.83358
1 0.9 0.675 0.796209555 17.95697
1 1 0.701 0.70147159 0.067274
1 1.2 0.75 0.565944624 24.54072
1 1.4 0.775 0.473864828 38.85615
1 1.6 0.81 0.407336589 49.71153
1 1.8 0.83 0.357071096 56.97939
1 2 0.845 0.317780354 62.39286
TR = 1.05
TR VR
'
From
Table
From
Equation % Error
1.05 0.7 0.64 1.112828194 73.87941
1.05 0.8 0.68 0.942651495 38.62522
1.05 0.9 0.71 0.816840904 15.04801
1.05 1 0.74 0.720204302 2.675094
1.05 1.2 0.77 0.581765455 24.44604
1.05 1.4 0.8 0.487557258 39.05534
1.05 1.6 0.419405385
1.05 1.8 0.367860496
1.05 2 0.327535613
TR = 1.10
TR VR
'
From
Table
From
Equation
1.4 0.7 1.24221212
1.4 0.8 1.05744281
1.4 0.9 0.91999765
1.1 1 0.73723404
1.1 1.2 0.596148029
1.1 1.4 0.500004921
1.1 1.6 0.430377018
1.1 1.8 0.377669042
1.1 2 0.336404031

15. 14 | P a g e
TR = 1.20
TR VR
'
From
Table
From
Equation % Error
1.4 0.7 1.24221212
1.4 0.8 1.05744281
1.4 0.9 0.91999765
1.2 1 0.767036082
1.2 1.2 0.621317533
1.2 1.4 0.521788333
1.2 1.6 0.449577376
1.2 1.8 0.394833997
1.2 2 0.351923762
TR = 1.40
TR VR'
From
Table
From
Equation % Error
1.4 0.7 1.24221212
1.4 0.8 1.05744281
1.4 0.9 0.91999765
1.4 1 0.813867863
1.4 1.2 0.660869611
1.4 1.4 0.556019408
1.4 1.6 0.479749366
1.4 1.8 0.421807498
1.4 2 0.37631191

16. 15 | P a g e
f) Equation for Departure
For simplicity of calculation we will consider the reduced equation at critical point,
𝒉∗
− 𝒉
𝑹𝑻𝒄
ℎ∗
− ℎ
𝑅𝑇𝑐
= ∫ 𝑍𝑐 [(
𝜕𝑃𝑟
𝜕𝑇𝑟
) − 𝑃𝑟] 𝑑𝑣𝑟 − 𝑇𝑟(1 − 𝑍)
𝑉𝑟
∞
= ∫ 𝑍𝑐 [(
𝜕
𝜕𝑇𝑟
) (
𝑇 𝑟
0.3333 𝑣 𝑟−0.08664
−
1
21.0541 𝑣 𝑟
2
+5.472384 𝑣 𝑟
) −
𝑇 𝑟
0.3333 𝑣 𝑟−0.08664
−
𝑉𝑟
∞
1
21.0541 𝑣 𝑟
2
+5.472384 𝑣 𝑟
] 𝑑𝑣𝑟 − 𝑇𝑟(1 − 𝑍)
= ∫ 𝑍𝑐 [(
1
0.3333 𝑣 𝑟−0.08664
−
𝑇 𝑟
0.3333 𝑣 𝑟−0.08664
+
1
21.0541 𝑣 𝑟
2
+5.472384 𝑣 𝑟
] 𝑑𝑣𝑟 − 𝑇𝑟(1 − 𝑍)
𝑉𝑟
∞
= 𝑍𝑐 [(
ln(0.3333 𝑣 𝑟−0.08664)
0.3333
−
𝑇𝑟 ln(0.3333 𝑣 𝑟−0.08664)
0.3333
+
ln(
5.472384
𝑣 𝑟
)+21.0541
5.472384
] − 𝑇𝑟(1 − 𝑍)
= 0.873 ln(0.333𝑣𝑟 − 0.08664) [1 − 𝑇𝑟] + 0.053 ln (
5.472384
𝑣 𝑟
) + 1.12 − 𝑇𝑟(1 − 𝑍)
(u*-u)/RTc
Now to derive departure from internal energy
𝑢∗
− 𝑢
𝑅𝑇𝑐
= − ∫ 𝑍𝑐 [𝑇𝑟 (
𝜕𝑃𝑟
𝜕𝑇𝑟
) − 𝑃𝑟] 𝑑𝑣𝑟
𝑉𝑟
∞
= ∫ 𝑍𝑐 [ 𝑇𝑟 (
𝜕
𝜕𝑇𝑟
) (
𝑇 𝑟
0.3333 𝑣 𝑟−0.08664
−
1
21.0541 𝑣 𝑟
2
+5.472384 𝑣 𝑟
) −
𝑇 𝑟
0.3333 𝑣 𝑟−0.08664
−
𝑉𝑟
∞
1
21.0541 𝑣 𝑟
2
+5.472384 𝑣 𝑟
] 𝑑𝑣𝑟
= ∫ 𝑍𝑐 [(
𝑇 𝑟
0.3333 𝑣 𝑟−0.08664
−
𝑇 𝑟
2
0.3333 𝑣 𝑟−0.08664
+
𝑇 𝑟
21.0541 𝑣 𝑟
2
+5.472384 𝑣 𝑟
] 𝑑𝑣𝑟
𝑉𝑟
∞
= 𝑍𝑐 𝑇𝑟 [(
ln(0.3333 𝑣 𝑟−0.08664)
0.3333
−
𝑇𝑟 ln(0.3333 𝑣 𝑟−0.08664)
0.3333
+
ln(
5.472384
𝑣 𝑟
)+21.0541
5.472384
]
= 0.873 𝑇𝑟 ln(0.333𝑣𝑟 − 0.08664) [1 − 𝑇𝑟] + 0.053 Tr ln (
5.472384
𝑣 𝑟
) + 1.12
𝒔∗
− 𝒔
𝑹
𝑠∗
− 𝑠
𝑅
= ∫ 𝑍𝑐 [(
𝜕𝑃𝑟
𝜕𝑇𝑟
) − (
1
𝑉𝑟
)] 𝑑𝑣𝑟 − ln(𝑧)
𝑉𝑟
∞
=∫ 𝑍𝑐 [(
1
0.3333 𝑣 𝑟−0.08664
−
1
𝑉𝑟
] 𝑑𝑣𝑟 − ln(𝑧)
𝑉𝑟
∞
= 𝑍𝑐 [(
ln(0.3333 𝑣 𝑟−0.08664)
0.3333
− (
1
ln 𝑣𝑟
) ]– ln (z)

17. 16 | P a g e
g) Accuracy of EOS for equation (C)
From table A1 for Nitrogen we get,
Zc=0.291
And at part (c) we calculated,
Zc=0.3471
Thus, Accuracy=
(0.3471−0.291)
0.291
=0.192783=19.2783%

18. 17 | P a g e
h) Derivation of Expressions:
For simplicity of calculation we will consider the reduced equation at critical point,
𝒂∗ − 𝒂
𝑹𝑻𝒄
:
We know,
𝑎∗
− 𝑎
𝑅𝑇𝑐
= − ∫ 𝑍𝑐 [(
𝜕𝑃𝑟
𝜕𝑇𝑟
) −
𝑇𝑟
𝑉𝑟
] 𝑑𝑣𝑟 + 𝑇𝑟𝑙𝑛(𝑍)
𝑉𝑟
∞
𝑜𝑟,
𝑎∗
− 𝑎
𝑅𝑇𝑐
= − ∫ 𝑍𝑐 [(
𝜕
𝜕𝑇𝑟
(
𝑇𝑟
0.3333 𝑣𝑟 − 0.08664
−
1
21.0541 𝑣𝑟
2 + 5.472384 𝑣𝑟
)) −
𝑇𝑟
𝑉𝑟
] 𝑑𝑣𝑟
𝑉𝑟
∞
+ 𝑇𝑟𝑙𝑛(𝑍)
Or,
𝑎∗−𝑎
𝑅𝑇𝑐
= − ∫ 𝑍𝑐 [
1
0.33∗𝑉𝑟−0.086
−
𝑇𝑟
𝑉𝑟
] 𝑑𝑉𝑟 + 𝑇𝑟𝑙𝑛(𝑍)
𝑉𝑟
∞
or,
𝑎∗−𝑎
𝑅𝑇𝑐
= −𝑍𝑐[(
ln(|165𝑉𝑟−43|)
0.33
− 𝑇𝑟𝑙𝑛|𝑉𝑟|] + 𝑇𝑟𝑙𝑛(𝑍)
After putting the value of Zc we get,
𝑎∗
− 𝑎
𝑅𝑇𝑐
= −0.291[(
ln(|165𝑉𝑟 − 43|)
0.33
− 𝑇𝑟𝑙𝑛|𝑉𝑟|] + 𝑇𝑟𝑙𝑛(𝑍)
𝒈∗ − 𝒈
𝑹𝑻𝒄
:
We know,
𝑔∗
− 𝑔
𝑅𝑇𝑐
= ∫ [𝑍𝑐𝑃𝑟 −
𝑇𝑟
𝑉𝑟
]𝑑𝑣𝑟 + 𝑇𝑟(𝑙𝑛𝑧 + 1 − 𝑧)
𝑉𝑟
∞
𝑔∗
− 𝑔
𝑅𝑇𝑐
= ∫ [𝑍𝑐 (
𝑇𝑟
0.3333 𝑣𝑟 − 0.08664
−
1
21.0541 𝑣𝑟
2 + 5.472384 𝑣𝑟
) −
𝑇𝑟
𝑉𝑟
] 𝑑𝑉𝑟 + 𝑇𝑟(𝑙𝑛𝑧 + 1
𝑉𝑟
∞
− 𝑧)
𝑔∗
− 𝑔
𝑅𝑇𝑐
= −
𝑇𝑟𝑙𝑛(|1375𝑉𝑟 − 361|)𝑍𝑐
0.33
+
ln (|
1368096
𝑉𝑟 + 5263525|) 𝑍𝑐
5.47
+ 𝑇𝑟(𝑙𝑛𝑧 + 1 − 𝑧)
After putting the value of Zc we get,
𝑔∗
− 𝑔
𝑅𝑇𝑐
= −
𝑇𝑟𝑙𝑛(|1375𝑉𝑟 − 361|) ∗ .291
0.33
+
ln (|
1368096
𝑉𝑟
+ 5263525|) ∗ .291
5.47
+ 𝑇𝑟(𝑙𝑛𝑧 + 1 − 𝑧)

19. 18 | P a g e
i) Speed of sound
𝑐 = √−𝑣2 √ (
𝜕𝑝
𝜕𝑣
)
𝑠
(
𝜕𝑝
𝜕𝑣
)
𝑠
=
𝜕
𝜕𝑣
{
𝑅𝑇
𝑣 − 𝑏
−
𝑎(𝑇)
𝑣(𝑣 + 𝑏)
}
=
𝑅𝑇(−1)
(𝑣 − 𝑏)2
+
𝑎(𝑇) (2𝑏 + 𝑏)
(𝑣 + 𝑏)2 𝑣2
= −
𝑅𝑇
(𝑣 − 𝑏)2
+
𝑎(𝑇) (2𝑏 + 𝑏)
(𝑣 + 𝑏)2 𝑣2
𝒄 = √−𝒗 𝟐{−
𝑹𝑻
(𝒗−𝒃) 𝟐 +
𝒂(𝑻) (𝟐𝒃+𝒃)
(𝒗+𝒃) 𝟐 𝒗 𝟐 }

20. 19 | P a g e
(j) Derive the Properties
Cp
We can directly derive Cp and Cv from Uj. Now we know Cp and Cv,
𝐶𝑝 = −
1
𝑈𝑗
[𝑇 [
𝑅𝑇
𝑣 − 𝑏
−
𝑎(𝑇)
𝑣(𝑣 + 𝑏)
−
𝑅𝑇
(𝑣 − 𝑏)2 +
𝑎(𝑇)
𝑣2(𝑣 + 𝑏)
+
𝑎(𝑇)
𝑣(𝑣 + 𝑏)2
] + 𝑣]
Cv
Also the relation between Cp and Cv is,
𝑐 𝑉 =
𝐶 𝑝
𝑘
So,
𝐶𝑣 = −
1
𝑈𝑗 𝐾
[𝑇 [
𝑅𝑇
𝑣 − 𝑏
−
𝑎(𝑇)
𝑣(𝑣 + 𝑏)
−
𝑅𝑇
(𝑣 − 𝑏)2 +
𝑎(𝑇)
𝑣2(𝑣 + 𝑏)
+
𝑎(𝑇)
𝑣(𝑣 + 𝑏)2
] + 𝑣]
1/v vp
β = (1/𝑣)(
𝜕𝑣
𝜕𝑇
)𝑝
(
𝜕𝑣
𝜕𝑡
) 𝑝 = −
(
𝜕𝑃
𝜕𝑇
)𝑣
(
𝜕𝑃
𝜕𝑣
)𝑇
=
𝑅
𝑣−𝑏
−
𝑎( 𝑇)
𝑣( 𝑣+𝑏)
−
𝑅𝑇
( 𝑣−𝑏)2+
𝑎( 𝑇)
𝑣2( 𝑣+𝑏)
+
𝑎( 𝑇)
𝑣( 𝑣+𝑏)2
β = −
1
𝑣
[
𝑅
𝑣−𝑏
−
𝑎(𝑇)
𝑣(𝑣+𝑏)
−
𝑅𝑇
(𝑣−𝑏)2+
𝑎(𝑇)
𝑣2(𝑣+𝑏)
+
𝑎(𝑇)
𝑣(𝑣+𝑏)2
]
k = v/p pv
Isentropic expansion coefficient:
𝑘 = −
𝑣
𝑃
(
𝜕𝑝
𝜕𝑣
) 𝑠
= -
𝑣
𝑃
[−
𝑅𝑇
(𝑣−𝑏)2 +
𝑎(𝑇)
𝑣2(𝑣+𝑏)
+
𝑎(𝑇)
𝑣(𝑣+𝑏)2]
𝑘 =
−𝑣
(
𝑅𝑇
𝑣−𝑏
−
𝑎(𝑇)
𝑣(𝑣+𝑏)
)
[
−𝑅𝑇
(𝑣−𝑏)2+
2𝑎
𝑇 (𝑣+𝑐)3]

21. 20 | P a g e
kT
We know,
𝐾 𝑇 = −
1
𝑣
𝛿𝑝
𝛿𝑣
)
𝑇
= -
1
𝑣
[−
𝑅𝑇
(𝑣−𝑏)2 +
𝑎(𝑇)
𝑣2(𝑣+𝑏)
+
𝑎(𝑇)
𝑣(𝑣+𝑏)2]
J
We know,𝑈𝑗 = (
𝜕𝑃
𝜕𝑇
) 𝑣
Also,
𝑑ℎ = 𝐶 𝑝 𝑑𝑇 + [𝑣 − 𝑇(
𝜕𝑣
𝜕𝑇
) 𝑝]
From isentropic process,
h= constant
so, 𝑑ℎ = 0
(
𝜕𝑇
𝜕𝑃
) ℎ =
𝑇 (
𝜕𝑣
𝜕𝑇
)𝑝−𝑣
𝑐𝑝
Now, (
𝜕𝑣
𝜕𝑇
) 𝑝 = -
(
𝜕𝑃
𝜕𝑇
)𝑣
(
𝜕𝑃
𝜕𝑣
)𝑇
𝑈𝑗 = (
𝜕𝑇
𝜕𝑃
) ℎ
=
𝑇[−
(
𝜕𝑃
𝜕𝑇
)𝑣
(
𝜕𝑃
𝜕𝑣
)𝑇
]−𝑣
𝑐𝑝
𝑈𝑗 =
−𝑇
𝑅
𝑣−𝑏
−
𝑎(𝑇)
𝑣(𝑣+𝑏)
−
𝑅𝑇
(𝑣−𝑏)2+
𝑎(𝑇)
𝑣2(𝑣+𝑏)
+
𝑎(𝑇)
𝑣(𝑣+𝑏)2
−𝑣
𝑐𝑝

22. 21 | P a g e
Summary:
In our project we firstly evaluated the Two parameters of the Redlich- Kwong-Soave equation. Then
converted the equation into reduced form. With the help of MATLAB and Excel we estimated tabulated
data and calculations.

23. 22 | P a g e
Appendix
MATLAB Code
clc;
clear;
x = 1;
zc = 1;
while x
v2 = zc^3 - zc^2 + (.42747-.08664-.08664^2)*zc + (-.42747*.08664);
if abs(v2) <= 0.00000025
x = 0;
clc;
fprintf('%d',zc);
else
zc = zc - 0.000025;
end
end

24. 23 | P a g e
Reference
1) http://en.wikipedia.org/
2) http://webbook.nist.gov/chemistry/fluid/
3) http://www.boulder.nist.gov/div838/theory/refprop/MINIREF/MINIREF.HTM
4) https://www.bnl.gov/magnets/staff/gupta/cryogenic-data-handbook/Section6.pdf
5) http://www.swinburne.edu.au/ict/success/cms/documents/disertations/yswChap3.pdf
6) https://www.e-education.psu.edu/png520/m10_p5.html
7) Advanced Engineering Thermodynamics-Adrian Bejan.
8) Thermodynamics, An Engineering Approach- Yunus A Cengel and M.B Boles
9) Provided class lectures and notes.