1. Carnot Cycle State 1 T hot State 2 T hot State 4 T cold State 3 T cold isothermal adiabatic isothermal q=0, w=C V  T q=0, w=-C V  T w= -nRT hot ln(V2/V1) q=-w w= -nRT cold ln(V4/V3) q=-w

2. Carnot Cycle: PV Diagram Volume Pressure 1 2 3 4 -nRT hot ln(V 2 /V 1 ) -C V  T -nRT cold ln(V 4 /V 3 ) C V  T

3. Carnot Cycle, Schematic View T hot T cold E 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.

4. 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, Prussia d. Aug. 24, 1888, Bonn "Heat cannot of itself pass from a colder to a hotter body."

5. Carnot Cycle: Step 1 State 1 T hot State 2 T hot State 4 T cold State 3 T cold isothermal adiabatic isothermal Isothermal reversible Expansion:  E=0Energy of an ideal gas depends only on temperature q= -w Heat converted to work! T hot

6. Carnot Cycle: Step 2 State 1 T hot State 2 T hot State 4 T cold State 3 T cold isothermal adiabatic isothermal Isothermal reversible Expansion: q=0No heat transferred in adiabatic process  T here defined: T hot - T cold = positive value Temp change for process is -  T : dT = negative Energy lost to expansion

7. Carnot Cycle: Steps 3 & 4 State 1 T hot State 2 T hot State 4 T cold State 3 T cold isothermal adiabatic isothermal Work done in compression

8. Carnot Cycle: Summary Step1Step2Step3Step4 w-nRT hot ln(V 2 /V 1 )-C V  T-nRT cold ln(V 4 /V 3 ) C V  T q-w0-w0  E0w0w w 1 + w 2 + w 3 + w 4 = -nRT hot ln(V 2 /V 1 ) - nRT cold ln(V 4 /V 3 ) q 1 + q 2 + q 3 + q 4 = nRT hot ln(V 2 /V 1 ) + nRT cold ln(V 4 /V 3 ) q = -w Heat flows through system, some work is extracted. T hot T cold E

9. 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/q hot We already know w = -(q hot + q cold ), therefor... 1)  = (q hot + q cold )/q hot = 1 + q cold /q hot Substituting in expressions for q hot and q cold 2)  = 1 + nRT cold ln(V 4 /V 3 ) / nRT hot ln(V 2 /V 1 )= 1+(T cold /T hot )(ln(V 4 /V 3 ) / ln(V 2 /V 1 ))

10.  = 1 + nRT cold ln(V 4 /V 3 ) / nRT hot ln(V 2 /V 1 )= 1+(T cold /T hot )(ln(V 4 /V 3 ) / ln(V 2 /V 1 )) This second expression is rather complicated-- but we have a relationship for the volumes in this process:  = 1 - (T cold /T hot ) Rearranging and plugging it in to the first equation on the page:

11.  = 1 - (T cold /T hot ) = 1 + (q cold /q hot ) 100% efficiency is only achieved when T cold =0 and/or T hot =  Practical impossibilities. (Foreshadowing the Third Law) Can we construct an engine more efficient than this one?

12. Impossible Machines Consider two engines E R and E’ operating between the same two reservoirs: T hot T cold ERER E’ Can the efficiencies of these two engines be different? We operate E R in reverse. We couple the operation of the two engines. We know: w R = q hot + q cold (forward direction) w’= q’ hot + q’ cold The composite engine then has a total work W= w’-w R Let’s couple the engines so that work from forward running engine drives the other one backwards with no work on the surroundings: w’ = w R (remember actual work of reverse process is -w R )

13. Impossible Machines Can the efficiencies of these two engines be different? Let’s assume  ’>  R :w’/q’ hot >w R /q hot But w’=w R soq’ hot < q hot Net heat transferred from hot reservoir = q’ hot - q hot 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?

14. The Carnot Cycle: Noticing state functions Assumptions in our proof: 1) The first law:Experimentally proven. 2) w’=w R A fully practical assumption 3)  ’>  R This assumption is disproven by contradiction Therefor we have proven:  ’   R Now we assumed that E’ was any engine, whereas, E R was reversible. So every engine is either of equal or less efficienct than a reversible engine (For the same two reservoirs)

15. Carnot Cycle: Noticing State Functions  = 1 - (T cold /T hot ) = 1 + (q cold /q hot ) Now--- we have this interesting relationship between temperature and heat for these systems: What Carnot noticed was that there was an implied state function here! (q hot /T hot ) = -(q cold /T cold ) or (q hot /T hot ) + (q cold /T cold ) = 0 This can also be written: This is a state function! Clausius called it the Entropy, S.

16. 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

17. Entropy is Not Conserved Vacuum 2 Atm T V 1 1Atm T 1 Atm T V 2 =2V 1 Two cases of expansion of an ideal gas: 1) Expansion in to a vacuum. 2) Reversible expansion (1)w=0,  E=0, q irreversible =0  S(surroundings)=0 To calculate  S(system) we look to the second process (they must be the same).

18. Entropy is Not Conserved In the second process we follow an isothermal, reversible path. We know that  T and thus  E=0 Now…q rev =  E - w= RT ln(2) so  S(system)= q rev /T= R ln(2) Vacuum 2 Atm T V 1 1Atm T 1 Atm T V 2 =2V 1

19. For the reversible process we’ve already calculated q rev = RT ln(V2/V1) = RT ln(2)  S(system) = q rev /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) = -q rev /T  S(total) = 0 For the total irreversible process,  S(surroundings) = 0 (nothing exchanged)  S (total) = 0 + R ln(2) > 0 Entropy is Not Conserved

20. Dissorder and Entropy Ludwig Boltzmann c. 1875 It turns out that disorder and entropy are intimately related. We start out by considering the spontaneity of this process. Why doesn’t the gas spontaneously reappear back in the box?

21. Dissorder and Entropy * Let’s 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 N>M for this problem. Number of ways of distributing M indistinguishable balls in N boxes is approximately:  ~ N M /M! (this approx. miscounts when multiple balls in same box) Boltzmann noted that an entropy could be defined as S= k ln(  )= R ln(  )/N A There are a number of reasons this is a good definition. One is it connects to thermodynamics.

22. So for a given state we have S= k ln(  )= R ln(  )/N A = R ln(N M /M!)/N A Let’s say we change state by increasing the volume. Well, for the same sized cells, N increases to N’. S’-S= (R/N A ) (ln(N’ M /M!) - ln(N M /M!)) =(R/N A ) ln(N’ M /N M ) So  S= (R/N A ) ln(N’ M /N M ) = M (R/N A ) ln(N’/N) And since N is proportional to volume:  S = M (R/N A ) ln(V2/V1) Dissorder and Entropy

23. Entropy of Materials Diamond S° 298 = 2.4 J/K Graphite S° 298 = 5.7 J/K How about water in its different phases: S° 298 (J/K mol) H 2 O(s,ice)44.3 H 2 O(l)69.91 H 2 O(g)188.72 Why does graphite have more entropy than diamond?

24. Entropy of Mixing What happens when you mix two ideal gases? What happens when you solvate a molecule?

25. Entropy and Chemical Reactions CO 3 2- (aq) + H + (aq)HCO 3 - (aq)148.1 J/K mol HC 2 O 4 - (aq) + OH - (aq)C 2 O 4 2- + H 2 O(l)-23.1 SS 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.

26. 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.