1

The

Ideal Gas

2

Ideal gas equation of state

Property tables provide very accurate information about the properties.

It is desirable to have simple relations among the properties that are sufficiently general and accurate.

Any equation that relates P-v-T are called Equation of state. Property relations that involve other properties of a substance at

equilibrium states are also referred to as equations of state. The simplest and best-known equation of state for substances in

gas phase is the ideal-gas equation of state.

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Ideal gas equation of state

Pv RTR is the Gas constantR= Ru / MRu is universal gas constant Ru = 8.314 KJ/Kmol . KM is Molar Mass of the gas (molecular weight)

The values of R and M are given in Table A-1 for several substance.

The vapor phase of a substance is called gas when it is above the critical temperature.

Vapor implies a gas that is not far from a state of condensation. In 1802, Charles and Lussac experimentally determined the

following (Ideal gas equation of state):

M is the mass of one mole (kmole) of substance in grams (kilograms)

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P-v-T relation for ideal gases( / )

( )

( / )u u

u

PV mRT v V m

PV NR T mR NMR NR

Pv R T v V N

Pv RT

The properties of an ideal gas for a fixed mass at two different states are related to each other by

m m1 2

1 1 2 2 1 1 2 2

1 2 1 2

and PV PV PV PV

RT RT T T

RT

PVm

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Ideal gasAn ideal gas is an imaginary substance that obeys the P-v-T relation.

The aforementioned relation approximates the behavior of real gas at low densities.

At low pressure and high temperature, the density of a gas decreases, and the gas behaves like an ideal gas.

In the range of practical interest, many familiar gases such as air, nitrogen, oxygen, hydrogen, helium, Aragon, neon, krypton, and even heavier gases such as carbon dioxide can be treated as ideal gases with negligible error (often less than 1 %).

Dense gases such as water vapor in steam power plants and refrigerant vapor in refrigerators, however, should not be treated as ideal gases.

Instead, the property tables should be used for these substances.

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Figure on the right shows the percentage of error involved in assuming steam to be an ideal gas, and the region where steam can be treated as an ideal gas.

Is steam an Ideal Gas? It depends !!!!!Is steam an Ideal Gas? It depends !!!!!

7

Is Water vapor an ideal gas?At pressure below 10 KPa, water

vapor can be treated as an ideal gas, regardless of its temperature, with negligible error (less than 0.1%).

At higher pressures, however, the ideal-gas assumption yields unacceptable errors.

In air-conditioning applications , where the pressure of the water vapor is very low (ideal gas relations can be used)

In steam power plant applications, they should not be used

8

Compressibility Factor ZReal gases deviate from ideal-gas behavior significantly at state near the saturation region and the critical point.

This deviation at a given temperature and pressure can accurately accounted for by introducing a correction factor called the compressibility factor

Pv RT

For real gases Z can be grater than or less than unity.

actual

ideal

vZ

v

PvZ

RT

1Pv

RT

Real Gas

Z > 1

Z < 1

Ideal GasZ = 1

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Compressibility Factor Z for Nitrogen

10

Compressibility Factor Z for H2

Similar charts have been

prepared for different gases and

they are found to be qualitatively

similar.

Can quantitative similarity

be achieved?

11

Generalized Compressibility chart

This can be done if the coordinates are modified such that PR =

P/Pcr and TR = T/Tcr where PR and TR are the reduced pressure and

reduced temperature.

RT

PZ

RP,pressureducedRe

The Z factor is the

same for all gases at

the same reduced

pressure and

temperature.

This is called the

principle of

corresponding states.

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Observations

B- It can be noticed that at many states the compressibility factor Z is approximately unity :

A- The deviation of a gas from

unity is greatest in the vicinity

of the critical point.

1. At very low pressure (PR<<1),

2. At high temperatures (TR>2) except when PR >>1.

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When P and v or T and v are given, generalized chart can still be used

crcr

actualR P/RT

vv

Pseudo-reduced

volume linesRT

PZ

RP,pressureducedRe

Pseudo reduced specific volume

Lines of constant vR are available on the compressibility chart (Figure A-30 pp. 867-869)

2- TR and vR

3- PR and vR

vR is pseudo reduced specific volume defined as:

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Example 2-11 Determine the specific volume of

refiregerant-134 at 1 MPa and 50 C, using:

A. Thermodynamic tablesB. The ideal gas lowC. The generalized compressibility chart.

Find the percentage error in the values obtained in B and C compared to the value obtained in part A.

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Equation of state

Pv RT

Equation of State predicts the P-v-T behavior of a gas quite accurately

16

Specific Heats

17

ImportanceWe know it takes more energy to warm up some materials than others

•For example, it takes about 10 times as much energy to warm up 1 kg of water, as it does to warm up the same mass of iron.

• it is desirable to have a property that will enable us to compare the energy storage capabilities for different substances.

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Specific Heat (heat capacity) Definition: It is the energy required to raise the

temperature of a unit mass by one degree Units

kJ/(kg 0C) or kJ/(kg K) cal/(g 0C) or cal/(g K) Btu/(lbm 0F) or Btu/(lbm R)

Simple mathematical definition:

TmCE

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Constant volume and Constant pressure specific heats, Cv and Cp

Cv can be viewed as energy

required to raise the temperature of a unit mass by one degree as the volume is maintained constant.

Cv can be viewed as energy

required to raise the temperature of a unit mass by one degree as the volume is maintained constant.

Cp can be viewed as energy

required to raise the temperature of a unit mass by one degree as the pressure is maintained constant.

Cp can be viewed as energy

required to raise the temperature of a unit mass by one degree as the pressure is maintained constant.

Cp > Cv

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Mathematical form of Cv

E=U+KE +PEE=U+KE +PE

ddddUU

UU

But dEBut dEmCmCvvdTdT

duduCCvvdTdT

vv T

uC

Consider constant volume system. Heat it from T1 to T2.Consider constant volume system. Heat it from T1 to T2.

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We’ll worry about the math later, but…

pp T

hC

h includes the internal energy and the work required to expand the system boundaries

Mathematical form of Cp

h u Pv

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CCpp is always bigger than C is always bigger than Cvv. This is . This is

because ibecause it takes more energy to warm up a constant pressure system due to the system boundaries expansion.

That is you need to provide the energy to

Observations

increase the internal energy do the work required to move the system

boundary

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Observations (continued)

Both are expressed in terms of u or h, and T, which are properties and thus Cv and Cp are also properties.

Because they are properties, they are independent of the type of process!!

vv T

uC

p

p T

hC

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Recall that :

)(Tuu Hence, Cv is at most function of T for an ideal gas.

vv T

uC

dT

duCv

Cv dependence on T for an ideal gas.

Joule found experimentally that the internal energy of an ideal gas is a function of temperature only

The partial derivative becomes ordinary derivative for an ideal gas. .

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h u Pv

pp T

hC

Recall that

dT

dhCp

Cp dependence on T for an ideal gas.

Hence, Cp is at most function of T for an ideal gas.

The partial derivative becomes ordinary derivative for an ideal gas.

)T(hRT)T(uh

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dTTCdu v )(

dTTCuuu v2

112

dTTCdh p )(

dTTChhhT

T p2

112

Internal energy and Enthalpy as functions of Cv and Cp for an ideal gas.

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To carry out these integrations, we need Cv and Cp as functions of T.

Analytical expressions are available in Table A-2c.

In this table, Cp is given as

Cp = a + bT + cT2 + dT3

The constants a, b, c, and d are tabulated for various gases.

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2

1

322

1)( dTdTcTbTadTCh p

432

41

42

31

32

21

22 TTdTTcTTb

aTh

This is inconvenient!! Only do it if you really need to be very accurate!!

Isn’t there an easier way?

Method 1

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T0=0 K was chosen to be an arbitrary reference. This choice has no effect on The u and h data are given in KJ/kg for air. Other gases in KJ/Kmol.

Method 2 These integrations of Eqs. 2-34

and 2-35 were tabulated in Table A-17 page 849.

0 0

T

o vTu u C dT

handu

dTChhT

T po

00

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Method 3 The variation of specific

heat for gases with complex molecules are higher and increase with temperature.

The variation of specific heats is smooth and can be approximated as linear over small temperature interval (a few hundred degrees or less)

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Method 3 Assume Cp and Cv is

constant over a short temperature range (a few hundred degrees or less).

The constant specific heats are evaluated at the average temperature (T1+T2)/2.

u2-u1=Cv,av(T2-T1)

h2-h1=Cp,av(T2-T1)

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Three Ways to Calculate u

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Example Air at 300 K and 200 kpa is heated at

constant pressure to 600 K. Determine the change in the internal energy of air per unit mass using:

Data from air tables (Table A-17) The functional form of the specific heat

(Table A-2c) The average specific heat value (Table A-

2b).

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The previous relations are not restricted to any kind of process.

The presence of constant volume specific heat in an equation should not lead to the concept that this equation is valid only for constant volume process.

The constant volume or constant pressure part of the name defines only how they are measured for each substance (see figure).

Once we have Cv or Cp as function of T, we can perform the integration for any process.

Important Observation

35

Recall that h=u+RTdh = du + RdTCpdT=CvdT+ RdT

Cp=Cv + R

Specific heat relations of an Ideal GasCCpp is modeled in the Appendix A-2(c) as a is modeled in the Appendix A-2(c) as a

function of temperature – so you could calculate function of temperature – so you could calculate dh, but what if you want to calculate du? dh, but what if you want to calculate du?

You’d need CYou’d need Cvv. . There is no corresponding Cv table !!

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v

p

C

Ck

Specific Heat Ratio

k also varies with temperature, but this k also varies with temperature, but this variation is very mild.variation is very mild.

k = 1.4 for diatomic gases (like air)k = 1.4 for diatomic gases (like air)

k = 1.667 for noble gasesk = 1.667 for noble gases

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The specific volume of incompressible substances remain constant during a process.

The Cv and Cp values of incompressible substances are identical and are denoted by C.

Cp = Cv = C

Specific heats of Solids and Liquids

38

du C dT CdTV

u C T C T T ( )2 1

Internal energy of Solids and Liquids

39

h u Pv dh du Pdv vdP

h u v P C T v P

But dv is 0 if the system is incompressible

0Small for solids

Enthalpy of Solids

TCuh avg

40

Enthalpy of Liquids

h u v P C T v P We have two cases:

Constant pressure process,

Constant temperature process,

TCuh avg0P

0T

Pvh