Chapter 13 (week 8) Thermodynamics and Spontaneous Processes Final Ex Ch 9, 10, ( , 11.5), , 13(Ex.5 &.8), 14(Ex.4 &.8), Entropy (S): measure of disorder Absolute Entropy S=k B ln  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 P=const and T const  S=  H/T;  S fus =  H fus /T M  S vap =  H vap /T B Gibbs Free Energy: G= H -TS  G=  H – T  P=const and T=const  G < 0 for Spontaneous Processes  H 0 always spontaneous  G = 0 for Equilibrium Processes(  S=  H/T)  G > 0 the Reverse is a Spontaneous Processes So for A  B spontaneously Equilibrium A  B G(A)=G(B) Equilibrium constant K(T)=exp{-  G°/RT}  G° is the standard Gibbs free energy for any rxn dD+bB  fF + eE In terms of the activity (a) K(T)=(a F ) f (a E ) e /(a D ) d (a B ) b for gases a E = P E (atm) in soln a E = [E](molar), for pure liquids/solids a E = 1

Heat transfer from the surroundings (T sur ) to the system (T sys ) System (Ideal Gas) Surroundings q If T sur ≠T sys energy flows form hot to cold spontaneously? By the 1 st Law Energy could flow from cold to hot! But by the 2 nd Law  S uni > 0 Entropy always increases in a spontaneous processes! So what is Entropy?

1.Reversible Isothermal Expansion Of an Ideal Gas Irreversible Expansion V 1 V V+dV V 2 dV q = nRTln(V 2 /V 1 )

Consider the Reversible Isothermal Expansion of an Ideal Gas: So T=const for the system.  U = q + w = 0 q = -w = (P ext  V) Heat must be transferred from the Surroundings to the System And the System does work on the surroundings If the process is reversible, P ext ~ P=nRT/V Which means done slowly and changes are infinitesimal  V  dV: The gas expands from V 1 to V 2 then  q ~ PdV= (nRT/V)dV = (nRT)(dV/V)=(nRT)dlnV since dlnV=dV/V Integrating from V 1 to V 2 q = nRT[lnV 2 - lnV 1 ]=nRTln(V 2 /V 1 ) (q) Heat transferred in the Reversible Isothermal Expansion of an Ideal Gas q = nRTln(V 2 /V 1 )

Carnot will use both Isothermal and Adiabatic q = 0,  U=w = nc v  T processes to define the entropy change  S Reversible P ext ~ P=nRT/V  T  dT and  V  dV nc v dT = - P dV nc v (dT/T)=- nRdV/V c v dln(T) =-R dln(V) c v ln(T 2 /T 1 )=Rln(V 1 /V 2 ) T2T2 T1T1 V1V1 V2V2 q=0

Carnot will use both Isothermal and Adiabatic q = 0 processes to define the entropy change  S For an Ideal gas  U = q + w =w  U=nc V  T = -P ext  V Reversible Process P ext ~ P=nRT/V  T  dT and  V  dV nc V dT= -PdV=- nRTdV/V nc v (dT/T)=-nRdV/V dx/x=dlnx so T=const q=0

Carnot will use both Isothermal and Adiabatic q = 0 processes to define the entropy change  S nc v (dT/T)=-nRdV/V c v dln(T) =-R dln(V) c v ln(T 2 /T 1 )=Rln(V 1 /V 2 ) (T 2 /T 1 ) C v = (V 1 /V 2 ) R (T 2 /T 1 ) = (V 1 /V 2 )  -1  =c p /c v Ratio of Specific Heats T2T2 T1T1 V1V1 V2V2

Carnot considered a cyclical process involving the adiabatic/Isothermal Compression/Expansion of an Ideal Gas: The so called Carnot cycle/Heat Engine This Yields (q h /T h )–(q l /T l ) = 0 and therefore  S = q/T and that S is a State Function. Since  S AC = q h /T h and  S CD =q l /T l are Isothermal and  S BC =  S AD = 0 since they are reversible and adiabatic(ad) ThTh TlTl qhqh ad qlql

Carnot considered a cyclical process involving the adiabatic/Isothermal Compression/Expansion of an Ideal Gas: The so called Carnot cycle Which Yields (q h /T h ) –(q l /T l ) = 0 and therefore define  S = q/T and so that S is a State Function Since q = nRTln(V 2 /V 1 ) then  S = q/T= nRln(V 2 /V 1 ) ThTh TlTl q h ;T h ad q l: : T l ad

Entropy Change for Reversible Isothermal process in an ideal Gas expansion:  S = q/T= nR ln(V 2 /V 1 ) q is the heat transferred reversibly! Consider one Molecule going from V 1 to V 2 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 U In 3-D U = (1/2m){(p x ) 2 + (p y ) 2 + (p z ) 2 }, for In 1-D motion it is ±p x = U 1/2 so   U  L Boltzmann Entropy

State =Royal Flush AKQJ10 1 suit  =4 micro states State=Straight flush <3 1 suit  =36=10*4-4 State =  of a kind  =624 52*3*2*1*48 4*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 5 Combo 5,598,960 POKER ENTROPY

W=  S =R ln   S

An Irreversible Process

For 3-D  ~ [(U   L] 3 = U  L 3 for one (mon-atomic) particle And now for N such independent particles   ~ U  L 3N But L 3 = V and since the process is isothermal U=(3/2)NkT = constant

  = # U    V 1 ) N and   = # U    V 2 ) N # is some arbitrary collection of constants. Going from Volume V 1 to V 2 and T=const Boltzmann Absolute Entropy: S=kln  S 1 = kln {# U   ( V 1 ) N } And S 2 = k ln {# U    V 2 ) N }

Absolute entropy S=kln   S=S 2 – S 1 = kln [#(U     V 2 ) N ] / [#(U     V 1 ) N ] S 2 –S 1 = kln(V 2 ) N / (V 1 ) N =kln(V 2 /V 1 ) N  S=Nkln(V 2 /V 1 )=nN 0 kln(V 2 /V 1 ) R=N 0 k and n =N/N 0  S= n Rln (V 2 /V 1 ) Same result as the Isothermal Expansion using  S=q rev /T consistent S=kln 

 S=S 2 – S 1 for Heat Transfer Processes at P=const and V=const Much easier to use  S =∫dq rev /T than S=kln  For Isothermal Processes:  S =(1/T) ∫dq rev =(1/T) q rev q rev =∫dq rev For heat transfer at V=const with no reactions or phase transitions dq rev = nc v dT and  S =∫ nc v dT/T = ∫ nc v d(lnT) If c v ≠f(T) then  S =nc v lnT 2 /T 1

 S=S 2 – S 1 for Heat Transfer Processes at P=const and V=const Much easier to use  S =∫dq rev /T than S=kln  For Isothermal Processes:  S =(1/T) ∫dq rev =(1/T) q rev q rev =∫dq rev For heat transfer at P=const with no reactions or phase transitions dq rev = nc p dT and  S =∫ nc p dT/T = ∫ nc p d(lnT) If c v ≠f(T) then  S =nc p lnT 2 /T 1

Electron Translation and Atomic Vibrations (phonons: sound particles like photons are light particles) Store Energy in Metals T S°(T) = 0 ∫ nc p dT/T Standard Entropy If no phase transition and P=const

Electron Translation and Atomic Vibrations(phonons) Store Energy in Metals T S°(T) = 0 ∫ nc p dT/T + n  S fus + n  S vap 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 gas  U = q + w =w  U=nc V  T = -P ext  V Reversible Process P ext ~ P=nRT/V  T  dT and  V  dV nc V dT = - P dV

Carnot will use both Isothermal and Adiabatic q = 0 processes to define the entropy change  S Reversible Process P ext ~ P=nRT/V  T  dT and  V  dV nc v dT = - P dV nc v (dT/T)=-nRdV/V c v dln(T) =-R dln(V) c v ln(T 2 /T 1 )=Rln(V 1 /V 2 ) T2T2 T1T1 V1V1 V2V2

c v ln(T 2 /T 1 )=-Rln(V 1 /V 2 ) A simpler form cab be obtained c v ln(T 2 /T 1 )= ln(T 2 /T 1 ) C v =Rln(V 1 /V 2 )= ln(V 1 /V 2 ) R (T 2 /T 1 ) C v = (V 1 /V 2 ) R Recall that c p = c v + R (T 2 /T 1 ) C v = (V 1 /V 2 ) R = (V 1 /V 2 ) C p -C v let  =c p /c v (T 2 /T 1 ) = (V 1 /V 2 )  -1 this relationship is very important for adiabatic processes in engines

S°/2 JK -1 mol -1