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Step 1: Isothermal Expansion (T constant, V1 à V2)

Since, Hence, Since V2 > V1 ,

so ΔQ1 > 0 è Heat is added to the gas

Carnot’s Process V2

V1=p1p2p

0 V

S1

S2 I

II

III

IV

v1 v4 v2 v3

Ideal Gas

ΔU1 = ΔW1 +ΔQ1 = 0

ΔQ1 = −ΔW1

dW = pV2

V2

∫1

2

∫ dV = −NkT dVV

= −NkT ln V2V1

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V1

V2

∫

ΔQ1 = −ΔW1 = NkTH lnV2V1

#

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&

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•  Every step is reversible •  Ideal Gas

(Ideal Engine)

Step 2: AdiabaKc expansion (Th à Tc , V2 à V3)

Th > Tc Since ΔQ2 = 0 (AdiabaKc), Recall, Ideal gas,

T, V independent

Carnot’s Process p

0 V

S1

S2 I

II

III

IV

v1 v4 v2 v3

Ideal Gas V3V2

=ThTc

!

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32

ΔW2 = ΔU2

dUdT

=Cv

ΔU =Cv (Tc −Th )

Cv =3Nk2

Step 3: Isothermal Compression (T constant , V3 à V4)

Since V4 < V3 ,

so ΔQ3 < 0 è Heat loss from the gas


0 V

S1

S2 I

II

III

IV

v1 v4 v2 v3

Ideal Gas V4V3=p3p4

ΔU3 = 0

ΔW3 +ΔQ3 = 0

ΔQ3 = −ΔW3 = NkTc lnV4V3

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Step 4: AdiabaKc compression (Tc à Th , V4 à V1)

Th > Tc Since ΔQ4 = 0,


0 V

S1

S2 I

II

III

IV

v1 v4 v2 v3

Ideal Gas

V1V4

=TcTh

!

"#

$

%&

32

ΔW4 = ΔU4 =Cv (Th −Tc )

Overall: Energy balance: Recall,

and Hence, Since,

è


0 V

S1

S2 I

II

III

IV

v1 v4 v2 v3

Ideal Gas ΔUtotal = ΔQ1 +ΔW1 +ΔW2 +ΔQ3 +ΔW3 +ΔW4

V3V2

=ThTc

!

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$

%&

32 V1

V4=TcTh

!

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32

V3V2

=V4V1

ΔQ1 = −ΔW1 = NkTH lnV2V1

#

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ΔQ3 = −ΔW3 = NkTc lnV4V3

#

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ΔQ1Th

+ΔQ3

Tc= 0

valid for any REVERSIBLE cyclic process

Overall:

Can be wriRen as: Entropy (extensive state funcKon)

à a measure of disorderess


0 V

S1

S2 I

II

III

IV

v1 v4 v2 v3

Ideal Gas ΔQ1Th

+ΔQ3

Tc= 0

valid for any REVERSIBLE cyclic process

δQrev

T!∫ = 0

dS = δQrev

T

Efficiency

SubsKtute….

…

Hence if ΔT é, η é But there’s always heat loss through path 3

So η < 1


0 V

S1

S2 I

II

III

IV

v1 v4 v2 v3

Ideal Gas

ΔW = ΔW1 +ΔW2 +ΔW3 +ΔW4

η =Th −TcTh

η =1− TcTh

Hess’s Law •  ΔH is a state funcKon, therefore, independent of path •  One can add enthalpies for a series of steps to get the desired heat of

reacKon

Thermochemistry

A+B

C

E

ReacKon coordinate

ΔE A+B

C

ΔH < 0 exothermic

ΔH > 0 endothermic

Example: Assume ideal mixing (well-‐mixed) Breakdown to elementary equaKon:

Thermochemistry 2MgO(s)+ SiO2 (s)→Mg2SiO4 (s)

2Mg(s)+ Si(s)+ 2O2 (g)→Mg2SiO4 (s)

Mg(s)+ 12O2 (g)→MgO(s)

Si(s)+O2 (g)→ SiO2 (s)

2Mg(s)+ Si(s)+ 2O2 (g)→Mg2SiO4 (s)

2MgO(s)+ SiO2 (s)→Mg2SiO4 (s)

ΔHo = -‐600.7 kJ

ΔHo = -‐910.6 kJ

ΔHo = -‐2174.2 kJ

ΔHo = -‐62.2 kJ

2x (-‐)

(-‐)

(+)

T dep. of ΔH Given ΔH(T1), Cp(A), Cp(B), Cp(C), Cp(D)

Thermochemistry

aA+ bB→ cC + dD

aA+ bB→ cC + dD

(T1) (T1)

(T2) (T2)

(1)

(2)

(3)

(4)

ΔH1 = ΔH4 +ΔH3 +ΔH2

ΔH4 = aCp(A)+ bCp(B)"# $%dT = − aCp(A)+ bCp(B)"# $%dTT1

T2

∫T2

T1

∫

ΔH3 =W = ΔH (T1)

ΔH2 = cCp(C)+ dCp(D)"# $%dTT1

T2

∫

Example: H2O (l) à H2O (g) at 1 atm, 373K ΔHvap = 9717 cal (1 cal = 4.2 joules) Liquid V = 0.018 L mol-‐1

Cp(l) ≈ 18 cal mol-‐1 K-‐1

Cp(g) ≈ 8 cal mol-‐1 K-‐1

Assume ideal behavior Calculate ΔH, ΔU, q, W when 1 mol of liquid water at 1 atm, 298 K(room temp) 600 K, 1 atm

Thermochemistry

Δ,rev" →""Heat slowly

Entropy of a reversible process is constant, dS = 0

Entropy of an irreversible process in an isolated system which evolves towards thermodynamic equilibrium always increases à maximum entropy is reached, when equilibrium is reached.

dS > 0 Second law, dS ≥ 0 ( = only for reversible process)

Second Law

ΔS = qT

1 2 q

T1 > T2

gives heat, ΔS ê, so –ve

gets heat, ΔS é, so +ve

ΔS1 =qT1

ΔS2 =qT2

ΔS1−2 = −qT1+qT2> 0

… is the science to use staKsKcal techniques to connect microscopic behavior with macroscopic properKes Microscopic scale : •  Approx. 1023 atoms/molecule •  Each has at least 6 degrees of freedom (x, y, z, px, py, pz) •  Completely intractable •  MICROSTATE – exact definiKon (all posiKons/momenta) of the microscopic

scale

Macroscopic scale: •  p, V, T, N, Cv, Cp etc. •  MACROSTATE

StaKsKcal Mechanics

DefiniKon: Ω (N, V, E) – number of microstates à number of states the system has for given N, V, E

Entropy (Microscopic InterpretaKon)

A B

p = 12!

"#$

%&N

N = # of balls

Vi

Vf One atom: N atoms: Vf > Vi è Ω(Vf) > Ω(Vi)

ω(Vi ) = βVi

Ω(Vi ) =ω(Vi )N = (βVi )

N

Ω(Vf ) = (βVf )N

ProporKonality factor

Higher probability Lower probability

ConnecKng S to Ω Ωtot = Ω1 Ω2 S = S1 + S2 Assume, S = f(Ω) è S α ln Ω S = k ln Ω where k = kb = Boltzmann constant

The entropy is a measure for the number of possible microstates of a system when a macrostate is given Arguments: S measures “disorder” Ω increases with “disorder”


S = k ln Ω Why is S logarithmically dependent on Ω? Ω measures phase space volume S measures our lack of knowledge going from microstate to macrostate Microscopic à Macrostate (S is the negaKve of informaKon) ** Phase space: Space of ___ dimensions, spanned by all posiKons and momenta, energy point corresponds to a microstate of the system. 2nd Law: (a)  There is no perpetual movement (perpetuum mobile) of the second kind –

i.e., it is an engine which transforms heat into work with 100% efficiency (b)  Each isolated macroscopic system wants to assume the most probable

state (largest # of microscopic realizaKon possibiliKes).


Losing informaKon

S (Vi) = k ln Ω(Vi) S (Vf) = k ln Ω(Vf)

Entropy

ΔS = S(Vf )− S(Vi )

= k lnΩ(Vf )− k lnΩ(Vi )

= k ln(βVf )N − k ln(βVi )

N

= kN lnVf

Vi

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ΔS = R lnVf

Vi

"

#$

%

&'

So, Constant P, Hence,

Entropy dS = dq

T

dS = dHT

dH = dU + pdV

dU = dq− pdV dS = dHT

dS = dHdT

1T!

"#

$

%&dT

dS =Cp

TdT

dSS(Ti )

S(Tf )

∫ =Cp

TdT

Ti

Tf

∫

ΔS =Cp lnTfTi

!

"#

$

%&

S(Tf ) = S(Ti )+Cp lnTfTi

!

"#

$

%&

ΔS

TfTi1

Change with Cp

S à 0 as T à 0K Third law of thermodynamics: “The entropy of a perfect crystal, at absolute zero kelvin, is exactly equal to zero” It is impossible to reach absolute “0” in a finite number of steps

T à 0k, Cp, Cv à 0 From 0 to Tf, When Tf à 0 For S(Tf) à 0, Cp à 0

Third Law

S(Tf ) = S(0)+Cp

TdT

0

Tf

∫

S(Tf ) =Cp

TdT

0

Tf

∫

S

T

Zero slope

Fundamental thermodynamic relaKon: Systems which experiences change in composiKon, must include composiKon variables… Chemical potenKal where ni, ni… are the number of moles of species i, j,…present in the system

Combining 1st Law and 2nd Law

∂G∂ni

"

#$

%

&'T ,p,nj ,...

= µi

∂G∂ni

"

#$

%

&'T ,p,nj ,...

=∂U∂ni

"

#$

%

&'T ,p,nj ,...

=∂H∂ni

"

#$

%

&'T ,p,nj ,...

=∂A∂ni

"

#$

%

&'T ,p,nj ,...

= µi

dU = TdS − pdV + µ∑ idni

dH = TdS +Vdp+ µ∑ idni

dA = −SdT − pdV + µ∑ idni

dG = −SdT +Vdp+ µ∑ idni

dU = TdS − pdV

Because all the natural variables of the internal energy, U are extensive quanKKes, from Euler’s homogeneous funcKon theorem,

Euler’s equaKon

DifferenKal of Euler’s equaKon, Compare:

Gibbs-‐Duhem relaKon

Euler’s EquaKon

dU = TdS − pdV + µ∑ idni

U = TS − pV + µ∑ iNi

dU = TdS − pdV + µidi∑ Ni + SdT −VdP + Nid

i∑ µi

SdT −VdP + Nidi∑ µi = 0

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