5.0 IDEAL GASES AND MIXTURES

At temperatures that are very much in excess of the critical temperature of a fluid, and

also as its pressure is considerably lowered, the vapor of the fluid tends to obey Boyle’s law and

Charle’s law exactly

Boyle’s law At constant temperature the pressure of a fixed mass of gas and its volume

are connected by the law

Pv = constant

i.e. pv= f(T) …………..5.1

Charle’s law

a. At constant pressure v α T for a fixed mass of gas

i.e. 2.5.........).........(PT

v

b. At constant volume pαT for a fixed mass of mass.

3.5.........).........(PT

P

Gay-Lussac also arrived at Charle’s law independently

5.1 THE EQUATION OF STATE FOR THE PREFECT GAS

Now consider a fixed mass of gas obeying both Boyle’s and charle’s laws. Change its

state from p1, v1, T1 to P2, V2, T2 by passing through an intermediate state P2, Vx, T1.

By Boyle’s law

P1V1 = P2Vx at T1

Vx = )4.5(2

11 p

vp

By Charle’s law 2

2

2

1

patT

V

T

Vx

2

21

2

11

2

21

5.54.5

)5.5(

T

VT

P

Vpandeqn

T

VTVx

2

22

T

VP = R where R is a constant for the particular gas.

6.5 RTpv

Suitably selecting the intermediate state, eqn 5.6 can also be obtained from the

combination of any other two of equations 5.1, 5.2, and 5.3. Perfect (or ideal) gas may be defined

as one which obeys exactly the equation pv=RT, where R is a constant having a particular value

for each substance and is called the characteristic Gas constant.

This equation is known as the equation of state for a perfect gas

7.5 mRTpv

Where m is the mass of the gas

The unit of R is Nm/kgK or kJ/kgK.

The molecular weight of a = )(32/1

tan

oxygenofmoleculeoneofmass

cesubstheofmoleculeoneofmass

The mole is a quantity of a substance whose mass is numerically equal to its molecular

weight.

In SI units the mole is the kilogram-mole

1 kilogram-mole of a substance has mass equal to M kg

Where M is molecular weight of the substance.

Eq. 5.5 then becomes

Pv= n MRT -------(5.8)

Where n is the number of moles

Now, Avogadro’s hypothesis states that equal volume of different ideal gases at the same

pressure and temperature contain the same number of molecules. This means that at the same

temperature and pressure 1 mole of any local gas will occupy a fixed volume.

Experiment has shown that the volume of one mole, of any perfect gas at 1 bar and

273.15k is appropriately 22.71m3

Consider equal volumes of ideal gases a and b at the same pressure and temperature.

Pv = nMaRaT

and pV = nMbRbT since equal volumes of different ideal gases contain the same number

of moles.

MaRa = MbRb

So the product MR is a constant for all gases and is called the universal gas constant and

is denoted by Ro.

MR = Ro

The unit of Ro is Nm/Kgmole-K or KJ/kgmole-K from of 5.6

9.5 onRT

pV

KkgmoleNmx

xx

nT

PVRo /3.8314

15.2731

71.22101 5

= 8.3143kJ/kgmole-K

5.2 Joule’s Law

Joule deduced experimentally that the change in internal energy of a gas is a function of

only the temperature change. The experiment, first performed by Gay-Lussac, had two copper

vessels A and B submerged in water (Fig. 5.1). Vessel A held air at 22atm while B, of the same

volume, was evacuated. The two vessels were connected with an initially closed value. The

system consisted of the insulated water container and the copper vessels. The contents of the

insulated system were in thermal equilibrium.

After the valve was opened and the gas had settled, the pressure was measured to be

1l atm. No water temperature change was observed and so Joule concluded that no change in air

temperature would have occurred.

dQ = 0 and dW = 0

since dQ = du + dW

du = 0.

The pressure of the air changed, the volume also changed and so

0

0

T

T

p

u

andv

u

Consequently Joule concluded that internal energy of the gas was not a function of the

volume or pressure. It must be a function of temperature.

Since the mass of the water and container was very much larger than that of the air; a

small change in air temperature might go undetected in water temperature. The experiment could

not have been accurate enough to make deductions, it is now known that a real gas should exhibit

a small drop in temperature after a free expansion. Nevertheless, Joule’s deduction can now be

proved mathematically for the perfect gas. The proof is beyond the present scope of work.

What is now known as Joule’s law will be stated as follows:

“The internal energy of a perfect gas is a function of only the temperature”.

U= U(T)

Enthalpy of a Perfect Gas

Since Pv = F(T) for a perfect gas

and U = U(T), it follows that h(=u + pv) is a function of only temperature for a perfect gas.

h = h(T)

5.3 The Joule-Thomson Experiment

As a result of criticisms of the Joule experiment described above, Joule and Thomson

(later Lord Kelvin) carried out the porous plug experiment to obtain more accurate results.

The experiment consisted of a well-insulated pipe with a porous plug that offered

considerable resistance to flow so that there could be relatively large drop in pressure from p1 to

p2 (fig. 5.2).

Fig. 5.2

It had been shown earlier that applying SFEE and neglecting potential and kinetic energies

h1 = h2

Joule and Thomson found that for air T2 < T1 and for hydrogen T2 >T1 and so found out

that for real gases h and u are functions of temperature. However for all the conditions for which

real gas behaviours approaches that of the perfect gas, T2 = T1; i.e h1=h2

5.4 Specific Heat of Perfect Gases

For constant volume process,

p

vT

hC

i.e. rate of change of enthalpy with temperature.

Since u and h are independent of p and v, the partial notation may be dropped for perfect

gases.

dT

dhCand

dT

duC pv

5.5 Relation between Cv, Cp, and R for a Perfect Gas

10.5

RCC

RdT

du

dT

dh

RTupvuh

vp

Molar Specific Heats:

and vv MCC

pp MCC

pv CandC are known as molar specific heats

11.5 RCC vp

Ratio of Specific Heats

13.51

/11

12.51

p

v

vp

v

p

V

P

C

R

C

R

CCR

C

C

C

C

Similarly

11

pv C

Rand

C

R

Units

KmolekgKJdT

MdhC

KkgKJdT

dhC

KmolekgKJdT

duMC

KkgKJdT

duC

p

p

v

v

/

/

/

/

5.6 Mean (on Average) Specific Heat for Gases

The mean specific heat for any process between T1 and Ti is defined by the relation

12

12

12

12

12

12

2

1

2

1

TT

hh

TT

DTCCp

and

TT

UU

TT

DTCCV

T

Tv

T

Tv

5.7 Properties of a Perfect Gas

For a perfect gas Cv and Cp are constant. For any change from state (1) i.e. p1, v1, T1 to

state (2) i.e. p2, v2, T2

U2-U1 = Cv (T2-T1) and h2-h1 = Cp (T2-T1)

Entropy of a Gas During a change from state, p1, v1, T1 to state p2, v2, T2

From the 1st law equation

dQ = du + dW

= du + pdv

From the definition of entropy

dQ = Tds

pdvduTds --------5.14

Now h = u + pv

dh = du + pdv + vdp

du = dh-pdv –vdp

vdpdhTds ------- 5.15

The entropy of a perfect gas may be obtained by impetrating eq. 5.14 or eq. 5.15.

from eq. 5.14

Tds = du + pdv

CvdT + RT v

dv

ds = V

Rdv

T

dTCv

1

2

1

221 11

V

VnR

T

TnCSS v

But 11

22

1

2

VP

VP

T

T

16.5...........ln1

11

11

ln11

ln1

1

2

1

212

1

2

11

2

1

2

11

2

1

2

1

2

11

2

1

2

11

2212

V

VC

P

PnCSS

V

VnC

P

PnC

V

VnRC

P

PnC

V

VR

V

VnC

P

PnC

V

VR

VP

VPnCSS

pv

pv

vv

vv

v

5.8 PROCESSES PERFORMED BY PERFECT GASES

a. Reversible Adiabatic (Isentropic) Process

If the change is isentropic

S2-S1 = 0

1

2

1

2

1

2

1

2

1

2

1

2

1

2

1

212

lnln

0lnln

0lnln

0ln1

V

V

P

P

V

V

P

P

V

V

C

C

P

P

V

VC

P

PnCSS

v

p

pv

18.5

17.5tan

1

2

1

1

1

2

2

1

1

1

2

2

1

1

1

2

2

1

1

22

1

11

2211

1

2

1

2

P

P

V

V

T

T

P

P

V

V

V

V

T

T

VRTVRT

VPVPNOW

tconsPV

V

V

P

P

b. Constant volume Process

dQ = du + dw

dw = 0

dQ = du

=CvdT

Q = Cv (T2-T1) -----5.19

Also dQ = Tds = CvdT

ds = Cv T

dT

20.5ln1

212

T

TCSS v

c. Constant Pressure Process

dQ = du + dw = cvdT + pdv

T

dTCds

dTCTdsdQ

TTCQ

TTCTTRC

VVpTTC

PdvdTCQ

p

p

p

pv

v

V

V

T

T

v

21.512

1212

1212

2

1

2

1

22.5ln 212

T

TCSS p

Fig. 5.5

Since Cp > Cv change in entropy in the constant pressure process is greater than entropy change

at constant volume.

A - Constant volume process; Entropy change (s2)v – s1

B constant Pressure Process; Entropy Change (s2)p – S1

WORK DONE IN ADIABATIC PROCESS (PERFECT GAS)

a. Non Flow Process

dQ = du + dw

= u2 – u1 + w = 0

W= u1 – u2 = cv (T1-T2)

W= cv (T1-T2) …….5.23

Cv = RTpvandR

1

24.51

1

1

1

1122

1122

2211

vpvpW

vpvp

R

vp

R

vpRW

b. Steady Flow Process

Using SFEE we have

- gZC

WQgZC

h hx 2

2

221

2

12

22

Where Wx is external shaft work. Q = 0 for adiabatic process, if kinetic and potential energy

terms are negligible.

R

PVT

TTR

W

RC

TTChhW

x

p

px

21

2121

1

1

25.51

1

1

1122

2211

2212

vpvpW

vpvpW

vpvpW

x

x

x

The work done in a reversible adiabatic process may also be calculated by means of integration.

Work done 2

1

v

v

pdvW

26.5

1

1

1

1

1

1

11

1

1

11

22211

1211

1

2

1

1

11

1

1

1

2

1

111

1

1

1

2

1

111

1

1

1

211

1

11

11

2

1

2

1\

2

1

TTRvpvpW

vpvpv

vvp

vv

vvp

vvvvp

Vvvp

r

vvp

dvVvpdvVC

yy

yy

y

yyy

yyyv

v

y

v

v

y

v

v

yy

POLYTROPIC PROCESS

When a gas undergoes a reversible process in which there is heat transfer, the process is called a

polytropic process. In this case pvn = c

The p-v-T relations are similar to those reversible adiabatic process

27.5

1

1

2

1

2

1

1

2

n

nn

p

p

v

v

T

T

Entropy Change Using the expression for entropy

andV

VR

T

TCSS v

1

2

1

212 lnln

Substituting

1

1

2

1

1

2

n

T

T

V

V

and

28.5ln

11

ln1

1

1

1

ln1

ln1

1,

1

1

212

1

2

1

2

1

212

T

T

n

nSS

T

T

nR

T

T

n

R

T

TRSS

RC

RC pv

Or

29.5ln

1

ln11

1

1

212

1

212

P

P

n

nSS

P

P

nn

nnSS

Or

30.5ln1

ln1

1

212

1

212

V

VRnSS

V

Vn

n

RnSS

Heat and Work in Polytropic Processes

Non Flow Process

Using First Law Equation

31.51

1

1

1

1

122

122

121212

vpvpWQ

vpvp

TTR

TTUUWQ

Steady Flow Process

Using SFEE

ggx Zc

Zc

hhWQ 1

2

12

2

212 0

22

Neglecting kinetic and potential energy termsd

32.51

1

1

1122

1122

12

1212

VPVPR

WQ

VPVPR

TTR

TTchhWQ

x

px

Work Transfer in Polytropic Process (non flow)

33.5

11

11

212211

1

1

111

2

1

n

TTR

n

vpvpW

v

v

n

vppdvW

v

v

n

Heat Transfer in Polytropic Process (non flow)

34.51

1

11

11

1

1

1

1

1

12

12

12

12

12

1212

12

TTCn

nQ

TTCn

n

n

nTTR

n

nTTR

nTTR

n

TTRTTC

WUUQ

v

v

v

The Isothermal Process

Pv=constant. In this case n=1, but the substitution of n=1 in the equations for work and heat in

the general polytropic case is invalid.

35,.5ln

ln

ln

2

21

2

211

1

21111

2

1

p

pRT

p

pvpW

v

vvpvppdvW

v

v

To find entropy

2

1

2

1

12

v

v

s

s

pdvuuTdsQ

But u = f(T). For isothermal process u2 – u1 = 0

36.5ln

ln

ln

1

212

1

2

1

2

1

1112

2

1

2

1

v

vRss

v

vR

v

v

T

vpss

pdsTdsQ

v

v

s

s

5.9 Mixtures of perfect Gases

Following two properties common for all ideal gas components overall.

Temperature = T2 = TA = TB ………TN

Overall volume = V = VA = VB ……..

Since each component occupies V as if present alone (molecular volume negligible) for any

component.

Partial volume = V1 = Volume occupied by i when compressed to the overall pressure P.

Then V = V1 = (Amagat – Leduc Law)

= mv = mA vA = mBvB = mivi

Partial pressure = pi = Pressure exerted when i occupies overall volume V.

Then P = p1

Since P’AV = P’AV (mAvA) = mARAT

i

i

A

B

A

AA

V

TRPiP

etcV

TRBP

V

TRAP

''

But ........1 BBAA VmVmmvm

mv

mi

vV

Tm

Rmpv

mvTRmvmm

Rm

v

TP

ii

ABAAii

/.......

This is the form PV = RT for ideal gas

Mixture behaves as ideal gas, where

m

RmR ii

mixture

Since )( iii vmmuUV

)(

)(

iii

iii

smmsSS

hmmhHH

and

m

CmC

CmdT

dTmCdU

dTCdu

viiv

vi

v

v

)

And

m

CmCp

pii