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Chapter 3 – The Second and Third Laws of Thermodynamics I. Second Law: Concepts.

A. The direction of spontaneous change.

1. Spontaneous process = one that does not require work to bring it about.

E.g. Expansion of a gas against a lower pressure.

The reverse does not happen naturally (i.e, spontaneously). E.g. Heat flows spontaneously from hot to cold, but not the other

direction.

E.g. the kinetic energy of the bouncing

ball is slowly but inexorably dissipated into the thermal motion of atoms in the floor, but never has the reverse been observed.

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E.g. The process on the right, in which all the heat flowing from the hot source into the engine can be converted to work, is not possible

Question: What determines spontaneity? What determines the irreversible direction of these and other processes? Verbal definition: Spontaneous direction of change leads to increased dispersal of energy in the system plus surroundings. Can we quantify this?

2. Entropy S and the Second Law.

ΔStot > 0

The total entropy of all parts of a total system (isolated system) increases in a spontaneous change.

3. Contrast with First Law.

• First Law describes what processes are permissible as far as conservation of energy is concerned.

• Second Law describes which of these permissible processes are likely to occur, that is, which can occur spontaneously.

4. What is Entropy, review microscopic or statistical definition.

S = kB ln W

kB = Boltzmann constant = gas constant for a single molecule rather than a mole.

kB = R/NA = 1.381 x 10-23J K-1 W = the number of different ways a system of atoms and molecules

can be arranged and still be in the same thermodynamic state.

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E.g. System of N particles that can have either spin up or spin down.

N+ = # up N- = # down N = N+ + N-

W(N+,N-) = # of ways of arranging N+ with spin up, etc.

=N!/N+!N-! W(0,100) = 1 way W(100,0) = 1 way W(50,50) = huge # of ways ~1030

S(50,50)= kB 100 ln 2 (maximum entropy)

Box of coins will spontaneously tend toward state of approximately the same number of ups and downs.

Conclusion: Isolated system will tend spontaneously toward a state containing the largest number of available microstates (molecular arrangements) at the fixed total energy.

Remember: S =U −U 0( )T

+kBlnQ

5. What is Entropy, thermodynamic or macroscopic definition?

Since entropy is about energy dispersal, and that in turn has a lot to do with thermal energy (disorganized), it is not surprising that changes in entropy are given by:

dS = dqrev/T or

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ΔS = dqrev

1

2

∫ /T (path integral along the reversible path

between two thermodynamic states.)

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i.e., Entropy change ΔS to go from state 1 to 2 by ANY path is the integral of dq/T along the reversible path that connects those two states.

Problem: Calculate ΔS when ideal gas expands isothermally from V1 to V2, whether reversibly or irreversibly.

Strategy: construct the reversible pathway for this process and integrate dqrev/T even if the process were to happen irreversibly.

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ΔS = dqrev1

2

∫ /T

reversible isothermal expansion

dw = - pdV = -(nRT/V)dV dq = - dw since dU =0 dq = +(nRT/V)dV

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ΔS = nR /V( )1

2

∫ dV

ΔS = nR ln V2/V1

(conditions here: revers or irrev/isothermal/perfect gas)

Note: for ΔS to be > 0, V2 must be > V1

Work example: Calculate the entropy change when one mole of ideal gas expands into a vacuum and doubles its volume at 298 K. (In other words, this is an irreversible process)

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6. System and Surrounding Entropy Change.

We know that dStot > 0 in spontaneous change, or

dS + dS' > 0 syst↑ ↑surr

A reversible change is a special idealized limiting case in which

dS + dS' = 0, or

dS = -dS'

In a reversible process the entropy change in system is exactly compensated by change in surroundings.

But in general:

dS ≥ -dS' ↑equality is for reversible case

so that dStot ≥ 0. We can calculate the entropy change in the surroundings as follows:

Since surroundings are the entire universe, it is a reservoir of constant volume, so energy dqsurr supplied to it as heat = dUsurr. So state function dUsurr= dqsurr

This in turn implies that dqsurr is independent of how the process

in the system occurred, reversible or irreversible. So:

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dSsurr =dqsurr,rev

Tsurr

=dqsurr

Tsurr

ΔSsurr =qsurr

Tsurr

(regardless of how process performed)

So for adiabatic change in the system (no heat transfer with surroundings)

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ΔSsurr = 0

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7. Show that transfer from a hot to a cold body is spontaneous

dStot= dS1 + dS2 = dq/T1 - dq/T2 = dq(1/T1 - 1/T2) ΔStot = q(1/T1 - 1/T2) > 0

B. Entropy changes in specific processes.

1. Isothermal expansion of an ideal gas: have already shown that, no matter how conducted (rev or irrev)

ΔS = nR ln V2/V1

2. Phase Transition at the Transition temperature.

Applies, e.g. to vap, fusion, etc. System

ΔStrans = ΔHtrans/Ttrans trans = vap, fus, etc.

Standard entropies of phase transitions, ΔtrsSo/(J Kʹ′ ʹ′ ʹ′1mol-1)

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Problem: The molar enthalpies of vaporization of Cl2 and benzene are 20.41 and 30.8 kJ/mol, respectively, and their boiling points are 239.1 and 353.2K, respectively. Calculate ΔSvap in J K-1mol-1 for both. Cl2: ΔSvap = (20.41kJ/mol)/239.1K = 85.36 J/Kmol C6H6: Δ Svap = (30.8 kJ/mol)/353.2K = 87.2 J/K mol Hmmm! Almost identical! Trouton’s rule: (a general observation)

ΔSvap ≈ 85 J K-1mol-1 for most liquids. Microscopic reason: change in degrees of freedom (W) going from liquid to gas is comparable for most liquids. Slightly more for well-ordered liquids vaporizing, like H2O(l) with its H-bonding network. ΔSvap(H2O) = 109.1 J K-1mol-1 Why larger? Can use Trouton’s Rule to estimate ΔHvap from boiling point of the

liquid: ΔHvap ≈ Tb x 85 J K-1mol-1 Just an estimate.

The standard entropies of vaporization of liquids ΔvapHo/(kJ mol-1) θb/°C ΔvapSo/(J K-1mol-1)

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3. Entropy change of heating a system at constant pressure (i.e., the usual scenario):

Always true that

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ΔS = dqrev1

2

∫ /T

Since constant p, dqrev = CpdT

Therefore,

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ΔS = CpdT /T∫ (at constant V, put in CV)

Over a short T range, assuming Cp is T-independent.

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S T2( ) = S T1( ) + Cp dT /TT1

T2

∫

= S T1( ) + Cp lnT2 /T1

Can also use these integrations to calculate the absolute entropy S(T) of a substance at a given T. Start integrating at T = 0.

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S T( ) = S T = 0( ) + Cp

0

Tf

∫ solid( ) /T dT( ) + ΔHfus /Tf + Cp liq( )Tf

Tb

∫ /T dT( ) + ΔHvap /T + Cp

Tb

T

∫ gas( ) /T dT( )

zero entropy of entropy of entropy of entropy of entropy of pt. heating solid to heating vaporizing heating entropy solid to liquid liquid from gas to melting trans. melting to desired T temp. boiling temp.

First integral has problems, because Cp(solid) may not be known at T near 0 K. Use Debye theoretical formula Cp = aT3 near T = 0.

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C. The Third Law. 1. Third law establishes what is the zero-point entropy, S(T=0 K). 2. Based on Nernst heat theorem: entropy changes of process occurring

near T = 0 are approaching zero.

ΔS → 0 as T → 0 K. Implication: S of all substances at T = 0 K are about the same.

3. Third law states: S of all perfect crystalline substances, whether

elements or compounds, are equal at T = 0 K. 4. For convenience, choose S=0 (at T=0K) for perfect crystalline

substance. 5. This matches Boltzmann formula:

S = kB ln W W = 1 for perfect crystal at 0 K (only one arrangement) no thermal motion S = kB ln 1 = k x 0 = 0

6. Any disorder in crystals raises S. 7. Note: some substances have a residual entropy at 0 K.

Synoptic Table 3B.1* Standard Third-Law molar entropies at 298 K

Som/(J K-1mol-1)

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8. We can use tabulated third-law standard state entropies to calculate the entropy change of chemical reactions and physical processes.

ΔSrxo = νS

mo

products

∑ − νSmo

reac tants

∑

D. Efficiencies of Thermal Processes.

1. Efficiency of heat engines (device for converting heat into work): a. Need to first compute maximum work possible. Heat engines work by converting a portion of the heat flow from

a hot to cold reservoir into work.

ΔS = ΔSc + ΔSh

ΔS = ⏐qc⏐/Tc - ⏐qh⏐/Th (must be > 0)

Therefore, must require ⏐qc⏐ > ⏐qh⏐ x Tc/Th

⏐wmax⏐ = useable energy (converted to work)

= ⏐qh⏐ - ⏐qc⏐ ↑=⏐qh⏐ Tc/Th ⏐wmax⏐ = ⏐qh⏐ (1 – Tc/Th) b. Now, efficiency ε is defined as

ε = max work/heat flow out of hot body = ⏐wmax⏐/⏐qh⏐

ε = 1 – Tc/Th

c. Note that most efficient conditions are when: Th > > Tc

ε = 1 only if Tc = 0K or Th = ∞ K.

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2. The Carnot engine (Ideal gas, 4 steps, all reversible):

A → B isothermal reversible expansion w = -nRTh ln VB/VA q = -w (since dU=0)

B → C adiabatic reversible expansion w = -CV(Th – Tc) q = 0

C → D isothermal reversible compression w = - nRTc ln VD/VC q = - w

D → A adiabatic reversible compression w = -CV (Tc - Th) q = 0

Since all steps reversible (or since S is state function) ΔSABCDA = 0 (around a closed loop) ΔSABCDA = nR ln VB/VA + 0 + nR ln VD/VC + 0 AB BC CD DA = nR ln VBVD/VAVC = 0 ↑= 1 Total work done by engine:

⏐w⏐ = +nRTh ln VB/VA + CV(Th – Tc) + nRTc ln VD/VC + CV(Tc – Th)

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Max work achieved when all steps reversible, but to achieve this engine must run infinitesimally slowly, and so is of little value.

3. Refrigerator:

a. Is a heat engine operating in reverse.

Instead of generating work as heat flows from hot to cold, work is input to get heat to flow from cold to hot.

not feasible feasible b. Coefficient of performance c:

c = ⏐qc⏐/⏐w⏐ = heat flow out of the refrig insides/work required to draw it out

c. Ideal refrigerator has coefficient c°.

c° = Tc/(Th – Tc) This is refrigerator operating reversibly. Note: performance is diminished as Th – Tc gets larger.

A practical working refrigerator cannot work reversibly because it would be too slow and have lower c value.

4. Heat Pumps:

Like an inside out refrigerator, gathers heat out of the outdoors which is already at a lower T than the indoors.

c ≈ 5 for commercial heat pumps

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II. Second Law in terms of system variables only. A. Free energy.

1. 2nd Law states ΔStot > 0. This is not always convenient because we must keep track of what's happening in the surroundings.

2. Develop a restatement of 2nd Law written in terms of system variables only.

3. Result: for processes at constant p, useful to define:

Gibbs free energy G = H – TS “free” means free to do work

Then, for constant p & T process

dG = dH - TdS

or ΔG = ΔH - TΔS

Now we restate 2nd Law: a process is spontaneous if:

ΔG < 0 (at constant p and T)

Thus, spontaneous processes will be downhill in G.

So now we see there are two factors involved in spontaneity:

system ΔH and system ΔS

Example: system ΔS of a spontaneous process could actually be negative, but only if ΔH is negative (exothermic) enough to outweigh TΔS term.

That process would be energetically favorable but entropically unfavorable from a system viewpoint.

By same token:

Process could be endothermic (uphill in H) (ΔH > 0) and still be spontaneous if TΔS is larger and positive. This explains how endothermic reactions can still be spontaneous.

These process are said to be entropically driven. Summary:

ΔG = ΔH - TΔS

spontaneity fct. = energetic aspect - entropic aspect

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4. For processes at constant V, useful to define:

Helmholtz free energy A = U - TS

Then, for a const V and T process

dA = dU - TdS

ΔA = ΔU - TΔS

Same meaning as for ΔG, but constant V & T conditions. 5. Interpretation of ΔA as Maximum Work possible:

dwmax = dA wmax = ΔA

⏐ΔA⏐ is maximum work than be derived from process when internal energy expended is ⏐ΔU⏐.

-TΔS can be thought of as the “heat tax” imposed by nature. 6. Interpretation of ΔG as the maximum non-expansion work (or

additional work) possible.

This would be work like passing ions across a cell membrane in biology, or passage of electrons in an electrochemical cell.

dwadd,max = dG wadd,max = ΔG 7. Standard Reaction Gibbs energies can be calculated from

enthalpies and entropies:

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ΔGro = ΔHr

o − TΔSro

and standard Gibbs energies of formation can be tabulated and then used to find Gibbs energies of reactions as:

€

ΔGro = ν

products

∑ ΔGfo − ν

reactants

∑ ΔGfo

Synoptic Table 3C.1* Standard Gibbs energies of formation (298 K) ΔfGo/(kj molʹ′ʹ′ʹ′ʹ′1)

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B. Combined 1st and 2nd Law:

1. Derive the fundamental equation:

dU = dq + dw reversible reversible

TdS -pdV

dU = dq + dw

irreversible irreversible

(TdS-δ) (-pdV+δ)

cancel

So dU = TdS - pdV regardless of reversible or irreversible; only requirement is that of closed system (no matter flow)

dU = TdS - pdV

2. Can now make other identifications:

Since U = function of S and V, write:

dU = (∂U/∂S)VdS + (∂U/∂V)SdV ↑ ↑ T -p (comparing term by term with above)

Now we have completely thermodynamic (non-statistical) definition of T:

T = (∂U/∂S)V

3. Properties of the Gibbs energy:

dG = d(H - TS)

= dH - TdS - SdT ↑d(U + pV) dU + pdV + Vdp

dG = dU + pdV + Vdp - TdS - SdT ↑____________↑ -dU = Vdp – SdT

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dG = Vdp - SdT Implications: G↓ as T↑ since S always > 0. G↑ as p↑ since V always > 0. Viewing G as a function of p and T:

dG = (∂G/∂p)Tdp + (∂G/∂T)pdT ↑ ↑ comparing with previous equation V -S

Now we're ready to establish the T and p dependence of G. 4. Temperature dependence of G:

Important because it will later show us how the equilibrium constant Keq of a reaction depends on T.

∂G∂T

"

#$$

%

&''p

= −S from above; see graph at right

= −((H−G) / T)

∂G∂T

"

#$$

%

&''p

−G / T = −H / T

T∂ G T( )∂T

"

#

$$

%

&

''p

= −H / T using product rule of diff

so we obtain Gibbs-Helmholtz Equation:

∂ G T( )∂T

"

#

$$

%

&

''p

= −H / T2

or for whole processes, such as chemical reaction:

∂ ΔG T( )

∂T

#

$

%%

&

'

((p

= −ΔH / T2 (constant T & p)

Important thing to notice: ΔG dependence on T depends on the enthalpy of reaction.

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For example: endothermic rxn has positive ΔH.

Therefore, ΔG/T↓ as T↑ spontaneity↑ as T↑

Keq↑ as T↑

Will see later that

Keq= e−ΔG

o/RT

ΔGo = −RTlnKeq

5. Pressure dependence of G:

Let's study this at fixed T dG = Vdp - SdT ↑=0

Gf−G

i= Vdppi

pf

∫

For solid and liquids, since V changes very slightly with pressure:

Gf−G

i≈ V dp = VΔp

pi

pf

∫ ≈ very small since V small.

Implies ΔGrxn very weakly p-dependent for solids and liquids. For gases, however:

Gf−G

i= nRT / p( )pi

pf

∫ dp (perfect gas)

Gf – Gi = nRT ln pf/pi Or in terms of molar Gibbs energies Gm:

Gm(pf) – Gm(pi) = RT ln pf/pi

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Implies ΔGrxn more strongly p-dependent for rxns involving gases.

0 = RT ln p2/p1 + ΔGrxn(p2) - 2RT ln p2/p1 - ΔGrxn(p1)

ΔGrxn(p2) = ΔGrxn(p1) + (2-1)RT ln p2/p1 ↑ ΔGrxn(p2) = ΔGrxn(p1) + (Δng)RT ln p2/p1

Δng = change in gas moles 6. Relate Gibbs energy of a gas to the standard state Gibbs energy: Since Gm(p2) = Gm(p1) + RT ln p2/p1

if p1 = 1 bar (standard pressure) = pθ p2 = our working pressure p

Gm(p) = Gm° + RT ln p/pθ This is a very important statement when we begin to discuss

chemical equilibrium, because of the relation of the equilibrium constant to the reaction quotient of gas pressures.

Keq=pNO2

2

pN2O4

p2

p1

N2O4(g)

ΔGrxn(1) 2NO

2(g)

N2O4(g)

ΔGrxn(2) 2NO

2(g)

p2

p1

nRT ln p2p1

nRT ln p2/p1

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