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The Laws of Thermodynamics

The Laws of Thermodynamics. The Zeroth Law of Thermodynamics. “If two systems are separately in thermal equilibrium with a third system, they are in thermal equilibrium with each other .”. This allows the design & the use of Thermometers !. The First Law of Thermodynamics.

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The Laws of Thermodynamics

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  1. The Laws of Thermodynamics

  2. TheZeroth LawofThermodynamics “If two systems are separately in thermal equilibrium with a third system, they are in thermal equilibrium with each other.”

  3. This allows the design & the use ofThermometers!

  4. The First Law of Thermodynamics Q = ∆Ē + W l For Infinitesimal, Quasi-Static Processes đQ = dĒ+ đW Both of these forms are expressions of the Conservation of Total Energy. So The Physical Meaning of the 1st Law of Thermodynamicsis that Total Energy is Conserved Heat absorbed bythe system Work done bythe system Change inthe system’sinternal energy

  5. Rudolf Clausius’ statement of the 1st Law of Thermodynamics “Energy can neither be created nor destroyed. It can only be changed from one form to another.” Rudolf Clausius (1850) • The Physical Meaning of the 1st Law of Thermodynamics •  Conservation of Total Energy!!!! • However, it says nothing about • The Direction of Energy Transfer!

  6. The 2nd Law of Thermodynamics: “The entropy of an isolated system increases in any irreversible process & is unaltered in any reversible process.” • This is sometimes called • The Principle of Increasing Entropy • DS³ 0 • It gives the Preferred (natural) • Direction of Energy Transfer • It thus determines whether a process can occur or not. Change in system entropy

  7. Mathematically, this means that in any process ΔS ≥ 0 or, at equilibrium, S → Smax. For (idealized) reversible processes only, ΔS = 0, dS = đQ/T. Examples of irreversible (real) processes: Temperature Equalization, Mixing of Gases, & Conversion of Macroscopic (ordered) KE to Thermal (random) KE The last two cases are obvious examples of the Association of Entropy with Disorder. The Second Law of Thermodynamics: “The entropy of an isolated system never decreases”. Change in Entropy 7

  8. General Featuresof the Entropy S It is a state function, so that ΔS between given macrostates is independent of the path. It is a quantitative measure of the disorderin a system. It gives a criterion for the directionof a process, since an isolated system will reach a state of maximum entropy. ΔS may be negativefor a portionof a composite system. An increase in S does notrequire an increase in temperature. For example, in the mixing of gases at the same temperature, or in the melting of a solid at the melting point. An increase in temperature does not necessarily imply an increase in S. For example, in the adiabatic compression of a gas. 8

  9. Some Historical Comments • Much of the early thermodynamics development was driven by practical considerations. • For example, building heat engines & refrigerators. • So, the original statements of the • Second Lawof Thermodynamics • may sound very different than those just mentioned.

  10. Various Statements of the2nd Law of Thermodynamics: • “No series of processes is possible whose sole result is the absorption of heat from a thermal reservoir and the complete conversion of this energy to work.” That is There are NO perfect engines!

  11. Two of Rudolf Clausius’ statements of the 2nd Law of Thermodynamics • “It will arouse changes while the heat transfers from a low temperature object to a high temperature object.” or: • “It is impossible to devise an engine which, working in a cycle, shall produce no effect other than the transfer of heat from a colder to a hotter body.”

  12. “It will arouse other changes while the heat from the single thermal source is taken out and is totally changed into work.” • “It is impossible to extract an amount of heatQHfrom a hot reservoir and use it all to do workW. Some amount of heatQCmust be exhausted to a cold reservoir.” Lord Kelvin’s (William Thompson’s) statement of the2nd Law of Thermodynamics The Kelvin-Planck statement of the2nd Law of Thermodynamics

  13. “A universe containing mathematical physicists will at any assigned date be in the state of maximum disorganizationwhich is not inconsistent with the existence of such creatures.” Sir Arthur Eddington’s version of the 2nd Law of Thermodynamics!

  14. The 2nd Law of Thermodynamics The Heat Flow Statement “Heat flows spontaneously from a substance at a higher temperature to a substance at a lower temperature. It neverflows spontaneously in the reverse direction.”

  15. Definition: Heat Engine • A system that can convert some of the random molecular energy of heat flow into macroscopic mechanical energy. QH HEAT absorbed by a Heat Engine from a hot body -W WORK performed by a Heat Engine on the surroundings -QC HEAT emitted by a Heat Engine to a cold body

  16. The Second LawApplied to Heat Engines Efficiency (W/QH) = [(QH - QC)/QH]

  17. A “Heat Engine” That Violatesthe Second Law

  18. Definition: Refrigerator A system that can do macroscopic work to extract heat from a cold body & exhaust it to a hot body, thus cooling the cold body further. Or, A system that operates like a Heat Engine in reverse. QC HEAT extracted by a Refrigerator from a cold body W WORK performed by a Refrigerator on the surroundings -QH HEAT emitted by a Refrigerator to a hot body

  19. The 2nd Law of ThermodynamicsClausius’ Statement for Refrigerators • “It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.” • Or: There are no perfect Refrigerators! • This statement about refrigerators also applies to air conditioners & heat pumps which use the same principles.

  20. The Second LawApplied to Refrigerators Efficiency (QC/W) = [(QC)/(QH - QC)]

  21. Conceptual Example:You Can’t “Beat” the 2nd Law of Thermodynamics Question Is it possible to cool your kitchen by leaving the refrigerator door open or cool your bedroom by putting a window air conditioner on the floor by the bed?

  22. Conceptual Example:You Can’t “Beat” the 2nd Law of Thermodynamics Question Is it possible to cool your kitchen by leaving the refrigerator door open or cool your bedroom by putting a window air conditioner on the floor by the bed? Answer:NO!!! Rather than cooling the kitchen, the open refrigerator will warm it up!! The air conditioner also will warm the bedroom.

  23. Entropy & The 2nd LawFor a System in Equilibrium with a Heat Reservoir at Temperature T The 2nd Law Heat flows from objects of high temperature to objects at low temperature because this process increases the disorder of the system. For a system interacting with a heat reservoir at temperature T & exchanging heat Q with it, the entropy change is:

  24. The 2nd Law of Thermodynamics can be used to generally classify Thermodynamic Processesinto Three Types: 1. Natural Processes • These are Always Irreversible Processes. • These are alsoSpontaneous Processes. 2. Impossible Processes • These violate either the 1st Law or the 2nd Law or both. 3. Reversible Processes • These are ideal processes & are never found in nature. • We’ll discuss each with examples next. (courtesy F. Remer)

  25. Increasing Entropy:A Gas With & Without Gravity Without gravity, entropy increases as the gas spreads out. When gravity is present, clumping increases the entropy. This effect has been used as a simple model of black hole formation. A gas with no gravity A gas with gravity

  26. Entropy Changes:Reversible Processes In General dS = (đQr/T) = (dĒ/T) + P(dV/T) = CV(T)(dT/T) + P (dV/T)

  27. Entropy Changes:Reversible Processes In General dS = (đQr/T) = (dĒ/T) + P(dV/T) = CV(T)(dT/T) + P (dV/T) Special Case: Any Classical Ideal Gas: dU = CVdT , PV = nRT . So, ΔS = CV(dT/T) + nR (dV/V) = CV ln(T2/T1) + nR ln(V2/V1)

  28. Entropy Changes:Reversible Processes In General dS = (đQr/T) = (dĒ/T) + P(dV/T) = CV(T)(dT/T) + P (dV/T) Special Case: Any Classical Ideal Gas: dU = CVdT , PV = nRT . So, ΔS = CV(dT/T) + nR (dV/V) = CV ln(T2/T1) + nR ln(V2/V1). Special Case: Ideal Monatomic Gas CV = (3/2) nR. So, ΔS = nR {ln[(T2/T1)3/2(V2/V1)]}, or ΔS = nR ln(T3/2V) + constant.

  29. Entropy Changes:Reversible Processes In General dS = (đQr/T) = (dĒ/T) + P(dV/T) = CV(T)(dT/T) + P (dV/T)

  30. Entropy Changes:Reversible Processes In General dS = (đQr/T) = (dĒ/T) + P(dV/T) = CV(T)(dT/T) + P (dV/T) Adiabatic Process: ΔS = ∫đQr/T = 0, since đQr = 0.

  31. Entropy Changes:Reversible Processes In General dS = (đQr/T) = (dĒ/T) + P(dV/T) = CV(T)(dT/T) + P (dV/T) Adiabatic Process: ΔS = ∫đQr/T = 0, since đQr = 0. Phase Change: ΔS = Q/T = Li/T.

  32. Entropy Changes:Reversible Processes In General dS = (đQr/T) = (dĒ/T) + P(dV/T) = CV(T)(dT/T) + P (dV/T) Adiabatic Process: ΔS = ∫đQr/T = 0, since đQr = 0. Phase Change: ΔS = Q/T = Li/T. Isochoric Process (ideal gas): ΔS = CV ln(T2/T1).

  33. Entropy Changes:Reversible Processes In General dS = (đQr/T) = (dĒ/T) + P(dV/T) = CV(T)(dT/T) + P (dV/T) Adiabatic Process: ΔS = ∫đQr/T = 0, since đQr = 0. Phase Change: ΔS = Q/T = Li/T. Isochoric Process (ideal gas): ΔS = CV ln(T2/T1). Isobaric Process(ideal gas): ΔS = CP ln(T2/T1).

  34. Reversible Processes Path a (isotherm) ΔS12 = nR ln(V2/V1), since T2 =T1. Path bi (isochore) ΔS13 = CV ln(T3/T1). Path bii (isobar) ΔS32 = CP ln(T2/T3) =CP ln(T1/T3). Paths b(i + ii) ΔS12 = CV ln(T3/T1) + CP ln(T1/T3) = (CP – CV ) ln(T1/T3) = nR ln(T1/T3) = nR ln(V2/V1). since for bii, V3/T3 = V2/T2 = V2/T1. Isochore P Isotherm a bi bii V Isobar 34

  35. Entropy Changes:Irreversible Processes Free Adiabatic Expansion of a Gas (into a vacuum). This is the Joule process, for whichQ, W & ΔEare all zero. Ideal gas: E = E(T), so that ΔT = 0. Since the final equilibrium state is that which would have been obtained in a reversible isothermal expansion to the same final volume, ΔS =nRln(V2/V1). Remember, that the entropy is a state function, so that its change depends only on the initial and final states, & noton the process. 35

  36. Entropy Changes:Irreversible Processes Free Adiabatic Expansion of a Gas (into a vacuum). This is the Joule process, for whichQ, W & ΔE are all zero. For a Real (non-ideal) Gas: ΔE = Δ(KE) + Δ(PE) = 0. Since the intermolecular PE increases with increasing volume, Δ(KE) decreases, so that the temperature decreases. 36

  37. Irreversible Processes Let the system have initial temperature T1and entropy S(T), and the reservoir have temperature T0and entropy S0. For the System: Q = ∫C(T)dT & ΔS = ∫(đQ/T) = ∫[C(T)/T]dT. 37

  38. Irreversible Processes Let the system have initial temperature T1and entropy S(T), and the reservoir have temperature T0and entropy S0. For the System: Q = ∫C(T)dT & ΔS = ∫(đQ/T) = ∫[C(T)/T]dT. For the Bath: Q0 = – Q = – ∫C(T) dT & ΔS0 = – Q/T0. 38

  39. Irreversible Processes Let the system have initial temperature T1and entropy S(T), and the reservoir have temperature T0and entropy S0. For the System: Q = ∫C(T)dT & ΔS = ∫(đQ/T) = ∫[C(T)/T]dT. For the Bath: Q0 = – Q = – ∫C(T) dT & ΔS0 = – Q/T0. Special Case:Constant C. C ∫(dT/T) = C ln(T0/T1), ΔS0 = – Q/T0 = – C(T0 – T1)/T0 = C[(T1/T0) – 1]. ΔSuniv =ΔS + ΔS0 = C [ln(T0/T1) + (T1/T0) – 1] = C(T1/T0)f(x), where f(x) = x lnx + 1 – x, and x = T0/T1. 39

  40. Irreversible Processes Let the system have initial temperature T1and entropy S(T), and the reservoir have temperature T0and entropy S0. For the System: Q = ∫C(T)dT & ΔS = ∫(đQ/T) = ∫[C(T)/T]dT. For the Bath: Q0 = – Q = – ∫C(T) dT & ΔS0 = – Q/T0. Special Case:Constant C. C ∫(dT/T) = C ln(T0/T1), ΔS0 = – Q/T0 = – C(T0 – T1)/T0 = C[(T1/T0) – 1]. ΔSuniv =ΔS + ΔS0 = C [ln(T0/T1) + (T1/T0) – 1] = C(T1/T0)f(x), where f(x) = x lnx + 1 – x, and x = T0/T1. From this result, we may show that ΔSuniv> 0 for T1 T0 . 40

  41. Philosophy!!!!The 2nd Law & Life on Earth The existence of low-entropy organisms like ourselves Has sometimes been used to suggest thatwe live in violation of the 2nd Law! Sir Roger Penrose has considered our situation in his work “The Road to Reality: a Complete Guide to the Laws of the Universe”(2005). 41

  42. In “The Road to Reality: a Complete Guide to the Laws of the Universe” (2005). Sir Roger Penrose points out that it is a common misconceptionto believe that the Sun’s energy is the main ingredient needed for our survival. However, he says, what is important is that the energy source be far from thermal equilibrium. For example, a uniformly illuminated sky supplying the same amount of energy as the Sun, but at a much lower energy, would be useless to us. Fortunately the Sun is a hot sphere in an otherwise cold sky. So, it is a low entropy source, which keeps our entropy low. 42

  43. The 2nd Law & Life on Earth The optical photons supplied by the Sun contain much more energy than the IR photons leaving us, since εph = hν. Since the energy the energy reaching us is contained in fewer photons, the Sun is a low entropy source. Plants utilize the low entropy energy, to reduce their entropy through photosynthesis. We keep our entropy low by breathing oxygen produced by plants, and by eating plants, or animals ultimately dependent on plants. 43

  44. The Third Law of Thermodynamics “It is Impossible to Reach a Temperature of Absolute Zero.” On the Kelvin Temperature Scale, T = 0 K is often referred to as “Absolute Zero”

  45. Another Statement of The Third Law of Thermodynamics: “The entropy of a true equilibrium state of a system atT = 0 Kis zero.” • Strictly speaking, this statement is true only ifthe quantum mechanicalground state is non-degenerate. If it is degenerate, the entropyatT = 0 Kis a small constant, not 0! This is Equivalent to: “It is impossible to reduce the temperature of a system toT = 0 Kusing a finite number of processes.”

  46. Some Popular Versions ofThe Lawsof Thermodynamics 1st Law: You can’t win. 2nd Law: You can’t break even. 3rd Law: There’s no point in trying.

  47. Other Popular Versions of The Laws of Thermodynamics • Version 1 • Zeroth Law:You must play the game. • First Law:You can't win the game. • Second Law:You can't break even in the game. • Third Law:You can't quit the game. • Version 2 • Zeroth Law:You must play the game. • First Law:You can't win the game; • you can only break even. • Second Law:You can only break even at absolute zero. • Third Law:You can't reach absolute zero.

  48. Version 3 • Zeroth Law:You must play the game. • First Law:You can't win the game. • Second Law:You can't break even except on a very cold day. • Third Law:It never gets that cold! • Version 4 • Zeroth Law:There is a game. • First Law:You can't win the game. • Second Law:You must lose the game. • Third Law:You can't quit the game.

  49. “Murphy's Law” of Thermodynamics: Things get worse under pressure!!

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