02-gaslaws

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State variables

Thanks to molecular chaos, ergodic motion, and the existence ofequilibrium states, the macroscopic state of the system is describedby a handful of macroscopic state variables (and not ~1023positionsand momenta!).

For a one component system, 3 state variables are usually needed tocompletely specify the state of the system, one of which must be anextensive variable. They may be

T, V, n T, P, n

T, V, (= n/V)

T, P, (why not?)

Extensive: V, n,

Intensive: P,, V,( )nVV /=

Equations of state

A relationship giving a macroscopic variable in terms of others iscalled an equation of state.

P = P(V,T,n) is a common equation of state.

The ideal gas law is of this form. P = nRT/V.

An intensive variable like P can be written as a function of only twoother intensive variables. For example, P = P(T,) where = n/V.

The ideal gas law can be written as P =RT.

It is the special properties of systems in equilibrium (molecularchaos, ergodic motion) that permit state variables to be functions ofa handful of other state functions, and equations of state to exist.

Other equationsof state

===

=

=

=

V

n

n

SSTSS

nTVSS

nPUVV

nTPUU

,),(

),,(

),,(

),,(

For a system that is a mixture ofc components, an extensive propertyis a function of c+2 state variables, one of which is extensive. Anintensive property is a function ofc+1 intensive state variables.

)52,components(3),,,,( 321 =+= cnnnTVSS

+===

=+=

211

1

1,,

,31,components2

),,(nnn

n

nx

n

SS

c

xTVSS

Partial derivatives

+=

+=

+=

=

dz

zyxfdzzyxf

z

f

dy

zyxfzdyyxf

y

f

dx

zyxfzydxxf

x

f

zyxff

dzyx

dyzx

dxzy

),,(),,(lim

),,(),,(lim

),,(),,(lim

),,(

0,

0,

0,

For an example, takef to be a function of three variables,x,y,z.

A partial derivatives is an ordinary derivative with respect to one ofthe variables, holding the other variables constant.

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Isotherms

andisobars

The equation of state is a 2-dimensionalsurface in a 3-dimensional space withaxes labeled by P,T,V.

0

2.5

5

7.5

100

200

400

0

1000

2000

0

2.5

5

7.5

10

V(L)

P(Pa)

T(K)

0 2 4 6 8 1 0

250

500

750

1000

1250

1500

1750

2000

V(L)

P(Pa)

isotherms

0 100 200 300 400 500

0

250

500

750

1000

1250

1500

V(L)

T(K)

isobars

slope of isotherms

nTPV

,

slope ofisobars

nPT

V

,

Understanding partial derivatives on amathematical level

The variables on the bottomare the independent variables. nTP

V

,

Regard Vas a function

of the independentvariables P,T,n.

nTP

V

,

tells you the slope of the isotherms.

Understanding partial derivatives on aphysical level

,1

or,,, nT

T

nT P

V

VP

V

=

which is normalized by Vto

make Ta positive intensive quantity, is a linear responsefunction orsusceptibility.

perturbation

response

A susceptibility gives the response to an external perturbation. In thiscase Ttells you how volume responds to a pressure change.

The susceptibility gives the linear response

is a linear response function orsusceptibility.

Ttells you how volume responds to a pressure change.

PPP

V

VV

VT

nT

=

=

,

1

P

V

P

V

nT

=

,

, for small Vand P, constant T

, linear response of volume to PPP

VV

nT

=,

V

P

,1

,nT

TP

V

V

=

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Gases at 1 atm are nearly ideal

At 1atm ( 1 bar) gas molecules are about 3.5nm (35) apart.

~3

Intermolecular interactions can be neglected.

Gases at 1 atm are nearly ideal

At 1atm ( 1 bar) gas densityis ~ 2.4 x 102 molecules/nm3.

Density of liquid water is33.4 molecules/nm3.

Real gases

Gases deviate from ideal behavior at high pressure or density, or at

low temperature. Deviations from PV=nRTare caused by

interactions between molecules.

1 mol of N2, CH4, H2at 300K, CO2 at 313K

Real gases

Gases deviate from ideal behavior at high pressure or density, or at

low temperature. Deviations from PV=nRTare caused by

interactions between molecules.

1 mol of N2

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Intermolecular interactions

The interaction

between molecules is

typically repulsive

at longer distance.

at very short

distances, attractive

Excluded volume

The volume available to molecules in a gas is reduced by the volume of

other molecules. This effect is known as excluded volume.

The actual volume available to a

molecule is Vnb, where b is the

volume excluded by 1 mole of gas

and n is the number of moles.

Replace Vby Vnb in the ideal gas law. The effect

of excluded volume is to raise the pressure.

nbV

nRTP

=

b describes molecule volume

m3/molx105

The b values youfind in tables are fitto experimentalequations of state.

Molecular attractions

The ideal gas law is corrected by an extra term that lessens

the pressure.2

=

=V

na

V

nRT

V

naRT

V

nP

Attractive interactions will cause the

gas molecules to pull together, strike

the walls with less force, and decrease

the pressure.

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Molecular attractions

The correction term is proportional to density (n/V) because theeffect is only important when the gas is sufficient dense for

molecules to feel each other. At low density, the correction

should vanish. The constant a reflects the strength of attractions

between molecules.

2

=

=V

na

V

nRT

V

naRT

V

nP

The van der Waals equation

The effects of attractions and repulsions are

combined in an empirical gas law, the van

der Waals equation. a and b are chosen to

match experimental data.

2

=V

na

nbV

nRTP

( ) nRTnbVV

naP =

+2The van der Waals equation

is sometimes written these

equivalent forms.

( ) RTbVV

aP =

+

2

Common two-parameter equations of state

2V

a

bV

RTP

=

( )BVVT

A

BV

RTP

+

=2

1

( ) ( )

++

=

VVVV

RTP

van der Waals

Redlich-Kwong

Peng-Robinson

Redlich-Kwong and Peng-Robinson better reproduceexperimental data (see text book).

A generic phase diagram in the P-Vplane

P

V

triple point

solid

liquid vapor

fluid

critical point

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Coexistence regions

P

V

triple point

solid

liquidvapor

fluid

critical point

liquid-vapor

coexistence

solid-vapor

coexistence

Tie linesare shown

Spontaneous phase separation in

coexistence regions

If prepared in a homogeneous state within a coexistence region, asubstance will spontaneously separate into two coexisting phases.

equilibrate

Continuous path from liquid tovapor above the critical point

P

V

triple point

solid

liquid vapor

fluid

critical point

Isotherms within the P-Vplane

V

triple point

solid

liquid

vapor

fluid

critical point

Super-critical,critical, and sub-critical isotherms,as you would

calculate fromvdW, Redlich-Kwong, or Peng-Robinsonequations of stateare shown

P

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Super-critical isotherm within the P-Vplane

V

triple point

solid

liquid

vapor

fluid

critical pointeverywhere alongthe super-criticalisotherm.

0,0 >

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Sub-critical isotherm within the P-Vplane:Maxwell equal-area construction

P

V

triple point

solid

liquid

vapor

fluid

critical point

ed

c

b

aAreaabc = Areacde

The region wheretwo phases are morestable (more likely)than one phase ispredicted by

This is called the

Maxwell equal-areaconstruction, whichwe will justify later.

Procedure to find critical parameters

P

V

triple point

solid

liquid

vapor

fluid

critical point

0

0

,

2

2

,

=

=

nT

nT

V

P

V

P

),(),,( TVPnTVPP ==

P function ofT,V,n.

2 eq.s in 2unknowns, T,V.

Solve to findTc, Vc.

Plug Tc, Vc.into P(T,V) to obtain Pc.

work out van der Waals critical point as example

Critical parameters of van der Waals gas

RT

aq

V

q

bVRTP

2where,

2

12

=

=

vdW equation

Set and0,

=

nTV

P0

,

2

2

=

nTV

P

c

cRT

aq

2=

See text book for analternative derivationof critical parameters2

V

a

bV

RTP

=

The value ofq at the critical point is .

Think ofq as an alternativetemperature variable


RT

aq

V

q

bV

RTP

V

a

bV

RTP

2where,

2

12

2

=

=

=

( )0

132

,

=

+

=

V

q

bVRT

V

P

nT

vdW equation

1st derivative = 0

( )23 bVqV =

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RT

aq

V

q

bVRTP

2where,

2

12

=

=vdW equation

1st derivative = 0 ( )23 bVqV =

( ) 032

43,

2

2

=

=

V

q

bVRTV

P

nT

2nd derivative = 0

( )342

3bV

qV =


RT

aq

V

q

bVRTP

2where,

2

12

=

=vdW equation


2nd derivative = 0 ( )342

3bV

qV =

divide these two equations

( )bVVcc = 2

3at critical point

bVc 3=


RT

aq

V

q

bVRTP

2where,

2

12

=

=vdW equation


2nd derivative = 0 ( )342

3bV

qV =

bVc 3=critical volumeplug back in to get qc and Tc

( ) ( )23 33 bbqb c =

4

27bqc =

,2

Sincec

cRT

aq =

4

272 b

RT

a

c

=

Rb

aTc

27

8=


RT

aq

V

q

bVRTP

2where,

2

12

=

=vdW equation

bVc 3=critical volume

,

4

27bqc =

Rb

aTc

27

8=critical

temperature

=2

2

1

c

c

ccc

V

q

bVRTP

( )( )

=232

427

3

1

27

8

b

b

bbRb

aRPc

( )( )

=22 32

427

13

1

27

8

b

a

227b

aPc =

=8

3

2

1

27

82b

a

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RT

aq

V

q

bVRTP

2where,

2

12

=

=vdW equation

bVc 3=critical volume

,4

27bqc =

Rb

aTc

27

8=critical

temperature

227b

aP

c =critical pressure

criticalcompressibilityfactor c

ccc

RT

VPZ =

Zc is independent ofthe substance! This

gives us a hint thatconstructing

dimensionlessquantities will

producesimplifications.

( )

=

Rb

aR

bb

a

27

8

327 2

8

3=

Reduced variables and corresponding states

We will reach the surprising conclusion that in the right units allsubstances obey nearly the same equation of state.

Construct dimensionless variables:

c

R

c

R

c

RV

VV

T

TT

P

PP === ,,

theinto,,Substitute cRcRcR VVVTTTPPP ===

van der Waals equation of state, ( ) RTbVV

aP =

+

2

The van der Waals equation in universal form

theinto,,Substitute cRcRcR VVVTTTPPP ===

van der Waals equation of state, ( ) RTbVV

aP =

+

2

( )

( ) cRcRcR

cR TRTbVV

VV

aPP =

+

2

( )( )

Rb

aRTbbV

bV

a

b

aP RR

R

R27

83

32722

=

+

RR

RR TV

VP

3

8

3

132

=

+In reduced variables, allvdW gases obey a universalequation of state.

Two-parameter equations of state

Any two-parameter equation of state can, in principle, betransformed into universal form in reduced units.

The reduction may be algebraically messy, but choosingintroducing dimensionless units in principle gives us enoughfreedom to cancel two parameters that depend on the chemical

nature of the substance.

As long as a group of substances can be described by some two-parameter equation of state*, their equation of state in reducedvariables will be universal.

*even if this equation of state may not have been discovered yet!

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Corresponding

states

Z

PR

The fact that data from

so many substancescollapses onto auniversal curve supportsthe corresponding statesanalysis.

Virial expansion:

Zas a function ofdensity or pressure

Z

PR

Except near thecritical point

TR = PR = 1Zis smooth function

ofT, P or.

The virial expansion

Analyze the deviation from ideality in terms of a series expansion ineither the density of the pressure.

L++++== 34

232 )()()(1

V

TB

V

TBV

TBRT

VPZ VVV

L++++== 342

32 )()()(1 PTBPTBPTBRT

VPZ PPP

Density virial expansion: (density)1

==V

n

V

Pressure virial expansion:

The virial expansion

Analyze the deviation from ideality in terms of a series expansion ineither the density of the pressure.

L++++== 34

232 )()()(1

V

TB

V

TB

V

TB

RT

VPZ VVV

L++++== 342

32 )()()(1 PTBPTBPTBRT

VPZ PPP

The virial coefficients,BnVorBnP , are functions ofTonly.

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PlotZas a

function ofdensity

L++++==3

42

32 )()()(1

V

TB

V

TB

V

TB

RT

VPZ VVV

L++++== 342

32 )()()(1 TBTBTBRT

VPZ VVV

V

1=

RT

VPZ=

0)( 12 TB V

21 TTT Boyle

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Taylor series expansion generates a localpolynomial approximation to a complicated function

( )+

+==

00

0

)()( xxdx

dfxfxf

xx

The Taylor series expansion is a polynomial is (x x0).

The derivatives off(x) atx0 are needed toconstruct the Taylor series expansion.

( ) +

=

303

3

0!3

1xx

dx

fd

xx

( ) +

=

202

2

0!2

1xx

dx

fd

xx

( ) LL +

+

=

n

xx

n

n

xxdx

fd

n0

0!

1

If (x x0) is small, (x x0)2 will be smaller, (x x0)

3 smaller still, andwe can hope that truncating the expansion will not cause much error.

Taylor expand sin(2x) aboutx0

[ ]( ) [ ]( )

[ ]( ) [ ]( )

[ ]( ) [ ]( ) L+++

++

++=

600

500

400

300

200000

)2sin(64!6

1)2cos(32

!5

1

)2sin(16!4

1)2cos(8!3

1

)2sin(4!2

1)2cos(2)2sin()2sin(

xxxxxx

xxxxxx

xxxxxxxx

( ) ( )

( ) ( ) LL +

++

+

+

+=

==

==

n

xx

n

n

xx

xxxx

xxdx

fd

nxx

dx

fd

xx

dx

fdxx

dx

dfxfxf

03

03

3

202

2

00

00

00

!

1

!3

1!2

1)()(

Expansion of sin(2x) aboutx0 = 1 through order 1

[ ]( )000 )2cos(2)2sin()2sin( xxxxx +

sin(2x)

Taylor approximation

x


[ ]( ) [ ]( )200000 )2sin(4!2

1)2cos(2)2sin()2sin( xxxxxxxx ++

sin(2x)


x

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[ ]( ) [ ]( )

[ ]( )300

2

00000

)2cos(8!3

1

)2sin(4!2


xxx

xxxxxxxx

+

++

sin(2x)


x


[ ]( ) [ ]( )

[ ]( ) [ ]( )4003

00

2

00000

)2sin(16!4

1)2cos(8

!3

1

)2sin(4!2


xxxxxx

xxxxxxxx

++

++

sin(2x)


x


[ ]( ) [ ]( )

[ ]( ) [ ]( )

[ ]( )500

4

00

3

00

2

00000

)2cos(32!5

1

)2sin(16!4

1)2cos(8

!3

1

)2sin(4!2


xxx

xxxxxx

xxxxxxxx

+

++

++

sin(2x)


x


[ ]( ) [ ]( )

[ ]( ) [ ]( )

[ ]( ) [ ]( )6005

00

4

00

3

00

2

00000

)2sin(64!6

1)2cos(32

!5

1

)2sin(16!4

1)2cos(8

!3

1

)2sin(4!2


xxxxxx

xxxxxx

xxxxxxxx

++

++

++

sin(2x)


x

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[ ]( ) [ ]( )

[ ]( ) [ ]( ) [ ]( )

[ ]( ) [ ]( )7006

00

500

400

300

200000

)2cos(128!7

1)2sin(64

!6

1

)2cos(32!5

1)2sin(16

!4

1)2cos(8

!3

1

)2sin(4!2


xxxxxx

xxxxxxxxx

xxxxxxxx

++

+++

++

sin(2x)


x


[ ]( ) [ ]( )

[ ]( ) [ ]( ) [ ]( )

[ ]( ) [ ]( ) [ ]( )8007

00

6

00

500

400

300

200000

)2sin(256!8

1)2cos(128

!7

1)2sin(64

!6

1

)2cos(32!5

1)2sin(16

!4

1)2cos(8

!3

1

)2sin(4!2


xxxxxxxxx

xxxxxxxxx

xxxxxxxx

+++

+++

++

sin(2x)


x

Virial coefficients by Taylor expansion ofZ =

L+

+

+

+=

===

3

0

3

32

0

2

2

0!3

1

!2

1)0()(

d

Zd

d

Zd

d

dZZZ

ExpandZ(,T) about = n/V = 0. (The Tdependence ofZis notshown explicitly on the next line.)

RT

VP

L++++= 342

32 )()()(1 TBTBTBZ VVV

Compare with

Virial coefficients by Taylor expansion ofZ =RT

VP

1)0(0

=

=

=RT

VPZ

0

2 )(

=

=

d

dZTB V

0

2

2

3!2

1)(

=

=

d

ZdTB V

L

0

3

3

4!3

1)(

=

=

d

ZdTB V

It would be more accurate towrite these as partialderivatives with respect to .

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Virial coefficients of the van der Waals gas

2

=V

na

nbV

nRTP

VRT

a

bV

V

RT

VP

=

van der Waals equation of state

RT

a

bTZ

=

1

1),(

compressibility factor

L++++

+=443322

1),( bbbRT

a

bTZ

Taylor expansion

1232 )(,)(,)(

==

= nnVVV

bTBbTBRT

abTB L

virial coefficients

V

1=

Second virial coefficient of the van der Waals gas

1232 )(,)(,)(

==

= nnVVV bTBbTBRT

abTB L

virial coefficients

B2V(T) vanishes at the Boyle temperature TBoyle.

bR

aTBoyle =

ForT< TBoyle attractions dominate.

ForT> TBoyle repulsions dominate.

When T= TBoyle attractions and repulsions cancel inB2V(T).

The Boyle temperature and idealityIt is often said that a gas behaves as if ideal at the Boyle temperature.

This is only in regards toB2V(T). Even at TBoyle the other virialcoefficients are not zero.

There are otherpossible measures of the cancellation betweenattractions and repulsions, for exampleZ =1.

1 mol of N2, CH4, H2 at300K, CO2 at 313K

Z= 1

02-gaslaws

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