Atkins-chapter05-second law the machinery-1

5.1 Fundamental equations
Chapter 5 The second law: the machinery
The combined equations of 1st and 2nd law: fundamental equations of thermodynamics Some typical applications of fundamental equations
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5.1.1 Combined forms of 1st and 2nd laws From the contents of previous chapters, we have learned two basic concepts of thermodynamics:
1st law: dU=δQ+δW dU= 2nd law: TdS=δQre TdS= closed system closed system in a reversible process
For a closed system taking a reversible process and no non-expansion work: nonwork:
δW=δWpV+δW’=δWpV=–pexdV=–pdV dV= and 1st law can be expressed as dU=TdS–pdV dU= TdS–
Because U is a state function and independence on the path of state preparation (the internal energy change in an irreversible process is same as the changes of U in a reversible process of same initial and final states)
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Derivation of fundamental equations
For a closed system taking a reversible process and no non-expansion work: nonwork: Basic equation: dU=TdS–pdV dU= TdS– Combined definition of enthalpy: H=U+pV H=U+pV dH=dU+d(pV)=TdS–pdV+pdV+Vdp=TdS+Vdp dH= dU+d(pV)=TdS– pdV+pdV+Vdp= For Helmholtz energy: A=U–TS A=U– dA=dU–d(TS)=TdS–pdV–TdS–SdT=–SdT–pdV dA= dU– d(TS)=TdS– pdV– TdS SdT SdT For Gibbs energy: G=A+pV=H–TS G=A+pV=H– dG=dA+d(pV)=–SdT–pdV+pdV+Vdp=–SdT+Vdp dG= dA+d(pV)=–SdT pdV+pdV+Vdp= SdT+Vdp
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5.1.2 fundamental equations of thermodynamics and Maxwell relations
Fundamental equations of thermodynamics: Closed system dU=TdS–pdV dU= TdS– Volume work only dH=TdS+Vdp dH= p,V,T changes dA=–SdT–pdV dA= SdT Phase equilibrium dG=–SdT+Vdp dG= SdT+Vdp Chemical equilibrium Extension usage of fundamental functions:
Irreversible and reversible processes have same changes of state function between same initial and final states When a system happened a phase transformation or chemical reaction inside, the system can still be treated as a closed system because no matter exchanges between system and surrounding.
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5.2 Maxwell relations and deductions of fundamental equations
Math. theorem: if Z =Z(x, y) is continuous and differentiable
Deduction of fundamental equations
dU = TdS－ pdV dH = TdS + Vdp dA = －SdT － pdV dG = －SdT + Vdp
? ?Z ? ? ?Z ? dZ = ? ? dx + ? ? ?y ? dy ? ? ?x ? y ? ?x
dU = TdS－ pdV
Maxwell relations
? ? ? ?Z ? ? ? ? ? ?Z ? ? = ? ? ? ?y ? ?x ? ? ? ?x ? ?y ? ? ? y ?x ? ? ? x ?Y ? ? ? ?
dH = TdS + Vdp
? ?T ? ? ?V ? ? ? ?p ? = ? ? S ? ? ?p ? ?S ?
? ?U ? ? =T ? ? ?S ?V
? ?T ? ? ?p ? ?? ? = ? ? ? ?S ?V ? ?V ? S
dA = －SdT － pdV
? ?H ? ? ? =T ? ?S ? p ? ?A ? ? ? = ?S ? ?T ?V ? ?G ? ? ? = ?S ? ?T ? p
? ?U ? ? ? = ?p ? ?V ? S ? ?H ? ? ? ?p ? = V ? ? ?S
? ?A ? ? ? = ?p ? ?V ? T
dG = －SdT + Vdp
? ?S ? ? ?V ? ? ? = ?? ? ? ?p ? ? ?T ? p ? ?T
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? ?S ? ? ?p ? ? ? =? ? ? ?V ?T ? ?T ?V
? ?G ? ? ? ?p ? = V ? ? ?T
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5.3 Applications of fundamental equations
Internal energy changes with temperature and volume in a volume-work-only process volume- work∵ U = U (T ,V ) ? ?U ? ? ?U ? ∴ dU = ? ? dT + ? ? dV ? ?T ?V ? ?V ? T ? ?U ? because of ? ? = CV ? ?T ?V ? ?U ? ? ?U ? ∴ dU = CV dT + ? ? dV ( for i . g . : ? ?V ? = 0, dU = CV dT ) ? ?V ? T ? ?T Based on the fundamental equation : dU = TdS ? pdV ? ?U ? ? ?S ? hence ? ? = T ? ?V ? ? p ? ?V ?T ? ?T ? ?S ? ? ?p ? According to the Maxwell relation : ? ? =? ? ? ?V ?T ? ?T ?V
Internal energy changes with temperature and volume in a volume-work-only process volume- work? ?U ? ? ?S ? ? ?p ? ? ? =T? ? ? p =T? ? ?p ? ?V ?T ? ?V ?T ? ?T ?V Thus ? ? ?p ? ? dU = CV dT + ?T ? is true for i.g and r.g ? ? p ? dV ? ? ?T ?V ? nR ? ?p ? ? ?p ? For i.g : pV = nRT T? ? ? = ? =p V ? ?T ?V ? ?T ?V That is : dU = CV dT For liquid or solid substances : So dU ≈ CV dT dV ≈ 0
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The internal energy changes can be measured by observing variables of CV, T, p, and V
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Enthalpy changes with T & p in a volumevolumework-only process workIn a closed system of of constant composition : H = H (T , p ) ? ?H ? ∵? ? = Cp ? ?T ? p ? ?H ? ? ?H ? dH = ? ? dp ? dT + ? ? ?T ? p ? ?p ?T (definition of heat capacity at const. p)
Enthalpy changes of a W’=0 process
So Thus ? ?H ? ? ?S ? ? ?V ? ? ? = T ? ? + V = ?T ? ? +V ? ?T ? p ? ?p ?T ? ?p ?T ? ? ?V ? ? dH = C p dT + ? ?T ? ? + V ? dp ?T ? p ? ? ? ? ? pV = nRT is true for i.g and r.g
? ?H ? ∴ dH = C p dT + ? ? dp ? ?p ?T Based on the fundamental equation : dH = TdS + Vdp ? ?H ? ? ?S ? hence ? ? = T ? ? +V ? ?p ?T ? ?p ?T
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In a case of i.g : Thus
nR ? ?V ? and ? ? = p ? ?T ? p
? ?V ? T? ? = V and ? ?T ? p
dH = C p dT
? ?V ? For solid or liquid : V ≈ 0, ? ? ≈ 0 hence dH ≈ C p dT ? ?T ? p
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? ?S ? ? ?V ? According to the Maxwell relation : ? ? = ? ? ? ?p ?T ? ?T ? p ?
Entropy changes with T & V if W’=0
In process of ΔN = 0, W ' = 0 : S = S (T ,V ) ? ?S ? ? ?S ? dS = ? ? dT + ? ? dV ? ?T ?V ? ?V ?T ? ?S ? ? ?p ? ∵ Maxwel relation : ? ? =? ? ? ?V ?T ? ?T ?V CV ? ?S ? ? ?S ? ? ?U ? ? ? ? = ? ? =? T ? ?T ?V ? ?U ?V ? ?T ?V ∴ dS = CV ? ?p ? dT + ? ? dV T ? ?T ?V nR ? ?p ? For case of i.g : ? ? = V ? ?T ?V
The difference of heat capacity of real gases
In a process of ΔN = 0, W ' = 0 : S = S (T , p )
( dA = ? SdT ? pdV )
? ?S ? ? ?S ? ? ?S ? ? ?H ? ? ?V ? dS = ? ? dT + ? ? dp = ? ? ? ? dT ? ? ? dp ? ?T ? p ? ?H ? p ? ?T ? p ? ?T ? p ? ?p ?T ? ? ?S ? ? ?V ? ? ? dG = ? SdT + Vdp, ? ? = ? ? ? ? ? ?p ?T ? ?T ? p ? ? ? ? = ? 1? ? ?S ? ? dH = TdS + Vdp, ? ? = ? ? ? ?H ? p T ? ? ? CV ? ?p ? dT + ? From the result of last section : dS = ? dV T ? ?T ?V Cp T ? ?V ? dT ? ? ? dp ? ?T ? p
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? 1? ? ?S ? ? dU = TdS ? pdV , ? ? = ? ? ?U ?V T ? ?
so
dS = CV d (ln T ) + nRd (ln V ) dS ≈ CV d (ln T )
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For case of liquid or solid : dV ≈ 0,
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Cp,m－CV,m of real gases
CV ? ?V ? ? ?p ? Because dS = dT ? ? dT + ? ? dp and dS = ? dV T T ? ?T ? p ? ?T ?V Cp The difference of two equations : ? C p ? CV ? ? T ? ? ?V ? ? ? ? ?p ? dp + ? ? dT ? ? ? ? dV ? = 0 ? ? ?T ? ? ?p ? ?T ?V ? ? ? and
The difference of heat capacity of real gases
In a closed system of constant composition, ? ?V ? ? ?p ? C p ? CV = T ? ? ? ? ? ?T ? p ? ?T ?V i.g : pV = nRT , nR nR ? ?V ? ? ?p ? T? ? ? ? =T ?T ? p ? ?T ?V p V ? C p ? CV = nR liquid or solid :
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? ? ?V ? ? T ? ?p ? C p ? CV = ? ? dp + ? ? dV ? ? ? ?T ? ? dT ?T ?V ?p ? ? ? In a closed system of constant composition, there are only two independent variables (one of these three variables depands others ) ? ?V ? ? ?p ? ? If dV = 0, then C p ? CV = T ? ? ? ? ? ? ?T ? p ? ?T ?V ? ? are same expressions ? ?p ? ? ?V ? ? or dp = 0, then C p ? CV = T ? ? ? ? ? ? ?T ?V ? ?T ? p ?
? ?V ? ? ? ≈ 0, ? ?T ? p
C p ≈ CV
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Summarizing: the measurement of indirect thermodynamic properties
For a closed system of no additional work: The measurement of ? ? ?p ? ? dU = CV dT + ?T ? ? ? p ? dV indirect ? ? ?T ?V ? thermodynamic ? ? ?V ? ? properties can be dH = C p dT + ? ?T ? + V ? dp ? undertaken by ? ? ?T ? p ? ? ? observation of state CV ? ?p ? dS = dT + ? ? dV properties of the T ? ?T ?V system ?V ?p
Gibbs energy changes against T &p
Fundamental equation dG=－SdT+Vdp dG=
? ?G ? ? ? = ?S ? ?T ? p
? ?G ? ? ? =V ? ? p ?T ? ?G ? ? ?G ? dG = ? ? dp ? dT + ? ? ?T ? p ? ?p ? T
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S (T ) = S (T0 ) + ∫
T
C p dT T
T0
? ? ? ? C p ? CV = T ? ? ? ? ?T ? p ? ?T ?V ?
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Gibbs - Helmholtz equation
? ? ? ? H ?TS ? G ? ? (G / T ) ? T ?G? ? = ? ? T ?p =? 2 ? ? ? ? T ?p = ? ?T ? 2 T T ? ?p T2
The efficiency of energy transformation
Nowadays, 90% of energy comes from chemical fuels, such as petroleum, coal, natural gas… gas… Two typical manner for transforming chemical energy into mechanical one Thermo engine
Fuels burn in a engine and transform chemical energy into heat, and then into mechanical work Exothermal chemical reaction happens in a thermo engine, its chemical energy release as heat ΔH=Qp, η<1-T1/T2 H=Q <1In case of automobiles, the efficiency of energy transformation is less than 40% and <15% of that be transferred into wheels We need a more effective way to transforming fuels and also save natural resources Transforming heat into work in turbine is more effective but hardly to reach 60%
?G
?T
dG = ? SdT + Vdp
H ? ? (G / T ) ? ? ?T ? = ? T2 ? ?p
U ?? ( A / T )? ? ?T ? = ?T2 ? ?V
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G = H ? TS
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Transform chemical energy in electrolytic ways
Electrolytic cell is a more effective way to use fuels
A chemical reaction happens in a manner of doing additional work ΔrG≤W’ Comparing thermo transformation, energy efficiency of cells is much higher than the thermo way η=–W/Qp=ΔrG/ΔrH=1–TΔrS/ΔrH W/Q G/Δ =1– S/Δ Since most of burning process are exothermic: ΔrH<0; if ΔrS>0, the maximum efficiency will be higher than 100% The practical fuel cells have a higher energy efficiency of 70% Higher effective, less resources consumption Fossil fuel is unrenewable, use as chemical material is unrenewable, more reasonable than just use as fuel
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Homework
P.120:
P 4.32
P.132:
P 5.2 P 5.11
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