JouleThomson.pdf
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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