the story of spontaneity and energy dispersal

The Story of Spontaneity and Energy DispersalYou never get what you want:

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SpontaneitySpontaneous process are those that

occur naturally.Hot body coolsA gas expands to fill the available volume

A spontaneous direction of change is where the direction of change does not require work to bring it about.

SpontaneityThe reverse of a spontaneous

process is a nonspontaneous processConfining a gas in a smaller volumeCooling an already cool object

Nonspontaneous processes require energy in order to realize them.

SpontaneityNote:

Spontaneity is often interpreted as a natural tendency of a process to take place, but it does not necessarily mean that it can be realized in practice.

Some spontaneous processes have rates sooo slow that the tendency is never realized in practice, while some are painfully obvious.

SpontaneityThe conversion of diamond to

graphite is spontaneous, but it is joyfully slow.

The expansion of gas into a vacuum is spontaneous and also instantaneous.

2ND LAW OF THERMODYNAMICS

The 2nd 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”

Statement formulated by Lord Kelvin

The 2nd Law of ThermodynamicsThe 2nd Law of Thermodynamics

recognizes the two classes of processes, the spontaneous and nonspontaneous processes.

HotReservoir

Heat Engine Work

HeatColdReservoir

What determines the direction of spontaneous change?The total internal energy of a system

does NOT determine whether a process is spontaneous or not.

Per the First Law, energy is conserved in any process involving an isolated system.

What determines the direction of spontaneous change?Instead, it is important to note that

the direction of change is related to the distribution of energy.

Spontaneous changes are always accompanied by a dispersal of energy.

Energy DispersalSuperheroes with

energy blasts and similar powers as well as the Super Saiyans are impossible characters.

They seem to violate the Second Law of Thermodynamics!

Power

Kamehame wave

Energy DispersalA ball on a warm

floor can never be observed to spontaneously bounce as a result of the energy from the warm floor

Energy DispersalIn order for this to

happen, the thermal energy represented by the random motion and vibrations of the floor atoms would have to be spontaneously diverted to accumulate into the ball.

Energy Dispersal It will also require the

random thermal motion to be redirected to move in a single direction in order for the ball to jump upwards.

This redirection or localization of random, disorderly thermal motion into a concerted, ordered motion is so unlikely as to be virtually impossible.

Energy Dispersal and SpontaneitySpontaneous change can now be

interpreted as the direction of change that leads to the dispersal of the total energy of an isolated system!

EntropyA state function, denoted by S.

While the First Law can be associated with U, the Second Law may be expressed in terms of the S

Entropy and the Second LawThe Second Law can be expressed in

terms of the entropy:

The entropy of an isolated system increases over the course of a spontaneous change: ΔStot > 0

Where Stot is the total entropy of the system and its surroundings.

EntropyA simple definition of entropy is that

it is a measure of the energy dispersed in a process.

For the thermodynamic definition, it is based on the expression:

EntropyFor a measurable change between two

states,

In order to calculate the difference in entropy between two states, we find a reversible pathway between them and integrate the energy supplied as heat at each stage, divided by the temperature.

Example

Change in entropy of the surroundings: ΔSsur If we consider a transfer of heat dqsur to the surroundings,

which can be assumed to be a reservoir of constant volume.

The energy transferred can be identified with the change in internal energy dUsur is independent of how change brought about (U is state

functionCan assume process is reversible, dUsur= dUsur,rev

Since dUsur = dqsur and dUsur= dUsur,rev, dqsur must equal dqsur,rev

That is, 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 dividing the heat transferred by the temperature at which the transfer takes place.

Change in entropy of the surroundings: ΔSsur

For adiabatic change, qsur = 0, so DSsur = 0

Entropy: A molecular look Boltzmann formula:

Entropy is a reflection of the microstates, the ways in which the molecules of a system can be arranged while keeping the total energy constant.

Entropy as a State Function To prove entropy is a state function we must show that ∫dS

is path independent Sufficient to show that the integral around a cycle is zero or

Sadi Carnot (1824) devised cycle to represent idealized engine

dSdqT 0

HotReservoir

ColdReservoir

Engine-w2

-w1w3

w4

qh

qc

Th

Tc

Step 1: Isothermal reversible expansion @ Th

Step 2:Adiabatic expansion Th to Tc

Step 3:Isothermal reversible compression @ Tc (sign of q negative)Step 4: Adiabatic compression Tc to Th

Carnot CycleStep 1: ΔU=0

Step 2: ΔU=w

Step 3: ΔU=0

Step 4: ΔU=-w

Carnot Cycle - Thermodynamic Temperature Scale

The efficiency of a heat engine is the ratio of the work performed to the heat of the hot reservoir

e=|w|/qh The greater the work the greater

the efficiency Work is the difference between

the heat supplied to the engine and the heat returned to the cold reservoir

w = qh -(-qc) = qh + qc

Therefore, e = |w|/qh = ( qh + qc)/qh = 1 + (qc/qh )

HotReservoir

HeatEngine Work

HeatColdReservoir

qh

-qc

w

Efficiency of Heat EnginesEfficiency is the ratio of the work done by

an engine in comparison to the energy invested in the form of heat for all reversible engines

All reversible engines have the same efficiency irrespective of their construction.

Carnot Cycle - Thermodynamic Temperature Scale

HotReservoir

HeatEngine Work

HeatColdReservoir

qh

-qc

w William Thomson (Lord Kelvin) defined

a substance-independent temperature scale based on the heat transferred between two Carnot cycles sharing an isotherm

He defined a temperature scale such that qc/-qh = Tc/Th

e = 1 - (Tc/Th ) Zero point on the scale is that

temperature where e = 1 Or as Tc approaches 0 e approaches

1 Efficiency can be used as a measure

of temperature regardless of the working fluid

Applies directly to the power required to maintain a low temperature in refrigerators

Efficiency is maximized

Greater temperature difference between reservoirs

The lower Tc, the greater the efficiency

Refrigeration

Coefficient of performance (COP or β or c)

COP describes the qc in this case as the minimum energy to be supplied to a refrigeration-like system in order to generate the required entropy to make the system work.

Entropy changes: ExpansionEntropy changes in a system are

independent of the path taken by the process

Total change in entropy however depend on the path:Reversible process: ΔStot = 0 Irreversible process: ΔStot > 0

Isothermal Isochoric Isobaric Adiabatic

ΔU 0 nCvΔT q+w w

q nRT ln or -wirrev

nCvΔT nCpΔT or –wirrev 0

wrev -nRT ln 0 -nRT ln

wirrev -pextΔV 0 -pextΔV =-nCvΔT=-pextΔV

ΔH 0 (for ideal gas) ΔU=ΔU + pΔV

=nCpΔT

ΔS = = 0

Entropy changes: Phase Transitions

Trouton’s rule: An empirical observation about a wide range of liquids providing approximately the same standard entropy of vaporization, around 85 J/mol K.

General equations for entropy during a heating process S as a function of T and V, at

constant P

S as a function of T and P, at constant V

Measurement of Entropy (or molar entropy)

Measurement of Entropy (or molar entropy)The terms in the previous equation

can be calculated or determined experimentally

The difficult part is assessing heat capacities near T = 0.

Such heat capacities can be evaluated via the Debye extrapolation

Measurement of Entropy (or molar entropy)In the Debye extrapolation, the

expression below is assumed to be valid down to T=0.

Third Law of ThermodynamicsAt T = 0, all energy of thermal motion has been

quenched, and in a perfect crystal all the atoms or ions are in a regular, uniform array.

The localization of matter and the absence of thermal motion suggest that such materials also have zero entropy.

This conclusion is consistent with the molecular interpretation of entropy, because S = 0 if there is only one way of arranging the molecules and only one microstate is accessible (the ground state).

Third Law of Thermodynamics

The entropy of all perfect crystalline substances is zero at T = 0.

Nernst heat theoremThe entropy change accompanying

any physical or chemical transformation approaches zero as the temperature approaches zero: ΔS 0 as T 0 provided all the substances involved are perfectly crystalline.

Third-Law entropiesThese are entropies reported on the

basis that S(0) = 0.

Exercises

HELMHOLTZ AND GIBBS ENERGIES

Clausius inequality

The Clausius inequality implies that dS 0.

“In an isolated system, the entropy cannot decrease when a spontaneous change takes place.”

Criteria for spontaneity

In a system in thermal equilibrium with its surroundings at a temperature T, there is a transfer of energy as heat when a change in the system occurs and the Clausius inequality will read as above:

Criteria for spontaneityWhen energy is transferred as heat at constant volume:

*dq = dU At either constant U or constant S:

Which leads to

Criteria for spontaneityWhen energy is transferred as heat at

constant pressure, the work done is only expansion work and we can obtain

At either constant H or constant S:

Which leads to

Criteria for spontaneityWe can introduce new

thermodynamic quantities in order to more simply expressand

Helmholtz and Gibbs energyHelmholtz energy,

A:A = U - TS

dA = dU – TdS

dAT,V ≤ 0

Gibbs energy, G:G = H - TS

dG = dH – TdS

dGT,p ≤ 0

Helmholtz energyA change in a system at constant

temperature and volume is spontaneous if it corresponds to a decrease in the Helmholtz energy.

Aside from an indicator of spontaneity, the change in the Helmholtz function is equal to the maximum work accompanying a process.

Helmholtz energy

, useful

Variation of the Gibbs free energy with temperature

Variation of the Gibbs free energy with pressure

Variation of the Gibbs free energy with pressure

Homework1. When 1.000 mol C6H12O6 (glucose) is oxidized to carbon

dioxide and water at 25°C according to the equation C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l), calorimetric measurements give ΔrHθ= -2808 kJ mol-1 and ΔrSθ = +182.4 J K-1 mol-1 at 25°C. How much of this energy change can be extracted as (a) heat at constant pressure, (b) work?

2. How much energy is available for sustaining muscular and nervous activity from the combustion of 1.00 mol of glucose molecules under standard conditions at 37°C (blood temperature)? The standard entropy of reaction is +182.4 J K-1 mol-1.

3. Calculate the standard reaction Gibbs energies of the following reactions given the Gibbs energies of formation of their components

a) Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)b) C12H22O11(s) + 12 O2(g) 12 CO2(s) + 11 H2O(l)

One for the roadLife requires the assembly of a large

number of simple molecules into more complex but very ordered macromolecules. Does life violate the Second Law of Thermodynamics? Why or why not?