Lecture 6: Intro to Entropy • Reading: Zumdahl 10.1, 10.3 • Outline: – Why enthalpy isn’t enough. – Statistical interpretation of entropy – Boltzmann’s Formula
Lecture 6: Intro to Entropy
• Reading: Zumdahl 10.1, 10.3
• Outline: – Why enthalpy isn’t 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. But this can’t be the whole story---consider an ice cube on a warm day.
• Consider the following reaction:H2O(s) H2O(l) H°rxn = +6.02 kJ
• Endothermic (ice feels cold)…..yet spontaneous (and it melts) !
Enthalpy and Spontaneous Rxns
• Consider the following problem: Mixing of a gas inside a bulb adiabatically (q = 0).
• q = 0, w = 0, E = 0, and H = 0 This process is NOT exothermic, but it still happens.
Statistical Interpretation of
Entropy• Imagine that we have a collection of 3 distinguishable particles who have a total energy of 3.
• 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.
• There are three “quanta” of energy () available for a total of energy of “3”
1
2
3
1 2 3
quanta
First Arrangement: All on one
• The first possible arrangement we consider is one in which all energy resides on one particle
1
2
3
1 2 3
1
2
3
1 2 3
1
2
3
1 2 3
There are three ways to do this
Second Arrangement: 2, 1, 0• Next arrangement: 2 on 1, 1 on
another, and the third has 0
1
2
3
1 2 3
1
2
3
1 2 3
1
2
3
1 2 3
1
2
3
1 2 3
1
2
3
1 2 3
1
2
3
1 2 3
Six ways to do this
Third Arrangement• The final possible arrangement is 1 on each particle.
1
2
3
1 2 3
Only one way to do this.
Which Arrangement?
• Which arrangement is most probable?
• Ans: The arrangement which the greatest number of possibilities
• In this case: “2, 1, 0”
1
2
3
1 2 3
1 way
1
2
3
1 2 3
6 ways
1
2
3
1 2 3
3 ways
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 (Why does a rope tangle?)
Determining Weight• Weight (): the number of microstates associated with a given configuration.
• We need to determine , without having to write down all the microstates.
W =A!
a0!( ) a1!( )...=
A!ai!
i∏
A = the number of particles in your system.ai is the number of particles with the same amount of energy.! = factorial, and means take the product.
Determining Weight (cont.)
• Consider 300 students where 3 students have 1 of energy, and the other 297 have none.
• A = 300 a1 = 3
a0 = 297€
=A!
ai!i
∏
€
=300!
3!( ) 297!( )
= 4.5 x 106
Weight and Entropy• The connection between weight () and entropy (S) is given by Boltzmann’s Formula:
S = k(lnk = Boltzmann’s constant = R/Na
= 1.38 x 10-23 J/K
• The dominant configuration will have the largest ; therefore, S is greatest for this configuration
Young Ludvig Boltzmann
A devoted fatherAnd husband.
His wife called him“My sweet, fat darling”
• Boltzmann at 58
Troubled by severe bouts of depression, and criticism of his scientific ideas
Boltzmann took his own life while on a family vacation inSwitzerland.
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: = Na!/(Na/2!)2
= 2Na
Example: Crystal of CO
• For a mole of CO: = Na!/(Na/2!)2
= 2Na
• Then, S = k ln() = k ln (2Na)
= Nak ln(2)
= R ln(2) = 5.64 J/mol.K
Another Example: Expansion
• What is S 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 =
• Final Weight =
• Then S = S2 -S1
= k ln(2) - kln()
= k ln(2/)
= k ln(2)
Expansion (cont.)
• Therefore, the S per particle = k ln (2)
• For a mole of particles:S = 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:S = k ln (final) - kln(initial)
= k ln(final/initial)
= k ln(final/initial) = Nk ln(final/initial) for N molec.
= Nkln(Vfinal/Vinitial)