Chem 340 - Lecture Notes 6 – Fall 2013 – Second law

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Chem 340 - Lecture Notes 6 – Fall 2013 – Second law

In the first law, we determined energies, enthalpies heat and work for any process from

an initial to final state. We could know if the system did work or gave out heat

(exothermic) or the reverse. However, we could not predict if it would actually happen.

In nature, and in all our experiences, we know there are processes that always go in

one way on their own, spontaneously. Heat always flows from a hotter to colder

system, gas always expands from a higher to lower pressure region, rocks roll downhill,

not up, Na metal always reacts with water to form NaOH and H2, salt dissolves in water.

These processes happen spontaneously, without outside intervention, unless we

counteract them (e.g. with work). The reverse processes do not happen, a heated

metal rod does not spontaneously get hot and one end and cold at another, the air in

the room never spontaneously moves all to one corner, sugar does not crystalize out of

your coke unless you evaporate the water or cool the solution, i.e. do some work on it.

These processes are irreversible, spontaneously go in one direction and not in reverse.

To reverse them you need to take energy from the surroundings – have them work/heat

on system. By contrast, reversible processes go in infinitesimal steps and can reverse

(at equilibrium). We learned that work done by the system is maximum for reversible

processes, so this concept relates to getting the most work out of a system - efficiency.

Thermodynamics needs a parameter, measure, ideally a state function, to tell us if a

process will occur spontaneously, or in what direction a process/reaction will proceed.

That is entropy, and is the heart of the second law of thermodynamics.

There are various ways to introduce the concept of entropy into thermodynamics, none

of them is possibly ideal. The idea of reversible and irreversible processes opens up a

path to a function.

Lets assume there is entropy = S and we want it to be a state function, so only depends

on initial and final state: S = ∫dS = Sf – Si , and over a cycle: ∫odS = 0

S cannot be U or H since those state functions do not indicate direction of a process

i.e. system does not just lose energy, e.g. expand ideal gas to vac., U=0, spont.

Similarly dS cannot be ᵭq or ᵭw since those variables are not state functions.

Second law statements tend to be abstract (and sort of negative) but they always hook

up heat and temperature into a combined change:

Clausius: “heat cannot flow form a colder to a hotter body without work added” or “It is

impossible to devise a continuously cycling engine that produces no effect other than the transfer

of heat from a colder to a hotter body.” – i.e. refrigerator needs engine and power (compressor)

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Kelvin: “one cannot use a cyclic process to transform heat

into work without also transferring heat from a hotter to a

colder reservoir” or “It is impossible to devise a continuously

cycling engine that produces no effect other than the extraction of

heat from a reservoir at one temperature and the performance of an

equal amount of mechanical work.”

(no perpetual motion machines – example, ship cannot draw energy

from ocean, without losing heat to something colder, text--ball at rest cannot bounce up)

Carathéodory statement: “In the neighborhood of every equilibrium state of a closed system

there are states that cannot be reached from the first state along any adiabatic path.” – oh well!

These statements combine heat and temperature (colder,

hotter) and seem to depend on machines. Thermodynamics

originally analyzed at heat engines like a steam engine, and

evaluated the work derived from the engine by providing

energy in form of heat. These models lead to entropy, S.

Also they tend to disperse energy (into both heat and work)

like text example bouncing ball dispersing energy into floor.

So to get at elusive entropy, lets turn

to a model engine, the Carnot cycle:

Heat engine works by expanding gas

diathermal contact cylinder with Thot,

and contracting by contact to Tcold

timing the contacts drives rod down and up turning the crankshaft

Model this with a cycle of

reversible steps, 1st ab,

isothermal expansion (in

contact Thot) follow by

bc, adiabatic expansion,

then reverse, lower Tcold

isothermal compression,

cd, follow by adiabatic

compression, da. Since

each step reversible and

has controlled parameters

can be modeled with the

(H,U,q,w) we know.

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Have an empirical look at the plot, what it means

First 2 steps, expansion, system does work, equivalent to area under curve abc

Second 2 steps, surroundings do work on system, equivalent to area under cda

Clearly the cycle does work on surroundings, i.e. design of engine, wtot = area inside

Design need mix isothermal and adiabatic, since both process occur on non-

intersecting curves, to get a cycle cannot use just one, must combine them

Cyclic, so 4 steps restore the system to its initial state.

isothermal, Th, U=0

adiabatic, q=0,U=w

isothermal, Tc, U=0

adiabatic, q=0,U=w

Since a cycle: ΔU = 0 wcy = w1 + w2 +w3 +w4 qcy = qh + qc = qab +qcd ΔU = 0 = wcy + qcy thus - wcy = qh + qc

Work done by system is greater than done on system, wtot<0 (since syst does work) Means qcy>0, or |qab=qh|>|qcd=qc| because heat flow: qh into syst, qc out from syst So to get work, need more heat form surroundings than give back If wcy is negative (work has flowed from the system across the boundary and appeared in the surroundings), any work produced by the system has been produced at the expense of the thermal energy of the surroundings. That is, heat (qh + qc) has flowed from the surroundings to the system at the same time. (Kelvin 2nd law)

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From steps 2 and 4: for wcy note ∫CVdT terms cancel, opp. sign integrals ln(VC/VB) = -ln(VA/VD) since both = (CV/R)ln(Tc/Th) rearrange: (VB/VA ) = (VC/VD ) wcy = R( Th −Tc ) ln(VB/VA) depends on the temperatures of the two reservoirs and on the ratio (VB/ VA) (i.e. compression ratio) for an ideal gas as the working substance in a reversible cycle.

Second law says heat engine must always lose heat to surroundings Process (a) normal, but (b) cannot happen ( perpetual motion of second kind) If engine (b), then Th=Tc and PV curve is a line, no area, wcy=0

no machine has 100% efficiency, but efficiency improves for Th>>Tc

but only ~1 for Th~∞, Tc~0 Example: calculate max work done by a reversible heat engine operating between 500 and 200K if 1kJ is absorbed at 500K Efficiency: e = 1 – T200/T500 = 1-2/5 = 0.6

Work: (step of expansion) w = .qab = 0.6 x 1000J = 600 J

Go back to efficiency equation:

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So have a state function, dS = dqrev/T, or S = ∫dqrev/T and it describes entropy technically, dqrev = ᵭqrev inexact differential, but if divide by T, exact (state fct) Because of definition, must calculate entropy for a reversible path, even if process is irreversible, need reversible path to get entropy between initial and final states

Categories: 1) reversible adiabatic, q = 0, so S = ∫dqrev/T = 0

2) cyclic process: S = ∫odqrev/T = 0 state function

3) isothermal expansion, U=0, qrev = -wrev =nRT ln(Vf/Vi) from w= -∫PdV

S = ∫dqrev/T =qrev/T = nR ln(Vf/Vi)

S > 0 for an isothermal expansion, ideal: no energy change but entropy change

Note any process ViVf, keep at const. T, has same S

4) reversibly change T, const. V U = qrev, or dqrev = CVdT

S = ∫dqrev/T = ∫ nCVmdT/T = nCVm ln(Tf/Ti) 5) const P see same thing, dqrev = CPdT

S = ∫dqrev/T = ∫ nCPmdT/T = nCPm ln(Tf/Ti)

Note: above assumes CV or CP constant over T range If change both V and T, ViTiVfTf, then break into 2 seg, const T, const V same for PiTiPfTf but Pf < Pi - expansion, need (-) sign for work (const T) part

Example: See extra sheets-S irrev – find rev. path

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Example: Showed on Friday for irreversible path (adiabatic expansion of gas T=500 K,P=100 atm P=1 atm against Pext=1 atm) that first needed to determine Tf and Vf,

which you could do from U=w, since q=0 for adiabatic)

Then: create a reversible path from ViTiVfTf and use this path to calculate S=qrev/T

For adiabatic, const V -T change, then const T- V change

S=qrev/T=R ln(24.8/0.411) + 3/2 R ln (500/302) = (34 + 6.3)JK-1 = 40 JK-1

Now if a reversible expansion know from previous chapter: Tf/Ti = (Vi/Vf)-1 = (Vi/Vf)

2/3

S=qrev/T= R ln(Vi/Vf) + 3/2 R ln (Tf/Ti) = -R ln(Vi/Vf) + 3/2 R ln (Vi/Vf)2/3 = 0

So Srev = 0 and qrev = 0 but Sirev > 0 Phase change –can happen at a const T since system needs heat (or loss of heat) to change phase, stays at Tb or Tm for boiling or melting (note, vapor press different)

S = ∫dqrev/T = qrev/T = Hvap/T

Same for fusion: S = ∫dqrev/T = qrev/T = Hfus/T Example for phase change: supercooled water, irreversible freezing at T < Tm,

for S calc need reversible cycle, heat to 0 C, freeze, cool to low T, separate page

Generalize, if know and i.e. expansion coefficient and compressibility, for

substance, then can determine S without equation of state for change: ViTiVfTf

S = ∫CV/T dT + ∫ dV = CV ln(Tf/Ti) + V Alternatively have change: PiTiPfTf

S = ∫CP/T dT - ∫V dP

For liquids and solids, CP and constant (assume small

S = CP ln(Tf/Ti) - V P Example for CO expansion, choose reversible path, separate sheet Similarly, for liquid Hg pressure increase, use formula above, separate sheet

Entropy was shown to be a state function, and relates to efficiency, and calculable for chemically interesting process (by finding rev path) Back to our goal – determine the direction of a process

Example: bring 2 metal rods together, one at T1, and other at T2, assume T1> T2, what will happen? You know that the hotter one (1) will spontaneously cool and the cooler one (2) will warm up. Does entropy indicate this?

Need a reversible path: slowly (rev) cool rod 1 by coupling to reservoir, const P

dqP = ∫CPdT or T change caused by q: T = 1/CP ∫dqP = qP/CP (assume CP const)

since done under const P, path defined, so not restrict how fast heat go, qP = H = qrev if system isolated heat only from 1 into 2, i.e. q = q1 + q2 = 0, entropy change is then

S = q1/T1 + q2/T2 = qP(1/T1 -1/T2) uses: qP = q1 = - q2

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--So look at normal process, heat flow from hot cold, T will get smaller, qP = q1 =(-) T1 > T2 so (1/T1 -1/T2)<0 and qP < 0 makes: dS > 0 --but if opposite, unnatural, if heat flow from cold to hot, then qP > 0 and dS < 0

Thus S (+) indicates natural direction of process, if isolated system

Example: Next consider a volume containing an ideal gas that suddenly collapses to half its previous volume, constant T with no forces, spontaneous partition into gas and vacuum, which we all recognize as unnatural.

Design a reversible analog to compute S, i.e. imagine dripping water on piston to slowly increase pressure and compress gas to half its volume. Since ideal, this is

isothermal (recall U not depend on V), so ViTi½ViTi and U = 0, so q = - w

S = ∫dqrev/T = qrev/T = -wrev/T = nR ln(½Vi/Vi) = -nR ln 2 < 0 (recall: ln½ = - ln 2) If reverse the process, gas spontaneous expands to fill the volume

S = ∫dqrev/T = nR ln2 > 0

Consistent: i.e. spontaneous, irreversible process is the one with S > 0

General: For any irreversible process in an isolated system, there is a unique direction

of spontaneous change: S > 0 for spontaneous change, S < 0 for the opposite or

nonspontaneous change. For these S = 0 for a reversible process only works if no energy flow: no heat or work on the system, isolated!

Contrast: U is neither created or destroyed, but S can be increased for an isolated

system, S > 0, but not destroyed – imbalance! For S < 0, need work

Aside: Interpretation S (+) when system becomes less ordered, more random e.g. expansion of gas, more volume, more places to be, less order molecules same with heating rods, heat (vibrations of motions of atoms) more dispersed

Statistical view: S = k ln where = # ways system can be in that state

S = k ln (/recall k = R/NA - and concept more states more probable

System then tends to a state of higher probability S (+)

Can think of number of states as /V2/V1

S = k lnV2/V1)N = Nk lnV2/V1) = nR lnV2/V1) were N = nNA

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The Clausius inequality recall: dU = dq + dw = dqrev + dwrev rearrange: dqrev - dq = dw - dwrev now the system can do maximum work for reversible process: |dw| ≤ |dwrev| but definition is work by system is (-), or -dw ≤ -dwrev or dw - dwrev ≥ 0 Substitute: dqrev - dq = dw - dwrev ≥ 0 or dqrev ≥ dq Now recall entropy: dS = dqrev/T dS ≥ dq/T the entropy is greater or equal to than heat change divided by temperature

Example (text) consider transfer of heat from hot to cold body (same example again) dS ≥ dqh/Th + dqc/Tc but dqh = - dqc so dS ≥ - dqc/Th + dqc/Tc or dS ≥ dqc(1/Tc - 1/Th) since dqc > 0 and Tc < Th then dS > 0

finally if system is isolated from surroundings, then dS ≥ 0, isolated system cannot reduce its entropy second law

Clausius extension for cyclic integral, dq/T around a path Imagine step 12 is reversible and then back, 21 is irreversible

∫o dq/T = ∫12 dqrev/T + ∫21 dqirev/T = - ∫21 dqrev/T + ∫21 dqirev/T

But dqrev > dqirev so ∫o dq/T ≤ 0 i.e. neg. (recall S = 0 = ∫o dqrev/T so S ≥ ∫o dq/T )

Now consider the surroundings, this is a large temperature bath, so const T, heat in has negligible effect, stays in equilibrium no matter what happens to system

Const V change: qsur = Usur or const P : qsur = Hsur H and U state fct, path indep

dSsur = dqsur/Tsur or macroscopic: Ssur = qsur/Tsur (acts rev, since state fct) this is the real heat of process, path indep, whereas for system need rev path

Example: compare process for reversible and irreversible compression of a gas,

evaluate S tot = Ssys + Ssur and compare (see separate sheet)

In example see S tot = Ssys + Ssur = 0 for reversible path, equilibrium no direction

but S tot = Ssys + Ssur > 0 for irreversible path, for spontaneous process.

Looking at this more globally, consider system and surrounding, S tot ≥ 0 or entropy of the universe is always increasing

Entropy of mixing: U = 0, q = -w, T=Pt=Vt=0, know spont. (ideal gas), Pi,Vi≠0

rev. path: Sa,b=na,bRln(Vt/Va,b) St=Sa+Sb = -ntR(xalnxa+xblnxb) >0 (lnxa<0)

In some way this defines time. If watch a process like mixing 2 gases, know Stot > 0, and we know we are watching it in a real time direction, but if it were filmed and

played backward, it would separate, but this tells us that S tot < 0, so not real time

We will use this Clausius inequality relationship to define Gibbs and Helmholtz free energies and those will tell us directions of process, using system properties alone

Absolute entropy and the Third law U and H were defined as changes, dU = dq + dw, or dH = dU + d(PV) and an absolute

value was not established, but relative values to a standard state were used: Hof

Entropy is different, can define an absolute, but need to know what is zero 3rd law

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Third law of thermodynamics: the entropy of a pure, perfectly crystalline substance (element of compound) is zero at 0 K.

Critical issue: idea of absolute temperature (from Kelvin and gas thermometer) and idea of a perfect crystal. In perfect crystal, all the molecules are identical, so interchanging

them give the same state, perfect crystal has only one state, so: S = k ln = k ln 1 = 0

How to calculate entropy – relates to heat and temperature, so need heat capacity, but for all the phases from perfect crystal to other solid phases, to liquid and to gas, allow for heats of phase changes (fusion, vaporization, and sol1sol2, etc.) along the way

Plot of Cp vs, T for O2 shows growth from 0 K, initially like T3, but change at each phase change

To get entropy, divide by T, integrate those curves,

and then add in H/T for phase changes, as below:

- entropy increases monotonic-ally, not lost, - heat capacity grows or drops with phase, each phase change:

Hph(+), S(+),

-vapor the big one: Hvap/T and then relatively const, all Cp > 0 gives trend: Svap>Sliq>Ssol

- molar entropy increases with molecule size, more vibrations (degree of freedom) - gases translate relatively freely, rotate, vibrate (3N-6 modes), liquids in hindered

motion (trans., rot. collide), solids only vibrate (like big coupled springs) - strongly bound solids higher energy vibrations, less accessible, lower entropy

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Example - copies Engel calculation of the entropy of O2 for all phases at T = 298K

Cp values measured from 12K, lower ones fit to T3 law, each phase change has H

See largest changes for vaporization (75.6 J/K), then heating (CpT) gas and liquid Standard states Although we have absolute values for entropy, U and H are

defined relative to standard states, so can also do that for entropy. Can use heat capacity to define entropy for T=298 K and P = 1 bar, but for gas phase, entropy highly dependent on pressure

At const T, ideal gas: Sm = R ln(Vf/Vi) = - R ln(Pf/Pi) if let Pi = Po = 1 bar Sm(P) = Sm

o – R ln(P/Po)

Graph shows Sm vs. P for gas, see max, Sm∞, P 0 Meaning is that V∞, P0, so infinite states Solids and liquids, P dependence small, can often neglect Entropy in Chemical reactions

Since entropy is a state function we can do the same approach as for H and U for reactions, Hess’s law works, just sum up reactants and subtract from sum of products

ΔSorxn= So

prod - Soreact since absolute values do not need

Recall, for H we used Hof which were referenced to elements in standard states, so

reference was same for reactants and products (mass law—atoms same)

But So use absolute values for reactant and product, reference to perfect crystal, 0 K Example: synthesis of glucose, C6H12O6(s), from carbon dioxide and water, T = 298 K: 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g)

Determine if spontaneous by testing if S tot = Ssys + Ssur > 0

Ssys = Sorxn = So(glu) + 6 So(O2) - 6 So(CO2) - 6 So(H2O)

= (209 + 6x205 – 6x214 – 6x70)JK-1mol-1 = - 262 JK-1 (for 1 mole rxn)

See Ssys (-) but need to consider Ssur – use heat transfer form reaction to surrounding

Hsys = Hof (glu) + 6 Ho

f (O2) - 6 Hof (CO2) - 6 Ho

f (H2O) = [-1273 + 6x0 – 6x(-294) – 6x(-286)]kJmol-1 = 2801 kJ (for 1 mole rxn)

Ssur = Hsur /T = Hsys/T = -2802 kJ/298 K = -9398 JK-1

S tot = Ssys + Ssur = (-262 -9398) JK-1 = 9.66 kJK-1 < 0

So reaction as written is not spontaneous at 298 K, logic is to form solid glucose from liquid and gas phase molecules, high disorder to more order, must do work on system

Also need to consider T dependence of S – Kirchoff’s law again - must account for phase changes in the entropy, integrate between them

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Refrigerator, Heat Pump, Reverse Carnot cycles:

Goal of Refrigerator is to pump heat form cold body to a hot one, since second law says this cannot happen without work. Efficiency then defined as coefficient of

performance c = r = |qc|/|w|

Energy deposited in hot body is |qh| = |qc|+|w| rearrange: |w| = |qh| - |qc| 1/c = (|qh| - |qc|)/|qc| = |qh|/|qc| - 1

From previous efficiency use : qh/qc= -Th/Tc

1/c = 1-Th/Tc or c = r = Tc/(Th-Tc)

if Tc decrease relative to Th, c = r decreases

Example: refrigerator operates freezer at Tf = 255 K, refrigertor at Tr = 277 K and room about Th = 295 K

Let Tf = Tc, maximum efficiency is r = 6.5, which implies

1 J work produces 6.5 J cooling, actual r ~ 1.5 The difference is loss of efficiency fo irreversible cycle

Alternate, heat pump, bring heat out of ground into house

(illustrated below), now c = hp = |qh|/|w|

Same picture here, if inside is Th = 295 K (~72 F) and

outside is Tc = 273 K (freezing), then c = hp = 13.4 but

this again is ideal efficiency (max) real is hp ~ 2-3 But if you use the ground, which deep enough is always at Tc ~ 285 K, it helps, and even with efficiency

hp ~ 2-3, get 2-3x advantage over direct electric heat

Entropy considerations,

Process is not spontaneous, however, can convert

work to heat at 100%, so hp calculated for minimum

work input to make S favor a spontaneous process for withdrawing qc and depositing qh+w Beyond this consider solar, wind, and fuel cells as alternate energy sources. Electrical energy can be directly converted to work without heat engine loss