Thermodynamics:the Second Law

자연과학대학 화학과 박영동 교수

The Direction of Nature and spontaneity

Thermodynamics: the Second Law

4.1 Entropy 4.1.1 The direction of spontaneous change 4.1.2 Entropy and the Second Law 4.1.3 The entropy change accompanying expansion 4.1.4 The entropy change accompanying heating 4.1.5 The entropy change accompanying a phase transition 4.1.6 Entropy changes in the surroundings 4.1.7 Absolute entropies and the Third Law of thermodynam-ics 4.1.8 The statistical entropy 4.1.9 Residual entropy 4.1.10 The standard reaction entropy 4.1.11 The spontaneity of chemical reactions 4.2 The Gibbs energy 4.2.12 Focusing on the system 4.2.13 Properties of the Gibbs energy

The direction of spontaneous change

the dispersal of matter

the chaotic dispersal of energy

Can we convert heat to work com-pletely?

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The Second Law

1. Reversible vs Irreversible processes

Reversible: A process for which a system can be restored to its initial state, without leaving a net influence on the system or its environment.

* idealized, frictionless;

* proceeds slowly enough for the system to remain in thermodynamic equilibrium.

Irreversible: Not reversible

* natural;

* proceeds freely, drives the system out of thermodynamic equilibrium;

* interacts with environment, can not be exactly reversed

Example: Gas-piston system under a constant temperature

* Slow expansion and compression

0)( 2112 wwpdvwq

* Rapid expansion and compression

)( ppex

( )ex ep p p ( )ex cp p p

)0( dU

2 1

1 2

2 2 2

1 1 1( ) 0

e c

e c e c

q w p dV p dV

p dV p dV p p dV

The Carnot cycle and en-tropy

1) 1 →2: Isothermal expansion

2 1 2 1,hotT T T V V

2) 2 → 3: Adiabatic expansion

3 2 3 2,cold hotT T T T V V

012 U

22

12 12 11

lnhot

Vq w pdV RT

V

023 q

23 23 ( )V cold hotw U c T T

The cycle operates with one mole ideal gas.

3) 3 → 4: Isothermal compression

1 4 1 4,hot coldT T T T V V

4 3 4 3,coldT T T V V

4) 4 → 1: Adiabatic compression

034 U

44

34 34 33

lncold

Vq w pdV RT

V

041 q

41 41 ( )V hot coldw U c T T

4. Work for adiabatic expansion

For an adiabatic process for an ideal gas,

2 2 1 1

V VC C

nR nRV T V T 2 2 1 1c cV T V T where, VC

cnR

1

1

"

( ) '

'

'

c

c

c c

c

c

T V const

pV V const

p V const

pV const

1 11 pV V V

V V V

CC C nR C nRc nR

c nR c nR C C C

pV const

2 2

1 1

2 2

1 1

2 1 2 1

1 2 1 2

1 1

1 1ln ln

ln ln

V V

VV

T VV V

T V

C C

nR nR

CnRTC dT pdV dV dT dV

V nR T V

C C T VdT dV

nR T V nR T V

T V T V

T V T V

from chap. 2

The net heat transfer and the net work over the Carnot cycle are:

2 4

1 3

ln lnhot cold

V Vw q RT RT

V V

32 4 2

3 1 1 4

1 ,VV V V

V V V V

2

1

( ) ln 0hot cold

Vw q R T T

V

Then,

So the system absorbs heat and performs net work in the Carnot cycle,which behaves as a heat engine.

VC

nRf i

i f

T V

T V

2

3

VC

nRcold

hot

T V

T V

2→3, 4

1

VC

nRhot

cold

T V

T V

4→1,

The Carnot cycle -summary

q w ∆U ∆H ∆S

1→2 RThln(V2/V1) -RThln(V2/V1) 0 0 Rln(V2/V1)

2→3 0 cV(Tc-Th) cV(Tc-Th) cP(Tc-Th) 0

3→4 -RTcln(V2/V1) RTcln(V2/V1) 0 0 -Rln(V2/V1)

4→1 0 cV(Th-Tc) cV(Th-Tc) cP(Th-Tc) 0

cycle R(Th-Tc)ln(V2/V1) -R(Th-Tc)ln(V2/V1) 0 0 0

For the Carnot cycle, we can also have:

3412 2 4

1 3

ln ln 0hot cold

qq V VqR

T T T V V

This relationship also hold for the reversed Carnot cycle.

This is called the Carnot’s Theorem:

revT

qdSSd

,0

The change of S is independent of path under a reversible process. S is a state function, means a system property. It is called entropy.

(4.1)

3. The Second Law and its Various Forms

To get the second law, we use the Clausius Inequality, i.e.,

0T

q

for a cyclic process,

which indicates during the cycle,

1) heat must be rejected to the environment;

2) heat exchange is larger at high temperature than at low temperature under reversible conditions;

3) the net heat absorbed is smaller under the irreversible condition than under the reversible condition.

Now, consider two cycles as shown in the plot.

For the cycle which contains onereversible process and one irreversible process,

01

2

2

1

irrevrev T

q

T

q (4.2)

For the cycle which have two reversible processes,

01

2

2

1

revrev T

q

T

q (4.3)

The difference between (4.2) and (4.3) gives

irrevrev T

qSd

T

q

1

2

1

2

1

2

Because states 1 and 2 are arbitrary, we have the second law,

T

qdS

(4.4)

It indicates that the heat absorbed by the system during a processhas an upper limit, which is the heat absorbed during a reversibleprocess.

Combine (4.1) and (4.4), we have

revT

q

T

q

The first law relates the state of a system to work done on it and heat it absorbs.

The second law controls how the systems move to the thermodynamicequilibriums, i.e., the direction of processes.

rev

w w

T T

or

16

The Second Lawof Thermodynamics

The Second Law of Thermodynamics establishes that all spontaneous or natural processes increase the en-tropy of the universe

DStotal = DSuniv = DSsys + DSsurr

In a process, if entropy increases in both the system and the surroundings, the process is surely spontaneous

Several simplified forms of the second law:

1) For an adiabatic process, (4.4) becomes

0dSIf the adiabatic process is reversible, then

0dS

It is also isentropic (S is constant).

2) For an isochoric process, (4.1) becomes

VV

revVV T

dTc

T

dTcdS

Because only state variables are involved, it holds for eitherreversible or irreversible processes.

(4.5) and (4.6) show that :

Irreversible work can only increase entropy; heat transfer can either increase or decrease entropy.

(4.5)

(4.6)

5. Thermodynamic Equilibrium

* Consider an adiabatic process, the second law becomes

0dS

or0SS

For an irreversible condition,

0dS

is the entropy at the initial state. 0s

When reaches the maximum,the state is in thermodynamicequilibrium because the entropy can not increase anymore.

s

Calculate the changes in entropy as a result of the transfer of 100 kJ of energy as heat to a large mass of water (a) at 0°C (273 K) and (b) at 100°C (373 K).

(a) ΔS at 0°C (273 K)

(b) ΔS at 100°C (373 K)

31100 10 J

366 J K273 K

revqS

T

31100 10 J

268 J K373 K

revqS

T

Heat engines

The engine will not operate spontaneously if this change in entropy is negative, and just becomes spontaneous as ΔStotal becomes positive. This change of sign occurs ΔStotal = 0, which is achieved when

the efficiency, η, of the engine, the ratio of the work produced to the heat absorbed, is

H L

hot cold

q qS

T T

coldL H

hot

Tq q

T

work produced1

heat absorbed

1

H L L

H H

cold

hot

q q q

q q

T

T

Refrigerators, and heat pumps

coldL Lcooling

in H L hot cold

Tq qCOP

w q q T T

hotH Hheating

in H L hot cold

Tq qCOP

w q q T T

coldL H

hot

Tq q

T

Refrigerator powerNo thermal insulation is perfect, so there is always a flow of energy as heat into the sample at a rate proportional to the temperature difference. The rate at which heat leaks in can be written as A(Th – Tc ), where A is a constant. Calculate the minimum power, P, re-quired to maintain the original temperature difference? Assume the refrigerator is operating at 100% of its theoretical efficiency. Express P in terms of A, Th, Tc.

The entropy change with isother-mal expansion

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2

1

2 2 2

1 1 1

2 2

1 1

1

1

ln ln

S dqT

p nRdw dV dV

T T VV p

nR nRV p

The entropy change with heating

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2

1

22

11

1

lnxx

S dqTC T

dT CT T

In case heat capacity is constant,

The entropy change with heating

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2

1

2

1

1

x

S dqTC

dTT

In case heat capacity is tempera-ture dependent,

The entropy change with a phase transition

When a solid (a), melts, the molecules form a more chaotic liquid, the disor-derly array of spheres (b). As a result, the entropy of the sample increases.

2

1

2

1

1

1 tr

tr tr

S dqT

Hdq

T T

At phase transition, the tempera-ture stays constant, T = Ttr.

Absolute entropies and the Third Law of thermodynamics

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S (0) = 0 for all perfectly or-dered crystalline materials.

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0( )

T pCS T dT

T

nonmetallic solids, Debye T 3 -law:At temperatures T << TD,

Cv,m = aT 3 ,

and

Sm (T) = Cv,m  

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Standard Molar EntropiesIn general, the more atoms in its molecules, the greater is the entropy of a substance

The statistical entropy

The 19 arrangements(W ) of four molecules in a system with three energy levels and a total energy of 4ε.

U = 4ε

U = 0 U = ε U = 2ε U = 3ε

W = 1 W = 4 W = 10 W = 20

U W0 11 42 103 204 355 566 847 1208 1659 220

10 286

A simple 4-particle system(in equal-spaced energy levels)

En-

erg

y

S = kB ln(W )W = C(n+3, n) = (n+3)(n+2)(n+1)/6

U = nε

A simple 4-particle system

W = C(n+3, n) = (n+3)(n+2)(n+1)/6U = nεS = kB ln(W )

U = 0 U = ε

A two-level system

Two–Level systema very simplified version

En-

erg

yIn case when ε >> kT, almost no particle can reach upper state.According to Boltzmann, the probability of a particle at level 1 at T isp1 = exp(-ε/kT ).

p1 = exp(-ε/kT ) 

Two–Level systema very simplified version

U(T) = Nε exp(-ε/kT )

Cv,m = (∂U/ ∂T) = Nε exp(-ε/kT ) (-ε/k) (-1/T 2)= N (ε 2/kT 2) exp(-ε/kT ) ≈ 0

1. This is why we do not consider heat capacity contribu-tions from subatomic particles such as electrons, etc.

2. This is why heat capacity approaches 0 when T ap-proaches 0 for every material.

ε 1.66E-19"=100kJ/mol"kB 1.36E-23 　 J/K

T, K U(T), J/mol Cv(T), J/mol0.001 0 0

1 0 0 10 0 0

100 9.00E-49 1.10E-481000 4.96E-01 6.06E-03

W and Energy Level Spac-ing

At a given temperature, the number of arrangements corre-sponding to the same total energy is greater when the energy levels are closely spaced than when they are far apart.

Residual entropy

for CO molecules, there are 2N possible arrangements at T= 0.Sm = NAk ln 2 = R ln 2 = 5.8 J K−1 mol−1

Ice at 0 K

W =N, S = k lnN = Nk ln, Sm = R ln = 3.4 J K−1 mol−1.

The spontaneity of chemical reac-tions

surr

HS

T

0univ surr

HS S S S

T

0univ surrS S S

at constant p, qp=ΔH.

surr

qS

T

q

0H T S

if we define G ≡ H – TS, Gibbs Free Energy,ΔG = ΔH –TΔS at constant T.

So, at constant T and p, ΔG = ΔH –TΔS ≤ 0.

Gibbs Energy, G

Spontaneity criterionΔSuniv > 0. at constant pressure and temperature, ΔGsystem < 0.

Calculate the change in molar entropy when one mole of argon gas is compressed from 2.0 dm3 to 500 cm3 and si-multaneously heated from 300 K to 400 K. Take CV,m = (3/2)R

p

V

T = 300 K

T = 400 K

Calculate the change in entropy when 100 g of water at 80°C is poured into 100 g of water at 10°C in an insulated vessel given that Cp,m = 75.5 J K−1 mol−1.

The enthalpy of vaporization of chloroform (trichloro-methane), CHCl3, is 29.4 kJ mol−1 at its normal boiling point of 334.88 K. (a) Calculate the entropy of vaporiza-tion of chloroform at this temperature. (b) What is the en-tropy change in the surroundings?

Suppose that the weight of a configuration of N molecules in a gas of volume V is proportional to VN. Use Boltz-mann’s formula to deduce the change in entropy when the gas expands isothermally.

Without performing a calculation, estimate whether the standard entropies of the following reactions are positive or negative: