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

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

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)

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

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 = 1.998 atm Pext = 1.998 atm Step 2 P = 1.999 atm Pext = 1.999 atm Step 1 (2) P = 2 atm Pext = 2 atm

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

IRREVERSIBLE EXPANSION Vi 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

REVERSIBLE COMPRESSION Vf 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

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

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

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

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 = 0.460 V

Standard Electrode Potential – Standard Reduction Potential, Eo Pt H2(g,1atm) H+(1M)  Cu2+(aq) Cu(s) Eo = 0.340 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 = 0.340 V Zn2+(1M) + 2e-  Zn(s) Eo = -0.763 V Eocell = Eo(cathode) – Eo(anode) = 0.340 – (-0.763) = 1.103 V For the reaction: Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

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 839 - 841