S = S(f) – S(i)  dQrev/T In 1855, Clausius proved the following (it is actually a corollary to “Clausius’ Theorem”): If a system changes between two equilibrium states, i and f, the integral dQrev/T is the same for any reversible path between i and f. Therefore there is a quantity, Clausius named “entropy” (S) that only depends on the endpoints, and not the path: S = S(f) – S(i)  dQrev/T dS = dQrev/T

S = S(f) – S(i)  dQrev/T dS = dQrev/T S = S(f) – S(i)  dQrev/T If reversible, adiabatic process, S = 0. If reversible, isothermal process, S = Q/T (i.e. S > 0 if heat in, S < 0 if heat out) For other processes, find a reversible process that takes the system from its initial to final state, e.g. bring into contact with an infinite series of heat baths: dQrev = C dT dS = C dT/T S =  C dT/T [If C = constant, S = C  dT/T = C ln (Tf/Ti)]

Problem: 2 kg of ice at 0oC melts and is then heated to 50 oC Problem: 2 kg of ice at 0oC melts and is then heated to 50 oC. What is its change in entropy?

Problem: 2 kg of ice at 0oC melts and is then heated to 50 oC Problem: 2 kg of ice at 0oC melts and is then heated to 50 oC. What is its change in entropy? a) melting: Tmelt = constant = 273 K. Therefore Smelt = Q/Tmelt = mLf/Tmelt. Smelt = (2 kg) (3.33 x 105 J/kg) / 273 K = 2.4 x 103 J/K b) warming: Tf = 323K, Ti = 273 K, c = 4186 J/kgK Swarm = dQ/T = mc dT/T = mc ln(Tf/Ti) = (2 kg) (4186 J/kgK) ln(323/273) Swarm = (2) (4186) (0.168) J/K = 1.4 x 103 J/K Stotal = Smelt + Swarm = 3.8 x 103 J/K

Temperature change: S =  C dT/T What if C is not a constant: Example: At low temperatures, the molar specific heat of metals is of the form: CP = aT + bT3 [a is determined by the number of electrons free to move around. b is determined by the spectrum of “phonons” (atomic vibrations).] Suppose the temperature changes from Ti to Tf at constant pressure (e.g. 1 atmosphere). Then for 1 mole: STi  Tf =  (a + bT2) dT = a(Tf-Ti) +(b/3)(Tf3-Ti3) e.g.: Ti = 0, Tf = T: S(T) = aT + bT3/3 + S(0) [“Third Law of Thermodynamics”: S(0) = 0]

dQrev = T dS dQirrev/T < dS dQirrev < T dS dS  dQ/T S  dQ/T What about the heat flow in an irreversible process: Clausius: dQirrev/T < S dQirrev/T < dS dQirrev < T dS So in general: dS  dQ/T S  dQ/T [=: reversible process; > irreversible process] Examples: 1) Q  T S in an isothermal process. 2) System goes through cycle: S = 0   dQ/T  0 [Clausius’ Theorem] 3) Isolated System: Q = 0  S  0 [Restatement of 2nd Law of Thermo.]

1) Q  T S in an isothermal process Example: Adiabatic free expansion of ideal gas: Q = 0 and T = constant. Since this process is irreversible, must have T S > Q = 0 : To find S, consider a reversible process that takes the gas from Vi  Vf at constant T: Isothermal expansion: Qrev = TS = nRTln(Vf/Vi )  S = nR ln(Vf/Vi) So TS > 0  ln(Vf/Vi) > 0  Vf > Vi . Can only have free expansions, cannot have “free compression” (even though allowed by energy conservation): some processes are forbidden by 2nd Law, even though permitted by 1st Law: [Example of “Arrow of Time”]

2) System goes through cycle: S = 0   dQ/T  0 Example 1: Carnot Cycle of ideal gas. Since reversible, expect  dQ/T = 0. On adiabatic steps, dQ = 0. Isothermal at Th: dQ = nRTh dV/V  dQ/T = nR ln (VB / VA) (>0) Similarly, isotherm at Tc:  dQ/T = - nR ln (VC / VD) (<0) But VB/VA = VC / VD   dQ/T = 0 

2) System goes through cycle: S = 0   dQ/T  0 Example 2: Isothermal expansion of Carnot Cycle replaced with adiabatic free expansion from VA to VB. Since irreversible, expect  dQ/T < 0. Now Qh = 0. The only step with heat flow is CD, with  dQ/T = nR ln (VD / VC) < 0 Therefore,  dQ/T < 0.  Rapid change A  B, not through well-defined P, T states

3) Isolated System: Q = 0  S  0 Example 1: Irreversible process (expect S > 0): Two energy reservoirs (i.e. each with C = ), brought into contact for a time interval in which heat Q flow from hot reservoir (at Th) to cold reservoir (at Tc). Sh = -Q/Th, Sc = +Q/Tc  Stot = Q (1/Tc – 1/Th) >0 (since Tc < Th).  [i.e. S > 0 means heat only flows spontaneously from hot to cold.]

3) Isolated System: Q = 0  S  0 Example 2: Irreversible process (expect S > 0): Heat flow from an object (O) originally at Th and with heat capacity C into an energy reservoir at Tc until it reaches Tc (i.e. equilibrium). |Q| = C (Th – Tc)  SRes = C (Th-Tc)/Tc = C (Th/Tc – 1) SO =  dQ/T = C dT/T = C ln (Tc/Th) = - C ln (Th/Tc) Stot = C [Th/Tc – 1 – ln(Th/Tc)] > 0 

Isolated System: S  0 1) Kelvin-Planck Statement of 2nd Law  Clausius Statement of 2nd Law  Sisolated  0 2) The universe is an isolated system [actually, the only perfectly isolated system] : 2nd Law: Suniverse  0 3) a) All microscopic processes (classical or quantum physics) are “time reversible”. If this can occur: Then so can this:

Isolated System: S  0 1) Kelvin-Planck Statement of 2nd Law  Clausius Statement of 2nd Law  Sisolated  0 2) The universe is an isolated system [actually, the only perfectly isolated system] : 2nd Law: Suniverse  0 3) a) All microscopic processes (classical or quantum physics) are “time reversible”. b) The only physical law that is not time reversible – that says that there are processes which can occur with time running forward that are not allowed to run with time running backwards, is the 2nd Law: the 2nd Law determines “the arrow of time”.

S  0 3b) The only physical law that is not time reversible – that says that even though some events can occur, they would not occur if time was run backwards, is the 2nd Law: the 2nd Law determines “the arrow of time”. Examples : a) Energy (in form of heat) spontaneously flow from hot to cold. To get it to go from cold to hot (e.g. in refrigerator), must do work. b) Gases will spontaneously expand to fill available volumes. To get the gas back into its original volume, must do work. c) Heat cannot be transformed completely into work, but work can be transformed completely into heat: i) friction: mechanical work makes an object’s temperature increase, which can flow into other objects. ii) Resistance in an electrical circuit: Electrical (work) power = IV = I2R – makes object hot (e.g. in electrical furnace, toaster, …). -- Reversing the current does not make it cold. -- no material has negative R. d) Next class: things break into pieces – they do not spontaneously put themselves back together: S  0  the universe is becoming more “disordered”. e) If you can tell whether a movie is running backwards because motions seem awkward – it is because they break the second law (i.e. have entropy decreasing). 4) The universe is evolving to a state in which everything is at the same temperature (and no work will be possible): “Heat Death of the Universe”

S  0 5) (Next class) The second law is only true “probabilistically”: As the number of particles in an isolated system* N  , the probability that S(isolated system) < 0 for any process decreases exponentially. Thermodynamic system: one with N >>> 1, e.g. N ~ 1023 As work in “nanomaterials” has progressed, researchers have looked for examples of S < 0 (e.g. negative electrical resistance) in nanoparticles. In 2014, reports of heat flowing briefly from cold glass nanoparticles (N ~ 107) to hotter surrounding gas. Next class: why “size matters” --------------------------------------------------------------------------------------------------------------------- *Isolated system: If the system you are considering is not isolated from its surroundings (e.g. W 0 and/or Q  0), then extend the system to include all the relevant surroundings (so that Q and W are internal to the extended system). Note that S may be < 0 for a process in a system which is not isolated, but Sj  0 if you sum up all the pieces of the extended system.