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

H2 and O2 combine spontaneously to form water (exothermic)… liquid water vaporizes spontaneously at room temperature; an endothermic process. The spontaneous change

When the valve is the gases mix spontaneously. opened …

• 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 dq dS = rev T qrev is the heat supplied reversibily

f dq ΔS = rev ∫i i T 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

dqsur,rev dqsur dSsur = = Tsur Tsur 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”.

qsur ΔSsur = Tsur 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. For an adiabatic change, ΔSsur = 0 3.2 Entropy

(b) The statistical view of entropy Boltmann formula for entropy S = k lnW • 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

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 3.2 Entropy

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 q q ∫ dS = q + c Th Tc 3.2 Entropy

(c) The entropy as a state function A Carnot Cycle q q ∫ dS = q + c Th Tc q T h = − h qc Tc dq ∫ rev = 0 Tsur 3.2 Entropy

(c) The entropy as a state function We need to show that the following equation applies to any materials dq ∫ rev = 0 Tsur Efficiency (η)

work performed w η = heat absorbed from hot source = qh q − q q η = h c =1− c qh qh q T T Because h = − h , η =1− c qc Tc Th 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 T η =1− c is true for any substance Th 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 q q ∑ rev = ∑ rev =0 all T perimeter T • dS is an exact differential • S is a state function 3.2 Entropy

(d) Thermodynamic temperature • Thermodynamic temperature scale

T = (1−η)Th • 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 3.2 Entropy

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

dwrev ≥ dw ⇒ −dwrev ≥ −dw

dU = dq + dw = dqrev + dwrev

dqrev − dq = dw − dwrev ≥ 0 ⇒ dqrev ≥ dq dq dq dS = rev ≥ ; the Clausius inequality T T dS ≥ 0 for an isolated system due to dq = 0 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 3.3 Entropy changes of specific processes

(a) Expansion V ΔS = nRln f Vi • The total change in entropy does depend on how the expansion takes place

• A reversible change ; dqsur = -dq

qsur qrev Vf ΔSsur = = − = −nRln T T Vi ΔStot = 0 • Free expansion ; w = 0 and q = 0

Vf ΔStot = nRln > 0 Vi 3.3 Entropy changes of specific processes

(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,

ΔtrsH ΔtrsS = Ttrs 3.3 Entropy changes of specific processes

(b) Phase transition • Trouton’s rule ; a wide range of liquids give approximately the same standard entropy of vaporization about 85 J/K⋅mol 3.3 Entropy changes of specific processes

Tf logarithm dependence S T = S T + dqrev ( f ) ( i ) ∫T T i At constant pressure,

Tf C dT S T = S T + p ( f ) ( i ) ∫T T i

When Cp is independent of T in the range

Tf T S T = S T + C dT = S T + C ln f ( f ) ( i ) p ∫T T ( i ) p i Ti 3.3 Entropy changes of specific processes

(d) The measurement of entropy

Tf C (s,T ) Δ H Tb C (1,T ) S T = S T + p,m dT + fus + p,m dT m ( f ) m ( i ) ∫ 0 T ∫T T f Tf

Δ H T C (g,T ) + vap + p,m dT ∫T T b Tb

3 Debye extrapolation ; Cp,m=aT near T = 0 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 3.4 The Third Law of Thermodynamics

(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, ΔS → 0 as T → 0 • 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 3.4 The Third Law of Thermodynamics

(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

Θ Θ Θ ΔrS = ∑ vSm − ∑ vSm Products Reactants

Θ Θ ΔrS = ∑vJSm (J) J SΘ ( H+, aq ) = 0