Chem131_Chapter3

7/28/2019 Chem131_Chapter3

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CHAPTER 3The Second Law of Thermodynamics

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Experimental observation: All spontaneous processes are irreversible Not all irreversible processes are spontaneous

SECOND LAW OF THERMODYNAMICS

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plug

vacuum

yes

3


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vacuum

no

4


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water freezing at -10 C

water/ice at 0 C

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ice melting at +10 C

water/ice at 0 C

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Some spontaneous processes have U and H > 0 (e.g., melting of ice at 10 C)

Some spontaneous processes have U and H < 0 (e.g., freezing of water at -10 C)

Issue: First law treats heat (q) and work (w) on an equal footing

In reality they are not interchangeable:there are restrictions on how much qcan be converted to w


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Equivalent statements:


(1) Kelvin: heat cannot, by a cyclic process, be taken from a reservoir and convertedinto work without simultaneously delivering heat from a higher to a lowertemperature reservoir.

Hot reservoir, TH

Heatengine

qH

-w

It is not possible to build a cyclicmachine that converts heat intowork with 100% efficiency!

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(2) Clausius: In a cyclic process, heat cannot flow from a lower to a higher temperaturereservoir without the expenditure of work on the system. Generalization: all naturalprocesses are irreversible.



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Energy of universe is constant, but increasingly unavailable for useful work

(3) It is impossible to achieve perpetual motion of the second kind, that is, to construct amachine that will convert heat into work isothermally.

(4) No heat engine operating between two temperatures can be more efficient than areversible heat engine operating between the same two temperatures.

(2) Clausius: In a cyclic process, heat cannot flow from a lower to a higher temperaturereservoir without the expenditure of work on the system. Generalization: all naturalprocesses are irreversible.


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A heat engine is a device that operates cyclically to convert heat to useful work.

HEAT ENGINES

Heating

Work

(1) A substance (normally a gas) is in acontainer closed by a piston

(2) The substance is heated and expands,

causing a piston to move (work).

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HEAT ENGINES

Cooling




(3) The substance is then cooled down to theoriginal temperature and the pistonreturns to its original position.

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HEAT ENGINES





(4) The cycle starts again.

Heating

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HEAT ENGINES

Cooling





(4) The cycle starts again.

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The engine:

(1) Absorbs some amount of heat qH from ahot reservoir (a hot body such as a boiler)

(2) Converts part of it to work (-w) done on thesurroundings by, e.g., pushing a piston

(3) The difference qH - (-w) = -qC is released

back to the surroundings (a cold bodysuch as a condenser ) as heat.

HEAT ENGINES


Hot reservoir, TH

Cold reservoir, TC

Heatengine

qH

qC

-w

Important:The reservoirs are large so thattheir temperature is not affected

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Efficiency: =work output per cycle

energy input per cycle=

w

qH

HEAT ENGINES

For a cyclic process U = 0

U= 0 = q+ w = qH+ qC+ w

But...

w = qH+ qC

=qH+ qC

qH

= 1 +qC

qH

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Carnots principle:

No heat engine can be more efficient than a reversible heat engine whenboth engines work between the same pair of temperatures TH and TC

HEAT ENGINES

super > revImagine:

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Carnots principle:


HEAT ENGINES

super > revImagine:

Hot reservoir, TH

Cold reservoir, TC

Reversibleheat engine

qH

qC

-w

rev =w

qH

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Carnots principle:


HEAT ENGINES

super > revImagine:

Hot reservoir, TH

Cold reservoir, TC

qH

qC

-wSuper

heat engine

QH

QC

-WReversibleheat engine

rev =w

qH

super =W

QH

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Carnots principle:


HEAT ENGINES

Imagine: QH = 1.3 qH

Hot reservoir, TH

Cold reservoir, TC

qH

qC

-wSuper

heat engine

QH

QC

-WReversibleheat engine

rev =w

qH

super =W

QH

10 cycles of the super engine = 13 cycles of the reversible engine

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Carnots principle:


HEAT ENGINES


Hot reservoir, TH

Cold reservoir, TC

qH

qC

w

QH

QC

-WReversibleheat pump

Superheat engine


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Carnots principle:


HEAT ENGINES


Hot reservoir, TH

Cold reservoir, TC

qH

qC

QH

QC

Reversibleheat pump

Superheat engine


W

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Carnots principle:


HEAT ENGINES

Hot reservoir, TH

Cold reservoir, TC

qH

qC

QH

QC

Reversibleheat pump

qH = QHafter 10 cycles of the super engine: super > revbut

|W| > |w|

Superheat engine

W

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Carnots principle:


HEAT ENGINES

Hot reservoir, TH

Cold reservoir, TC

qH

qC

QH

QC

Reversibleheat pump

qH = QHafter 10 cycles of the super engine: super > revbut

|W| > |w|

net work

Superheat engine

W

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Carnots principle:


HEAT ENGINES

Hot reservoir, TH

Cold reservoir, TC

qH

qC

QH

QC

Reversibleheat pump

net work

U= 0 = q+ w = qH+ qC+ wnet work must come from some energy input

qH = QHbut (in 10 cycles of the super-engine):

Superheat engine

W

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Carnots principle:


HEAT ENGINES

Hot reservoir, TH

Cold reservoir, TC

qH

qC

Superheat engine

QH

QC

Reversibleheat pump

net work


the energy input must come from the cold reservoir...

W

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HEAT ENGINES

Hot reservoir, TH

Cold reservoir, TC

qH

qC

Superheat engine

QH

QC

Reversibleheat pump

net work



Heat cannot, by a cyclic process, be taken from a reservoir and converted into workwithout simultaneously delivering heat from a higher to a lower temperature reservoir.

NOT POSSIBLE BECAUSE VIOLATES THE KELVIN STATEMENT

W

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HEAT ENGINES



NOT POSSIBLE BECAUSE VIOLATES THE KELVIN STATEMENT

super > rev

Heat cannot, by a cyclic process, be taken from a reservoir and converted into workwithout simultaneously delivering heat from a higher to a lower temperature reservoir.

(any engine) (reversible engine)

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HEAT ENGINES

Hot reservoir, TH

Cold reservoir, TC


qB,HReversibleheat pump

Two reversible heat engines

A = A

qA,H

B = B

qB,H

A B

qB,CqA,C

qA,H

B A

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HEAT ENGINES

Hot reservoir, TH

Cold reservoir, TC




A = A

qA,H

B = B

qB,H

B A

qB,CqA,C

qA,H

A B

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HEAT ENGINES

Hot reservoir, TH

Cold reservoir, TC




A = A

qA,H

B = B

qB,H

B A

qB,CqA,C

qA,H

A = B

All reversible engines operating between two given temperaturesmust have the same efficiency

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HEAT ENGINES

(any engine) (reversible engine)

irrev rev

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HEAT ENGINES

All reversible engines operating between two given temperaturesmust have the same efficiency

rev = f(TH, TC)

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CARNOT CYCLE

Hot reservoir, TH

Cold reservoir, TC

qC

qHisotherm, TH

V2V1 V

P1

P2

P

1

2

qI = wI = RTH lnV2

V1> 0

heat adsorbed by the system from the reservoir = work done by the system

1) Reversible, isothermal expansion

UI = 0

I

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CARNOT CYCLE

Hot reservoir, TH

Cold reservoir, TC

qC

qH

V2V1 V

P1

P2

P

1

2

V3V4

isotherm, TC

4 3

heat is released by the system = work is done on the system

UIII = 0

3) Reversible, isothermal compression

III

qIII = wIII = RTC lnV4

V3< 0

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CARNOT CYCLE

Hot reservoir, TH

Cold reservoir, TC

qC

qH

V2V1 V

P1

P2

P

1

2

4 3

V3V4

adiabat

II

2) Reversible adiabatic expansion

qII = 0

wII < 0 work done by the system

UII = +wII = CV(TC TH)

37

C O C C


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CARNOT CYCLE

Hot reservoir, TH

Cold reservoir, TC

qC

qH

V2V1 V

P1

P2

P

1

2

4 3

adiabatIV

UIV = +wIV = CV(TH TC)

work done on the systemwIV > 0

qIV = 0

4) Reversible adiabatic compression

V3V4

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CARNOT CYCLE


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CARNOT CYCLE

Hot reservoir, TH

Cold reservoir, TC

qC

qHisotherm, TH

V2V1 V

P1

P2

P

1

2

isotherm, TC

4 3

adiabatadiabat

U around the cycle is zero

IV

IIIII

V3V4

I

39

CARNOT CYCLE


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CARNOT CYCLE

Hot reservoir, TH

Cold reservoir, TC

qC

qHisotherm, TH

V2V1 V

P1

P2

P

1

2

isotherm, TC

4 3

adiabatadiabat

The net work is done by the gas on the surroundings

IV

IIIII

V3V4

w = (wI + wII + wIII + wIV) = RTH lnV2

V1

+ RTC ln

V4

V3

I

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CARNOT CYCLE


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CARNOT CYCLE

Hot reservoir, TH

Cold reservoir, TC

qC

qHisotherm, TH

V2V1 V

P1

P2

P

1

2

V3V4

isotherm, TC

4 3

adiabatadiabat


T2

T1

= V1V2R/CV

adiabatic processfor a perfect gas THV

R/CV2

= TCVR/CV3

THVR/CV1

= TCVR/CV4

IV

IIIII


V1

+ RTC ln

V4

V3

I

41

CARNOT CYCLE


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CARNOT CYCLE

isotherm, TH

V2V1 V

P1

P2

P

1

2

isotherm, TC

4 3

adiabatadiabat


THVR/CV2

= TCVR/CV3

THVR/CV1

= TCVR/CV4

V2

V1= V3

V4

w = RTH lnV2

V1+RTC lnV2

V1

IV

IIIII

V3V4


V1

+ RTC ln

V4

V3

I

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CARNOT CYCLE


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CARNOT CYCLE

isotherm, TH

V2V1 V

P1

P2

P

1

2

isotherm, TC

4 3

adiabatadiabat


THVR/CV2

= TCVR/CV3

THVR/CV1

= TCVR/CV4

V2

V1= V3

V4

w = R(TH TC) lnV2

V1

IV

IIIII

V3V4


V1

+ RTC ln

V4

V3

I

43

CARNOT CYCLE


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CARNOT CYCLE

isotherm, TH

V2V1 V

P1

P2

P

1

2

isotherm, TC

4 3

adiabatadiabat

=wqI

= R(THTC) ln(V2/V1)

RTH ln(V2/V1)

=

TH TC

TH

IV

IIIII

V3V4

w = R(TH TC) lnV2

V1

q= qI

We know:

=

work output per cycle


w

qH

I

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CARNOT CYCLE


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CARNOT CYCLE

isotherm, TH

V2V1 V

P1

P2

P

1

2

isotherm, TC

4 3

adiabatadiabat

=wqI

= R(THTC) ln(V2/V1)

RTH ln(V2/V1)

IV

IIIII

V3V4

w = R(TH TC) lnV2

V1

q= qI

We know:

=

work output per cycle


w

qH

=TH TC

TH< 1

I

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CARNOT CYCLE


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CARNOT CYCLE

isotherm, TH

V2V1 V

P1

P2

P

1

2

isotherm, TC

4 3

adiabatadiabat

IV

IIIII

Ucycle = 0

qcycle = wcycle

qcycle = qI+ qIII

= TH

TCTH

=

wqI

= qI+

qIIIqI

qI+ qIII

qI=

TH TC

TH

V3V4

We know:

I

46

CARNOT CYCLE


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CARNOT CYCLE

isotherm, TH

V2V1 V

P1

P2

P

1

2

isotherm, TC

4 3

adiabatadiabat

IV

IIIII

= TH

TCTH

=

wqI

= qI+

qIIIqI

V3V4

qIII

TC+

qI

TH= 0

qI+ qIII

qI=

TH TC

TH

Ucycle = 0

qcycle = wcycle

qcycle = qI+ qIII

We know:

I

47

CARNOT CYCLE


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CARNOT CYCLE

isotherm, TH

V2V1 V

P1

P2

P

1

2

isotherm, TC

4 3

adiabatadiabat

IV

IIIII

V3V4

I

qIII

TC+

qI

TH= 0

Carnots cycle

48

ENTROPY


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ENTROPY

49

ENTROPY


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ENTROPY

ycle

qrev

T= 0

Sum of Carnot cycles

qrev

T= 0

In the limit of an infinite number ofCarnots cycles

state functionqrev

T=

50

ENTROPY


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ENTROPY

ycle

qrev

T= 0

Sum of Carnot cycles

qrev

T= 0

In the limit of an infinite number ofCarnots cycles

dS=qrev

TS=

state 2

state 1

qrev

T= S2 S1

51

ENTROPY


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ENTROPY

qrev is not an exact differential is not a state function

qrev

Tis an exact differential is a state function

dS=qrev

TS=

state 2

state 1

qrev

T= S2 S1

52

ENTROPY


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Calculation of entropy change - Pure substance

ENTROPY

dU= q+ w

q= dU+ Pext dV

qrev = dU+ P dV S(V, T)

dS= qrev

T

53

ENTROPY


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dS= qrev

T

ENTROPY

dS=S

T

VdT+

SV

TdVS(V, T)

dS=

dU

T +

P

T dVqrev = dU+ P dV


54

ENTROPY


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dS=

dU

T +

P

T dV

ENTROPY

dS=

S

T

V

dT+

S

V

T

dV


dU =

U

T

V

dT+

U

V

T

dV

but...

dS=1

T

U

T

V

dT+1

T

U

V

T

dV+P

TdV

55

ENTROPY


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ENTROPY

dS=

CV

T dT+PT +

1

TU

V

T

dV

dS=1

T

U

T

V

dT+1

T

U

V

T

dV+P

TdV


56

ENTROPY


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ENTROPY

dS=CV

TdT+

P

T+1

T

U

V

T

dV

dS=

ST

V

dT+

SV

T

dV

S

T

V

=

CV

T

S

V

T

=

P

T+1

T

U

V

T

dV


57

ENTROPY


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ENTROPY

S

T

V

=

CV

T

S

V

T

=

P

T+1

T

U

V

T

dV


2f(x, y)

xy=

2f(x, y)

yx

for a state function

V

S

T

V

T

=

T

S

V

T

V

58

ENTROPY


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ENTROPY

S

T

V

=CV

T

SV

T

=PT+ 1

T

UV

T

dV


V

ST

V

T

= T

SV

T

V

1

T

V

U

T

V

T

=

T

P

T+1

T

U

V

T

V

59

ENTROPY


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ENTROPY

1

T

V

U

T

V

T

=

T

P

T+1

T

U

V

T

V


1

T

2U

VT=1

T

P

T

P

T2

1

T2

U

V

+1

T

2U

TV

1

T

P

T+1

T

U

V

T

=1

T

P

T

V

60

ENTROPY


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ENTROPY

S

V

T

=

P

T+1

T

U

V

T

dV

S

V

T

=

P

T

V


1

T

P

T+1

T

U

V

T

=1

T

P

T

V

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HOW TO CALCULATE ENTROPY CHANGES


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S for an isothermal expansion of a perfect gas from V1 to V2 at temperature T


dS=

ST

V

dT+

PT

V

dVdS=

ST

V

dT+

SV

T

dV

dS=

P

T

V

dV

PV= nRT

Perfect gas P

T

V

=

nR

V

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dS=

ST

V

dT+

PT

V

dVdS=

ST

V

dT+

SV

T

dV

P

T

V

=nR

VS= nR

V2

V1

dV

V

S= nR lnV2

V1

independent of the path


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S for a process occurring at constant volume


dS=

ST

V

dT+

PT

V

dVdS=

ST

V

dT+

SV

T

dV

dS=

S

T

V

dT

S

T

V

=

CV

T

From last lecture...

dS=

S

T

V

dT =CV

TdT

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dS=

ST

V

dT+

PT

V

dVdS=

ST

V

dT+

SV

T

dV

dS=

S

T

V

dT =CV

TdT S=

T2

T1

CVdT

T

S= CV lnT2

T1

(assuming the CV is independent of T)

S for a process occurring at constant volume


65

FEW OTHER DEFINITIONS & RELATIONS


66/92

dS=dH

T

V dP

T

dS=dU

T

+P

T

dV

dU= dH PdV V dP

dS=dH

T

PdV

T

V dP

T+PdV

T


dS=qrev

T

dH= dU+ PdV+ V dPH= U+ PV

But...

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dS=dU

T

+P

T

dV

dU= dH PdV V dP


dS=qrev

T

H= U+ PV

But...

dH= dU+ PdV+ V dP

dS=dH

T

V dP

T

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dS=dH

T

V dP

T

CP =

H

T

P

&

dH= HT

P

dT+ HP

T

dP

But...

S=CP

TdT+

V

T+1

T

H

P

T

dP

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dS= S

TP

dT+ S

PT

dP

S=

CP

T dT+

V

T +

1

TH

P

T

dP

S(T, P)

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dS= S

TP

dT+ S

PT

dP

S=

CP

T dT+V

T +

1

TH

P

T

dP

S

P

T

= V

T+1

T

H

P

T

S

TP

=

CP

T

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72/92

dS= S

T

P

dT+ S

P

T

dP

S

T

P

=

CP

T

S

P

T

= V

T+1

T

H

P

T

not measurable

S

P

T

=

V

T

P

= V

=1

V

V

T

P

thermal expansion

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dS=

ST

P

dT+

SP

T

dP


PV= nRT

Perfect gas

dS=

S

P

T

dP =

V

T

P

dP

V

T

P

=

nR

P

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74/92

dS=

ST

P

dT+

SP

T

dP

VT

P

=

nR

P


S= nR

RT/V2RT/V1

dP

P

dS=

nR

dP

P

S= nR lnV2

V1

independent of the path

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75/92

S= CP lnT2

T1

dS=CP

TdT

dS=S

T

P

dT+ S

P

T

dP

S for a process occurring at constant pressure

dS=

S

T

P

dTS

T

P

=

CP

T

S=

T2

T1

CPdT

T

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76/92

We can use either S(T,V) or S(T,P)

dS=

S

T

V

dT+

S

V

T

dV

dS= CV

TdT+

PT

V

dV

dS=CV

T

dT+nR

V

dV

S for a perfect gas from (V1, T1, P1) to (V2, T2, P2)

P

TV

=

nR

V

S= CV lnT2

T1+ nR ln

V2

V1

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q = qrev = 0

S for an adiabatic reversible expansion of a gas

Pext

V

T1

T2

P1, V1

P2, V2

S=?

dS=qrev

T

S = 0

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78/92

S for an adiabatic irreversible expansion of a gas

dS=qrev

T

Pext

V

T1

T2

P1, V1

P2, V2

S=?

q = qirrev = 0

S= 0

S must be calculated over a reversible path that connects theinitial and final states of the irreversible adiabatic expansion

78

THERMODYNAMIC TEMPERATURE SCALE


79/92

Up to now, ideal gas temperature scale

depends on the substance

We should not have to rely on any particular substance or eq. of state to define atemperature

Carnots cycle (independent of the substance)

rev = 1 +qC

qH= f(C, H)

qC

qH

=C

H

Choose a reference temperature (triple point of water)

= tr |q|

|qtr|

T = 273.16K|q|

|qtr| tr = 273.16

79

REVERSIBILITY & IRREVERSIBILITY


80/92

Suniv = Ssys +Ssurr

Tsys = Tsurr

Reversible processes

system

surroundings

dSuniv = dSsys + dSsurr =qsys

Tsys

+qsurr

Tsurr

Suniv = 0

=qrev

Tsys+qrev

Tsurr

qrev

80



81/92

reversibleisotherm, T

V

P

1

2

1 3

reversibleadiabat

2irreversible adiabat

Irreversible adiabatic processes (closed system)

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82/92


V

P4

3 reversibleadiabat

1


reversibleadiabat

S12 +S23 +S34 +S41 = 0

S23 = S41 = 0

reversible adiabats

S12 = S34 = q34

TdU =

(q+ w) = q34 + w = 0

q34 > 0

Not possible!

it violates the 2nd law of thermodynamics (Kelvins statement)

w = q34


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V

P4

3 reversibleadiabat

1


reversibleadiabat

S12 +S23 +S34 +S41 = 0

S23 = S41 = 0

reversible adiabats

S12 = S34 = q34

T

w = q34

dU =

(q+ w) = q34 + w = 0

S12 0


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84/92


V

P4

3 reversibleadiabat

1


reversibleadiabat

S12 +S23 +S34 +S41 = 0

S23 = S41 = 0

reversible adiabats

S12 = S34 = q34

T

w = q34

dU =

(q+ w) = q34 + w = 0

S12 = 0 S12 > 0


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S > 0

Irreversible adiabatic processes (isolated system)

S > 0

The universe can be considered as an isolated system

Suniv = Ssys +Ssurr > 0 irreversible process

Suniv 0

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MOLECULAR CONNECTION


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molecular kinetic energyT

all energies [kinetic, potential (electronic,vibrational, rotational, nuclear, ...)]

U

?S ?dS=dqrev

T

Higher S more probable states

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probability ?

1/4

probability ?

1/4

probability ?

1/4

probability ?

1/4

=

1/2probability


A more uniform distribution much more probable than a very uneven one

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probability

1/4

probability

1/4

=1/2

probability


macroscopic states

A macroscopic state is associated with many microscopic states

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probability

1/4

probability

1/4

=1/2

probability


What is a microscopic state?

Depends on the system

For an ideal gas

If you know at an instant of time where every molecule is and what its velocity is youhave the full microscopic state

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Some macroscopic states have many more microscopic states consistent with it(e.g, same P, T, components)

Let p = probability of a thermodynamic state

p =num er o m crostates cons stent w t a g ven t ermo ynam c state

total number of microstates

k = Boltzmanns constant

true for anything

entropy is associated with themicroscopic states of highest probability

S= k lnp

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S= k lnp

order / disorderS

more disorder more probable more microstates

S= k ln p2

p1S=

21

qrevT

difficult !!!

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= lnp

order / disorderS

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Chem131_Chapter3