Unit-2: Thermodynamic Properties of Real Gases

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Course: Thermodynamics - Free Course
Book: Unit-2: Thermodynamic Properties of Real Gases
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Date: Friday, 22 November 2024, 11:40 AM

Contents

PVT behaviour of fluids; Mathematical representation of PVT behaviour; Generalized compressibility factor correlation; Generalized equations of state.  


T-v diagram for the heating process of water at constant pressure




T-v diagram of contant-pressure phase-change processes







PVT Diagram of Pure Components






The PVT surface above represents a substance which contracts upon freezing. Most substances do so, the notable exception being water, bismuth and antimony which expands upon freezing. The expansion of water upon freezing has an enormous impact on the nature of the Earth.










PT Diagram







Passing from the solid phase directly into the vapor phase is called sublimation.

For substances that have a triple-point pressure above the atmospheric pressure such as solid CO2 (dry ice), sublimation

is the only way to change from the solid to vapor phase at atmospheric conditions.


PT Diagram of Water



Triple point: P = 611.7 N/m2; T = 0.01oC

Critical point: P= 220.64 x 105 N/m2; T = 374oC

Normal boiling point: 100oC; Normal freezing point: 0oC (1 atm)


PV Diagram






PV Diagram of Water







TV Diagram of Water














Ideal Gas Equation










van der Waals Equation




van der Waals Equation



  • The parameter b is related to the size of each molecule. The volume that the molecules have to move around in is not just the volume of the container V, but is reduced to ( V - b )
  • The parameter a is related to intermolecular attractive force between the molecules, and 1/V is the density of molecules. The net effect of the intermolecular attractive force is to reduce the pressure for a given volume and temperature.
  • When the density of the gas is low (i.e., when 1/V is small and b is small compared to V)  the van der Waals equation reduces to that of the ideal gas law.

van der Waals Constants

Substance

a

(J. m3/mole2)

b

(m3/mole)

Pc

(MPa)

Tc

(K)

Air

1.1358

3.64x10-5

3.77

133

Carbon Dioxide (CO2)

1.3643

4.27x10-5

7.39

304.2 

Nitrogen (N2)

1.1361

3.85x10-5

3.39

126.2 

Hydrogen (H2)

1.0247

2.65x10-5

1.30

33.2 

Water (H2O)

1.5507

3.04x10-5

22.09

647.3 

Ammonia (NH3)

1.4233

3.73x10-5

11.28

406 

Helium (He)

1.00341

2.34x10-5

0.23

5.2 

Freon (CCl2F2)

1.0780

9.98x10-5

4.12

385 


Observe that inert gases like Helium have a low value of a as one would expect since such gases do not interact very strongly, and that large molecules like Freon have large values of b.









Comparison of ideal and van der Waals gas







Comparison of actual behavior during isothermal compression to prediction of van der Waals equation.





An isotherm of the van der Waals equation for T < Tc.





Redlich-Kwong Equation


Peng-Robinson Equation


Berthelot Equation


Dieterici Equation


Example: Trial and Error solution of cubic Equation
















Derivation for a and b of van der Waals







Example : van der Waals equation - Derivation for a and b in another way van der Waals equation is given as





Dieterici Equation















Compressibility Factor


Virial Equation


Corresponding State Principle










Equation of State for a Single Phase






Equation of State for Liquids




Quality of Vapor (Dryness Fraction)








Steam Tables





Single-phase tables
















Example - Volume from Steam Tables










From the corresponding state principle by considering Z as a function of (Tr, Pr) estimate the volume of 1 kmol of oxygen at 200 K and 6 x 106 N/m2.  (Pc = 5.05 x 106 N/m2; Tc = 155 K). Z values for various Tr, Pr are as given below:


Z

Tr               \( \downarrow \)P\( \rightarrow \)

0.8

1.0

1.2

1.5

1.1

0.7649

0.6880

0.5984

0.4580

1.2

0.8330

0.7858

0.7363

0.6605

1.3

0.8764

0.8438

0.8111

0.7624

1.4

0.9062

0.8827

0.8595

0.8256

1.5

0.9278

0.9103

0.8933

0.8689


First by considering the interpolation at fixed values of Pr, get Z; then at fixed Tr value get the Z.

That way, you will get Z as 0.807


Integration Calculations



 

Surprise Test