Lecture 6: Introduction to Entropy

1. Lecture 6: Introduction to Entropy • Reading: Zumdahl 10.1, 10.3 • Outline: • Why enthalpy isn’t enough to explain spontaneity • Statistical interpretation of entropy • Boltzmann’s Formula

2. Enthalpy and Spontaneous Rxns • Early 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) DH°rxn = +6.02 kJ This process is endothermic (ice feels cold), …..yet spontaneous (it melts all by itself) !

3. Enthalpy and Spontaneous Rxns • Consider the following problem: Mixing of a gas inside a bulb adiabatically (q = 0). q = 0, w = 0, DE = DH=0 This process is NOT exothermic, but it still happens. Why???

4. Statistical Interpretation of Entropy • Imagine that we have a collection of 3 distinguishable particles who have a total energy of 3e (e is a unit of energy) • Let’s ask the question, “How will this fixed amount of energy distribute itself over the particles?”

5. Our system consists of three distinguishable particles. There are three “quanta” of energy (e) available for a total of energy of “3e”

6. First Arrangement: All on one • The first possible arrangement we consider is one in which all energy resides on one particle There are three ways to do this

7. Second Arrangement: 2, 1, 0 • Next arrangement: 2e on 1, 1e on another, and the third has none. Six ways to do this

8. Third Arrangement • The final possible arrangement is 1e on each particle. Only one way to do this.

9. Which Arrangement? • Q: Which arrangement is most probable? • Ans: The arrangement which the greatest number of possibilities • In this case: “2, 1, 0”

10. 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?)

11. 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. W 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.

12. 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 = 4.5 x 106

13. 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

14. Young Ludvig Boltzmann

15. A devoted father and husband. His wife called him “My sweet, fat darling”

16. Troubled by severe bouts of depression, and criticism of his scientific ideas, • Boltzmann at 58

17. Troubled by severe bouts of depression, and criticism of his scientific ideas,… Boltzmann took his own life while on a family vacation in Switzerland. • Boltzmann at 58

19. 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

20. 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 Recall: ln (Ab) = bln(A)

21. Another Example: Expansion of a volume • 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.

22. Original Weight = W Final Weight = 2W So, DS = S2 - S1 = kln(2W) - kln(W) = kln(2W/W) = kln(2) Recall: lnA - lnB = ln(A/B)

23. 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/moleK

24. Note in the previous example that weight(W was directly proportional to volume (V): Generalizing: DS = k ln (Wfinal) - kln(Winitial) = k ln(Wfinal/Winitial) = k ln(Wfinal/Winitial) = Nk ln(Wfinal/Winitial) for N molecules Finally, DS = Nk ln(Vfinal/Vinitial)