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• Ice, Water, Steam Quiz• Internal Energy, Heat & Work problem Set

Energy Cycles

Edward A. Mottel

Department of Chemistry

Rose-Hulman Institute of Technology

04/18/23

Energy Cycles

Reading Assignment:• Chang, Chapter 6.6

This lecture involves the concept of thermodynamic energy cycles and calculation of the heat energy released by these cyclic processes.

The importance of different pathways is exemplified by the Carnot cycle and Hess' Law.

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Energy Cycles

A series of energy steps following a defined pathway in which the final step returns the system to the original state conditions.

Pathways• Isobaric• Isothermal (constant external pressure)• Isothermal (varying external pressure)• Adiabatic

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Sublimation of Water at 0 C

sublimationH2O (s, 0 C) H2O (g, 0 C)

H2O (g, 100 C)

H2O (l, 100 C)

H2O (l, 0 C)

Which thermodynamic terms are associatedwith each step?

heat capacityof steam

enthalpy of vaporization

heat capacityof water

enthalpyof fusion

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Determine the Enthalpy ofSublimation of Water at 0 C

Assume the heat capacity of water vaporis constant from 0 C to 200 C.

H2O (s, 0 C) H2O (g, 0 C)

H2O (l, 100 C)

H2O (l, 0 C) H2O (g, 100 C)

sublimation

+50 cal/g+209 J/g

-540 cal/g-2259 j/g

-100 cal/g-418 J/g

-80 cal/g-335 J/g

+670 cal/g+2803 J/g

04/18/23

Ice, Water, Steam Quiz

Select various masses of ice, water and steam at temperatures consistent with the phases.

Determine the final temperature of the mixture and the number of grams of each phase present in the final mixture.

Confirm your answer with the program ICEWATER in the class folder.

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Isothermal ExpansionConstant External Pressure

1.5 atm1.5 atm1.5 atm

One liter of a compressed gas in a cylinder causes a piston to expand against a constant

external pressure of 1.5 atm until the total volume of the gas in the cylinder is three liters.

The initial and final temperatureof the gas in the cylinder

is the same.

Sketch a graph of this process

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Isothermal ExpansionConstant External Pressure

0 1 2 3 40

1

2

Volume (L)

Pe

xt (

atm

)

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Isothermal ExpansionConstant External Pressure

0 1 2 3 40

1

2

Volume (L)

Pe

xt (

atm

)

Work may berepresentedas the area

under the curve.

work = -Vf

Vi

P dV = - PVf

Vi

dV = - P (Vf - Vi)

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Energy Units

Latm can be converted to more common energy units (e.g., J or cal) by using the value of R as a conversion factor.

Determine the work done in calories and joules

work = - (1.5 atm) (3.0 L - 1.0 L) = -3.0 Latm

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Energy Units

Determine the work done in calories and joules

work = - (1.5 atm) (3.0 L - 1.0 L) = -3.0 Latm

= -3.0 Latm ×1.987 calmol-1K-1

0.08206 Latmmol-1K-1= -73 cal

= -3.0 Latm ×8.314 Jmol-1K-1

0.08206 Latmmol-1K-1= -304 J

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Engines

An engine is a machine which can perform work.

The expansion of a gas in a piston can do work.

Describe the activities of an expanding gas at constant external pressure.

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EnginesConstant External Pressure

1.5 atm

Expansion Stroke

1.5 atm1.5 atm

Compression Stroke

1.5 atm 1.5 atm

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Engines

0 1 2 3 40

1

2

Volume (L)

Pe

xt (

atm

) expansioncycle

As described this does not represent a practical enginebecause the final state does not equal the initial state.

compressioncycle

How much work is done in this overall process?

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Isothermal ExpansionDecreasing External Pressure

4.5 atm

One liter of a compressed gas in a cylinder causes a piston to expand against a decreasing

external pressure until the total volume of the gas in the cylinder is three liters.

2.25 atm1.5 atm

The initial and final temperatureof the gas in the cylinder

is the same.

Sketch a graph of this process

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Isothermal ExpansionDecreasing External Pressure

PV = nRT Work may berepresentedas the area

under the curve.

work = -Vf

Vi

P dV

0 1 2 3 40

1

2

Volume (L)

Pe

xt (

atm

)

3

4

The pressure term is rewritten in terms of volume.

= - Vf

Vi

dVnRTV

= - nRT ln Vf

Vi

Isothermal ExpansionDecreasing External Pressure

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0 1 2 3 40

1

2

Volume (L)

Pe

xt (

atm

)

3

4Isothermal Expansion

Decreasing External Pressure

Is there anynet heat flow

in this process?

E = q + w the internal energy

of a phase is a functionof its temperature

E = qv = m Cv T

isothermal

= 0

Is heat flowinginto or out of the system?

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0 1 2 3 40

1

2

Volume (L)

Pe

xt (

atm

)

3

4Isothermal Expansion

Decreasing External Pressure

The decreasing external pressure piston performsmore work (greater efficiency) than a piston workingagainst a constant external pressure equal to Pfinal.

ConstantExternal Pressure

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0 1 2 3 40

1

2

Volume (L)

Pe

xt (

atm

)

3

4Isothermal Expansion

Decreasing External Pressure

If the process is reversed,how much work

is done?

If the isothermal expansion process is reversedby isobaric compression,how much work is done?

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Gas Expansion Room temperature

gas

colder gas

When a gas expands againsta low restraining pressure

why does it cool?

E = q + w

adiabatic expansionq = 0

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Gas ExpansionCarbon Dioxide Fire Extinguisher

Why is it dangerous to point acarbon dioxide fire extinguisher

at a person?

liquid CO2

gaseous CO2

Enthalpy

H = E + PV

Heat energycontent ofa molecule

internal molecular motion

electronic energy

pressure-volume workrequired to maintain

the volume of the moleculeOnly definedfor a constantpressure process

Enthalpy

Absolute enthalpy values cannot be measured,but changes in enthalpy can be measured

relative to an arbitrary reference state

Enthalpy Values

An element in itsmost stable phase

at 25 ºC and1 atm pressure

An element, moleculeor ion at 25 ºC and

1 atm pressure

Each element hasonly one reference state

Enthalpy of formationvalues are listed in a

Hfº tableHfº = 0

ReferenceState

StandardState

Hfºchange

(final value - initial value)

enthalpy standard state25 ºC and 1 atm

formation

Is enthalpy a function of temperature?

Enthalpy of Formation

What do the initial and final values refer to?

Enthalpy of FormationExamples

Hfº at 25 ºC and 1 atm for one mole of

O2(g) O2(g)oxygen:

Hfº = 0

O3(g) 3/2 O2(g)ozone:

Hfº = +142.7 kJ·mol-1

CH4(g) C(gr) + 2 H2(g)methane:

Hfº = -74.81 kJ·mol-1

Which of the reactions is exothermic?

Write the equation and determine theenthalpy of formation at 25 ºC and 1 atm for

one mole of

HCHO(g)

formaldehyde:

Hfº =

C(dia)

diamond:

Hfº =

KBr(s)

potassium bromide:

Hfº =

Write the equation and determine theenthalpy of formation at 25 ºC and 1 atm for

one mole of

HCHO(g) C(gr) + H2(g) + 1/2 O2(g)

formaldehyde:

Hfº = -108.57 kJ·mol-1

C(dia) C(gr)

diamond:

Hfº = +1.897 kJ·mol-1

KBr(s) K(s) + 1/2 Br2(l)

potassium bromide:

Hfº = -392.17 kJ·mol-1

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Hess' Law

The enthalpy change of a chemical reactionis equal to the difference of the enthalpy of the

products and the enthalpy of the reactants.

Reactants Products

Elements

Hrx

Hf,productsHf,reactants

Hrx = Hf,products - Hf,reactants

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Thermodynamic ApplicationsComparison of Liquid Fuels

methanol ethanol isooctane