Enthalpy and Entropy - ocw.snu.ac.kr
Transcript of Enthalpy and Entropy - ocw.snu.ac.kr
SNU NAOE
Y. Lim
Enthalpy and Entropy
SNU NAOE
Y. Lim
How to know βπ’, ββ?
β’ u and h is not measurable directly.
β’ u and h is associated with heat capacity
β’ The heat capacity is measurable from experiment!
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Heat Capacity Specific heat capacity Molar heat capacity
πΆ = π/βπ π = π/βπ π = π/βπ
Extensive Property Intensive Property Intensive Property
J/K J/KΒ·g (cal/gΒ·K) J/KΒ·mol
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Heat Capacity and U, H
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π
Constant v
π
π
π =?πππππ = ππ£
ππ’ = πΏπ + πΏπ€ = πΏπ β πππ£0
ππ£ =ππ’
πππ£
=ππ’
ππ
ππ’ = ππ£ππ
βπ’ = ΰΆ±ππ£ππ
cv
π
π =?
πβ = π π’ + ππ£ = πΏπ + πΏπ€ + πππ£ + π£ππ = πΏπ + π£ππ0
Constant P
π
π
πππππ = ππ ππ =πβ
πππ
=πβ
ππ
πβ = ππππ
ββ = ΰΆ±ππππ
cp
For ideal gas, u=f(T) only
=ππ’
πππ£
=π
βπ π£
= βπ’
= ββ
=π
βπ π=
πβ
πππ
For ideal gas, h=u+Pv=u+RTh=f(T) only
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Correlation for ββ from experiment
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πππ = π΄ + π΅π + πΆπ2 + π·πβ2 + πΈπ3 πΎ
(Ideal gas condition)good approximation with at low P
Cp (ice-water-steam)
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cP and cv
β’ For solids and liquidsβ’ ππ β ππ£
β’ For ideal gas,
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ππ β‘πβ
πππ
=π π’ + ππ£
πππ
=ππ’
πTP
+ππ π
πTP
=ππ’
πTP
+ π
ππ’
πTP
=ππ’
ππ=
ππ’
πTπ£
ππ = ππ£ + π
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cv of Gases
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cv β3/2 R
Diatomic
cv β5/2 R
Polyatomic
cvβ6/2 R +f(T)
Linear
cv β5/2 R +f(T)
For an ideal gas, cv=3/2R
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ββ with Phase change
ββ = ΰΆ±ππππ = ΰΆ±π1
ππ
ππ,π ππππ ππ + ββππ’π +ΰΆ±ππ
ππ
ππ,πππ ππ + ββπ£ππ +ΰΆ±ππ
π2
ππ,πππ ππ
http://industries-news.blogspot.kr/2010/06/fundamentals-of-refrigeration-and-air.html
Latent heat : heat required to change the phase
Sensible heat : heat required to increase temperature
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Enthalpy of reaction
β’ Standard enthalpy of formation, ππππ
β Change of enthalpy when one mole of substance is formed from its pure element in nature at standard condition, 25β, 1 atm
β’ Standard enthalpy of reaction, ππππππ
β Change of enthalpy by a chemical reaction
β’ ππππππ < 0: exothermic reaction
β’ ππππππ > 0: endothermic reaction
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Example 2.11
β’ In: 10 mol/s of hexane @25β
β’ Out: 100β vapor
β’ Q=? (435 kJ/s in the textbook)
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18PR
429
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Example 2.15
β’ Propane in an adiabatic chamber at 25β.β’ (a) React with stoichiometric amount of oxygen
(T=4910K in textbook)
β’ (b) React with stoichiometric amount of air(T=2370K in textbook)
β’ (c) React with stoichiometric amount of air and the product contains 90% CO2 and 10% CO (T=2350K in textbook)
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18PR 18PR 18SRK
3283C 3469K 3557K 3557
18PR
2364K
18PR
2211K
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HHV vs LHV?
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Understanding of Entropy, S
β’ Difficult to understand the concept.
β Think about Enthalpy
β’ H is hard to understand, because it is close to the mathematical notion, not natural discovery such as T or P.
β Same for Entropy!
β’ It was born as a notion, and cannot be explained clearly until 19th century.
β’ It can be defined by two kinds of view:β The classical thermodynamics definition
β The statistical mechanics definition.
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Statistical understanding of S
β’ S is a measure of the number of possible configurations(π΄) of the individual molecules.
β If you have a system with 4 molecule in a room,
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The number of possible configurations
4C0=1
4C1=4Entropy is high
Entropy is low
When you observe the system,the probability you can see the state=1/16=6.25%
the probability you can see this state=6/16=37.5%
πΊ = ππ© π₯π§π΄
4C2=6
4C3=4
4C4=1
Where ππ© is theBoltzmann constant
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Statistical understanding of S
β’ When there are 100 molecules,
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The probability you can observe this state:7.9*10-29%
All the system moves to maximize its entropy
Low entropy(Possible but very rare)
High entropy
ΞS>0
Actually there are 6.022 x 1023
gas molecules in 0.224m3
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Second Law
β’ Entropy can be defined by two different points of view, but it is proved as same.
β ππ =πππππ£
π
β π = ππ΅ lnπΊ
β’ 2nd law says:
β ππ β‘πΏππππ£
πβ βπ β‘
πΏππππ£
π=
ππππ£
π
β’ βπ π’πππ£ = βπ π π¦π + βπ π π’ππ β₯ 0
β For a reversible process, βπ π’πππ£ = 0
β For an irreversible process, βπ π’πππ£ > 0
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(ππ π = ππππ π‘πππ‘)
If you want to know how it can be same, you need read more references:
Textbook (Chapter 3.3, 3.11) andArieh Ben-Naim, A Farewell to Entropy : StatisticalThermodynamics Based on Information, Chapter 5, World Scientific Publishing Company, 2008.(you can find an e-book from SNU library)
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Why entropy is important?
β’ Actually, all the real processes are irreversible, but we cannot know the exact path of the real process.
β’ We can βassumeβ an ideal reversible (dS=0) process for each equipment and this is the best case for the equipment.
β’ Then we can compare the real result with the ideal best case, then can say its βefficiencyβ
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Entropy for a reversible process
β’ Reversible, isothermal
β’ Ξπ π π¦π =ππππ£
π
β’ Ξπ π π’ππ = βππππ£
π
β’ Ξπ π’πππ£ = Ξπ π π¦π + Ξπ π π’ππ = 0
β’ Ξπ π π¦π β 0, π π π’ππ β 0
β’ Reversible and Adiabatic
β’ Ξπ π π¦π =ππππ£
π= 0
β’ Ξπ π π’ππ = βππππ£
π= 0
β’ Ξπ π’πππ£ = Ξπ π π¦π + Ξπ π π’ππ = 0
β’ Ξπ π π¦π = 0, π π π’ππ = 0
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Ξ΄qrev
Isentropic
=Isentropic
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How to get Ξs from T, P, v without qrev?
β’ Entropy is a state function
β’ For a real(irreversible) process from state 1 to 2, we can construct a hypothetic reversible process from 1 to 2 to get βπ
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βπ ππππ(πππ’π) = π 2 β π 1= βπ πππ£(πππ)
π (bar)
π£ π3\mol
1
2Irreversible
Reversible
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Entropy of irreversible Process
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SuddenIsothermalExpansion A
ππ΄,πππ‘π’ππ = βπ€π΄,πππ‘π’ππ= 1 πππ β 0.08 β 0.04 Ξ€π3 πππ= 4000 Ξ€π½ π ππ β ππππ
ππ΄,πππ£ = βπ€πππ£ = +ΰΆ±π£1
π£2
πππ£ = ΰΆ±π£1
π£2 π1π£1π£
ππ£
= π1π£1 lnπ£2π£1
= 5545 Ξ€π½ π ππ
π (bar)
π£ π3\mol
1
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2
0.04 0.08
π (bar)
π£ π3\mol
1
2
A1
2
0.04 0.08
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How to get Ξs for an irreversible process?
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P1
T1T2
P2
βπππππ= βππππ
βπππππππππππππ
A
B
State 1
State 2
Step A: reversible isothermal expansion
ππ’ = 0 = πΏππππ£ + πΏπ€πππ£
πΏππππ£ = βπΏπ€πππ£ = πππ£
βπ π΄ = ΰΆ±πΏππππ£π
= ΰΆ±πππ£
π
For Ideal gas, Pv=RT
βπ π΄ = ΰΆ±π£1
π£2 π
π£ππ£ = π ln
π£2π£1
= βπ lnπ2π1
Step B: reversible isobaric expansion
ππ’ = πΏππππ£ + πΏπ€πππ£ = πΏππππ£ β πππ£
πβ = π π’ + ππ£ = ππ’ + πππ£ + πππ
πΏππππ£ = πβ = ππππ
βπ π΅ = ΰΆ±πΏππππ£π
= ΰΆ±π1
π2 ππ
πππ
βπ = βππ¨ + βππ© = ΰΆ±π»π
π»π ππ
π»π π» β πΉ π₯π§
π·π
π·π
If cp is constant,
βπ = ππ lnπ2π1β π ln
π2π1