Real Gases

Deviation from ideal gas Law:-Gas particles have volume

- Attraction exists between gas particles (liquefication)

Ideal gas law applies only at low gas density:

high T, low p

TRpVn

VvolumemolarV

TRn

VpTRnpV

mm

isotherm

Van der Waal’s Equation

Ideal gas law needs modification:

- Let’s consider the molecular volume:

b: size of 1 mole of gas molecules

For 1 mole of gas: Vm,total = Vm,space + b

bVbVV

VV

mtotalmidealm

spacemidealm

,,

,,

- Let’s consider the attraction forces between

molecules:

- Due to attraction: - velocity decreased, - colliding less with wall, - pressure drops

- How big is this drop in pressure?

attraction force

1f2f3f

1cFattr

1V2V

attraction force

1f2f3f

2cFattr

1V2V

- How big is this drop in pressure?

21 ccFattr

ccc 212cFattr

V

nc

2

2

V

nFattr

22222

2 111

mVnVnVV

n 2

1

mattr V

F

- How big is this drop in pressure?

drop in pressure proportional to attraction force

drop in pressure indirectly proportional to square of Vm

2mV

constppressureindrop

2m

ideal

realideal

idealreal

idealreal

V

constpppp

ppp

ppp

pp

RTVp idealmideal ,

RTbVV

ap m

m

2

n

VVm

RTnbnVV

nap

2

2

VdW Equation

of state

b: volume of 1 mole of gas molecules.

a: represents strength of attraction.

Maxwell construction

a1 = a2

Many other equations of state Mostly empirical

Virial Equation

midealmideal V

RTV

RTp

1

,

to correct, develop in a series (additional correction terms:

...1

...1

32

32

mmm

mmm

V

C

V

B

VRTp

V

CRT

V

BRT

VRTp

B,C, …: second, third, … virial coefficients: exp. determined

Compression factor z

describes deviation from ideal gas behavior.

idealm

m

V

Vz

,

p

RTVTRVp mm

RT

Vpz m

For ideal gas, z=1),,( TVpfz m

At a given temperature, z depends on the nature of gas

Why?

Let’s find for a VdW gas at p → 0

The slope of the curve z=f(p) in the previous diagram.

Tp

z

RTbVV

ap m

m

2

2mm V

a

bV

RTp

2mm V

a

bV

RTp

mm

m

m

mm

m

VTR

a

bV

Vz

RT

V

V

a

bV

RT

RT

Vpz

2

m

m

mm

m

VTR

a

VbVTR

a

bV

Vz

1

1

1...11

1 32

xforxxxx

1mV

bx

...11

132

mmm

m

V

b

V

b

V

b

Vb

mmmm VTR

a

V

b

V

b

V

bz

...1

32

TmTm

T

m

m

V

z

RTVRTz

p

z

V

TRp

TR

Vpzpat

11

,1,1,0

...1

132

mmm V

b

V

b

VTR

abz

TR

ab

RTp

z

T

1lim

0p

slope at p → 0

TR

ab

RTp

z

T

1lim

0p

> 0

= 0

< 0

bR

aT

TR

ab

TR

ab

TR

ab

RT

Boyle

BoyleBoyle

Boyle

0

01

0

0

0

patpositiveslopeTR

abif

patzeroslopeTR

abif

patpositiveslopeTR

abif

b: volume of 1 mole of gas molecules.

represents repulsive forces

a: represents strength of attraction

Size effect > attraction

Size effect = attraction

Size effect < attraction

Explain the trend!

Critical Point

Vm range in which L and G phases coexist shrink to a single point

Above Tc, L and G phases can not be distinguished from each other:

Density (L) = Density (G) No Interface

b

apthatshow

Rb

aT

b

a

b

RT

b

a

b

RT

V

a

bV

RT

bVVbVVbV

V

a

bV

RT

V

a

bV

RT

V

p

V

a

bV

RT

V

a

bV

RT

V

p

c

ccc

cmcm

c

cmcmcmcmcm

cmcm

c

cmcm

c

Tm

cmcm

c

cmcm

c

Tm

27

27

8

27

2

427

2

40

2

323332

620

62

20

2

32323,

2,

,,,,,

4,

3,

4,

3,

2

2

3,

2,

3,

2,

002

2

TmTm V

p

V

p

2mm V

a

bV

RTp

Experimental measurements of pc and Tc are more

accurate than that of Vc, a and b are calculated from the

experimental values of pc and Tc .

!

64

27

8

22

bandaofunitstheDerive

p

TRa

p

RTb

thatshow

c

c

c

c

Is it possible to find an equation of state that doesn’t

contain material specific constants?

The law of corresponding states

Reduced properties:

cm

mrm

cr

cr V

VV

T

TT

p

pp

,,

mcmrmcrcr VVVTTTppp ,,

b

apc 27

Rb

aTc 27

8 bV cm 3,

mrmrr VbVTbR

aTp

b

ap 3

27

8

27 ,2

mrmrr VbVTbR

aTp

b

ap 3

27

8

27 ,2

2mm V

a

bV

RTp

2,,

3

13

8

rmrm

rr VV

Tp

•Two gases at the same Tr and pr have the same Vr.

•These two gases are in “corresponding states’’.

•Ar (302 K and 16 atm) ↔ ethylene (381 K and 18 atm)