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The Story of Spontaneity and Energy DispersalPowerPoint Presentation

The Story of Spontaneity and Energy Dispersal

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### The Story of Spontaneity and Energy Dispersal

Efficiency is maximized Greater temperature difference between reservoirs The lower Tc, the greater the efficiency

You never get what you want: 100% return on investment

Spontaneity

- Spontaneous process are those that occur naturally.
- Hot body cools
- A 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.

Spontaneity

- The reverse of a spontaneous process is a nonspontaneous process
- Confining a gas in a smaller volume
- Cooling an already cool object

- Nonspontaneous processes require energy in order to realize them.

Spontaneity

- Note:
- Spontaneous 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.

Spontaneity

- The 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 Thermodynamics

- The 2nd Law of Thermodynamics summarizes the recognition of the spontaneous and nonspontaneous processes.

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 Dispersal

- Superheroes with energy blasts and similar powers as well as the Super Saiyans are impossible characters.
- They violate the Second Law of Thermodynamics!

Energy Dispersal

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

Energy Dispersal

- In 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 be random thermal motion to be redirect 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 Spontaneity

- Spontaneous change can now be interpreted as the direction of change that leads to the dispersal of the total energy of an isolated system!

Entropy

- A 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 Law

- The 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.

Entropy

- A 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:

Entropy

- For 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.

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 function
- Can assume process is reversible, dUsur= dUsur,rev

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

- dUsur is independent of how change brought about (U is state function

Change in entropy of the surroundings: ΔSsur

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

Reservoir

Th

qh

-w1

w3

Engine

w4

-w2

qc

Cold

Reservoir

Tc

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

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 Tcto Th

Reservoir

w

Engine

Work

qh

Heat

-qc

Cold

Reservoir

Heat

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 )
- William Thomson (Lord Kelvin) defined a substance-independent temperature scale based on the heat transferred between two Carnot cycles sharing an isotherm

Reservoir

w

Engine

Work

qh

Heat

-qc

Cold

Reservoir

Heat

Carnot Cycle - Thermodynamic Temperature Scale- 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 (Atkins, p. 98-99)

Efficiency of Heat Engines

- Efficiency 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.

Exercises (10 pts each)

Entropy changes: Expansion

- Entropy 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

Enropy 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)

- 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 Thermodynamics

- At 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 theorem

- The 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 entropies

- These are entropies reported on the basis that S(0) = 0.

Exercise

- Calculate the entropy of transition between orthorhombic sulfur and monoclinic sulfur at the transition temperature, 369 K. The transition enthalpy is -402 J/mol.
- Calculate the change in entropy that 1.5 moles of krypton suffers, assuming that it behaves ideally, when it is heated. Its volume changes during heating from 50L to 150L, all at 0.5 atm.

One for the road

- Life 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?

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 spontaneity

- At either constant U or constant S:
- Which leads to

Criteria for spontaneity

- When 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 spontaneity

- We can introduce new thermodynamic quantities in order to more simply express and

Helmholtz and Gibbs energy

- Helmholtz energy, A:
A = U - TS

dA = dU – TdS

dAT,V≤ 0

- Gibbs energy, G:
G = H - TS

dG= dH– TdS

dHT,p≤ 0

Helmholtz energy

- A 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.

Exercise

- When 1.000 molC6H12O6(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-1and ΔrSθ = +182.4 J K-1 mol-1at 25°C. How much of this energy change can be extracted as (a) heat at constant pressure, (b) work?
- 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.
- Calculate the standard reaction Gibbs energies of the following reactions given the Gibbs energies of formation of their components
- Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
- C12H22O11(s) + 12 O2(g) 12 CO2(s) + 11 H2O(l)

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