1. Work, Energy & Heat 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, or matter with surroundings Isothermal process: Temperature of the system stays the same Adiabatic: No heat (q) exchanged between system and surroundings

2. 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

3. 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)

4. 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

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

6. 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.

7. 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

8. 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

12. A system from a state 1 (or 2) to a new state 2 (or 1). Regions A, B, C, D, and E correspond to the areas of the 5 segments in the diagram. 1. If the process is isothermal reversible expansion from state 1 to state 2, the total work done by the system is equal to A. Area C + Area E B. Area C only C. Area E only D. Area A + Area C + Area E E. Area A only 3. If the process in question 1 was carried out irreversibly against a constant pressure of 2 atm, the total work done by the system is equal to A. Area C + Area E B. Area C only C. Area A + Area C + Area E D. Area E + Area D E. Area E only 2. If state 2 undergoes irreversible compression to state 1 against an external pressure of 5 atm, the work done by the surroundings on the system is equal to A. Area A + Area C + Area E B. Area C only C. Area A + Area C E. Area E only F. Area A only 6. Which one of the following is not true? A. There is no heat flow between system and surrounding for a reversible adiabatic process B. Work done by the gas in a reversible expansion is a maximum C. Work done by the gas in a reversible expansion is a minimum D. Work done by the gas in a reversible compression is a minimum E. Work done by the gas in a reversible expansion is not the same as the work done against a constant external pressure 4. For a reversible adiabatic expansion of a gas, which one of the following is correct? A. Heat flows to maintain constant temperature B. The gas suffers a maximum drop in temperature C. The gas suffers a minimum drop in temperature D. The work done is a positive quantity E. There is zero change in internal energy 5. The heat capacity (Cp) for a solid at low temperatures is approximately represented by Cp = AT 3, where A is a constant. Using the equation for Cp, the change in entropy (  S) for heating a solid from 0K to 1K is A. A/4C. A/2 B. A/3D. A

13. 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 Statistical Entropy

14. 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 