Chapter 2

PROPERTIES OF PURE

SUBSTANCES

2

Objectives

• Introduce the concept of a pure substance.

• Discuss the physics of phase-change processes.

• Illustrate the P-v, T-v, and P-T property diagrams and P-v-T

surfaces of pure substances.

• Demonstrate the procedures for determining thermodynamic

properties of pure substances from tables of property data.

• Describe the hypothetical substance “ideal gas” and the

ideal-gas equation of state.

• Apply the ideal-gas equation of state in the solution of typical

problems.

• Present some of the best-known equations of state.

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PURE SUBSTANCE

• Pure substance: A substance that has a fixed chemical

composition throughout.

• Air is a mixture of several gases, but it is considered to

be a pure substance (as along as it in the same phase).

Nitrogen and gaseous air are

pure substances. A mixture of liquid and gaseous

water is a pure substance, but a

mixture of liquid and gaseous air

is not.

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PURE SUBSTANCE PROPERTIES

DIAGRAM

• To study the phase change process for a

pure substance, we take water as an

example.

• Water, in liquid phase is filled in a piston

cylinder without any weight on the piston.

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PHASE-CHANGE PROCESSES OF PURE SUBSTANCES

At 1 atm

(≈1bar) and

20°C, water

exists in the

liquid phase

(compressed

liquid).

At 1 atm

pressure and

100°C, water

exists as a

liquid that is

ready to

vaporize

(saturated

liquid).

As more heat is

transferred, part of

the saturated liquid

vaporizes

(saturated liquid–

vapor mixture).

At 1 atm

pressure, the

temperature

remains

constant at

100°C until

the last drop

of liquid is

vaporized

(saturated

vapor).

As more

heat is

transferred,

the

temperatur

e of the

vapor starts

to rise

(superheat

ed vapor).

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• Compressed liquid (subcooled liquid): A substance that it

is not about to vaporize.

• Saturated liquid: A liquid that is about to vaporize.

T-v diagram for the heating process of water

at constant pressure.

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• Saturated vapor: A vapor that is about to condense.

• Saturated liquid–vapor mixture: The state at which the liquid and vapor phases coexist in equilibrium.

• Superheated vapor: A vapor that is not about to condense (i.e., not a saturated vapor).

T-v diagram for the heating process of water

at constant pressure.

8

Saturation Temperature and Saturation Pressure • The temperature at which water starts boiling depends on the pressure;

therefore, if the pressure is fixed, so is the boiling temperature.

• Water boils at 100C at 1 atm pressure.

• Saturation temperature Tsat: The temperature at which a pure substance

changes phase at a given pressure.

• Saturation pressure Psat: The pressure at which a pure substance changes

phase at a given temperature.

The liquid–

vapor

saturation

curve of a

pure

substance

(numerical

values are for

water).

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• Latent heat: The amount of energy

absorbed or released during a phase-

change process.

• Latent heat of fusion: The amount of

energy absorbed during melting. It is

equivalent to the amount of energy

released during freezing.

• Latent heat of vaporization: The amount

of energy absorbed during vaporization

and it is equivalent to the energy released

during condensation.

• The magnitudes of the latent heats

depend on the temperature or pressure at

which the phase change occurs.

• At 1 atm pressure, the latent heat of

fusion of water is 333.7 kJ/kg and the

latent heat of vaporization is 2256.5 kJ/kg.

• The atmospheric pressure, and thus the

boiling temperature of water, decreases

with elevation.

10

PROPERTY DIAGRAMS FOR PHASE-

CHANGE PROCESSES • The variations of properties during phase-change processes are best

studied and understood with the help of property diagrams such as the T-v, P-v, and P-T diagrams for pure substances.

T-v diagram of

constant-pressure

phase-change

processes of a pure

substance at various

pressures

(numerical values

are for water).

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Why properties diagrams are important?

• From the diagram we can

show any process that a

system go through.

• For example: 1kg water is

heated in constant pressure

from state 1 to state 2. Then

the system is heated in

constant volume until state 3.

• The above process can be

shown in Pv diagram or Tv

diagram

• The value of a system

characteristics are also can

be defined at each state

involved in the process

3

1

2

12

PROPERTY TABLES • For most substances, the relationships among thermodynamic properties are too

complex to be expressed by simple equations.

• Therefore, properties are frequently presented in the form of tables.

• Some thermodynamic properties can be measured easily, but others cannot and

are calculated by using the relations between them and measurable properties.

• The results of these measurements and calculations are presented in tables in a

convenient format.

Enthalpy—A Combination Property

The

combination

u + Pv is

frequently

encountered

in the analysis

of control

volumes.

The product pressure

volume has energy units.

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Saturated Liquid and Saturated Vapor States

• Table A–4: Saturation properties of water under temperature.

• Table A–5: Saturation properties of water under pressure.

A partial list of Table A–4.

Enthalpy of vaporization, hfg (Latent

heat of vaporization): The amount of

energy needed to vaporize a unit mass

of saturated liquid at a given

temperature or pressure.

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Examples Pressure of saturated

liquid

A rigid tank contains 50kg of

saturated liquid water at 90oc.

Determine the pressure in the tank

and the volume of the tank.

15

P saturation at 90 ºC= 70.183 Kpa

vf= 0.001036 m3/kg,

Total volume= mass x specific volume= 50 kg x 0.001036 m3/kg

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Example: Temperature of saturated

vapor

A piston cylinder device contains 0.06m3

of saturated water vapor at 350kPa

pressure. Determine the temperature

and the mass of the vapor inside the

cylinder

17

T saturation at 350 kPa = 138.86 C,

Specific volume at saturated vapor vg= 0.52422 m3/kg

Mass of the water vapor= Total volume (m3)/specific volume (m3/kg)

= 0.06/0.52442 kg =0.11 kg

• A mass of 200 g of saturated liquid water is completely vaporized at a

constant pressure of 100 kPa. Determine (a) the volume change

and (b) the amount of energy transferred to the water.

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From Table A5, we get vf= 0.001043 m3/kg, vg= 1.6941 m3/kg, hf=417.51 kJ/kg, hg=2675.0 kJ/kg,

so, Vchange= (vg-vf)xmass= (1.6941-0.001043)x0.2 m3=

Hchange= (hg-hf)xmass= (2675.0-417.51)x0.2 kJ

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Saturated Liquid–Vapor Mixture Quality, x : The ratio of the mass of vapor to the total mass of the mixture.

Quality is between 0 and 1 0: sat. liquid, 1: sat. vapor.

The properties of the saturated liquid are the same whether it exists alone or in

a mixture with saturated vapor.

The relative

amounts of

liquid and

vapor phases

in a saturated

mixture are

specified by

the quality x.

A two-phase system can be

treated as a homogeneous

mixture for convenience.

Temperature and

pressure are dependent

properties for a mixture.

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The v value of a

saturated liquid–

vapor mixture lies

between the vf and vg

values at the

specified T or P.

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Superheated Vapor In the region to the right of the

saturated vapor line and at

temperatures above the critical

point temperature, a substance

exists as superheated vapor.

In this region, temperature and

pressure are independent

properties.

Compared to saturated vapor,

superheated vapor is characterized by

An excellent tricks to memorize the above

relation:

• As you go right, the values of the

properties will increase except pressure.

22

Compressed Liquid Compressed liquid is characterized by

Hence, a compressed liquid

may be approximated as a

saturated liquid at the given

temperature as below.

The compressed liquid properties

depend on temperature much more

strongly than they do on pressure.

An excellent tricks to memorize the above relation:

• As you go left, the values of the properties will

decrease except pressure.

23 P-v diagram of a pure substance.

T-v diagram of a pure substance.

Very important note:

• Compressed liquid region (use table

A7)

• Saturated liquid-vapor region (use

table A4 or A5)

• Super heated vapor region (use table

A6

Very important!

Steps to determine unknown properties:

1) Find regions from given properties or

phase description.

2) Then go into the relevant table.

3) Then determine the unknown

properties (either directly or by

interpolation or by equations).

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PRESSURE AND VOLUME OF

SATURATED MIXTURE

• A rigid tank contains 10kg of water at

90 ºC. If 8kg of the water is in the liquid

form and the rest is in the vapor form,

determine (a) the pressure in the tank

and (b) the volume of the tank

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Examples: Saturated liquid-vapor

mixture states on T-v and P-v diagrams.

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T, °C P, kPa v, m3/kg u, kJ/kg h, kJ/kg Quality,

x

Phase description

i 140 0.05 Saturated liquid vapor

mixture

ii 550 Saturated liquid

iii 125 750

iv 300 0.2579

v 400 1450

vi 220 Saturated vapor

vii 190 2050

By using data from Table A4, Table A5, Table A6 or Table A7

Complete this table for H2O

Solutions are in next slide

Solutions

i) From Table A4 (because the substance is in saturated stage), we find; P= 361.53 kPa,

vf= 0.001080 m3/kg, vg= 0.50850 m3/kg, uf=588.77 kJ/kg, ufg= 1960.9 kJ/kg, hf=589.16 kJ/kg,

hfg=2144.3 kJ/kg

vfg= vg-vfg= 0.50850-0.001080 m3/kg = 0.50782 m3/kg,

Here, given v= 0.05, we know v= vf+ x.vfg; so, x= (v-vf)/vfg = (0.050-0.001080)/0.50782 = 0.096,

h= hf+x.hfg= 589.16+0.096x2144.3 kJ/kg= 795.02 kJ/kg,

ii) By using table A5 (because water is saturated and pressure is given), Tsaurated= 155.48 ºC,

vf= 0.001097 m3/kg, uf= 655.32 kJ/kg, hf= 655.77 kJ/kg

For saturated liquid, we know that v=vf, u= uf, and h= hf and x=0,

Therefore, v= 0.001097 m3/kg, u=655.32 kJ/kg, h= 655.77 kJ/kg

iii) By checking table A4( because we are not sure yet which table we should use) we see that the

saturation pressure corresponding to the given temperature (i.e. 125 C) is 232.23 kPa, but our given

pressure is 750 kPa, that mean P> Psaturation , therefore the substance is compressed liquid.

So, we should use table A4, we find

vf= 0.001065 m3/kg, uf= 524.83 kJ/kg, hf= 525.07 kJ/kg,

For compressed liquid (it can be assumed that the properties are same as the saturated liquid at the given

temperature), i.e. v=vf, u=uf and h=hf

Therefore, v= 0.001065 m3/kg, u= 534.83 kJ/kg, h= 525.07 kJ/kg,

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iv) We can check the table A4 (we yet not know which table we should use), we see that at 300 ºC,

vg= 0.021659,

But the given v= 0.2579 m3/kg is greater that vg, therefore the substance is superheated vapour

(check lecture slides for understanding),

So, We will use table A6, we should now find the values which match with T=300 C and v=0.2579

m3/kg,

We get, P= 1 MPa, u= 2793.7 kJ/kg, h= 3051.6 kJ/kg

v) We will check table A5, we see that uf= 604.22 kJ/kg, ug= 2553.1 kJ/kg, but given u= 1450 kJ/kg,

therefore the given u is greater that uf and less that ug, and so, the substance is saturated liquid-

vapour mixture,

So we should continue to use table A5, we find T= 143.61 C, ufg= 1948.9 kJ/kg, vf= 0.001084 m3/kg ,

vfg=0.46133 m3/kg, hf= 604.66 kJ/kg, hfg=2133.4 kJ/kg

We know, x=(u-uf)/ufg, so, x=(1450-604.22)/1948.9 = 0.433,

v=vf+x.vfg= 0.001084+0.433x0.46133 m3/kg= 0.1997 m3/kg,

h=hf+x.hfg= 604.66+0.433x2133.4= 1528.42 kJ/kg,

vii) Given T= 190 C and h= 2050 kJ/kg

We can check table A4, we see that the hf=807.43 kJ/kg, hg=2785 kJ/kg, So, hf<h<hg, therefore the

substance is saturated liquid vapour mixture,

So we will continue to use table A4, the values are, P=1255.2 kPa, vf= 0.001141, vfg= 0.1552,

uf=806.00, ufg= 1783, hfg=1977.9, we can find from h=hf+x.hfg, so x= 0.6282

v=vf+x.vfg=0.09863 m3/kg, u=uf+x.ufg= 1926 kJ/kg, Done !

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29

THE IDEAL-GAS EQUATION OF STATE

• Equation of state: Any equation that relates the pressure, temperature,

and specific volume of a substance.

• The simplest : ideal-gas equation of state. This equation predicts the P-v-T

behavior of a gas quite accurately within some properly selected region.

R: gas constant

M: molar mass (kg/kmol)

Ru: universal gas constant

Ideal gas equation

of state

Different substances have different

gas constants.

Note , P is absolute pressure

T is absolute temperature

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The ideal-gas

relation often is not

applicable to real

gases; thus, care

should be exercised

when using it.

Mass = Molar mass Mole number

Ideal gas equation at two

states for a fixed mass

Real gases

behave as an ideal

gas at low

densities (i.e., low

pressure, high

temperature). Pv= RT

PV=mRT

PV= NRuT

Different forms of equation of state

P1V1=mRT1

P2V2=mRT2

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When the Ideal gas equation is used

• The pressure is small compared to the

critical pressure

P< Pcr

• The temperature is twice the critical

temperature and pressure is less than 10

times of the critical pressure.

T=2Tcr and P<10Pcr

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Is Water Vapor an Ideal Gas?

• YES: at pressures below 10 kPa.

• NO: in actual application (higher pressure environment) such as Steam Power Plant

33

4-7 COMPRESSIBILITY FACTOR—A MEASURE

OF DEVIATION FROM IDEAL-GAS BEHAVIOR Compressibility factor Z

A factor that accounts for

the deviation of real gases

from ideal-gas behavior at

a given temperature and

pressure.

Z bigger than unity, the more the gas deviates

from ideal-gas behavior.

Gases behave as an ideal gas at low densities

(i.e., low pressure, high temperature).

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Summary

• Pure substance

• Phases of a pure substance

• Phase-change processes of pure substances

Compressed liquid, Saturated liquid, Saturated vapor, Superheated vapor

Saturation temperature and Saturation pressure

• Property diagrams for phase change processes

The T-v diagram, The P-v diagram, The P-T diagram, The P-v-T surface

• Property tables

Enthalpy

Saturated liquid, saturated vapor, Saturated liquid vapor mixture,

Superheated vapor, compressed liquid

Reference state and reference values

• The ideal gas equation of state

Is water vapor an ideal gas?