Quiz
Thermodynamics Equations
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1S= [blank_start]k[blank_end] ln W
ΔS = [blank_start]qrev[blank_end]/T
dS= d qrev/[blank_start]T[blank_end]
For all liquids, [blank_start]ΔvapH[blank_end]/T[blank_start]b[blank_end] = [blank_start]+85[blank_end] [blank_start]J K-1 mol-1[blank_end]
[blank_start]ΔS[blank_end]= [blank_start]nR[blank_end] [blank_start]ln[blank_end] [blank_start]Vf[blank_end]/[blank_start]Vi[blank_end] = nR ln [blank_start]Pi/Pf[blank_end]
[blank_start]ΔSm[blank_end] = [blank_start]Cp,m[blank_end] ln [blank_start]Tf[blank_end]/[blank_start]Ti[blank_end]- Ak
- Bqrev
- CT
- DΔvapH
- Eb
- F+85
- GJ K-1 mol-1
- HnR
- Iln
- JVf
- KVi
- LPi/Pf
- MCp,m
- NTf
- OTi
- PΔSm
- QΔS
2For a perfect gas, [blank_start]Cp[blank_end] = [blank_start]Cv[blank_end] + [blank_start]n[blank_end]R
If monatomic: Cv,m = [blank_start]3/2[blank_end] R Cp,m=[blank_start]5/2[blank_end] R
If polyatomic: Cv,m [blank_start]>[blank_end] 3/2 R Cp,m>5/2 R
At temps close to 0, [blank_start]Sm[blank_end] = [blank_start]1/3 Cv[blank_end],m- ACp
- BCv
- Cn
- D3/2
- E5/2
- F>
- G1/3 Cv
- HSm
3ΔSsurr = q[blank_start]surr[blank_end]/T
qsurr=[blank_start]-q[blank_end]system
At constant pressure, Δ[blank_start]S[blank_end]surr= [blank_start]-[blank_end]Δ[blank_start]H/T[blank_end]- Asurr
- B-q
- C-
- DH/T
- ES
4ΔrSθ = [blank_start]Σv[blank_end]Smθ ([blank_start]products[blank_end]) [blank_start]-[blank_end] ΣvSmθ ([blank_start]reactants[blank_end])
ΔStotal = ΔS[blank_start]system[blank_end] + [blank_start]ΔSsurroundings[blank_end]. The process is spontaneous if [blank_start]ΔStotal is positive[blank_end].
Stθ = [blank_start]Sθ298[blank_end] + C[blank_start]p[blank_end] [blank_start]ln T[blank_end]/[blank_start]298[blank_end]- AΣv
- Bproducts
- C-
- Dreactants
- Esystem
- FΔSsurroundings
- GΔStotal is positive
- HSθ298
- Ip
- J298
- Kln T
5Gibbs energy:
G= [blank_start]H-TS[blank_end]
ΔG=ΔH[blank_start]-TΔS[blank_end]. It is to do with the system, for a spontaneous reaction, ΔG must be [blank_start]under 0[blank_end].
ΔG = the maximum [blank_start]NON EXPANSION work[blank_end] that can be obtained from [blank_start]a process[blank_end] at a constant [blank_start]T[blank_end] and [blank_start]p[blank_end]. It doesn't change the [blank_start]volume[blank_end].- AH-TS
- B-TΔS
- Cunder 0
- DNON EXPANSION work
- Ea process
- FT
- Gp
- Hvolume
6Pressure changes:
For solids or liquids, ΔG = [blank_start]V[blank_end][blank_start]Δ[blank_end][blank_start]p[blank_end]
For a gas, [blank_start]ΔG[blank_end]= [blank_start]nRT[blank_end] ln [blank_start]Pi[blank_end]/[blank_start]Pf[blank_end]
ΔrGθ= Σv[blank_start]G[blank_end]mθ (products) - ΣvGmθ ([blank_start]reactants[blank_end])
ΔrG changes with temperature: ΔrG= [blank_start]ΔrH[blank_end] - [blank_start]TΔrS[blank_end], if we assume ΔrH and S change very little with temperature.- AV
- BΔ
- Cp
- DnRT
- EΔG
- FPi
- GG
- Hreactants
- ITΔrS
- JΔrH
- KPf
7K= ([blank_start]Pc[blank_end]/Pθ)^[blank_start]c[blank_end] / ([blank_start]Pa/Pθ[blank_end])^a (Pb/Pθ0^b
SO Δr[blank_start]G[blank_end] = [blank_start]-RT[blank_end] ln K
[blank_start]lnK[blank_end] = [blank_start]-[blank_end]Δr[blank_start]H[blank_end]θ/[blank_start]R[blank_end] [blank_start](1/T)[blank_end] + [blank_start]ΔrS[blank_end]θ[blank_start]/R[blank_end]
ΣθΔ- APc
- Bc
- CPa/Pθ
- D-RT
- EG
- FlnK
- G-
- HH
- IR
- J(1/T)
- KΔrS
- L/R
Informações
- Descrição
- Quiz sobre Thermodynamics Equations, criado por M S em 12-04-2016.
- Autor
- M S
- Idioma
- English
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- 0
- Etiquetas
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