Lecture 5: Intro to Entropy
• Reading: Zumdahl 10.1, 10.3
• Outline:
– Why enthalpy is not enough
– Statistical interpretation of entropy
– Boltzmann’s Formula
Enthalpy and Spontaneous Rxns
• Early on in the development of
thermodynamics, it was believed that if a
reaction was exothermic, it was
spontaneous.
• Consider the following reaction:
H2O(s)
H2O(l)
DH°rxn = +6.02 kJ
• Endothermic…..yet spontaneous!
Enthalpy and Spontaneous Rxns
• Consider the following problem: Mixing of
a gas inside a bulb adiabatically (q = 0).
• q = 0, w = 0, DE = 0,
and DH = 0
….but it still happens
Statistical Interpretation of
Entropy
• Imagine that we have a collection of 3
distinguishable particles who have a total
energy of 3e.
• Let’s ask the question, “How will this fixed
amount of energy distribute itself over the
particles?”
Statistics and Entropy (cont.)
• Our system consists of
three distinguishable
particles.
quanta
3
2
1
1
2
3
• There are three “quanta”
of energy (e) available for
a total of energy of “3e”
First Arrangement: All on one
• The first possible arrangement we consider is one
in which all energy resides on one particle
3
3
3
2
2
2
1
1
1
1
2
3
1
2
3
There are three ways to do this
1
2
3
Second Arrangement: 2, 1, 0
• Next arrangement: 2e on 1, 1e on another, and the
third has 0e
3
3
3
2
2
2
1
1
1
1
2
1
3
2
3
3
3
3
2
2
2
1
1
1
1
2
3
1
2
3
Six ways to do this
1
2
1
2
3
3
Third Arrangement
• The final possible arrangement is 1e on each particle.
3
2
1
1
2
3
Only one way to do this.
Which Arrangement?
3
3 ways
2
1
1
2
• Which arrangement is
most probable?
3
3
6 ways
2
1
1
2
3
• Ans: The arrangement
which the greatest
number of possibilities
3
1 way
2
• In this case: “2, 1, 0”
1
1
2
3
The Dominant Configuration
• Configuration: a type of energy distribution.
• Microstate: a specific arrangement of energy
corresponding to a configuration.
• Which configuration will you see? The one with the
largest # of microstates. This is called the dominant
configuration.
Determining Weight
• Weight (W): the number of microstates associated
with a given configuration.
• We need to determine W, without having to write
down all the microstates.
A!
A!
W

W
a0 !a1!...  ai !
i
A = the number of particles in your system.
ai is the number of particles with the same amount of energy.
! = factorial, and P means take the product.
Determining Weight (cont.)
• Consider 300 students
where 3 students have
1e of energy, and the
other 297 have none.
• A = 300
a1 = 3
a0 = 297
A!
W
 ai!
i
300!

3!297!

= 4.5 x 106

Weight and Entropy
• The connection between weight (W) and entropy (S)
is given by Boltzmann’s Formula:
S = k lnW
k = Boltzmann’s constant = R/Na
= 1.38 x 10-23 J/K
• The dominant configuration will have the
largest W; therefore, S is greatest for this
configuration
Example: Crystal of CO
• Consider the depiction of
crystalline CO. There are
two possible arrangements
for each CO molecule.
• Each arrangement of CO
is possible.
• For a mole of CO:
W = Na!/(Na/2!)2
= 2Na
Example: Crystal of CO
• For a mole of CO:
W = Na!/(Na/2!)2
= 2Na
• Then, S = k ln(W)
= k ln (2Na)
= Nak ln(2)
= R ln(2)
= 5.64 J/mol.K
Another Example: Expansion
• What is DS for the
expansion of an ideal gas
from V1 to 2V1?
• Focus on an individual
particle. After expansion,
each particle will have
twice the number of
positions available.
Expansion (cont.)
• Original Weight = W
•
Final Weight = 2W
• Then DS = S2 -S1
= k ln(2W) - kln(W)
= k ln(2W/W)
= k ln(2)
Expansion (cont.)
• Therefore, the DS per
particle = k ln (2)
• For a mole of particles:
DS = k ln (2Na)
= Nak ln(2)
= R ln(2)
= 5.64 J/mol.K
Expansion (cont.)
• Note in the previous example that weight was
directly proportional to volume.
• Generalizing:
DS = k ln (Wfinal) - kln(Winitial)
= k ln(Wfinal/Winitial)
= k ln(Wfinal/Winitial)
= Nk ln(Wfinal/Winitial) for N molec.
= Nkln(Vfinal/Vinitial)