please pick up ice, water, steam quiz internal energy, heat & work problem set
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Please Pick Up
• Ice, Water, Steam Quiz• Internal Energy, Heat & Work problem Set
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Energy Cycles
Edward A. Mottel
Department of Chemistry
Rose-Hulman Institute of Technology
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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
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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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04/18/23
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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04/18/23
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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04/18/23
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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04/18/23
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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04/18/23
Gas ExpansionCarbon Dioxide Fire Extinguisher
Why is it dangerous to point acarbon dioxide fire extinguisher
at a person?
liquid CO2
gaseous CO2
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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
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Enthalpy
Absolute enthalpy values cannot be measured,but changes in enthalpy can be measured
relative to an arbitrary reference state
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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
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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?
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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?
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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 =
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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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04/18/23
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