1. THERMODYNAMICS thermo – heat dynamics – changes
Laws of Thermodynamics:
0th, 1st, 2nd, 3rd 

2. Note: Thermodynamics ……..
·     is not concerned about rate of changes
(kinetics) but the states before and after
the change
·     does not deal with time

3. Classical Thermodynamics
Marcoscopic observables
T, P, V, …
Statistical Thermodynamics
Microscopic details dipole
moment, molecular size, shape

4. Joule’s experiment T  mgh adiabatic wall (adiabatic process)
T  mgh
adiabatic wall
(adiabatic process)
U (energy change) = W (work) = mgh
Thermometer w h

5. Conclusion: work and heat has the same effect
Page 59 w
T  time interval of
heating
U = q (heat)
Conclusion: work and heat has the same effect
to system (internal energy change) *
FIRST LAW: U = q + W

6. U: internal energy is a state function [ = Kinetic Energy (K.E.)
+ Potential Energy (P.E.) ]
q: energy transfer by temp gradient
W: force  distance
E-potential  charge
surface tension  distance
pressure  volume
First Law:
The internal energy of an isolated system is constant

7. Convention:
Positive: heat flows into system
work done onto system
Negative: heat flows out of system
work done by system
Pressure-volume Work
P1 = P2 V2 V1 M

8. If weight unknown, but only properties of system
are measured, how can we evaluate work? M M V2 P P V1
Assume the process is slow and steady,
Pint = Pext

9. Free Expansion: Free expansion occurs when the external pressure
is zero, i.e. there is no opposing force

10. Quasi equilibrium process:
Reversible change: a change that can be reversed by an infinitesimal modification of a variable.
Quasi equilibrium process:
Pint = Pext + dP (takes a long time to complete)
infinitesimal at any time
quasi equilibrium process

11. what we have consider was isobaric expansion
Page 64-66 P V
Example:
P1 = 200kPa = P2
V1 = 0.04m3 V2 = 0.1m3 *
what we have consider was isobaric expansion
(constant pressure) other types of reversible
expansion of a gas: isothermal, adiabatic

12. Isothermal expansion remove sand slowly at the same time
Page 65-66
Isothermal expansion
remove sand slowly
at the same time
maintain temperature
by heating slowly * P T V1 V2 V
area under curve

13. Ex. V1 = 0.04m3 P1 = 200kPa V2 = 0.1m3

14. Adiabatic Reversible Expansion
For this process
PV = constant for ideal gas
(proved later) V1
P T1
(slightly larger
than 1) V2
P T2

15. |Wa|>|Wb|>|Wc|>|Wd|
Ex. V1 = 0.04m3, P1 = 200kPa, V2 = 0.1m3,
 = 1.3
200kPa
const a d b V P
d: pressure drop without
volume change
|Wa|>|Wb|>|Wc|>|Wd|
isothermal
adibatic

16. State Function Vs Path Function
State function: depends only on position in
the x,y plane
e.g.: height (elevation)
300 1 B 2
200
100m Y X A

17. Path Function: depends on which path is taken to reach destination
from 1  2, difference of 300m (state function)
but path A will require more effort.
Internal energy is a state function,
heat and work are path functions P 3 2
481.13K
192.45K 1
200kPa
0.04
0.1
V/m3

18. 5 moles of monoatomic gas

19. Joule’s second experiment
Energy UU(V,T) ???
thermometer V2 V1
adiabatic wall
At time zero, open valve

20. After time zero, V1 V1+V2 T=0  Q=0 W=0 no Pext  U=0 thermometer
No temperature
change
adiabatic wall
After time zero, V1 V1+V2
T=  Q=0
W=0 no Pext  U=0

21. U=U(T) Energy is only a function of temperature for ideal gas*
0 (for ideal gas)
CV is constant volume heat capacity

22. *Enthalpy Define H = U + PV state function intensive variables
  state function intensive variables
locating the state
Enthalpy is also a state function
H = U + PV + VP

23. H = QP constant pressure heating
At constant pressure
H = U+PV
= U - W
= Q
H = QP constant pressure heating
H is expressed as a functional of T and P
for ideal gases

24. Thermochemistry
Heat transferred at constant volume qV = U
Heat transferred at constant pessure qP = H
Exothermic H = -
Endothermic H = +

25. Standard states, standard conditions
do not measure energies and enthalpies
absolutely but only the differences, U or H
The choice of standard state is purely a
matter of convenience

26. What is the standard state ?
The standard states of a substance at a specified
temperature is its pure form at 1 atmosphere

27. 25oC, 1 atmosphere:
the most stable forms of elements assign
“zero enthalpy”
Ho298 = 0 used for chemical reactions

28. Standard enthalpy of formation
Standard enthalpy change for the formation
of the compound from its elements in their
reference states.
Reference state of an element is its most stable
state at the specified temperature & 1 atmosphere
C (s) + 2H2 (g)  CH4 (g) Hfo = -75 kJ
289K, 1 atm
From the definition, Hfo for elements  0

29. Hess’s Law
The standard enthalpy of an overall reaction is the sum of the
standard enthalpies of the individual reactions into which a
reaction may be divided.
Standard reaction enthalpy is the change in enthalpy when the
reactants in their standard states change to products in their
standard states.

30. Hess’ Law H1 H1 = H2 + H3 + H4 state function
Hess’s law is a simple application of the first law of thermodynamics

31. e.g. C (s) + 2H2 (g)  CH4 (g) H1o = ?
298K, 1 atm
C (s,graphite) + O2 (g)  CO2 (g)
Ho = kJ
H2 (g) + ½O2(g)  H2O (l)
Ho = kJ
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l)
Ho = kJ
H1o = (-285.8) - (-890.4)
= -75 kJ/mole

32. Heat of Reaction (Enthalpy of Reaction)
Enthalpy change in a reaction, which may be obtained from Hfo of products and reactants
Reactants  Products
I stoichiometric coefficient, + products,
- reactants
E.g. CH4 (g) + Cl2 (g)  CH3Cl (g) + HCl (g)

33. Hfo/ kJ
CH3Cl
HCl Cl CH
Hro = ( ) - ( ) = kJ
Reactants Products
elements elements

34. * Page 83 2A + B  3C + D 0 = 3C + D - 2A - B Generally, 0 = J J J
J denotes substances,
J are the stoichiometric numbers *

35. Bond energy (enthalpy)
Assumption – the strength of the bond is
independent of the molecular environment
in which the atom pair may occur.
C (s,graphite) + 2H2 (g)  CH4(g)Ho = kJ
H2 (g)  2H (g) Ho = kJ
C (s,graphite)  C (g) Ho= kJ
 C (s,graphite) + 2H2 (g)  C (g) + 4H (g)
Ho = 2(435.3) = kJ
C (g) + 4H (g)  CH4 (g) H =
= kJ
CH Bond enthalpy = /4 kJ = kJ

36. Temperature dependence of Hr
HrT
CP,P R
Hro
CP,R T

37. H2 = Ho + CP(T2-T1) assume CP,i constant wrt T
What is the enthalpy change for vaporization (enthalpy of vaporization) of water at 0oC?
H2O (l)  H2O (g)
Ho = (-286.1) = kJmol-1
H2 = Ho + CP(T2-T1) assume CP,i
constant wrt T
H (273) = Ho(298) + CP(H2O,g) - CP (H2O,l)( )
= ( )(-25)
= 43.0 kJ/mole

38. Reversible vs Irreversible
Non-spontaneous changes vs Spontaneous
changes
Reversibility vs Spontaneity
First law does not predict direction of changes,
cannot tell which process is spontaneous.
Only U = Q + W

39. Second Law of Thermodynamics
Origin of the driving force of physical and chemical change
The driving force: Entropy
Application of Entropy:
Heat Engines & Refrigerators
Spontaneous Chemical Reactions
Free Energy

40. Second Law of Thermodynamics
No process is possible in which the sole result is the absorption
of heat from a reservoir and its complete conversion into work
Hot Reservoir q w
Engine

41. Direction of Spontaneous Change
More Chaotic !!!

42. Entropy (S) is a measurement of the randomness
Spontaneous change is usually accompanied by a
dispersal of energy into a disorder form, and its
consequence is equivalent to heating
Entropy (S) is a measurement of the randomness
of the system, and is a state function!
S  Q S  1 / T

43. Page 122
Entropy S
For a reversible process, the change of entropy is defined as
(thermodynamic definition of the entropy) *
Another expression of the Second Law:
The entropy of an isolated systems increases in the course of a
spontaneous change:
Stot > 0
where Stot is the total entropy of the isolated system

44. Entropy S
The entropy of an isolated systems increases in the course of a
reversible change:
Stot = 0
where Stot is the total entropy of the isolated system

45. Carnot’s Theoretical Heat Engine
Heat flows from a high temperature reservoir
to a low temperature body. The heat can be
utilized to generate work.
e.g. steam engine.

46. The efficiencies of heat engines
Hot Reservoir q w
Engine
DS = - |q|/Th < not possible! contrary to the second law

47. The Nernst Theorem The entropy change accompanying any physical or
chemical transformation approaches zero as the
temperature approaches zero.
DS  0 as T0

48. Third Law of Thermodynamics If the entropy of each element in its most
state is taken as zero at the absolute zero of temperature, every substance has a positive entropy. But at 0K, the entropy of substance may equals to 0, and does become zero in perfect crystalline solids.
Implication: all perfect materials have the same
entropy (S=0) at absolute zero temperature
Crystalline form: complete ordered,
minimum entropy

49. Statistical Interpretation of S
S = 0 at 0K for perfect crystals
S = k ln 
  Boltzmann number of arrangements
postulate of
entropy Boltzmann constant

50. Entropy Change of Mixing
one distinguish
arrangement
number of
arrangement
increased

51. In general mixing NA, NB
Entropy change of mixing
Stirling’s approximation: ln N!  N ln N + 0(N)
for large N

52. Extensions of 2nd Law
TdS dq Clausius Inequality
For adiabatic process,
Entropy will always attain maximum in
adiabatic processes.
A similar function for other processes?

53. Define Helmholtz free energy
A = U - TS Thermodynamic
State Function
dA = dU - TdS - SdT
Substitute into Clausius Inequality
for isothermal, isochoric (constant volume)
process,

54. change in Helmholtz free energy = maximum isothermal work
Example of isothermal, isochoric process:
combustion in a bomb calorimeter
O2 +
fuel
Temp. bath
O2, CO2,
H2O
Higher P
heat given
out

55. Gibbs Free Energy
Define Gibbs free energy
G = H - TS Thermodynamic
state function
= U + PV -TS
dG = dU +PdV + VdP -TdS -SdT

56. constant pressure, constant temperature
G will tend to a minimum value
equilibrium
spontaneous change
More applications since most processes are
isothermal, isobaric
chemical reactions at constant T, P
Reactants  Products
endothermic H is positive
exothermic H is negative

57. Change of Gibbs free energy with temperature
(constant pressure)

58. In P
1/T

59. Example: What is the change in the boiling
point of water at 100oC per torr
change in atmospheric pressure?
Hvap = 9725 cal mol-1
Vliq = l mol-1
Vvap = l mol-1