1. The Distribution of Energy Consider 6 particles – Each particle has three energy levels E 3 = 2a E 2 = a E 1 = 0 a a E Let N = Total number of particles = 6 N 1 = Number of particles in state E 1 N 2 = Number of particles in state E 2 Total energy of the system = 4a N = N 1 + N 2 + N 3 = 6 N 1 E 1 + N 2 E 2 + N 3 E 3 = 4a

2. The Boltzmann Distribution – Effect of Temperature 30K 300K 3000K 30000K 300000K

3. The system partition function is for indistinguishable particles. For the two-level system:

4. The Distribution of Molecules on Two Energy Levels Box at restBox weakly shakenBox vigorously shaken

5. N 2 /N 1 = exp[-  E/k B T] N 2 /N 1 approaches 1 as T   A Two-Level System

6. As T approaches infinity, N 1 and N 2 approaches 50% N 1 /N = 1/(1+exp[-  E/k B T]) N 2 /N = 1/(1+exp[  E/k B T])

10. The First Law: Some Terminology System: Well defined part of the universe Surrounding: Universe outside the boundary of the system Heat (q) may flow between system and surroundings Closed system: No exchange of matter with surroundings Isolated System: No exchange of q, w, or matter with surroundings Isothermal process: Temperature of the system stays the same Adiabatic: No heat (q) exchanged between system and surroundings

11. THE CONCEPT OF REVERSABILTY Irreversible processes: Hot Cold Warm Temperature equilibration Mixing of two gases Expansion into a vacuum P = 0 Evaporation into a vacuum P = 0

12. THE CONCEPT OF REVERSABILTY REMOVE PINS Irreversible Expansion pins P = 2 atm P ext = 1 atm (1) (2) P = 2 atm P = 1.999 atm Step 1 P = 1.998 atm Step 2 Infinite number of steps Reversible Expansion P = 1 atm P ext = 1 atm

13. THE CONCEPT OF REVERSABILTY Reversible processes: PoPo P o +  P PoPo Tiny weight Condensation (pressure minimally increases by adding tiny weight) Evaporation (pressure minimally decreases by removing tiny weight)

14. P V V1V1 V2V2 P ext w = -P ext (V 2 – V 1 ) P, V 1 P, V 2 IRREVERSIBLE EXPANSION P ext V1V1 V2V2

15. REVERSIBLE EXPANSION P ext = Pressure of gas. If the gas is ideal, then P ext = nRT/V How does the pressure of an ideal gas vary with volume? P V This is the reversible path. The pressure at each point along curve is equal to the external pressure.

16. REVERSIBLE EXPANSION P V ViVi VfVf PiPi PfPf A B The reversible path The shaded area is IRREVERSIBLE EXPANSION P V ViVi VfVf PiPi PfPf A B The shaded area is P ext = P f Reversible expansion gives the maximum work

17. REVERSIBLE COMPRESSION P V VfVf ViVi PfPf PiPi A B The reversible path The shaded area is P Reversible compression gives the minimum work IRREVERSIBLE COMPRESSION V VfVf ViVi PfPf PiPi A B The shaded area is P ext = P f

18. The Carnot Engine P1V1TP1V1T Isothermal expansion q P2V2TP2V2T w = -nRT ln(V 2 /V 1 ) How can we take it back? Hot reservoir T h qhqh Heat engine w Cold reservoir T c qcqc

19. The Carnot Engine- what is the efficiency? A B Isothermal reversible expansion  U = 0, w = -nRT h ln(V 3 /V 1 ) q = +nRT h ln(V 3 /V 1 ) (T h ) C Adiabatic reversible expansion dU = C p dT = δw q = 0 (T c ) D Isothermal reversible compression  U = 0, w = -nRT c ln(V 2 /V 4 ) q = +nRT c ln(V 2 /V 4 ) P1P1 P2P2 P3P3 P4P4 Pressure V1V1 V2V2 V3V3 V4V4 (T c ) Adiabatic reversible compression dU = C p dT = δw q = 0 total work

20. Statistical Interpretation of Entropy Consider the following expansion: V/2 When stopcock is opened: At any given moment Probability that a given molecule is in the LHS bulb = 50 % (or ½) Probability that a given molecule is in the RHS bulb = 50 % (or ½) Probability that two molecules stay in LHS bulb = ½  ½ = ¼ (25%) Probability that N molecules stay in LHS bulb = (½) N Microstates: A particular way of arranging molecules among the positions accessible to them while keeping the total energy fixed.

21. Consider four molecules in two compartments: 1 way4 ways6 ways4 ways1 way Total number of microstates = 16 The most probable (the “even split”) If N   the “even split” becomes overwhelmingly probable

22. Boltzmann S = k B lnW Consider spin (or dipole restricted to two orientations) orW = 2, and S = k B ln21 particle 2 particles 3 particles 1 mole W = 4, and S = k B ln4,,, W = 8, and S = k B ln8 W =2 NA, and S = k B ln(2 NA ) = N A k B ln2 = Rln 

23. Some entropy problems The spectroscopic entropy of a diatomic molecule is given by the following equation. What is meant by residual entropy? (a)Calculate the spectroscopic entropy of CO at 298.15 K and 1 atm. (b)Write down the expression for calculating the (calorimetric) absolute molar entropy from 0 K to 298.15 K. Note that CO undergoes a solid-solid phase transition at 61.6 K. The experimentally determined calorimetric value is ~170 JK -1 mol -1.. (c) Calculate the residual entropy of CO. Make the correction for residual to the calorimetric value. Compare the two entropies (spectroscopic and calorimetric). The molar heat capacity of C 2 H 4 (g) can be expressed by the following equation over the temperature range 300 K < T < 1000 K. Calculate  S if one mole of ethene is heated from 300 K to 600 K at constant volume. There are no phase transitions.

24. Unit 3 – Real & Ideal Gases Non-ideality of gases - Fugacity  o o T P P RTTGPTGV P G ln,            Standard molar Gibbs energy 1 bar Can generalize for a real gas ....1 2 32  PTBPTB RT VP PP   o o f TPf TGPTG, ln,  The thermodynamic function fugacity. Note: f (P,T)  P as P  0   ......exp, 2 32  PTBPTB P P f TPf PP oo Gibbs energy must be taken relative to some standard state Standard state of real gas is taken to be the corresponding ideal gas at 1 bar i.e. must “adjust” the real gas to ideal behavior Real gas (T,P) Ideal gas (T,P) Real gas (T, P  0) = Ideal gas (T, P  0) G1G1 G2G2 G3G3 oo P P f f RT P P dPV P RT Gln ' 0 ' 1          

25. Fugacity coefficient P f  Ideal gas,  = 1. Fugacity coefficient measures extent of non-ideality ' 0 ' 1 lndP P Z P    (Z-1) / P P (bar) 0 P  1 0 ideal Real