Work, Energy & HeatThe 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

Cold

Warm

WarmTemperature equilibration

Mixing of two gases

Expansion into a vacuumP = 0

Evaporation into a vacuumP = 0

THE CONCEPT OF REVERSABILTY

Reversible processes:

Po

Po + P

Po

Tiny weight

Condensation(pressure minimally increases by adding tiny weight)

Evaporation(pressure minimally decreases by removing tiny weight)

P

VV1 V2

Pext

w = -Pext(V2 – V1)

P, V1 P, V2

IRREVERSIBLE EXPANSION

Pext

Pext

V1

V2

THE CONCEPT OF REVERSABILTY

REMOVE PINS

Irreversible Expansion

pins

P = 2 atm

Pext = 1 atm

(1)

(2)

P = 2 atm

Pext = 2 atm

P = 1.999 atm

Pext = 1.999 atm

Step 1

P = 1.998 atm

Pext = 1.998 atm

Step 2Infinite number of steps

Reversible ExpansionP = 1 atm

Pext = 1 atm

P = 1 atm

Pext = 1 atm

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?

P

V

This is the reversible path. The pressure at each point along curve is equal to the external pressure.

1

2ln12

1

2

1

2

1 V

VnRTdV

VnRTdV

V

nRTPdVdVPw

V

V

V

V

V

Vext

REVERSIBLE EXPANSION

P

VVi Vf

Pi

Pf

A

B

The reversible path

i

f

V

VnRTw ln

The shaded area is

IRREVERSIBLE EXPANSION

P

VVi Vf

Pi

Pf

A

B

fif VVPw

The shaded area is

Pext = Pf

Reversible expansion gives the maximum work

REVERSIBLE COMPRESSION

P

VVf Vi

Pf

Pi

A

BThe reversible path

i

f

V

VnRTw ln

The shaded area is

P

Reversible compression gives the minimum work

IRREVERSIBLE COMPRESSION

VVf Vi

Pf

Pi

A

B

fif VVPw

The shaded area is

Pext = Pf

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 EB. Area C onlyC. Area E onlyD. Area A + Area C + Area EE. 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 toA. Area C + Area EB. Area C onlyC. Area A + Area C + Area ED. Area E + Area DE. 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 toA. Area A + Area C + Area EB. Area C onlyC. Area A + Area CE. Area E onlyF. 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 processB. Work done by the gas in a reversible expansion is a maximumC. Work done by the gas in a reversible expansion is a minimumD. Work done by the gas in a reversible compression is a minimumE. 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 temperatureB. The gas suffers a maximum drop in temperatureC. The gas suffers a minimum drop in temperatureD. The work done is a positive quantityE. 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/2B. A/3 D. A

P/atm

1.0 10 Volume/L

2

5

A B

C

D E

State 1

State 2

Consider four molecules in two compartments:

1 way 4 ways 6 ways 4 ways 1 way

Total number of microstates = 16

The most probable(the “even split”)

If N the “even split” becomes overwhelmingly probable

Statistical Entropy

Boltzmann S = kB lnW

Consider spin (or dipole restricted to two orientations)

or W = 2, and S = kB ln21 particle

2 particles

3 particles

1 mole

W = 4, and S = kB ln4, , ,

W = 8, and S = kB ln8

W =2NA, and S = kB ln(2NA) = NA kB ln2 = Rln

ElectrochemistryGalvanic Cells (Electrochemical Cells) The combination of two half cells

e-

e-

0.460 V

Cu2+Cu2+

[KNO3 (aq)] Salt Bridge

NO3- K+

Ag+NO3

-

Cu(s) + 2Ag+(aq) Cu2+(aq) + Ag(s)

LSH - Anode 1.00 M Cu(NO3)2(aq)

Site of OxidationCu(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

Cu Ag

Standard Electrode Potential – Standard Reduction Potential, Eo

Pt H2(g,1atm) H+(1M) Cu2+(aq) Cu(s) Eo = 0.340 V

SHE Always Anode

Standard cell potential Eocell = Eo(right) – Eo(left)

Site of Reduction

Cathode Anode

Standard reduction half-potential: Cu2+(1M) + 2e- Cu(s) Eo = 0.340 VZn2+(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)

Gibb’s Free Energy and Electrode Potential

oor nFG oor nFG

Go = -n F Eocell

G = -n F Ecell Reactants/products not in standard states

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