S = S(f) S(i) dq rev /T. ds = dq rev /T

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Donald Richardson
5 years ago
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1 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 dq rev /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) dq rev /T ds = dq rev /T

2 ds = dq rev /T S = S(f) S(i) dq rev /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: dq rev = C dt ds = C dt/t S = C dt/t [If C = constant, S = C dt/t = C ln (T f /T i )]

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

4 Problem: 2 kg of ice at 0 o C melts and is then heated to 50 o C. What is its change in entropy? a) melting: T melt = constant = 273 K. Therefore S melt = Q/T melt = ml f /T melt. S melt = (2 kg) (3.33 x 10 5 J/kg) / 273 K = 2.4 x 10 3 J/K b) warming: T f = 323K, T i = 273 K, c = 4186 J/kg K S warm = dq/t = mc dt/t = mc ln(t f /T i ) = (2 kg) (4186 J/kg K) ln(323/273) S warm = (2) (4186) (0.168) J/K = 1.4 x 10 3 J/K S total = S melt + S warm = 3.8 x 10 3 J/K

5 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: C P = at + bt 3 [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 T i to T f at constant pressure (e.g. 1 atmosphere). Then for 1 mole: S Ti Tf = (a + bt 2 ) dt = a(t f -T i ) +(b/3)(t f3 -T i3 ) e.g.: T i = 0, T f = T: S(T) = at + bt 3 /3 + S(0) [ Third Law of Thermodynamics : S(0) = 0]

6 dq rev = T ds What about the heat flow in an irreversible process: Clausius: dq irrev /T < S dq irrev /T < ds dq irrev < 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 2 nd Law of Thermo.]

7 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 V i V f at constant T: Isothermal expansion: Q rev = T S = nrtln(v f /V i ) S = nr ln(v f /V i ) So T S > 0 ln(v f /V i ) > 0 V f > V i. Can only have free expansions, cannot have free compression (even though allowed by energy conservation): some processes are forbidden by 2 nd Law, even though permitted by 1 st Law: [Example of Arrow of Time ]

8 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 T h : dq = nrt h dv/v dq/t = nr ln (V B / V A ) (>0) Similarly, isotherm at T c : dq/t = - nr ln (V C / V D ) (<0) But V B /V A = V C / V D dq/t = 0

9 2) System goes through cycle: S = 0 dq/t 0 Example 2: Isothermal expansion of Carnot Cycle replaced with adiabatic free expansion from V A to V B. Since irreversible, expect dq/t < 0. Rapid change A B, not through welldefined P, T states Now Q h = 0. The only step with heat flow is C D, with dq/t = nr ln (V D / V C ) < 0 Therefore, dq/t < 0.

10 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 T h ) to cold reservoir (at T c ). S h = -Q/T h, S c = +Q/T c S tot = Q (1/T c 1/T h ) >0 (since T c < T h ). [i.e. S > 0 means heat only flows spontaneously from hot to cold.]

11 3) Isolated System: Q = 0 S 0 Example 2: Irreversible process (expect S > 0): Heat flow from an object (O) originally at T h and with heat capacity C into an energy reservoir at T c until it reaches T c (i.e. equilibrium). Q = C (T h T c ) S Res = C (T h -T c )/T c = C (T h /T c 1) S O = dq/t = C dt/t = C ln (T c /T h ) = - C ln (T h /T c ) S tot = C [T h /T c 1 ln(t h /T c )] > 0 T h /T c ln(t h /T c ) T h /T c

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

13 Isolated System: S 0 1) Kelvin-Planck Statement of 2 nd Law Clausius Statement of 2 nd Law S isolated 0 2) The universe is an isolated system [actually, the only perfectly isolated system] : 2 nd Law: S universe 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 2 nd Law: the 2 nd Law determines the arrow of time.

14 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 2 nd Law: the 2 nd 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 = I 2 R 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

15 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 ~ 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 ~ 10 7 ) 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 S j 0 if you sum up all the pieces of the extended system.

Reversible Processes. Furthermore, there must be no friction (i.e. mechanical energy loss) or turbulence i.e. it must be infinitely slow.

Reversible Processes. Furthermore, there must be no friction (i.e. mechanical energy loss) or turbulence i.e. it must be infinitely slow. Reversible Processes A reversible thermodynamic process is one in which the universe (i.e. the system and its surroundings) can be returned to their initial conditions. Because heat only flows spontaneously