Transcript
Page 1: 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

Page 2: 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) !

Page 3: 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

• 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.

Page 4: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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?”

Page 5: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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”

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2

3

1 2 3

quanta

Page 6: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

First Arrangement: All on one

• The first possible arrangement we consider is one in which all energy resides on one particle

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2

3

1 2 3

1

2

3

1 2 3

1

2

3

1 2 3

There are three ways to do this

Page 7: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

Second Arrangement: 2, 1, 0• Next arrangement: 2 on 1, 1 on

another, and the third has 0

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3

1 2 3

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1 2 3

1

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1 2 3

1

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1 2 3

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1 2 3

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1 2 3

Six ways to do this

Page 8: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

Third Arrangement• The final possible arrangement is 1 on each particle.

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2

3

1 2 3

Only one way to do this.

Page 9: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

Which Arrangement?

• Which arrangement is most probable?

• Ans: The arrangement which the greatest number of possibilities

• In this case: “2, 1, 0”

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1 2 3

1 way

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1 2 3

6 ways

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1 2 3

3 ways

Page 10: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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?)

Page 11: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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.

Page 12: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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

Page 13: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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

Page 14: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

Young Ludvig Boltzmann

Page 15: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

A devoted fatherAnd husband.

His wife called him“My sweet, fat darling”

Page 16: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

• Boltzmann at 58

Troubled by severe bouts of depression, and criticism of his scientific ideas

Page 17: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

Boltzmann took his own life while on a family vacation inSwitzerland.

Page 18: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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

Page 19: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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

Page 20: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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.

Page 21: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

Expansion (cont.)

• Original Weight =

• Final Weight =

• Then S = S2 -S1

= k ln(2) - kln()

= k ln(2/)

= k ln(2)

Page 22: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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

Page 23: Lecture 6: Intro to Entropy Reading: Zumdahl 10.1, 10.3 Outline: –Why enthalpy isn’t enough. –Statistical interpretation of entropy –Boltzmann’s Formula

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)


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