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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

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The Story of Spontaneity and Energy Dispersal

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- 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.

- 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.

- 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.

- 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.

- “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 summarizes the recognition of the spontaneous and nonspontaneous processes.

Hot

Reservoir

Engine

Work

Heat

Cold

Reservoir

Heat

- 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.

- 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.

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

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

- 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.

- 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.

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

- 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

- 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.

- 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:

- 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.

- 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

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

Hot

Reservoir

Th

qh

-w1

w3

Engine

w4

-w2

qc

Cold

Reservoir

Tc

- 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

Step 1: ΔU=0

Step 2: ΔU=w

Step 3: ΔU=0

Step 4: ΔU=-w

Hot

Reservoir

w

Engine

Work

qh

Heat

-qc

Cold

Reservoir

Heat

- 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

Hot

Reservoir

w

Engine

Work

qh

Heat

-qc

Cold

Reservoir

Heat

- 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 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.

- 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

- 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.

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

- 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

- In the Debye extrapolation, the expression below is assumed to be valid down to T=0.

- 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).

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

- 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.

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

- 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.

- 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?

- The Clausius inequality implies that dS 0.
- “In an isolated system, the entropy cannot decrease when a spontaneous change takes place.”

- 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:

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

- 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

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

- Helmholtz energy, A:
A = U - TS

dA = dU – TdS

dAT,V≤ 0

- Gibbs energy, G:
G = H - TS

dG= dH– TdS

dHT,p≤ 0

- 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.

, useful

- 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)