The Joule-Thomson expansionThe Joule-Thomson experiment is deceptively easy todescribe and correspondingly difficult to perform. Theexperiment consists of forcing a gas at constant backingpressure through a porous plug and measuring the change intemperature. The Joule-Thomson

coefficient is equal to

�

µJT

=!T

!P

The Joule-Thomson expansion...

The Joule-Thomson coefficient is

�

µJT

=!T

!PThe most straightforward way to measure the coefficient is tomeasure the temperature of the gas entering and leaving theporous plug and divide the change in temperature by thechange in pressure:

�

!T!P"

# $

%

& ' (

)T)P

=T2*T

1

P2* P

1

The Joule-Thomson expansion...

The Joule-Thomson coefficient is

�

µJT

=!T

!PAs usual, we need to decide what is being held constant whenwe take the partial derivative. In this case, since q=0, thechange in internal energy of the gas is just w, so that

�

!U =U2"U

1= P

1V1" P

2V2

�

U1

+ P1V1

=U2

+ P2V2

H1

= H2

or,

so that the expansion is isenthalpic.

�

µJT

=!T!P"

# $

%

& ' H

Some thermodynamic identities

�

µJT

=!T!P"

# $

%

& ' H

= (

!T!H"

# $

%

& ' P

!P!H"

# $

%

& ' T

= (

!H!P

"

# $

%

& ' T

!H!T

"

# $

%

& ' P

= (

!H!P

"

# $

%

& ' T

CP

�

dH = dU + PdV +VdP! dU = dH " PdV "VdP

dS =dU

T"P

TdV

#

$ %

& %

! dS =dH

T"V

TdP

Some thermodynamic identities

�

µJT

=!T!P"

# $

%

& ' H

= (

!T!H"

# $

%

& ' P

!P!H"

# $

%

& ' T

= (

!H!P

"

# $

%

& ' T

!H!T

"

# $

%

& ' P

= (

!H!P

"

# $

%

& ' T

CP

�

dH = dU + PdV +VdP! dU = dH " PdV "VdP

dS =dU

T"P

TdV

#

$ %

& %

! dS =dH

T"V

TdP

This gives us anexact differentialinvolving dH and dP.

Some thermodynamic identities...

�

dS =dH

T!V

TdP

�

dH =!H!T

"

# $

%

& '

P

dT +!H!P

"

# $

%

& '

T

dP = CPdT +

!H!P

"

# $

%

& '

T

dP

dS =

CPdT +

!H!P

"

# $

%

& '

T

dP

T(V

TdP

=CP

TdT +

!H!P

"

# $

%

& '

T

T(V

T

)

*

+

+

+

+

,

-

.

.

.

.

dP

Some thermodynamic identities...

�

dS =dH

T!V

TdP

�

dH =!H!T

"

# $

%

& '

P

dT +!H!P

"

# $

%

& '

T

dP = CPdT +

!H!P

"

# $

%

& '

T

dP

dS =

CPdT +

!H!P

"

# $

%

& '

T

dP

T(V

TdP

=CP

TdT +

!H!P

"

# $

%

& '

T

T(V

T

)

*

+

+

+

+

,

-

.

.

.

.

dP

This gives us anexact differentialinvolving dT and dP.

Some thermodynamic identities...

�

dS =dH

T!V

TdP

�

dH =!H!T

"

# $

%

& '

P

dT +!H!P

"

# $

%

& '

T

dP = CPdT +

!H!P

"

# $

%

& '

T

dP

dS =

CPdT +

!H!P

"

# $

%

& '

T

dP

T(V

TdP

=CP

TdT +

!H!P

"

# $

%

& '

T

T(V

T

)

*

+

+

+

+

,

-

.

.

.

.

dP

This gives us anexact differentialinvolving dT and dPthat will allow us tosolve for (∂H/∂P)T.

Some thermodynamic identities...

�

dS =CP

TdT +

1

T

!H!P

"

# $

%

& '

T

(V)

* +

,

- . dP

�

!!P

CP

T

"

# $

%

& '

"

# $

%

& '

T

=!!T

1

T

!H!P

"

# $

%

& '

T

(V)

* +

,

- .

"

#

$

%

&

'

"

#

$ $

%

&

' '

P

1

T

!!P

!H!T

"

# $

%

& '

P

"

# $

%

& '

T

= (1

T2

!H!P

"

# $

%

& '

T

(V)

* +

,

- . +1

T

! 2H!P!T

"

# $

%

& ' (

!V!T

"

# $

%

& '

P

)

*

+

,

-

.

0 = (1

T2

!H!P

"

# $

%

& '

T

(V)

* +

,

- . (1

T

!V!T

"

# $

%

& '

P

!H!P

"

# $

%

& '

T

(V)

* +

,

- . = (T

!V!T

"

# $

%

& '

P

!H!P

"

# $

%

& '

T

=V (T!V!T

"

# $

%

& '

P

This is an exact differential

Some thermodynamic identities...

!H!P

"#$

%&'T

= V ( T!V!T

"#$

%&'P

µJT= !

"H"P

#$%

&'(T

CP

This expression can be used to predict the Joule-Thomson coefficient from any equation of state whichallows the calculation of (∂V/∂T)P, so long as the heatcapacity at constant pressure is also known. (Recall thatthe heat capacity can be measured from the speed of sound.)

An exampleSuppose the compressibility Z=PV/RT of the gas follows avirial equation of the type

�

Z T,V ( ) =1+B T( )

V +

C T( )

V 2

+!

The Joule-Thomson coefficient may be written

�

µJT

=1

CP

T!V!T

"

# $

%

& ' P

(V"

# $

%

& '

where we now need (∂V/∂T)P from the equation of state.

An example...

�

PV

RT=1+

B T( )

V +

C T( )

V 2

+!

P =RT

V + RT

B

V 2

+ RTC

V 3

+!

dP =R

V dT !

RT

V 2

dV +R

V 2

B + T"B

"T

#

$ %

&

' (

#

$ %

&

' ( dT ! 2RT

B

V 3

dV +!

dP = 0

R

V +

R

V 2

B T( ) + T"B

"T

#

$ %

&

' (

#

$ %

&

' (

#

$ %

&

' ( dT =

RT

V 2

+ 2RTB T( )

V 3

#

$ %

&

' ( dV

Finding (∂V/∂T)P from the equation of state:

An example...

�

R

V +

R

V 2

B T( ) + T!B

!T

"

# $

%

& '

"

# $

%

& '

"

# $

%

& ' dT =

RT

V 2

+ 2RTB T( )

V 3

"

# $

%

& ' dV

R

V +

R

V 2

B + T!B

!T

"

# $

%

& '

"

# $

%

& '

"

# $

%

& '

RT

V 2

+ 2RTB

V 3

"

# $

%

& '

=!V

!T

"

# $

%

& '

P

!V

!T

"

# $

%

& '

P

=

V + B + TdB

dT

"

# $

%

& '

"

# $

%

& '

"

# $

%

& '

T + 2TB

V

"

# $

%

& '

Having set dP=0 we can solve for (∂V/∂T)P:

An example

�

!V

!T

"

# $

%

& '

P

=

V + B + TdB

dT

"

# $

%

& '

"

# $

%

& '

"

# $

%

& '

T + 2TB

V

"

# $

%

& '

We can expand the denominator using the binomialexpansion, provided we work at sufficiently large molarvolumes (i.e., low pressures):

�

1

T 1+ 2B

V

!

" #

$

% &

=1

T1' 2

B

V + 2

B

V

!

" #

$

% &

2

' 2B

V

!

" #

$

% &

3

+!

(

)

*

+

,

-

An exampleDoing so, the derivative becomes

�

!V

!T

"

# $

%

& '

P

= V + B + TdB

dT

"

# $

%

& '

"

# $

%

& '

(

) *

+

, - .1

T1/ 2

B

V + 2

B

V

"

# $

%

& '

2

/ 2B

V

"

# $

%

& '

3

+!

(

)

*

+

,

-

Keeping only the leading terms,

�

!V

!T

"

# $

%

& '

P

=1

TV + B + T

dB

dT

"

# $

%

& ' ( 2B

)

* +

,

- .

=1

TV ( B + T

dB

dT

"

# $

%

& '

)

* +

,

- .

An exampleSubstituting this into the equation for µJT gives

�

µJT

=1

CP

T!V

!T

"

# $

%

& '

P

(V

)

* +

,

- .

=1

CP

V ( B + TdB

dT

"

# $

%

& ' (V

)

* +

,

- .

=1

CP

TdB

dT

"

# $

%

& ' ( B

)

* +

,

- .

in the low-pressure limit.

An example

The point at which the two terms on the right are equalspecifies the inversion temperature of the gas. Above thistemperature, µ is negative (the gas warms as it expands);below this temperature, it is positive.A more realistic equation of state predicts that the inversiontemperature is a function of pressure and that a gas may inprinciple have more than one inversion temperature at agiven pressure.

�

µJT

=1

CP

TdB

dT

!

" #

$

% & ' B

(

) *

+

, -

An alternative approachWe could use the Euler permutation to evaluate (∂V/∂T)P.Again taking

P =RT

V+ RT

B

V2+ RT

C

V3+!

We can easily evaluate (∂P/∂T)V and (∂P/∂V)T :

!P!V

"#$

%&'T

= (RT

V2( 2RT

B

V3( 3RT

C

V4+!

!P!T

"#$

%&'V

=R

V+RB

V2+RT

V2

dB

dT+RC

V3+RT

V3

dC

dT+!

An alternative approach... From these expressions we can again evaluate the Joule-Thomson coefficient:

µJT=1

CP

!T

"P"T

#$%

&'(V

"P"V

#$%

&'(T

!V

)

*

++++

,

-

.

.

.

.

=1

CP

!T

R

V+RB

V2+RT

V2

dB

dT+RC

V3+RT

V3

dC

dT+!

!RT

V2! 2RT

B

V3! 3RT

C

V4+!

!V

)

*

+++

,

-

.

.

.

=1

CP

V + B + TdB

dT+C

V+T

V

dC

dT+!

1+ 2B

V+ 3

C

V2+!

!V

)

*

+++

,

-

.

.

.

Another exampleIt we take the van der Waals equation of state

P =RT

V ! b!a

V2

We again evaluate (∂P/∂T)V and (∂P/∂V)T :

!P!V

"#$

%&'T

= (RT

V ( b( )2+ 2

a

V3

!P!T

"#$

%&'V

=R

V ( b

Another example... From these expressions we can again evaluate the Joule-Thomson coefficient:

µJT=1

CP

!T

"P"T

#$%

&'(V

"P"V

#$%

&'(T

!V

)

*

++++

,

-

.

.

.

.

=1

CP

!T

R

V ! b#$%

&'(

!RT

V ! b( )2+ 2

a

V3

!V

)

*

+++++

,

-

.

.

.

.

.

=1

CP

1

1

V ! b( )! 2

V ! bRT

a

V3

!V

)

*

++++

,

-

.

.

.

.

Another example... From these expressions we can again evaluate the Joule-Thomson coefficient. Once again taking the limit at largemolar volume,

limV!"

µJT=1

CP

limV!"

#T

R

V # b$%&

'()

#RT

V # b( )2+ 2

a

V3

#V

*

+

,,,,,

-

.

/////

=1

CP

limV!"

#V#2ab2 + 4abV # 2aV 2 + bRTV 2( )#2ab2 + 4abV # 2aV 2 + RTV 3

*

+,,

-

.//

=1

CP

2a

RT# b*

+,-./

…and an important limiting caseWhen the gas is ideal, a and b are zero…

µJT=1

CP

1

1

V ! b( )! 2

V ! b

RT

a

V3

!V

"

#

$$$$

%

&

''''

=1

CP

1

1

V ! 0( )! 2

V ! 0

RT

0

V3

!V

"

#

$$$$

%

&

''''

=1

CP

1

1

V! 0

!V

"

#

$$$

%

&

'''

= 0