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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?