Download Lecture 4 - TCD Chemistry

Chapter 4
Joule Thomson Effect
Chemical Potential
Fugacity
Chemical Thermodynamics : Georg Duesberg
Chapter11111
Chapter 1 1: Slide
: Slide1
1
Joule – Thomson Effect
Refrigeration developed by Carl von Linde in 19th
Century, in response to a request from Guinness
in Dublin for a new cooling technique.
Based upon the fact that gases cool as they
expand: Joule-Thomson effect (William
Thomson, later Lord Kelvin, born in Belfast),
The Linde refrigerator combines the JT process
with a counter-flow heat exchanger.
The gas is re-circulated and it cools on expansion
through the throttle. The cooled gas cools the highpressure gas, which cools still further as it
expands. Eventually liquefied gas drips from the
throttle.
Chemical Thermodynamics : Georg Duesberg
2
Heat capacities
From first law of thermodynamics:
dU  dQ  dW with dW  PdV
dU  dQ  pV
H  U  pV
 U 
 U 
dU  
dT



 dV
 T V
 V T
 H 
 H 
 dp
dH  
 dT  
 T  p
 p T
 U  Heat capacity at
CV  

 T V constant volume
 H 
CP  

 T  P
(isochore)
 U 
dU  CV dT  
 dV
 V T
Heat capacity at
constant pressure
(isobar)
 H 
 dp
dH  C p dT  
 p T
Change of Internal Energy at constant temperature
for ideal gases


No interactions between molecules.
the internal energy of gas depends on temperature only.
•
•
Z = 1; PV = RT
U = U (T)
Heat capacity at
constant volume
 U 
 U 
dU  
 dT  
 dV
 T V
 V T
 U 
CV  

 T V
 U 
T  

 V T
dU  CV dT   T dV
[Note from VdW: a(n2/V2)= πT ]
Chemical Thermodynamics : Georg Duesberg
Internal Pressure
of a gas
(0 for ideal gas)
Internal Pressure, πT


For ideal gas, πT = 0 because U
independent of molecular separation
for volume changes at constant T
• Implies ideal gas law
For real gas,
If dU >0 as dV increases with T
constant, attractions between
molecules dominate and πT > 0
If dU <0 as dV increases with T
constant, repulsions between
molecules dominate and πT <0
[Note from VdW: a(n2/V2)= πT ]
Chemical Thermodynamics : Georg Duesberg
Joule Experiment: Measure πT
pex = 0, therefore w = 0
dT=0, therefore q = 0
DU = q +w = πT dV= 0
πT must =0 since dV>0
V2


 w   p ex dV


V1

“No change in temperature occurs when
air is allowed to expand in such a manner
as not to develop mechanical power” J. P.
Joule, Phil. Mag., 26, 369 (1845)
Joule experiment incapable of detecting
small effects since the water calorimeter he
used had a large heat capacity
Chemical Thermodynamics : Georg Duesberg
pex = 0
THE JOULE-THOMPSON EXPERIMENT
A further test of intermolecular forces in real gases.
 H 
 H 
DH  
 DT  
 DP  0
 T  P
 P T
Imagine a sample of gas pushed through a porous
plug, in an isolated tube (adiabatic system). The
temperature is measured on each side of the plug.
Analysis
w = piVi - pfVf
Since DU = Uf - Ui = w (because q = 0),
Uf + pfVf = Ui + piVi
Hf = Hi i.e. DH = 0
This is a constant enthalpy (isenthalpic) process.
Chemical Thermodynamics : Georg Duesberg
 H 

  CP
 T  P
What is
 H 
???  

 P T
Thermodynamic Consequences of State Functions
 H 
 H 
Enthalpy, H(p,T)
DH  
 DT  
 DP  0
 T  P

To evaluate (dH/dp)T we apply Euler chain rule:
 H

 p


 P T
 H 
???  

 P T


T
 T 


 H   T 
 P  H

 
 

 T 
 T  p  P  H


 H  p
 H 

  CP
 T  P
 T 
    JT
 P  H
Cp= (dH/dT)p and one defines (dT/dP)H as µ as Joule Thomson Coefficient
then (dH/dp)T =- Cpµ, is the Isothermal Joule Thomson Coefficient
 H 
 JT CP  

 P T
Chemical Thermodynamics : Georg Duesberg
Joule Thompson Experiment


Chemical Thermodynamics : Georg Duesberg
Modern methods measure µCP (Isothermal)
not µ
Isothermal J-T coefficient, µT = (dH/dp)T
= -Cpµ
Practical applications: Gas liquification,
Isotope separation, artificial snow
Chapter99999 1 : Slide
9
µ - Physical consensus


µ is a function of p and T (see Fig. for CO2)
µ can be either (+) or (-)
• Positive µ means dT is negative when dp is
negative
Gas cools on expansion
• Negative µ means that means dT is
positive when dp is negative
Gas warms on expansion
• Transition between positive and negative
µ is called the Joule-Thompson inversion
temperature (Ti =2TB )
Gases have both high and low inversion
temperatures
Chemical Thermodynamics : Georg Duesberg
Joule-Thomson Process for the vdW Gas
The JT process corresponds to an isenthalpic expansion:
 H 
 H 
DH  
 DT  
 DP  0

T

P

P

T
 H

P
 T 
   
 H
 P  H

 T
 H 

  CP

T

P
 H 
CP DT  
 DP

P

T
DH  TDS  VDP

 H 
 V 
T



 V
T
 P T
 T  P
 H 
 S 



  T  V
Cp
Cp

 P T
 P T

p
 S 
 V 
   
 (Maxwell relation)
 P T
 T  P
This is a pretty general (model-independent) result. By applying this result to the vdW equation,
one can qualitatively describe the shape of the inversion curve (requires solving cubic
equations...).
 V 
 
 T  P
May be obtained form the equation of state
Joule-Thomson Process for the vdW Gas
We’ll consider the vdW gas at low densities:

N 2a 
 P  2 (V  Nb )  Nk BT
V 

V  Nb
N 2a

(...)P
PV 
 PNb  Nk BT
V
T
N 2 a  V 
 V 
P
  2 
  Nk B
 T  P V  T  P
 V 
T
 V
T  P
 T 
    
Cp
 P  H
N 2a
P  2
V
Nk B
 V 



2
N
a

T

P P 
V2
2

Na
 Nb 2 a  b
k BT
 RT
CP
CP
Joule-Thomson Process for the vdW Gas
Cooling:
 JT 
2a
b  0
k BT
Heating:  JT 
2a
b  0
k BT
If b = 0, T always decreases in the JT process: an increase of Upot at the expense of K.
If a = 0, T always increases in the JT process (despite the work of molecular forces is 0):
The upper inversion temperature:
(at low densities)
TINV
2a
27


TC  2TB
k Bb
4
•Tc =8a/27Rb
•the critical temperature of the vdW gas
•TB = a/Rb
Thus, the vdW gas can be liquefied by
compression only if its T < 27/4TC.
Chemical Thermodynamics : Georg Duesberg
The JT Process in Real Gases
In real gases, molecules interact with each
other.
At low densities, the intermolecular forces
are attractive. When the gas expands
adiabatically, the average potential energy
increases, at the expense of the kinetic energy.
Thus, the temperature decreases because of
the internal work done by the molecules
during expansion.
Upot
vdW gas
x
expansion
H  U kin  U pot  PV
 JT  0
The JT Process in Real Gases
At high densities, the effect is reversed: the free
expansion results in heating, not cooling. The overall
situation is complicated: the sign of DT depends on initial T
and P.
heating
cooling
Upot
x
expansion
All gases have two inversion temperatures: in the
range between the upper and lower inversion
temperatures, the JT process cools the gas, outside
this range it heats the gas.
isenthalpic curves
(H =const) for
ideal and
real gases
For T < TINV, the
drop in pressure
(expansion) results
in a temperature
drop.
Chemical Thermodynamics : Georg Duesberg
 JT  0
Gas
boiling T
(P=1 bar)
inversion T
@ P=1 bar
CO2
195
(2050)
CH4
112
(1290)
O2
90.2
893
N2
77.4
621
H2
20.3
205
4He
4.21
51
3He
3.19
(23)
Liquification of Gases
Most gas liquifiers combine the expansion engines with JT process: the expansion
engine helps to pre-cool the gas below the inversion T. The expansion engines are a
must for He and H2 liquefiers (the inversion T is well below RT).
For air, the inversion T is above
RT. In 1885, Carl von Linde
liquefied air in a liquefier based
solely on the JT process: the gas
is recirculated and, since T is
below its inversion T, it cools on
expansion through the throttle.
The cooled gas cools the highpressure gas, which cools still
further as it expands. Eventually
liquefied gas drips from the
throttle.
Chemical Thermodynamics : Georg Duesberg
Linde
refrigerator
Virial Equation of State and Joule Thompson
Virial coefficients:
Z = p Vm / RT = (1 + B’p + C’p2 + ...)
Z = p Vm / RT = (1 + B/Vm + C/Vm2 + ...)
For Low pressures (0-15 bar) B is
sufficient
for moderate pressures (<100 bar) B
and C describe Z well
High Pressure better use other EOS
Also allow derivation of exact correspondence
between virial coefficients and intermolecular
interactions.
dB(T )
 JT p 0  B(T )  T
Chemical Thermodynamics : Georg Duesberg
dT
The Perfect gas and T, a, kT, 
Property
Internal
Pressure
Expansion
Coefficient
Acr
 U 


V

T
T  
1  V 
a  
V  T  p
1  V 


V  p T
Isothermal
Compressibility
kT  
Joule-Thomson
Coefficient
  
 T 


p
 H
Chemical Thermodynamics : Georg Duesberg
Value for Perfect Gas
T=0
a =1 / T
kT=1 / p
=0
Info:
Strength/nature of
interactions between
molecules
The higher T, the less
responsive is its volume
to a change in
temperature
The higher the p, the
lower its compressibility
Another indication of
molecular interactions.
Chemical Potential
Chemical Thermodynamics : Georg Duesberg
Chapter1919191919
Chapter 1 : Slide 119:
Slide 19
Chemical potential (m)
• Energy status of molecules in a system (e.g. Benzene in
water)
– Internal energies
• Chemical Bonds, vibrations, flexations, rotations.
– External Energies
• Whole molecule transitions, orientations
• Interactions of molecule with surroundings
• Energy status is a function of:
– Temperature
– Pressure
– Chemical composition
• “Average energy per molecule”
Chemical Thermodynamics : Georg Duesberg
Chemical Potential – definition
• No way to directly measure chemical potential.
• Can only determine differences in , based on the
tendencies of a chemical to move from one situation to
another.
• Need a reference point, like sea level or absolute zero.
• often: select pure liquid chem. as reference
For a pure substance
  Gm
Chapter 1 : Slide 21
Chemical Thermodynamics : Georg Duesberg
Chemical Potential
dG = (μiβ – μiα) dni
In a spontaneous processes, dG < 0 at constant T and P. dni moves
from phase α to phase β to have negative change in free energy.
The spontaneous transfer of a substance takes place from a region with
a higher μi to a lower μi. The process continues to equilibrium where
dG = 0, and μi and μi become equal.
Phase α
Phase β
If two populations of a chemicals (for example, the chemical
coexists in two separate phases): each will have its own value of
1 and 2
1: liquid
2: gas
1
2
Start: liquid benzene (1) and very little vapor benzene (2)
Initial disequilibrium: 1  2 (1 > 2)
Open stopcock. Benzene volatilizes. Net movement of benzene to the right.
1 decreases, 2 increases until they are equal – till there is no net movement
of benzene
Chemical Thermodynamics : Georg Duesberg
Chemical Potential
Partial molar Gibbs energy
Chemical potential for component 1
 G 


 1
 n1 T , p ,n
j
If T and p are kept constant (i.e. dT and dp = 0) then
 G 
 G 


dG  
dn1  
dn2  ...
 n1 T , p ,n j
 n2 T , p ,n j
Or:
dG  1dn1  2 dn2  ...
G  1n1  2 n2  ...
The total free energy is the sum of the contributions from all the
different components present:
G( P, T , ni )  ni i
Chemical Thermodynamics : Georg Duesberg
Chapter 6 : Slide 24
Chemical potential = free energy added to system with each added
increment of i
For a single component system containing 1 mole of substance:
Gm  
 G 
T , P, ni
i  
 ni 
Where  = chemical potential (kJ/mol)
DG = free energy (kJ)
ni = moles of component (i)
i  Gi  H i  T  Si
Chemical Thermodynamics : Georg Duesberg
Properties of the Gibbs energy
G = H - TS
dG = dH –TdS - SdT
H = U + pV
dH = dU +pdV + Vdp
dqrev =TdS and dwrev = - pdV
Both are not state functions but the sum
dG = dU + pdV + Vdp –TdS - SdT dq + dw can be treated to be reversible
because they only depend on the state
functions
dG = TdS – pdV + pdV + Vdp –TdS - SdT
dG = Vdp - SdT
dU = TdS –pdV
G = f ( p, T )
26
Chemical Thermodynamics : Georg Duesberg
Properties of the Gibbs energy
dG = Vdp - SdT
 G 

  S
 T  p
G
Slope = -S
 G 

  V
 p T
G
Slope = V
T (constant p)
S is positive (-S is negative)
so G is decreasing with
increasing T
Chemical Thermodynamics : Georg Duesberg
P (constant T)
V is positive so G is
increasing with
27
increasing p
Dependence of G on p
It would be useful to determine the Gibbs energy at one
pressure knowing its value at a different pressure.
dG = Vdp - SdT
We set dT = 0 (we make sure that the temperature is steady
and integrate:

Gf
Gi
pf
dG   Vdp
pi
pf
DG   Vdp
pi
pf
G( pf )  G( pi )   Vdp
pi
28
Chemical Thermodynamics : Georg Duesberg
Dependence of G on p
pf
G( pf )  G( pi )   Vdp
pi
Liquids and Solids.
Only slight changes of volume with pressure mean that we can effectively
treat V as a constant.
G( pf )  G( pi )  VDp
G( pf )  G( pi )  V ( pf  pi )
Often V Dp is very small and may be neglected i.e. G for solids and liquids
under normal conditions is independent of p. Exception: Geography
29
Chemical Thermodynamics : Georg Duesberg
Dependence of G on p
pf
G( pf )  G( pi )   Vdp
pi
Ideal Gases.
For gases V cannot be considered a
constant with respect to pressure.
For a perfect gas we may use:
dp
G ( pf )  G ( pi )  nRT 
pi p
pf
 G ( pi )  nRT ln
pi
pf
30
Chemical Thermodynamics : Georg Duesberg
Dependence of G on p
Ideal Gases.
pf
G( pf )  G( pi )  nRT ln
pi
We can set pi to equal the standard pressure, p ( = 1 bar).
Then the Gibbs energy at a pressure p is related to its
standard Gibbs energy, G, by:
pf
G ( pf )  G  nRT ln 
p

Chemical Thermodynamics : Georg Duesberg
Fugacity (lat.) = “urge to flee” – same
units as pressure
The fugacity is a parameter which enables us to apply the perfect gas
expression to real gases….
In the thermo dynamical treatment of the Chemical Potential
Chemical Thermodynamics : Georg Duesberg
Dependence of G on p
Real Gases.
For real gases we modify the
expression for a perfect gas and
replace the true pressure by a new
parameter, f, which we call the
fugacity.
Gm ( pf )  Gm

f
 RT ln 
p
The fugacity is a parameter we
have simply invented to enable us
to apply the perfect gas expression
to real gases.
33
Chemical Thermodynamics : Georg Duesberg
Dependence of G on p
Real Gases.
We may then write
f2
DG  RT ln
f1
We may show that the ratio of fugacity to pressure is called the fugacity
coefficient:
f

p
Where  is the fugacity coefficient
 is related to the compression factor Z and the Viral coefficients:
ln   
p
0
Z 1
dp
p
Chemical Thermodynamics : Georg Duesberg
f  pe
B ' p C ' P 2 ..)
Viral coefficients
A real gas is in its standard state when its fugacity is equal to 1 bar and it is
behaving as if it were an ideal gas at some specified temperature (the
diagram shown below is exaggerated to make this point):
fugacity
hypothetical
standard state
ideal gas fi
= Pi
real gas
f = 1 bar
not the standard state
P = 1 bar
In the limit of zero pressure real
gases behave more and more like
ideal gases and the fugacities of
real gases approach their partial
pressures: lim fi --> Pi
Pi --> 0
pressure
The standard state of a pure liquid or solid at a fugacity of 1 bar and some
specified temperature. Since the molar volumes of solids and liquids are
generally small and relatively insensitive to pressure, the activities of solids and
liquids at pressures that are not too far removed from 1 bar remain close to
unity.
Chemical Thermodynamics : Georg Duesberg
Pressure region
Z
f
a

I (very Low)
1
P
0
0
II (moderate)
<1
<P
>0
>0
III (high)
>1
>P
<0
<0
Chemical Thermodynamics : Georg Duesberg
Dependence of G on T
Using the same procedure as for the
dependence of G on p we get:
 G 

  S
 T  p
 dG   SdT
To go any further we
need S as a function of T
?
Instead we start with: G = H - TS
-S = (G – H)/T
37
Chemical Thermodynamics : Georg Duesberg
Dependence of G on T
GH
S 
T
G H
 S
T T
Let G/T = x
H
x  S
T
H
  (G / T ) 

  2
T
 T  p
This is the Gibbs-Helmholtz
Equation
DH
 (DG / T ) 

  2
T
T

p
H
 x 

  2
T
 T  p
38
Chemical Thermodynamics : Georg Duesberg
Dependence of G on T
Two expressions:
 DG 

  DS
 T  p
DH
 (DG / T ) 

  2
T
T

p
Gibbs-Helmholtz Equation
Changes in entropy or, more commonly, changes in enthalpy can be used
to show how changes in the Gibbs energy vary with temperature.
For a spontaneous (DG < 0) exothermic reaction (DH < 0) the change in
Gibbs energy increases with increasing temperature.
Less negative
Slope = -DH/T2 = positive for exothermic reaction
DG/T
Very negative
T (constant p)
39
Chemical Thermodynamics : Georg Duesberg
Chemical Thermodynamics : Georg Duesberg
• Comparing the chemical potential of the real gas to the chemical
potential of an ideal gas at the same pressure
(P)
))
 (  RT ln (P
P
real  id     RT ln f



• The chemical potential of a real gas is
written in terms of its fugacity
J   o  RT ln f J
• The activity coefficient (J) relates the
activity to the concentration terms of interest.
• In gaseous systems, we relate the fugacity (or activity) to the ideal
pressure of the gas via
 J PJ  f J
Chemical Thermodynamics : Georg Duesberg
Change in Internal Energy @ Constant P
• Suppose we want to know how the internal energy, U, changes with temperature at
constant pressure
dU   dV  C dT
 U    ( dV  C dT)
  

 T  
T

T
V
T
V
p
p
 U 
 V 
    T    CV
 T  p
 T  p
– But the change in volume with temperature
 V 
1  V 
a    or    aV
 T 
V  T 
at constant pressure is the related to the
 U 
thermal compressibility of the gas, a
so     a V  C
 T 
p
p
T
V
p
• Large a means sample responds strongly to changes in T
• For an ideal gas, T = 0 so
 U 
 U 
   C V    definition 
 T  p
 T  p
Thermodynamic Consequences of State Functions
Enthalpy, H(p,T)
•
•
Like U, H is a state function
For closed system with constant composition
 H 
 H 
dH    dp    dT
 T  p
 p  T
but
•
•
 H 
   Cp
 T  p
 H 
so dH    dp  C p dT
At constant volume
 p  T
 H   p 
 H 
        Cp
 T  V  p  T  T V
By the chain rule, we can find (dp/dT)V
 y   x    z 
Chain Rule : if z  z(x, y) then        1
 x  z  z  y  y  x
Re call p(V, T)
 p 
1
  
  T  V   T   V 
   
 V  p  p  T

By analogy to a earlier, the isothermal
compressibility, kT,
kT  
1  V 
 
V  p  T
 V 
or    k T V
 p  T
so
 p 
1
  
 T  V  T 
  kT V
 V  p


 y 
by reciprocal identity ,   
 x  z

 T 
  
 V  p
1



1
 x  ,
  
 y  z 
 V 
 V  and we know  T   aV
p
 
 T  p
 T 
1
so   
 V  p a V
 p 
a
This makes   
 T  V k T
 H  a
 H 
and     
 Cp
 T  V  p  T k T
Thermodynamic Consequences of State Functions
Enthalpy, H(p,T)
• To evaluate (dH/dp)T apply chain rule and reciprocal identity again
 H 
1
  
 p  T  T   p 
   
 H  p  T  H
Euler chain rule
 H 
 T 
 H 
        reciprocal relation 2x
  T  p  p  H
 p  T
• But, Cp= (dH/dT)p and if one defines (dT/dP)H as µ, then (dH/dp)T =- Cpµ
• Thus,
 H  a
 H 

 
 Cp
 
 T  V  p  T k T
 H 
a
 Cp
   C p 
 T  V
kT
or

 H 
a 
   C p 1   
 T  V
kT

 U 
 U 
dU  
dT



 dV
 T V
 V T
 U 
 U   U 

  Cv  
 

 T  p
 V T  T  p

1  V 


V  T  P
 U 
 U 

C


V




v
 T  p
 V T
k 
 H 

  (1   JT / k )C p
 T V
45
1  V 


V  P T
A Word about kT
• We know that if we increase the pressure the volume decreases so if
dp is positive, dV is negative
– Since kT(1/V)(dV/dp)T, kT is always positive
• For and ideal gas V=nRT/p so
(dV/dp)T= -nRT/p2 and kT(1/V)(-nRT/p2 )
Simplifying kT,
kT (nRT /V)(1/p2 )
kT (p)(1/p2 )
kT (1/p )
• This means that as the pressure increases, the compressibility
of a gas decreases