Transcript Document

The Distribution of Energy
Consider 6 particles – Each particle has three energy levels
E3 = 2a
a
a
E
E2 = a
E1 = 0
Let N = Total number of particles = 6
N1 = Number of particles in state E1
N = N1 + N2 + N3 = 6
N2 = Number of particles in state E2
N2 = Number of particles in state E2
Total energy of the system = 4a
N1E1 + N2E2 + N3E3 = 4a
The Boltzmann Distribution – Effect of Temperature
1.20
300000K
1.00
Pj = Nj/N
0.80
3000K
0.60
30000K
30K
0.40
300K
0.20
0.00
0
500
1000
1500
2000
2500
Energy
3000
3500
4000
4500
5000
Pj  N ,V ,T  
aj
A

e
 E j  N ,V  / k BT
Q N ,V ,T 
The system partition function is
QN ,V ,     e  E  N ,V  
i
qv ,  N
i
N!
for indistinguishable particles.
For the two-level system:
N2
 eE
N1
2
 E1  / k B T
N1  N
1
1 e
 E / k BT
N2  N
1
1 e
 E / k BT
The Distribution of Molecules on Two Energy Levels
Box at rest
Box weakly shaken
Box vigorously shaken
A Two-Level System
N2/N1 approaches 1 as T  
1.00
0.90
0.80
N2/N1 = exp[-E/kBT]
0.70
N2/N1
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0.00
20000.00
40000.00
60000.00
Tempertaure (K)
80000.00
100000.00
120000.00
A Two-Level System
1.00
0.90
N1/N = 1/(1+exp[-E/kBT])
0.80
0.70
Ni/N
0.60
As T approaches infinity,
N1 and N2
approaches 50%
0.50
0.40
0.30
N2/N = 1/(1+exp[E/kBT])
0.20
0.10
0.00
0
10000
20000
30000
40000
50000
60000
Temperature (K)
70000
80000
90000
100000
Energy can be evaluated form the partition function:
For the system:
  ln Q 
  ln Q 

E  
 k BT 2 




T

 N ,V

 N ,V
For the atom/molecule:

  
 k BT 2 




T

 N ,V

 N ,V
  ln q 
  ln q 
E  N  , and for 1 mol: E  N A   U
  ln q 
and so we see that U  N A k B T 2 

 T V
Heat capacity and pressure can also be evaluated from the partition function:
 U 
Cv  
and


T

 N ,V
  ln Q 
P  k BT 


V

 N ,
You should be able to calculate any of these properties for any partition function
I give you. Here follows some results for a diatomic molecule. You should
understand (and derive completely) how these results were obtained.
3
The translational partition function (3D):
 2Mk BT  2
qtrans V ,T   
 V
2
 h

The rotational partition function (large T):
8 2 Ik BT
qrot T  
h2
The vibrational partition function:
e  / 2T
qvib T  
1  e  / T
vib
vib
Translational
U
3
RT
2
CV
3
R
2
Rotational
RT
R
Vibrational



R vib   /vib

T
e
 1
 2
vib
e 
  vib 
R

 T  1  e 
2
vib

/T
vib

/T 2
Total internal energy is the sum of the individual internal energies and total heat
capacity is the sum of the individual heat capacities. Note, we have ignored the
electronic contribution.
Note q V ,T   qtrans qrot qvib qelec and
 total   trans   rot   vib   elec
The First Law: Some Terminology
System: Well defined part of the universe
Surrounding: Universe outside the boundary of the system
Heat (q) may flow between system and surroundings
Closed system: No exchange of matter with surroundings
Isolated System: No exchange of q, w, or matter with surroundings
Isothermal process: Temperature of the system stays the same
Adiabatic: No heat (q) exchanged between system and surroundings
THE CONCEPT OF REVERSABILTY
Irreversible processes:
Hot
Warm
Cold
Warm
Temperature equilibration
Mixing of two gases
P=0
P=0
Expansion into a vacuum
Evaporation into a vacuum
THE CONCEPT OF REVERSABILTY
Pext = 1 atm
Pext = 1 atm
REMOVE PINS
(1)
pins
Irreversible Expansion
P = 1 atm
P = 2 atm
Step 2
Step 1
Infinite number of steps
(2)
Reversible Expansion
P = 1 atm
P = 2 atm
P = 1.999 atm
P = 1.998 atm
THE CONCEPT OF REVERSABILTY
Reversible processes:
Tiny weight
Po
Po
Po + P
Condensation
Evaporation
(pressure minimally
increases by adding
tiny weight)
(pressure minimally
decreases by removing
tiny weight)
IRREVERSIBLE EXPANSION
Pext
Pext
V2
V1
P
P, V1
P, V2
Pext
w = -Pext(V2 – V1)
V1
V
V2
REVERSIBLE EXPANSION
Pext = Pressure of gas. If the gas is ideal, then Pext = nRT/V
How does the pressure of an ideal gas vary with volume?
This is the reversible path. The pressure
at each point along curve is equal to the
external pressure.
P
V
V2
V2
V1
V1
w   Pext dV   PdV  
V2 1
V
nRT
dV  nRT
dV  nRT ln 2
V1 V
V
V1
REVERSIBLE EXPANSION
A
Pi
IRREVERSIBLE EXPANSION
The reversible path
P
A
Pi
Pext = Pf
P
B
B
Pf
Pf
Vi
Vf
The shaded area is
w  nRT ln
Vf
Vi
V
Vi
The shaded area is
w  Pf Vi V f 
Reversible expansion gives the maximum work
Vf
V
REVERSIBLE COMPRESSION IRREVERSIBLE COMPRESSION
Pext = Pf
B
B
Pf
The reversible path
P
Pf
P
A
Pi
A
Pi
Vf
Vi
The shaded area is
w  nRT ln
Vf
Vi
V
Vf
Vi
The shaded area is
w  Pf Vi V f 
Reversible compression gives the minimum work
V
The Carnot Engine
Heat engine
Hot reservoir
Th
Cold reservoir
Tc
qh
qc
w
q
Isothermal expansion
P1
V1
T
P2
V2
T
How can we take it back?
w = -nRT ln(V2/V1)
The Carnot Engine- what is the efficiency?
Isothermal reversible expansion U = 0, w = -nRTh ln(V3/V1)
q = +nRTh ln(V3/V1)
A
Pressure
P1
(Th)
(Th) B
P2
P3
Adiabatic reversible expansion
(Tc)
dU = CpdT = δw
q=0
total work
D
C (T )
c
P4
V1
V2
V3
V4
Adiabatic reversible compression
dU = CpdT = δw
q=0
Isothermal reversible compression U = 0, w = -nRTc ln(V2/V4)
q = +nRTc ln(V2/V4)
Statistical Interpretation of Entropy
Consider the following expansion:
When stopcock is opened:
V/2
V/2
At any given moment
Probability that a given molecule is in the LHS bulb = 50 % (or ½)
Probability that a given molecule is in the RHS bulb = 50 % (or ½)
Probability that two molecules stay in LHS bulb = ½  ½ = ¼ (25%)
Probability that N molecules stay in LHS bulb = (½)N
Microstates: A particular way of arranging molecules among the
positions accessible to them while keeping the total energy fixed.
Consider four molecules in two compartments:
Total number of microstates = 16
1 way
4 ways
6 ways
4 ways
1 way
The most probable
(the “even split”)
If N   the “even split” becomes overwhelmingly probable
Boltzmann
S = kB lnW
Consider spin (or dipole restricted to two orientations)
1 particle
2 particles
3 particles
1 mole
or
,
W = 2, and S = kB ln2
,
,
W = 4, and S = kB ln4
W = 8, and S = kB ln8
W =2NA, and S = kB ln(2NA) = NA kB ln2 = Rln
Some entropy problems
The spectroscopic entropy of a diatomic molecule is given by the following equation.
 2Mk T  3 2 V e5 / 2 
 /T
S
Te
B
  ln
 ln 
 ln 1  e vib / T  vibvib/ T
 ln g e1

2
R
NA 
 rot
e
1

 h



What is meant by residual entropy?
(a)Calculate the spectroscopic entropy of CO at 298.15 K and 1 atm.
(b)Write down the expression for calculating the (calorimetric)
absolute molar entropy from 0 K to 298.15 K. Note that CO undergoes
a solid-solid phase transition at 61.6 K. The experimentally determined calorimetric value is ~170 JK -1mol-1..
(c) Calculate the residual entropy of CO. Make the correction for residual
to the calorimetric value. Compare the two entropies (spectroscopic and calorimetric).
The molar heat capacity of C2H4(g) can be expressed by the following equation over the temperature range 300 K < T < 1000 K.
CV T  / R  16.4105 
6085.929K 822826K 2

T
T2
Calculate S if one mole of ethene is heated from 300 K to 600 K at constant volume. There are no phase transitions.
Unit 3 – Real & Ideal Gases
Non-ideality of gases - Fugacity
 G

 P

  V
T
 G T , P   G o T  RT ln
P
P
o
The thermodynamic function fugacity.
Note: f (P,T)  P as P  0
1 bar
Standard molar Gibbs energy
Can generalize for a real gas
f P, T 
f
o

P
P
o

PV
 1  B2 P T P  B3 P T P 2  ....
RT

G T , P   G o T  RT ln
f P, T 
fo

exp B2 P T P  B3 P T P 2  ......
Gibbs energy must be taken relative to some standard state
Standard state of real gas is taken to be the corresponding ideal gas at 1 bar
i.e. must “adjust” the real gas to ideal behavior
Real gas (T,P)
G2
G1
Ideal gas (T,P)
G3
Real gas (T, P  0) = Ideal gas (T, P  0)
G1 
 RT

P
f
  '  V dP '  RT ln o  RT ln o

P
f
P 0  P
P
Fugacity coefficient
g f
Ideal gas, g = 1. Fugacity coefficient measures extent of non-ideality
P
(Z-1) / P
ln g
0
P
P Z 1

dP '
'
0 P
P (bar)
Real
g
1
ideal
0
P (bar)
Unit 2 Summary
You need to know the Boltzmann equation in terms of probability and fractional
occupancy in a two level system. In general, the Boltzmann equation gives the
probability, Pj, that a randomly chosen system (from an ensemble of systems in
thermal contact) will be in state j with energy Ej(N,V):
Pj  N ,V ,T  
aj
A

e
 E j  N ,V  / k BT
Q N ,V ,T 
The system partition function is Q  N ,V ,     e
 E i  N ,V 

i
qv ,  N
N!
for
indistinguishable particles.
For the two-level system:
N2
 eE
N1
2
 E1  / k B T
,
N1  N
1
1 e
 E / k B T
,
N2  N
1
1 e
 E / k BT
You should be able to obtain partition functions for simple energy level systems.
The Boltzmann equation seeks to find the maximum number of configurations.
For a system with large N, there is a configuration with so great a weight that is
overwhelms the rest. The system will almost always be found in it, and it will
determine the properties of the system. The Boltzmann equation gives the
probability of realizing this configuration as a function of energy and
temperature.
Energy can be evaluated form the partition function:
For the system:
  ln Q 
  ln Q 

E  
 k BT 2 

 T  N ,V
   N ,V
For the atom/molecule:
  ln q 
  ln q 

  
 k BT 2 



 T  N ,V

 N ,V
E  N  , and for 1 mol: E  N A   U
  ln q 
and so we see that U  N A k BT 2 

 T V
Heat capacity and pressure can also be evaluated from the partition function:
 U 
Cv  
and

 T  N ,V
  ln Q 
P  k BT 

 V  N ,
You should be able to calculate any of these properties for any partition function
I give you. Here follows some results for a diatomic molecule. You should
understand (and derive completely) how these results were obtained.
3
The translational partition function (3D):
 2Mk BT  2
qtransV ,T   
 V
 h2 
The rotational partition function (large T):
qrot T  
8 2 Ik BT
h2
The vibrational partition function:
qvib T  
e  / 2T
1  e / T
vib
vib
The vibrational partition function depends on
how we define the zero of energy. If we define it
as the bottom of the internuclear potential, we
obtain the above expression. However if we
define zero as the v = 0 state, then we obtain the
following expression:
e2
e1 = -De
v=2
v=1
E = 0 for v = 0
E=0
1/2h
qvib T  
1
1 e
vib / T
The expression for the rotational partition function above works for high
temperatures. Room temperature measurements have to calculated numerically
using the following equation.
qrot T   1  3e 2 B / k BT  5e 6 B / k BT  
From these partition functions the following can be obtained.
Translational
U
3
RT
2
Rotational
RT
Vibrational



R vib   /vib

T
e
 1
 2
vib
CV
3
R
2
R
e 
 
R vib 
 T  1  e 
2
vib

/T
vib

/T 2
Total internal energy is the sum of the individual internal energies and total heat
capacity is the sum of the individual heat capacities. Note, we have ignored the
electronic contribution.
Note q V ,T   qtrans qrot qvib qelec and
 total   trans   rot   vib   elec
WORK, ENERGY & HEAT
Work is the transfer of energy due to unbalanced forces:
w   Pext V
If Pext is not constant during the process:
w  
Vf
Vi
For a reversible process (compression or expansion of an ideal gas):
V
w  nRT ln 2
V1
Pext dV
The first law of thermodynamics:
dU  q  w
Path & state functions: know the
U  q  w
difference.
Some important processes:
ISOLATED SYSTEM:
q  0 , w  0 , U  0
ISOTHERMAL (no change in T):
U  0
T  0
V
wrev  q rev  nRT ln 2
V1
q 0
ADIABATIC (no change in q):
U  wrev   CV T dT
T2
T1
U  qV (constant volume)
PRESSURE-VOLUME WORK:
H  U  PV
constant pressure
H  U  PV
q P  H
The temperature of a gas decreases in a reversible adiabatic expansion:
}
3
 T2  2 V1
  
(Know the derivation)
V2
 T1 
ENTHAPLY
At constant pressure: H  U  PV , and q P  H
To calculate the enthalpy from T = 0K to T = T, must take into account any phase
transitions.
H T   H 0  
T fus
T
 CP T dT   fus H   CP T dT , etc. In other words H   CP dT
S
0
L
T fus
Note also that C P 
H
, and C P  CV  nR
T
Make sure you know how to calculate enthalpy changes for chemical reactions.
You can use Hess’s law and enthalpies of formation to calculate enthalpies of
reaction. Know how to calculate enthalpy changes at different temperatures
using the following equation:
H T2   r H T1   
T2
T1
CP  products   CP react' s 
ENTROPY
Entropy (state function): dS 
q rev
T
dS  0 Equilibrium (reversible process)
dS  0 Spontaneous process in isolated system
Know the Second Law.
Isothermal expansion:
S  R ln
V2
, For mixing of two gases:
V1
N
mix S   R  yi ln yi (Learn the derivation)
j 1
The Third Law: Every substance has finite positive entropy, but at 0 K the
entropy may become 0, and does so in the case of a perfectly crystalline
substance.
Determining absolute entropies from calorimetric data:
S
T fus C P
T dT   fus H  Tvap C PL T dT  vap H  T C Pg T dT
0
T fus
Tvap
T
T fus
T
Tvap
T
S T   


These entropies are usually called “calorimetric entropies.” Note we have
assumed the Third Law holds true for calorimetric entropies.
Statistical Entropy:
S  k B ln W
From partition functions:
  ln Q 
 qV ,T  
  ln q 
S  k B ln Q  k BT 
 Nk B  Nk B ln 
 Nk BT 


 T  N ,V
 N 
 T V
Note that the above expression relates entropy to the system partition function
(Q) and the molecular partition function (q). Make sure you know how to go
from the expression for system Q to molecular q, using Stirling’s approximation
(lnN! = NlnN – N).
The molar entropy of an ideal monoatomic gas: S 
3


 2mk B T  2 V g e1 
5

R  R ln 
 h 2 
2
NA 


The molar entropy of a diatomic gas is given by:
3


 2Mk BT  2 V e 5 / 2 
 /T
S
Te

 ln 
 ln
 ln 1  e  vib / T  vib
 ln g e1

2



R
NA
2 rot
e vib / T  1

 h

You should be able to derive entropy expressions using simple partition
functions. You should be able to determine entropies of monoatomic and
diatomic molecules from spectroscopic parameters. The entropies are generally
called “statistical entropies.”


We have assumed the Third Law holds perfectly when calculating calorimetric
entropies, i.e. S = 0 at T = 0 K. Spectroscopic entropies are therefore more
accurate. Realize that at 0 K some molecules (usually those possessing small
dipoles) may get “locked” into two or more possible degenerate states, e.g. CO
and linear NNO, W = 2; CH3D, W = 4. You should be able to calculate residual
entropies, and add the values to the calorimetric values to get entropies in closer
agreement to statistical ones.
Helmholtz Energy:
A   U  T  S  0
A  0
A  0
A  0
For constant T and V
Spontaneous (irreversible)
Not spontaneous
Equilibrium
Gibbs Energy:
G  H  TS
G  0
G  0
G  0
For constant T and P
Spontaneous (irreversible)
Not spontaneous
Equilibrium
Maxwell Relations
Maxwell relations allow us to derive some important thermodynamic equations. You
need to be able to derive all of them. Below I have reproduced Table 22.1 from the
textbook. Please learn the differential expressions and the derivations for the
corresponding Maxwell relations.
Thermodynamic Energy
Differential expression
U
dU = TdS – PdV
H
dH = TdS + VdP
A
dA = -SdT – PdV
G
dG = -SdT + VdP
Corresponding Maxwell
relations
 T 
 P 

   
 V  S
 S V
 T 
 V 

 


P

 S  S  P
 S 
 P 

 


V

T  T V
 S 
 V 
   


P
 T
 T  P
Two important ones are the temperature and the pressure dependency of the Gibbs
energy.
 G 

   S This gives:
 T  P
proof)
 G 

 V
 P T
This gives:
H
 G / T 

   2 Gibbs-Helmholtz equation (learn the

T

P
T
G  RT ln
P2
(ideal gas)
P1
P
 S 
 V 
 
 , which gives S  nR ln 2 (ideal
P1
 P T  T  P
Other very important one are:  
gas)
 A 
 A 

   P and     S so that using cross derivatives, we find that
 V T
 T V
 P 
 S 
V
  
 . We end up with S  nR ln 2 for an ideal gas.
V1

T

V
 V 
T
Beware, I may ask to solve the equations using other equations of state.
You need to know how internal energy varies with volume.
 U 
 P 

  P  T 
 (learn the proof)
 V T
 T V
You need to know how enthalpy varies with pressure.
 H 
 V 

 V T
 (learn the proof)
 P T
 T  P
For the above two, you may be asked to solve the partial derivative using an equation of
state other than the ideal gas equation (see your homework for examples).
The Non-Ideality of Gases
At P1 = 1 bar G T , P   G o T   RT ln P , where G o T  = standard molar Gibbs energy.
For non-ideal gases G T , P   G o T   RT ln
Using the virial equation for non-ideality:
f P ,T 
fo
, where f(P,T) = fugacity of gas.


f P ,T  P
B3 P T P 2



exp
B
T
P

 .....

2P
o
o
2
f
P


P
Z 1
f
PV
Note, the fugacity coefficient g  , and ln g  
dP , where Z 
P
P
RT
0
(Z = compressibility factor)
 U 
 P 
4. The relationship is 
  P  T 

 V T
 T V
R
 P 
For a van der Waals gas 
, and so
 
 T V V  b
Integrating
U
u
a
 U 

  2
 V T V
1
dV which leads to
Videal V 2
1
a
   U ideal    U ideal
V
V
V
dU  a 
ideal
 1
U T ,V   a
 Videal
For ethane a = 5.5818 dm6 bar mol-2, and
Uideal = 14.55 kJ mol-1
14.53
Molar Energy / kJ/mol
Using the van der Waals equation
RT
a
P
 2 , we can find P from
V b V
various molar volumes. We can then
determine U as a function of P and plot the
two. I have used the molar volume range
suggested in problem 22-4.
14.54
14.52
14.51
14.5
14.49
14.48
70
270
470
P /bar
R
RT
 P 
 U 
(a) For an ideal gas 
and so 
0
 
  P 
V
 T V V
 V T
a
 U 
(b) For a van der Waals gas, see above. 
  2
 V T V
RT
A
(c) For Redlich-Kwong gas, where P 
 1/ 2
V  B T V V  B 
R
A
 P 
 3/ 2
  
 T V V  B 2T V V  B
670
R
A
RT
A
RT
AT
 U 
 3/ 2

 1/ 2

 3/ 2

  P 
V  B 2T V V  B 
V  B T V V  B  V  B 2T V V  B 
 V T
A
AT
A
A
 1/ 2
 3/ 2
 1/ 2
 1/ 2
T V V  B  2T V V  B  T V V  B  2T V V  B 
2A
A
2A  A
3A
 1/ 2
 1/ 2
 1/ 2
 1/ 2
2T V V  B  2T V V  B  2T V V  B  2T V V  B 
(d) For the virial equation of state Z  1  B2 P T V 1  B3P T V 2 , up to the third virial
coefficient. Solving for pressure gives:
RT
RT
RT
 B2 P T  2  B3 P T  3
V
V
V
R
R dB RT
R dB RT
 P 

   B2 P T  2  2 P 2  B3P T  3  3P 3
dT V
dT V
V
V
 T V V
P
R dB RT
R dB RT 
R
 U 

   P  T   B2 P T  2  2 P 2  B3P T  3  3P 3 
dT V
dT V 
V
V
 V T
V
RT
RT
RT RT
RT dB2 P RT 2
RT dB3P RT 2

 B2 P T  2  B3P T  3 
 B2 P T  2 
 B3P T  3 
V
V
dT V 2
dT V 3
V
V
V
V
dB2 P RT 2 dB3P RT 2

dT V 2
dT V 3
RT 2 dB2 P RT 2 dB3P RT 2  dB2 P 1 dB3P 
 U 
And so 
 3
 2 


 
dT
V dT 
V
V  dT
 V T V 2 dT

5. We can determine how enthalpy varies with P, since H = U + PV. You simply add the
corresponding PV values to the U values determined In the last problem. Remember to convert
the PV units into joules. Basically 1 L = 1000 cm3 = 10-3 m3, and 1 bar = 105 Pa. Therefore the
conversion factor is 102. But we need kJ not J, and so divide further by 1000. The final
conversion factor is 0.1.
 H 
 V 
We need 
 V T
 for the remaining parts of the problem.
 P T
 T  P
 H
(a) For an ideal gas 
 P

 V 
RT
R
T   0
  V  T 
 
P
P
T
 T  P
(b) For the virial equation of state Z  1  B2 P T P  B3P T P 2 , up to the third virial coefficient.
Solving for volume gives:
V 
RT
RT
 B2 P T RT  B3P T RTP 
 B2 P T RT  B3 P T RTP
P
P
 V

 T

dB
R
dB
   B2 P T R  2 P RT  B3P T RP  3P RTP
dT
dT
P P
 H 
 V 
RT
 B2 P T RT  B3P T RTP

 V T
 
P
 P T
 T  P
dB
dB
R

 T   B2 P T R  2 P RT  B3P T RP  3P RTP 
dT
dT
P

dB
 H 
dB

   2 P RT 2  3P RPT 2
dT
dT
 P T
3.