1. An Introduction to Statistical Thermodynamics

2. ( ) Gas molecules typically collide with a wall or other molecules about once every ns. Each molecule has a different speed, s i. For N molecules, the root-mean- square (rms) speed, c, is: If n moles of gas, with a molar mass of M, are in a volume of V, then the pressure is: Kinetic Model of Gas Total mass

3. As PV=nRT for an ideal gas, we can solve for c as a function of T: Kinetic Model of Gas There is a proportionality between the rms speed of molecules and the square root of temperature. Note: There is a distribution of speeds, which is described by the Boltzmann distribution.

4. Boltzmann Distribution of Speeds in a Gas Dashed lines show the rms speed, c

5. Statistical Approach Statistical thermodynamics uses a mathematical description of the distribution of particle positions and speeds to calculate the state variables for a system and to obtain values of the state functions, such as U and S. The position of each molecule is given in “phase space” by the co-ordinates of x, y and z. The momentum (a vector that is the product of mass and velocity) of each molecule is represented in “phase space” by three vectors: p x, p y and p z. Neither position nor momentum can both be known with certainty but are only known within a range in phase space.

6. Position Phase Space y x z ViVi Defines occupied volume

7. Momentum Phase Space pypy pzpz pxpx Momentum

8. Statistical Approach The positions of molecules in phase space can be used to determine the bulk state variables for the system. Obvious correlation between positions and volume, V. Pressure, P can be related to the rms speed, c, which is related to momentum of all the molecules in the system. Temperature, T is related to the square of the rms speed, which is likewise related to the momentum. Several different arrangements of molecules in phase space can correspond to the same bulk state (P, V, T). The number of arrangements of molecules for a particular state of the system is called the thermodynamic probability, .

9. Analogy for Thermodynamic Probability “State” of the system = sum of dice = 7  = 6 (if distinguishable)  = 3 (if indistinguishable) + + + + + +

10. Analogy for Thermodynamic Probability “State” of the system = sum of dice = 2 +  = 1 Conclude that a “state” of 7 is more likely to be found as it has a higher probability, . In a “real” system, each state (P i, V i, T i ) will have a certain  i. As one mole contains ~10 23 molecules,  tends to be a very large number! The equilibrium state of a system is most likely and hence must have the largest . (indistinguishable)

11. Relationship between  and S In an isolated system, or thermodynamic Universe, molecules will spontaneously move from a low  state to a high  state. Thus the  for the system approaches a maximum. Previously we saw that in a thermally-isolated system (or t.d. Universe), S approaches a maximum. CONCLUDE that there must be some proportionality between  and S.

12. U1U1 Ω1(U1)Ω1(U1) U2U2 Ω2(U2)Ω2(U2) Two systems with differing internal energies (U 1 and U 2 ) and differing Ω are placed into contact. Together they define a thermodynamic universe. We know that the two systems will reach thermodynamic equilibrium over time. Total energy for the t.d. universe: U = U 1 + U 2 Ω for the universe is Ω = Ω 1 Ω 2 The equilibrium macroscopic state will correspond to the maximum Ω (i.e. most likely state). What will be the values for Ω 1 and Ω 2 at equilibrium?

13. U2U2 U1U1 Ω1Ω2Ω1Ω2 Find the maximum Ω by setting the derivative = 0. But, U 1 + U 2 is fixed, so that dU 1 = -dU 2 Thus,Dividing by Ω 1 Ω 2 : U 1 +U 2 = const.

14. We note that If then d (ln Ω 1 ) = d (ln Ω 2 ) Substituting, we find the condition for thermal equilibrium that is satisfied when Ω is a maximum: We say that System 1 and 2 are at the “same temperature” (T 1 = T 2 ) and have the same thermal energy, kT. Comparing to the condition above, we write that: k is the Boltzmann constant.

15. Classical Definition of Temperature and the Boltzmann Equation Recall the differential form of the First Law of Thermodynamics: dU = dQ + dW = TdS - PdV We recall that such that Comparing the two definitions for temperature, we see that: S = k lnΩ Now find the partial derivative w.r.t. S while holding V constant: Giving us a new definition of T

16. Ludwig Boltzmann (1844-1906)

17. W = thermodynamic probability, 

18. Relates the microscopic state of a system (molecular level) to its macroscopic state. Voted one of the ten most beautiful equations in a survey by Physics World! Significance: An increase in the entropy of a system corresponds to it going from a less probable state to a more probable state. Boltzmann Equation

19. Calculating the Thermodynamic Probability Imagine there are N=10 positions in phase space, and you are putting j molecules into these positions. If the molecules are indistinguishable, there is only one way to arrange 10 molecules, so  = 1! Here we are assuming single occupancy of sites. In position/momentum space, multiple occupancy is allowed!

20. How many ways are there to arrange four indistinguishable molecules on ten sites? If the particles were distinguishable, there would be 10x9x8x7 ways. That is, the first particle would have 10 sites to choose from; the second particle would have 9 sites, etc. Generalising for j particles on N sites: If the particles are indistinguishable, the number or arrangements are reduced by j!

21. Problem Revisited: What is  S when the volume of one mole of an ideal gas is doubled isothermally? Classical result gave us:  S = +nR ln2 Statistical Thermodynamics Initially, there are  i ways of arranging the N particles: One way: Double V

24. ToTo T1T1 C q W ToTo T1T1 C q T1T1 q W Note that T 1 < T 1 q*

25. Question: How much less work are we doing as a result of the “extra” heat flow?

27. For a thermodynamic Universe, we saw that: S Universe   The Universe can consist of several parts (systems and reservoirs) each with their own entropy, so: S Universe = S 1 + S 2 + S 3 +... Probabilities are multiplicative, so: S Universe = S 1 + S 2 + S 3 +...   1  2  3 …. But, since S 1   1 and S 2   2, etc., then: S 1 + S 2 + S 3 +...   1 +  2 +  3 +... How can this be? Boltzmann thought about this problem…. Boltzmann Equation

28. We note a property of logarithms: ln  1 + ln  2 + ln  3 +...  ln (  1  2  3 ….) The conditions are therefore satisfied if S  ln . Then, S 1 + S 2 + S 3 +...  ln  1 + ln  2 + ln  3 +... but also S 1 + S 2 + S 3 +...  ln(  1  2  3 ….), as required. Boltzmann derived a relationship in which k is the constant of proportionality: S = k ln  with k = 1.38 x 10 -23 JK -1. Voted one of the ten most beautiful equations in a survey by Physics World!