chemical thermodynamics 2013/2014

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Chemical Chemical Thermodynamics Thermodynamics 2013/2014 2013/2014 2 nd Lecture: Zeroth Law, Gases and Equations of State Valentim M B Nunes, UD de Engenharia

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Chemical Thermodynamics 2013/2014. 2 nd Lecture: Zeroth Law, Gases and Equations of State Valentim M B Nunes, UD de Engenharia. The Gaseous State. - PowerPoint PPT Presentation

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Page 1: Chemical Thermodynamics 2013/2014

ChemicalChemical ThermodynamicsThermodynamics2013/20142013/2014

2nd Lecture: Zeroth Law, Gases and Equations of StateValentim M B Nunes, UD de Engenharia

Page 2: Chemical Thermodynamics 2013/2014

2

The Gaseous The Gaseous StateState

When studying Thermodynamics we don’t need any theory about molecular structure but, in order to understand the subject, its good to have in mind molecular models for the states of matter. The gaseous state it is often taken as an example to describe the principles of Thermodynamics.

In this Lecture we will review some of the basic concepts about the gaseous state, namely the concepts of pressure and temperature, and the basic laws of gases.

Page 3: Chemical Thermodynamics 2013/2014

3

Ideal GasesIdeal Gases

A good vision of an ideal gas is an assembly of molecules, with sizes much smaller than average distances travelled between collisions, in continuous random (chaotic) movement, with velocities that increase with temperature, and that only interact in elastic collisionsFor a given system with volume, V, number of moles, n, and at pressure, p, and temperature, T, we say that it is at a given thermodynamic state. All this properties are not independent from each other. In other words they are related by an Equation of State.Before proceeding to those equations of state let us examine the concepts of pressure and temperature.

Page 4: Chemical Thermodynamics 2013/2014

PressurePressure

4

The pressure is a force exerted by an unit area. In the International System (SI) the unit for pressure is the Pascal (Pa).

1 Pa = 1 N.m-2

1 bar = 100 kPa (exact value)

1 atm = 101325 Pa (exact value)

1 atm = 760 mmHg (exact value)

hggr

hr

A

mg

A

FP

2

2

Page 5: Chemical Thermodynamics 2013/2014

TemperaturTemperature e

5

Although the temperature is a fundamental physical concept, it is very difficult to define it.

Temperature is the property that indicate us in what direction the heat flows!

The definition of temperature results from the so called “zeroth law of thermodynamicszeroth law of thermodynamics”: if two systems are separately in thermal equilibrium with a third, then they must also be in thermal equilibrium with each other

A, B and C share the same temperature.

Page 6: Chemical Thermodynamics 2013/2014

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Boyle´s LawBoyle´s Law

At constant n and T, we obtain

VppV

1or constant

Isothermals of the gas.

Page 7: Chemical Thermodynamics 2013/2014

7

Charles LawCharles Law

At constant n and p, we obtain

TVLord Kelvin was the first one to recognize the outstanding importance of this Law, giving origin to the absolute scale of temperatures.

15.273/º/ CtKT

Isobaric of the gas

Page 8: Chemical Thermodynamics 2013/2014

8

Avogadro's LawAvogadro's Law

At constant p and T, we obtain

nV

Page 9: Chemical Thermodynamics 2013/2014

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Perfect Gas Equation of StatePerfect Gas Equation of State

Combining the laws of gases we can easily obtain the perfect gas equation of state:

nRTpV Were R is the perfect gas constant. In the SI, R ≈ 8,314 J.K-

1.mol-1 . In non SI units it may be expressed by R ≈ 0,082 atm.L.K-1.mol-1

At STP (standard temperature and pressure), t=0 ºC and p=1 atm, the molar volume of a perfect gas, Vm = V/n is:

-1L.mol 4.22mV

Page 10: Chemical Thermodynamics 2013/2014

Surface of Possible StatesSurface of Possible States

10

Page 11: Chemical Thermodynamics 2013/2014

11

pVT surface of an ideal pVT surface of an ideal gasgas

Page 12: Chemical Thermodynamics 2013/2014

12

Mixtures: Dalton´s Mixtures: Dalton´s LawLawThe pressure exerted by a mixture of perfect gases is the sum of the pressures exerted by the individual gases occupying the same volume

A

A

BA

AA

BA

yP

P

VRTnn

VRTn

P

P

PPP

/

/

Dalton´s Law writes as follow:

PyP AA

Page 13: Chemical Thermodynamics 2013/2014

Real GasesReal Gases

13

For non-perfect gases there are present intermolecular forces, attractive and repulsive, between atoms and molecules, giving origin to deviations from ideality

A quantitative measure o non-ideality is the compressibity factor, Z:

RT

pVZ m Ideal gas

Z = 1

Page 14: Chemical Thermodynamics 2013/2014

14

Real Gas (COReal Gas (CO22) pV isotherms) pV isotherms

Experimental isotherms for carbon dioxide

CDE – vapor pressure of CO2 at 20 ºC

The isotherm at 31,04 ºC corresponds to the critical isotherm

* Critical point

In the absence of intermolecular forces there will be no condensate states of matter (solids and liquids)!

Page 15: Chemical Thermodynamics 2013/2014

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Virial Equation of StatesVirial Equation of States

At higher temperatures some of the isotherms are similar to the ideal gas. Perfect gas equation can be expanded, in terms of virial equations:

.....12

mm

m

V

C

V

B

RT

pVZ

Leiden form

.....1 2'' pCpBZ Berlin form

B and B’ – 2nd virial coefficientC and C’ – 3rd virial coefficient 2

2

' and 'RT

BCC

RT

BB

Page 16: Chemical Thermodynamics 2013/2014

16

Virial coefficients: dependence of temperatureVirial coefficients: dependence of temperature

Page 17: Chemical Thermodynamics 2013/2014

17

van der Waals equationvan der Waals equation

van der Waals corrected the perfect gas equation by assuming that molecules occupy some space, and there are long–range intermolecular forces.

Page 18: Chemical Thermodynamics 2013/2014

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Molar volume calculationMolar volume calculation

The VDW equation is cubic in volume, so it gives three roots for the volume in the VLE zone.

023

p

abVp

aV

p

RTbV mmm

Molar volume of liquid

Molar volume of gas

Without physical significance!

Page 19: Chemical Thermodynamics 2013/2014

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The critical isothermThe critical isotherm

For the critical isotherm there is an inflexion point. In mathematical terms we have:

02

Cc TmTm V

p

V

p

Solving these equations we have

432

2

32

62

2

mmm

mmm

V

a

bV

RT

V

p

V

a

bV

RT

V

p

Solving altogether we obtain the critical values:

227

27

8 3

b

ap

Rb

aTbV CCC

Page 20: Chemical Thermodynamics 2013/2014

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The Principle of Corresponding StatesThe Principle of Corresponding States

If we define now a new set of parameters (reduced parameters):

c

mr

cr

cr V

VV

T

TT

p

pp

Substituting in the VDW equation we obtain:

2

3

13

8

rr

rr VV

Tp

This means that two different gases at the same reduced temperature and reduced volume should exert the same reduced pressure, and they are in corresponding states.

Page 21: Chemical Thermodynamics 2013/2014

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Z factor for several gasesZ factor for several gases

Page 22: Chemical Thermodynamics 2013/2014

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Equations of stateEquations of state