Chapter 3. The Second Law

Chapter 3. The Second Law

2011 Fall Semester Physical Chemistry 1

(CHM2201)

Contents

The direction of spontaneous change 3.1 The dispersal of energy 3.2 The entropy 3.3 Entropy changes accompanying specific processes 3.4 The Third Law of thermodynamics

Concentrating on the system 3.5 Helmholtz and Gibbs energies 3.6 Standard molar Gibbs energies

Combining the First and Second Laws 3.7 The fundamental equation 3.8 Properties of internal energy 3.9 Properties of the Gibbs energy

The Second Law of thermodynamics

No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work

The spontaneous change

•  A spontaneous process is one that can occur in a system left to itself; no action from outside the system is necessary to bring it about.

•  A non-spontaneous process is one that cannot take place in a system left to itself.

•  If a process is spontaneous, the reverse process is non-spontaneous, and vice versa.

The spontaneous change Water falling (higher to lower potential energy) is a spontaneous process.

H2 and O2 combine spontaneously to form water (exothermic)… liquid water vaporizes

spontaneously at room temperature; an endothermic process.

Conclusion: enthalpy alone is not a sufficient criterion for prediction of spontaneity.

The spontaneous change

When the valve is opened …

the gases mix spontaneously.

•  There is no significant enthalpy change. •  Intermolecular forces are negligible. •  So … why do the gases mix?

3.1 The dispersal of energy

Key points 1.  During a spontaneous change in an isolated

system the total energy is dispersed into random thermal motion of the particles in the system

We look for the direction of change that leads to dispersal of the total energy of the isolated system

3.2 Entropy

Key points 1.  The entropy acts as a signpost of spontaneous

change 2.  The Clausius definition for the entropy 3.  The Boltzmann formula for the entropy 4.  Entropy is a state function (Carnot cycle) 5.  Entropy increases in a spontaneous change (The

Clausius inequality)

3.2 Entropy (a) Thermodynamic definition of entropy

dS = dqrevT

qrev is the heat supplied reversibily

ΔS = dqrevTii

f∫

To calculate the difference in entropy, we find a reversible path between two states, and integrate the energy supplied as heat at each stage of the path divided by T

The Second Law of thermodynamics : The entropy of an isolated system increases in the course of a spontaneous change : ΔStot > 0

3.2 Entropy (a) Thermodynamic definition of entropy

1.  The surroundings consist of a reservoir of constant V. 2.  The energy supplied as heat can be identified with the

change in internal energy, ΔUsur. 3.  U is a state function and independent of the path. 4.  So we can drop “rev”.

dSsur =dqsur,revTsur

=dqsurTsur

ΔSsur =qsurTsur

ΔSsur = 0For an adiabatic change,

Regardless of how the change is brought about in the system, reversibly or irreversibly, we can calculate the change of entropy of the surroundings by using the above equation.

3.2 Entropy (b) The statistical view of entropy

S = k lnWBoltmann formula for entropy

•  k = 1.38×10-23 J/K: the Boltzmann constant •  W : the number of microstates : the ways in

which the molecules of a system can be arranged while keeping the total energy constant

•  When W = 1, S = 0 •  When molecules can access more

microstates for a given energy (e.g. as the system volume increases), the entropy increases

•  Molecules in a system at high T can occupy a large number of the available energy levels, so a small additional transfer of energy as heat will lead to a relatively small change in the number of accessible energy

3.2 Entropy (c) The entropy as a state function

Entropy is a state function

1.  We show that the above equation is true for a ‘Carnot cycle’ involving a perfect gas

2.  We show that the result is true whatever the working substance is

3.  We show that the result is true for any cycle

dqrevTsur∫ = 0


A Carnot Cycle

1.  A ⟶ B : Reversible isothermal expansion at Th ; The entropy change is qh/Th

2.  B ⟶ C : Reversible adiabatic expansion : No energy leaves the system as heat so the change in entropy is zero

3.  C ⟶ D : Reversible isothermal compression at Tc ; the change in entropy of the system is qc/Tc

4.  D ⟶ A : Reversible adiabatic compression ; No energy enters the system as heat so the change in entropy is zero

dS∫ =qqTh+qcTc


A Carnot Cycle

dS∫ =qqTh+qcTc

qhqc= −

ThTc

dqrevTsur∫ = 0


We need to show that the following equation applies to any materials

dqrevTsur∫ = 0

η = work performedheat absorbed from hot source =

wqh

Efficiency (η)

η =qh − qcqh

=1−qcqh

Because qhqc= −

ThTc

, η =1− TcTh

3.2 Entropy (c) The entropy as a state function •  Suppose two reversible engines are

coupled together and run between the same two reservoirs

•  The working substances and details of construction of the two engines are entirely arbitrary

•  Suppose that engine A is more efficient than B

•  Choose a setting of the controls that causes engine B to acquire energy as heat qc from the cold reservoir and to release a certain quantity of energy as heat into the hot reservoir

•  Because engine A is more efficient, not all the work that A produces is needed for this process

3.2 Entropy (c) The entropy as a state function •  The net effect of the processes is the

conversion of heat into work without there being a need for a cold sink : Contrary to the Kelvin statement of the Second Law

•  Initial assumption that engines A and B can have different efficiencies must be false

η =1− TcTh

is true for any substance

3.2 Entropy (c) The entropy as a state function •  Any reversible cycle can be

approximated as a collection of Carnot cycles

•  The integral around an arbitrary path is the sum of the integrals around each of the Carnot cycle

•  This approximation becomes exact as the individual cycles are allowed to become infinitesimal

•  dS is an exact differential •  S is a state function

qrevT

=all∑ qrev

T=

perimeter∑ 0

3.2 Entropy (d) Thermodynamic temperature •  Thermodynamic temperature scale

•  The temperature of the engine can be inferred from the measured efficiency

•  Kelvin scale is defined by using water at its triple point as the notional hot source and defining that temperature as 273.16K exactly

•  If the efficiency of an engine is 0.20, the temperature of the cold sink is 0.80×273.16K = 220K

•  This result is independent of the working substance of the engine

T = (1−η)Th

3.2 Entropy (e) The Clausius inequality •  More work is done when a change is reversible than when it

is irreversible

dS = dqrevT

≥dqT

; the Clausius inequality

dU = dq+ dw = dqrev + dwrev

dS ≥ 0 for an isolated system due to dq = 0

dwrev ≥ dw ⇒−dwrev ≥ −dw

dqrev − dq = dw− dwrev ≥ 0⇒ dqrev ≥ dq

In an isolated system the entropy cannot decrease when a spontaneous change occurs!!!

3.3 Entropy changes of specific processes

Key points 1.  The isothermal expansion of a perfect gas 2.  The entropy change at the transition

temperature 3.  Entropy change can be estimated in terms of the

heat capacity

(a) Expansion

•  The total change in entropy does depend on how the expansion takes place

•  A reversible change ; dqsur = -dq

•  Free expansion ; w = 0 and q = 0


ΔS = nR lnVf

Vi

ΔSsur =qsurT

= −qrevT

= −nR lnVf

ViΔStot = 0

ΔStot = nR lnVf

Vi> 0

(b) Phase transition

•  The transition of a substance accompanies a change in entropy

•  For example, when a substance vaporizes, a compact condensed phase changes into a widely dispersed gas and the entropy increases

•  At the transition temperature, any transfer of energy as heat between the system and the surrounding is reversible because the two phases in the system are in equilibrium

•  Because q = ΔtrsH at constant pressure,


Δ trsS =Δ trsHTtrs

(b) Phase transition •  Trouton’s rule ; a wide range of liquids give approximately the

same standard entropy of vaporization about 85 J/K⋅mol


(c) Heating


S Tf( ) = S Ti( ) + dqrevTTi

Tf∫

S Tf( ) = S Ti( ) + CpdT

TTi

Tf∫

S Tf( ) = S Ti( ) + CpdTTTi

Tf∫ = S Ti( ) + Cp lnTfTi

At constant pressure,

When Cp is independent of T in the range

logarithm dependence

(d) The measurement of entropy


Debye extrapolation ; Cp,m=aT3 near T = 0

Sm Tf( ) = Sm Ti( ) + Cp,m (s,T )

T0

Tf∫ dT + ΔfusHTf

+ Cp,m (1,T )

TTf

Tb∫ dT

+ ΔvapHTb

+ Cp,m (g,T )

TTb

T∫ dT

3.4 The Third Law of Thermodynamics

Key points 1.  The Nernst heat theorem 2.  The Third Law allows us to define absolute

entropies of substances

•  At T = 0, all energy of thermal motion has been quenched •  In a perfect crystal all the atoms or ions are in a regular uniform array

(a) The Nernst heat theorem

•  Nernst heat theorem : The entropy change accompanying any physical or chemical transformation approaches zero as the temperature approaches zero. Provided all the substances involved are perfectly ordered,

•  If we arbitrarily ascribe the value zero to the entropies of elements in their perfect crystalline form at T = 0, then all perfect crystalline compounds also have zero entropy at T = 0

•  The Third Law of Thermodynamics : The entropy of all perfect crystalline substances is zero at T = 0

•  S = klnW •  Residual entropy : imperfectness


ΔS→ 0 as T→ 0

(b) Third-Law entropies

•  Entropies reported on the basis that S(0) = 0 are “Third-Law entropies”

•  The standard (Third-Law) entropy SΘ(T) : the entropy of the substance in its standard state at T

•  The standard reaction entropy


ΔrSΘ = vSm

Θ

Products∑ − vSm

Θ

Reactants∑

ΔrSΘ = vJSm

Θ(J)J∑

SΘ( H+, aq ) = 0

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Chapter 3. The Second Law