Chapter 13 (week 8) Thermodynamics and Spontaneous Processes

Final Ex Ch 9, 10, (11-1-11.3, 11.5), 12.-.7, 13(Ex .5 & .8), 14(Ex .4 & .8), 15.-15.3• Entropy (S): measure of disorder

Absolute Entropy S=kBln Number of Available Microstate (N,U,V)

• Second Law of Thermodynamics: Heat cannot be transferred from cold to hot without work S ≥ q/T Inequality of Clausius

• Phase Transitions: @ P=const and T constS=H/T; Sfus=Hfus/TM Svap=Hvap/TB

• Gibbs Free Energy: G= H -TS G= H – TS@ P=const and T=const G < 0 for Spontaneous Processes

H< and S>0 always spontaneous G = 0 for Equilibrium Processes(S=H/T) G > 0 the Reverse is a Spontaneous Processes

So for AB spontaneously G(A)<G(B)@ Equilibrium AB G(A)=G(B)

• Equilibrium constant K(T)=exp{-G°/RT} G° is the standard Gibbs free energy for any rxn

dD+bB fF + eEIn terms of the activity (a) K(T)=(aF)f(aE)e/(aD)d(aB)b for gases aE = PE(atm)in soln aE = [E](molar), for pure liquids/solids aE = 1

Heat transfer from the surroundings (Tsur) to the system (Tsys)

System(Ideal Gas)

Surroundings

q

If Tsur ≠Tsys energy flows form hot to cold spontaneously?By the 1st Law Energy could flow from cold to hot!But by the 2nd Law Suni > 0 Entropy always increases in a spontaneous processes! So what is Entropy?

1. Reversible Isothermal ExpansionOf an Ideal Gas

IrreversibleExpansion

V1 V V+dV V2

dV

q = nRTln(V2/V1)

Consider the Reversible Isothermal Expansion of an Ideal Gas:So T=const for the system. U = q + w = 0

q = -w = (Pext V)

Heat must be transferred from the Surroundings to the System And the System does work on the surroundings

If the process is reversible, Pext ~ P=nRT/VWhich means done slowly and changes are infinitesimalV dV: The gas expands from V1 to V2 then

q ~ PdV= (nRT/V)dV = (nRT)(dV/V)=(nRT)dlnV since dlnV=dV/V

Integrating from V1 to V2 q = nRT[lnV2 - lnV1]=nRTln(V2/V1)

(q) Heat transferred in the Reversible Isothermal Expansion of an Ideal Gas

q = nRTln(V2/V1)

Carnot will use both Isothermal and Adiabatic q = 0,

U=w = ncvT processes to define the entropy change S

Reversible Pext ~ P=nRT/V

T dT and VdVncv dT = - P dVncv(dT/T)=- nRdV/V

cvdln(T) =-R dln(V) cvln(T2/T1)=Rln(V1/V2)

T2

T1

V1 V2

q=0

Carnot will use both Isothermal and Adiabatic q = 0

processes to define the entropy change S

For an Ideal gasU = q + w =w

U=ncV T = -Pext VReversible Process Pext ~ P=nRT/V

T dT and VdV

ncVdT= -PdV=-nRTdV/V

ncv(dT/T)=-nRdV/Vdx/x=dlnx so

T=const

q=0

Carnot will use both Isothermal and Adiabatic q = 0

processes to define the entropy change S

ncv(dT/T)=-nRdV/V

cvdln(T) =-R dln(V)

cvln(T2/T1)=Rln(V1/V2)

(T2/T1)Cv = (V1/V2)R

(T2/T1) = (V1/V2) -1

=cp/cv

Ratio of Specific Heats

T2

T1

V1 V2

Carnot considered a cyclical process involving the adiabatic/Isothermal

Compression/Expansion of an Ideal Gas: The so called Carnot cycle/Heat Engine

This Yields (qh/Th)–(ql/Tl) = 0 and therefore S = q/T and that S is a State Function. Since SAC= qh/Th and SCD=ql/Tl are Isothermal and SBC=SAD= 0 since they are reversible and adiabatic(ad)

Th

Tl

qh

ad qlad

Carnot considered a cyclical process involving the

adiabatic/Isothermal Compression/Expansion of an Ideal Gas:

The so called Carnot cycle

Which Yields (qh/Th) –(ql/Tl) = 0 and therefore define S = q/T and so that S is a State Function

Since q = nRTln(V2/V1) then S = q/T= nRln(V2/V1)

Th

Tl

qh;Th

ad ql:: Tl

ad

Entropy Change for Reversible Isothermal process in an idealGas expansion:

S = q/T= nR ln(V2/V1) q is the heat transferred reversibly! Consider one Molecule going from V1 to V2

Using the Boltzmann Entropy S=kln Find S!The number of ways that molecule could occupy the Volume is proportional to the size V so V If it’s 1 Dimensional motion, L

The number of possible momentum states is obtained from the internal energy UIn 3-D U = (1/2m){(px)2 + (py)2 + (pz)2}, for

In 1-D motion it is ±px = U1/2 so UL

Boltzmann Entropy

State =Royal FlushAKQJ10 1 suit=4 micro states

State=Straight flush12345 <3 1 suit=36=10*4-4

State = of a kind=62452*3*2*1*484*3*2*1

State=full house=3744

State= flush =5108

State =straight=10,200

State=3 of a kind=36=54,912

State = pairs= 123,552

State= 1 pair=1,098,240

State= high card =1,303,540

Total possible 5Combo 5,598,960

POKER ENTROPY

W=S =R ln

S

An Irreversible Process

For 3-D

~ [(UL]3 = UL3 for one (mon-atomic) particle

And now for N such independent particles

~ UL3N

But L3= V and since the process is isothermal

U=(3/2)NkT = constant

= # UV1) N and = # UV2) N

# is some arbitrary collection of constants.

Going from Volume V1 to V2 and T=const

Boltzmann Absolute Entropy: S=kln

S1 = kln {# U(V1) N}

And

S2 = k ln {# UV2) N}

Absolute entropy S=kln

S=S2 – S1 = kln [#(UV2) N] / [#(UV1) N]

S2–S1= kln(V2)N/ (V1)N =kln(V2/V1)N

S=Nkln(V2/V1)=nN0kln(V2/V1) R=N0k and n =N/N0

S= n Rln (V2/V1)

Same result as the Isothermal Expansion using

S=qrev/T consistent S=kln

S=S2– S1 for Heat Transfer Processes at P=const and V=const

Much easier to use S =∫dqrev/T than S=kln

For Isothermal Processes: S =(1/T) ∫dqrev=(1/T) qrev

qrev=∫dqrev

For heat transfer at V=const with no reactions or phase transitions

dqrev = ncvdT and S =∫ ncvdT/T = ∫ ncvd(lnT)

If cv≠f(T) then S =ncv lnT2/T1

S=S2– S1 for Heat Transfer Processes at P=const and V=const

Much easier to use S =∫dqrev/T than S=kln

For Isothermal Processes: S =(1/T) ∫dqrev=(1/T) qrev

qrev=∫dqrev

For heat transfer at P=const with no reactions or phase transitions

dqrev = ncpdT and S =∫ ncpdT/T = ∫ ncpd(lnT)

If cv≠f(T) then S =ncplnT2/T1

Electron Translation and Atomic Vibrations (phonons: sound particles like photons are light particles) Store Energy in Metals

T

S°(T) =0∫ ncpdT/T

Standard EntropyIf no phase transitionand P=const

Electron Translation and Atomic Vibrations(phonons) Store Energy in Metals

T

S°(T) =0∫ ncpdT/T + nSfus + nSvap

Standard Entropy with phase transition and P=const

Carnot will use both Isothermal and Adiabatic q = 0

processes to define the entropy change S

For an Ideal gasU = q + w =w

U=ncV T = -Pext VReversible Process Pext ~ P=nRT/V

T dT and VdV

ncV dT = - P dV

Carnot will use both Isothermal and Adiabatic q = 0

processes to define the entropy change S

Reversible Process Pext ~ P=nRT/V

T dT and VdV

ncv dT = - P dVncv(dT/T)=-nRdV/V

cvdln(T) =-R dln(V)

cvln(T2/T1)=Rln(V1/V2) T2

T1

V1 V2

cvln(T2/T1)=-Rln(V1/V2)

A simpler form cab be obtained

cvln(T2/T1)= ln(T2/T1)Cv =Rln(V1/V2)= ln(V1/V2)R

(T2/T1)Cv = (V1/V2)R

Recall that cp = cv + R

(T2/T1)Cv = (V1/V2)R = (V1/V2)Cp-Cv let =cp/cv

(T2/T1) = (V1/V2) -1 this relationship is very important foradiabatic processes in engines

S°/2JK-1

mol-1