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Carnot Cycle

w -nRThotln(V2/V1) q-w

State 1 Thot

State 2 Thot

isothermal

q0, wCV??T

q0, w-CV??T

adiabatic

adiabatic

isothermal

State 4 Tcold

State 3 Tcold

w -nRTcoldln(V4/V3) q-w

Carnot Cycle PV Diagram

-nRThotln(V2/V1)

1

CV ?T

4

2

-CV ?T

Pressure

3

-nRTcoldln(V4/V3)

Volume

Carnot Cycle, Schematic View

Thot

E

Tcold

The engine operates between two reservoirs to and

from which heat can be transferred. We put heat

into the system from the hot reservoir and heat

is expelled into the cold reservoir.

Questions about Thermodynamic Cycles

How much of the heat put in at high temperature

can be converted to work? Can two engines with

the same temperature difference drive one

another? What does entropy have to do with it?

Clausius b. Jan. 2, 1822, Prussiad. Aug. 24,

1888, Bonn "Heat cannot of itself pass from a

colder to a hotter body."

Carnot Cycle Step 1

State 1 Thot

State 2 Thot

isothermal

adiabatic

adiabatic

Thot

isothermal

State 4 Tcold

State 3 Tcold

Heat converted to work!

Isothermal reversible Expansion ?E0 Energy

of an ideal gas depends only on

temperature q -w

Carnot Cycle Step 2

State 1 Thot

State 2 Thot

isothermal

adiabatic

adiabatic

isothermal

State 4 Tcold

State 3 Tcold

Energy lost to expansion

Isothermal reversible Expansion q0 No heat

transferred in adiabatic process DT here

defined Thot - Tcold positive value Temp

change for process is -DT dT

negative

Carnot Cycle Steps 3 4

State 1 Thot

State 2 Thot

Work done in compression

isothermal

adiabatic

adiabatic

isothermal

State 4 Tcold

State 3 Tcold

Carnot Cycle Summary

Step1 Step2 Step3 Step4 w -nRThotln(V2/V1) -C

V ?T -nRTcoldln(V4/V3) CV ?T q -w 0 -w 0 ?E

0 w 0 w

Thot

w1 w2 w3 w4 -nRThotln(V2/V1) -

nRTcoldln(V4/V3) q1 q2 q3 q4

nRThotln(V2/V1) nRTcoldln(V4/V3) q

-w Heat flows through system, some work is

extracted.

E

Tcold

The Carnot Cycle Efficiency

The efficiency of a heat engine is simply

total work accomplished/total fuel (heat)

input The heat is input only from the hot

reservoir so efficiency ? -w/qhot We

already know w -(qhot qcold),

therefor... 1) ? (qhot qcold)/qhot

1 qcold/qhot Substituting in expressions for

qhot and qcold 2) ? 1 nRTcoldln(V4/V3) /

nRThotln(V2/V1) 1(Tcold/Thot)(ln(V4/V3)/ln(V2/V1

))

? 1 nRTcoldln(V4/V3) / nRThotln(V2/V1)

1(Tcold/Thot)(ln(V4/V3)/ln(V2/V1)) This second

expression is rather complicated-- but we have a

relationship for the volumes in this process

Rearranging and plugging it in to the first

equation on the page

? 1 - (Tcold/Thot)

? 1 - (Tcold/Thot) 1 (qcold/qhot)

100 efficiency is only achieved when

Tcold0 and/or Thot? Practical

impossibilities. (Foreshadowing the Third

Law) Can we construct an engine more

efficient than this one?

Impossible Machines

Consider two engines ER and E operating between

the same two reservoirs

Can the efficiencies of these two engines be

different?

Thot

ER

We operate ER in reverse. We couple the operation

of the two engines. We know wR qhot qcold

(forward direction) w qhot qcold

E

Tcold

The composite engine then has a total work W

w-wR Lets couple the engines so that work

from forward running engine drives the other one

backwards with no work on the surroundings w

wR (remember actual work of reverse process is

-wR)

Impossible Machines

Can the efficiencies of these two engines be

different?

Lets assume ?gt ?R w/qhotgtwR/qhot But

wwR so qhotlt qhot Net heat transferred

from hot reservoir qhot- qhot In other words,

the heat withdrawn from the hot reservoir is

negative! From above, we also so that this

implies that the heat withdrawn from cold

reservoir is positive! Without doing any work we

extract heat from the cold reservoir and place it

in the hot reservoir!

What is wrong with this argument?

The Carnot Cycle Noticing state functions

Assumptions in our proof 1) The first

law Experimentally proven. 2) wwR A fully

practical assumption 3) ?gt ?R This

assumption is disproven by contradiction

Therefor we have proven ?? ?R Now we

assumed that E was any engine, whereas, ER was

reversible. So every engine is either of equal

or less efficienct than a reversible

engine (For the same two reservoirs)

Carnot Cycle Noticing State Functions

Now--- we have this interesting relationship

between temperature and heat for these systems

? 1 - (Tcold/Thot) 1 (qcold/qhot)

What Carnot noticed was that there was an implied

state function here! (qhot/Thot)

-(qcold/Tcold) or (qhot/Thot) (qcold/Tcold)

0 This can also be written

This is a state function! Clausius called it the

Entropy, S.

Conservation of Entropy ???

- The question arose, is entropy conserved? After

all, energy is. - But a great deal of experimental experience

indicated that - ?S(system) ?S(surroundings) ? 0
- This is the Second Law of Thermodynamics.
- Heat never spontaneously flows from a cold

body to a hot one. - A cyclic process can never remove heat from a

hot body and achieve complete conversion into

work

Entropy is Not Conserved

1 Atm T

1Atm T

2 Atm T V1

Vacuum

V22V1

Two cases of expansion of an ideal gas 1)

Expansion in to a vacuum. 2) Reversible

expansion (1) w0, ?E0, qirreversible0

?S(surroundings)0 To calculate ?S(system) we

look to the second process (they must be the

same).

Entropy is Not Conserved

1 Atm T

1Atm T

2 Atm T V1

Vacuum

V22V1

In the second process we follow an isothermal,

reversible path. We know that ?T and thus

?E0 Nowqrev ?E - w RT ln(2) so

?S(system) qrev/T R ln(2)

Entropy is Not Conserved

For the reversible process weve already

calculated qrev RT ln(V2/V1) RT ln(2)

?S(system) qrev/T R ln(2) One way to make

sure this is reversible is to make sure the

outside temperature is only differentially

hotter. In this case, ?S(surroundings)

-qrev/T ?S(total) 0 For the total

irreversible process, ?S(surroundings) 0

(nothing exchanged) ?S (total) 0 R ln(2)

gt 0

Dissorder and Entropy

It turns out that disorder and entropy are

intimately related.

Ludwig Boltzmann c. 1875

We start out by considering the spontaneity of

this process. Why doesnt the gas spontaneously

reappear back in the box?

Dissorder and Entropy

Lets break the box into N cells and consider the

gas to be an ideal gas composed of M molecules.

We ask What is the probability that M

molecules will be in the box labeled

This obviously depends on both M and N. We assume

NgtM for this problem. Number of ways of

distributing M indistinguishable balls in N boxes

is approximately ? NM/M! (this approx.

miscounts when multiple balls in same

box) Boltzmann noted that an entropy could be

defined as S k ln(?) R ln(?)/NA There are a

number of reasons this is a good definition. One

is it connects to thermodynamics.

Dissorder and Entropy

So for a given state we have S k ln(?) R

ln(?)/NA R ln(NM/M!)/NA Lets say we change

state by increasing the volume. Well, for the

same sized cells, N increases to N. S-S

(R/NA) (ln(NM/M!) - ln(NM/M!)) (R/NA)

ln(NM /NM) So ?S (R/NA) ln(NM/NM) M (R/NA)

ln(N/N) And since N is proportional to

volume ?S M (R/NA) ln(V2/V1)

Entropy of Materials

Why does graphite have more entropy than diamond?

Graphite S298 5.7 J/K

Diamond S298 2.4 J/K

How about water in its different phases

S298 (J/K mol) H2O(s,ice) 44.3 H2O(l) 69.91

H2O(g) 188.72

Entropy of Mixing

What happens when you mix two ideal gases?

What happens when you solvate a molecule?

Entropy and Chemical Reactions

?S

CO32- (aq) H(aq) HCO3- (aq) 148.1 J/K

mol HC2O4- (aq) OH- (aq) C2O42-

H2O(l) -23.1

Its hard to predict the change in entropy without

considering solvent effects

Calculation of ?S for a reaction is similar to

that for enthalpy. Entropies of elements are

not zero though.

Fluctuations

All the previous arguments relate to processes

involving large numbers of molecules averages

over long periods of time. When the system is

small and observation time is short, then

fluctuations around the maximum entropy

solution can be found.

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