Multiple Carnot engines can be used to construct any reversible cyclic engineconstruct any reversible cyclic engine

Provides proof that the Carnot efficiency is the best possible.

Multiple Carnot engines can be used to construct any possible reversible engineconstruct any possible reversible engine

What is the highestreservoir temperature

PV isothermsTh reservoir temperature

for this heat engine?Th

What is the lowestreservoir temperature

h

reservoir temperaturefor this heat engine?

Tc

What is the efficiencyof this heat engine?

c

Tc

T Tch

h

TT

T

S in Phase ChangesConsider a phase transition at temp Ttrans. This is a reversible process with entropy change. In fact, it is the first really reversible process that we have seen.Consider a system and the surroundings at the transition temperature (the temperature at which the two phases are in equilibrium: when P= 1 atm, this is called the normal transition temperature)E H O ( ) H O ( ) T 0 C 273 K E.g., H2O (l) H2O (s) T = 0 C or 273 K orH2O (l) H2O (g) T = 100 C or 373 K – both at 1 atm How does the boiling temperature change as the pressure is reduced? I e hiking in the mountains?I.e., hiking in the mountains?What does this look like on a phase diagram?

d transHS

P

g

Now Htrans = qp and so

for one mole. This is the first real example of a reversible process!

transtrans

trans

HST

T

example of a reversible process!

S in Phase Changes

This table gives the entropy changeat the normal transition temperature.at the normal trans t on temperature.How would we determine the absolute entropy at anarbitrary temperature when there may have been severalphase transitions at lower temperatures?phase transitions at lower temperatures?

S in Phase Changes

0T T

PC dTdqS T S

Third Law (S(0) =0)

0 0

0S T ST T

S in Phase ChangesCalculating the Standard Molar Entropy of N2

T T C dTdq 0 0

0 PC dTdqS T ST T

Solid - solid

0 (Third Law)

298.15 0 192.06m mJThus S K S

mol K

Let us find S for a more complex process Let us get some liquid

S in Phase ChangesLet us find S for a more complex process. Let us get some liquid water at -5 C (268 °K) and throw it onto a sidewalk also at -5 C. What happens?? It freezes - - - fast! ( liquid was a metastable state)Is this process spontaneous? YES Is it reversible? NO! Is this process spontaneous? YES Is it reversible? NO! How do we find S for the process? We invent a reversible path, and calculate S along that path, because S is a state function!Let’s invent a three part path. Calc ΔS for Part 1p p

H2O (liq) -5 C H2O (s) -5 C 273

11

qST

Calc ΔS for Part 1

1 heating 3 cooling

268 T

1 pq dH C dT

H2O (liq) 0 CH2O (s) 0 C

2 h t iti (f i )

  273 27321

1268 268

pC H OqS dTT T

H2O (liq) 0 C 2 phase transition (freezing) But fusion is tabulated! 1

273 Jln 1.392268 mol K

S Cp

Phase Transitions273

11

qS

Part 11

268 T1 pq dH C dT Also, we know

273 27321

1268 268

pC H OqS dTT T

Thus

268 268

We know Cp for liquid water is ~ cons’t and = 75.29 J/mol K.We carry out all these steps with one mole of material y por else we carry along an “n”.So finally

273 J I th i t?1

273 Jln 1.392268 mol K

S Cp

Is the sign correct?

S in Phase Changes

H2O (liq) -5 C H2O (s) -5 C

1 heating

Part 2Phase transition (freezing) at normal transition temperature H = H

H2O (s) 0 C

1 heating 3 cooling

transition temperature. Hfreezing = -Hfusion(the tabulated value)

J6008H2O (liq) 0 C 2 ( )2 phase transition (freezing) But fusion is tabulated!

2

6008 J22.01273 K mol K

molS

Part 3 268 268

233

273 273

pC H O sqS dTT T

C (H O ( ) ) 37 66 J/ l K d bt i 268 JΔS 37 66ln 0 696 Cp (H2O (s) )= 37.66 J/mol K, and we obtain 3ΔS =37.66ln = -0.696273 mol K

S in Phase Changes: Example

1273 Jln 1.392268 mol K

S Cp

Freezing (-5°C)H2O(l) -5°C H2O(s) -5°C

St 1 h ti Step 3 cooling J Step 1 heating

H2O(l) 0°C H2O(s) 0°C

Step 3 cooling

2

J6008 J22.01273 K mol K

molS

H2O(l) 0 CStep 2

Phase transition (freezing)But fusion is tabulated!

2 (s)

3268 J37.66 ln 0.696273 mol K

S

  J1 392 22 01 0 696 21 31S S S S Putting it all together (Freezing at -5°C),

1 2 3 1.392 22.01 0.696 21.31mol KoverallS S S S

The entropy change is negative!!!Does this process happen spontaneously? YES!! Why??Does this process happen spontaneously? YES!! Why??Spontaneous Stotal 0 So we need to consider the surroundings!!

Phase Transitions

J1 392 22 01 0 696 21 31S S S S

Putting this all together,

1 2 3 1.392 22.01 0.696 21.31mol KoverallS S S S

Spontaneous Δ Stotal 0. So we need to consider the surroundings!W d t l l t ΔS

processsurroundings

HS

T

We need to calculate ΔSsurroundings.

268 6008

surroundings

process fusion

TJH H

m l process fusion mol

T

We need to remember how we found Δ H at arbitrary temperature

o

To

pT

H T H C dT

Phase Transitions

268

268 2 2273

oH H Cp H O Cp H O s dT

6008 75.29 37.66 268 273J5820

5820mol

This was for fusion, but we need freezing. Hence ΔH = 5820 J/mol Thus

5820 J21.72268 mol K

processsurroundings

HS

T

Hence ΔHprocess 5820 J/mol. Thus

268 mol KT

J J21.31 21.72mol K mol Ktot sys surroundingsS S S mol K mol K

J0.41 0mol K

y g

The total entropy change is positive,and all is well!!

Phase Transitions

That was painful and tedious at best!We would like a better way to deal with the surroundingsWe would like a better way to deal with the surroundingsin a constant pressure process.It will be through the state function G, known as theGibbs Fr En r This ls ll s us t discuss quilibriumGibbs Free Energy. This also allows us to discuss equilibriumthe composition of reaction mixtures.First, however, we need to digress briefly to deal with theentropy of mixing.

Entropy of Mixing

We will mix two ideal gasesslowly and reversibly using animaginary path.The ideal gases do not interact, soThe ideal gases do not interact, so

expand the V1 gas slowly into V2

Entropy of Mixing

We will mix two ideal gasesslowly and reversibly using animaginary path.The ideal gases do not interact, soThe ideal gases do not interact, so

expand the V1 gas slowly into V2

ConcepTest 1Will S increase, decrease or remain unchanged upon mixing?

a. Increaseb D sb. Decreasec. Remain unchanged

Entropy of MixingWe will mix two ideal gases

slowly and reversibly using animaginary path.The ideal gases do not interact, so

expand the V1 gas slowly into V2.expand the V1 gas slowly into V2.

Similarly expand the V gas into V

1 21 1

1

ln V VS n RV

Similarly, expand the V2 gas into V1

1 22 2

2

ln V VS n RV

ΔStot is the sum of the two:1 2 1 2

1 2 1 21 2

ln lnV V V VS S S n R n RV V

tot

Express this result in mole fractions of the final mixture.WHY BOTHER???

Entropy of Mixing

1 2 1 21 2 1 2

1 2

ln lnV V V VS S S n R n RV V

We use this result soon to findSi P d T1 1

11 2 1 2

n Vxn n V V

We use this result soon to findequilibrium concentrationsof reactants and products. CHEMICAL POTENTIAL

Since same P and T,

1 2 1 2 CHEMICAL POTENTIAL

1 21 1 1 1 2 1 1 2ln ln ln and similarly for nV Vn R n R x n n x R x

V

Thus ln lnS n n R x x x x

No dependence on P! 1 1 1 1 2 1 1 2

1

y fV

1 2 1 1 2 2Thus ln lnmixingS n n R x x x x

Is S as given by this relation positive or negative?