Steps of Carnot Cycle

qh absorbed (+) from hot reservoir Thw done by system (–)

q=0w done by system (–)w = Cv(Tc-Th)S = 0

qc discarded (–) into cold reservoir Tcw done on system (+)

q=0w done on system (+)w=Cv(Th-Tc)S = 0

0hU q w

2

1

2

1

ln

ln (positive)

h hVq w nRTV

VS nRV

0hU q w

4

3

4

3

ln

ln (negative)

c cVq w nRTV

VS nRV

HOW??P,V,TAll change

Ideal Gas Equation of State

Isochores – const V

So let’s tip this surface so thatwe see the V,T projection

Reversible Adiabatic Expansionstep 2 of Carnot Cycle (we want w)

T

V

P2, V2, Th

P3, V3, Tc

, since adiabatic and 0U q w w q

So we could find U to get wU is a state function, and we canchoose any path that makes it easy.Here is one!

P2, V3, Th

a

b

0a b

v h c

v c h

U U UC T T

C T T

Is there a problem that P2, V3, Th is not on the surfaceof the ideal gas eq of state?No! It is just not an equilibrium state, and unstable!

Irreversible Processes: adiabatic expansion

Clausius inequality: dSsys+dSsurr≥0 or dSsys≥-dSsurr≥dq/T

q=0 so dSsurr=0 and dSsys ≥ 0But dSsys>0 for an irrev spon process!!!

The system did work and T<0. The work came from conversion of thermal energy of the system.

wrev>wirrev

Q: Where did the rest of the work go for an irreversible process?

A: Into entropy of the system!!

S in Phase Changes

Phase transitions are reversible because the two phases are in equilibrium at Ttrs

So trsS = trsH/Ttrs

Note: this is the molar entropy (JK-1mol-1)

solid → liquid → gas

dec Entropy inc

exo trsH endo

neg trsS pos

Consider a phase transition at temp Ttrans. This the first really reversible process that we have seen.Consider a system and the surroundings at Ttrans (temperature at which the two phases are in equilibrium: when P= 1 atm, called the normal transition temperature)

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?

P

T

g

Now Htrans = qp and so

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

transtrans

trans

HST

S in Phase Changes

Phase Diagram

Ttrans,norm

P1atm

S in Phase Changes

Trouton’s Law: vapS ~ 85 kJ/mol

Why is water > 85 kJ/mol?

Why is methane <85kJ/mol?

First: want to calculate entropy, S, of the system at temp, T:

S= Sf-Si=

Sf= Si+

For constant pressure and as long as only expansion work is being done: dqrev= CPdT

So: Sf= Si+

or: Sf= Si+ CPln(Tf/Ti) if CP is independent of T

NOTE: The same expresssion applies for constant volume by subsituting CV for CP

∫i

f dqrevT

∫i

f dqrevT

S in Phase Changes

∫i

f CPdTT

Second: must consider phase changesLet’s go from T=0 to T>Tb

So: Sf= S(T=O)+ + fusH/Tf + + vapH/Tb +

S in Phase Changes

∫0

Tf CPdTT ∫

Tf

TbCPdTT ∫

Tb

T CPdTT

Second: must consider phase changesLet’s go from T=0 to T>Tb

So: Sf= S(T=O)+ + fusH/Tf + + vapH/Tb +

S in Phase Changes

∫0

Tf CPdTT ∫

Tf

TbCPdTT ∫

Tb

T CPdTT

Difficult to measure Cp at low TS near T=0 we will use the

Debye extrapolation: CP=aT3

Will be addressed in the 3rd Law

S in Phase Changes

Let us find S for a more complex process. Throw liquid water at -5 C onto a sidewalk also at -5 C. What happens?? It freezes - - - fast! Is it reversible? NO!Is this process spontaneous? How do we find S for the process? Because S is a state function, we invent a reversible path, and calculate S along that path! Let’s invent a path!!!

S in Phase Changes: Example

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

Step 1 heating

H2O(l) 0°CStep 2

Phase transition (freezing)But fusion is tabulated!

H2O(s) 0°C

Step 3 cooling

1 pq dH C dT   273 27321

1268 268

pC H OqS dTT T

 1

273 Jln 1.392268 mol K

S Cp

Step 1

S in Phase Changes: ExampleFreezing (-5°C)

H2O(l) -5°C H2O(s) -5°C

Step 1 heating

H2O(l) 0°CStep 2

Phase transition (freezing)But fusion is tabulated!

H2O(s) 0°C

Step 3 cooling

CP (H2O(l)) = 75.32 J/mol K