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
Warm
Temperature equilibration
Cold
Warm
Mixing of two gases
P = 0
Expansion into a vacuum
P = 0
Evaporation into a vacuum

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

4. IRREVERSIBLE EXPANSION
Pext V1 V2 P V V1 V2
Pext
w = -Pext(V2 – V1)
P, V1
P, V2

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

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

7. IRREVERSIBLE EXPANSIONVi Vf Pi Pf A B
The reversible path
The shaded area is
IRREVERSIBLE EXPANSION P V Vi Vf Pi Pf A B
The shaded area is
Pext = Pf
Reversible expansion gives the maximum work

8. REVERSIBLE COMPRESSIONVf Vi Pf Pi A B
The reversible path
The shaded area is
IRREVERSIBLE COMPRESSION V Vf Vi Pf Pi A B
The shaded area is
Pext = Pf
Reversible compression gives the minimum work

12. A system from a state 1 (or 2) to a new state 2 (or 1)
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
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
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
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 = AT3, 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/4 C. A/2
B. A/3 D. A
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

13. Statistical Entropy Consider four molecules in two compartments: 1 way
4 ways
6 ways
Total number of microstates = 16
The most probable
(the “even split”)
If N   the “even split” becomes overwhelmingly probable

14. Consider spin (or dipole restricted to two orientations)
Boltzmann
S = kB lnW
Consider spin (or dipole restricted to two orientations)
1 particle or
W = 2, and S = kB ln2
2 particles
W = 4, and S = kB ln4 , , ,
3 particles
W = 8, and S = kB ln8
1 mole
W =2NA, and S = kB ln(2NA) = NA kB ln2 = Rln

17. Electrochemistry Galvanic Cells (Electrochemical Cells)
The combination of two half cells
Cu(s) + 2Ag+(aq)  Cu2+(aq) + Ag(s) e-
0.460 V
Cu2+
[KNO3 (aq)] Salt Bridge
NO3- K+
Ag+ Cu Ag
LSH - Anode
1.00 M Cu(NO3)2(aq)
Site of Oxidation
Cu(s)  Cu2+(aq) + 2e-
RSH - Cathode
1.00 M AgNO3 (aq)
Site of Reduction
Ag+(aq) + e-  Ag(s)
Measure the electromotive force of the cell (cell potential or potential difference, Ecell)
Cell diagram: Cu(s) Cu2+(aq)  Ag+(aq)  Ag(s) Ecell = V

18. Standard Electrode Potential – Standard Reduction Potential, Eo
Pt H2(g,1atm) H+(1M)  Cu2+(aq) Cu(s) Eo = V
SHE Always Anode
Site of Reduction
Standard cell potential Eocell = Eo(right) – Eo(left)
Cathode
Anode
Standard reduction half-potential: Cu2+(1M) + 2e-  Cu(s) Eo = V
Zn2+(1M) + 2e-  Zn(s) Eo = V
Eocell = Eo(cathode) – Eo(anode) = – (-0.763) = V
For the reaction: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

19. DG = -n F Ecell DGo = -n F Eocell Eocell = (RT/nF) lnK
Gibb’s Free Energy and Electrode Potential
DG = -n F Ecell
Reactants/products not in standard states
DGo = -n F Eocell
Reactants/products in standard states
Can combine reduction half equations to obtain Gibb’s free energy change.
Electrode Potential and the equilibrium constant
Eocell = (RT/nF) lnK
Must specify temperature – does K change with T?
Electrode Potential and concentration
Ecell = Eocell - (RT/nF) lnQ
Nernst equation – note Ecell = 0 when Q = K
Know how to do concentration cell calculations – page