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CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-1
Chapter 4
Thermodynamic Variablesand Relations
Assoc. Prof. Dr. Mazlan Abdul WahidFaculty of Mechanical EngineeringUniversiti Teknologi Malaysiawww.fkm.utm.my/~mazlan
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-2
Thermal Expansion Coefficient
Volumetric Thermal Expansion Coefficient
PV T
V
V
∂∂
= 1α
( )...,, XPTfV =α
PL T
L
L
∂∂
= 1α
LV αα 3= 321 LLL ααα ==Isotropic Material
Linear Thermal Expansion Coefficient
(SI Units = K-1)
(SI Units = K-1)
2
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-3
Coefficient of CompressibilityVolumetric Coefficient of Compressibility
TV P
V
V
∂∂
−= 1β
( )...,, XPTfV =β
Approximately:Note negative sign in definition.
E = Elastic ModulusEV /3====ββββ
jiViVj EE // ====ββββββββ
(SI Units = atm-1)
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-4
Heat Capacity
At Constant Pressure:
δQrev,P= CPdTP (SI Units = J/mole-K)
CP = f(T,P,X,…)
Empirical Fit: CP(T) = a ++++ bT ++++ c/T2
At Constant Volume:
δQrev,V = CVdTV (SI Units = J/mole-K)
CV = f(T,P,X,…)
In General: CP > CV & CP − CV = TVα2/β
3
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-5
Internal EnergydU = δQ + δW +δW/
δQrev = TdS
δWrev = −PdV
1st & 2nd Laws: dU = TdS - PdV +δW/
Coefficient relations:
TS
U
V
=
∂∂
PV
U
S
−=
∂∂
sV V
T
S
P
∂∂−=
∂∂
Maxwell relation:
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-6
EnthalpyDefining an energy & state function: H = U + PVDifferentiating: dH = dU + PdV + VdPSubstituting for dU: dH = TdS-PdV+dW/+PdV+VdP1st & 2nd Laws: dH = TdS + VdP +δW/
Good for isobaric processes: dP = 0
Coefficient relations:
TS
H
P
=
∂∂
VP
H
S
=
∂∂
PS S
V
P
T
∂∂=
∂∂Maxwell relation:
4
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-7
Helmholtz Free EnergyF (DeHoff) or A (Arbeiten)
Defining an energy & state function: F = U − TSDifferentiating: dF = dU − TdS − SdTSubstituting for dU: dF = TdS-PdV+dW/-TdS-SdT1st & 2nd Laws: dF = - SdT − PdV +δW/
Good for isothermal processes: dT = 0Coefficient relations:
ST
F
V
−=
∂∂
PV
F
T
−=
∂∂
VT T
P
V
S
∂∂=
∂∂Maxwell relation:
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-8
Gibbs Free EnergyG (DeHoff) or F (Others)
Defining an energy & state function: G = H - TSDifferentiating: dG = dU+PdV+VdP−−−−TdS-SdTSubstituting for dU: dG = TdS-PdV+dW/+PdV+VdP-TdS-SdT1st & 2nd Laws: dG = − SdT + VdP +δW/
Good for isothermal/isobaric processes: dT=0, dP=0
Coefficient relations:
ST
G
P
−=
∂∂ V
P
G
T
=
∂∂
PT T
V
P
S
∂∂−=
∂∂Maxwell relation:
5
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-9
State Functions (Table 4.4)• State Variables
– Temperature T– Pressure P– Volume V
• Energy Functions– Internal Energy U– Enthalpy H– Helmholtz Free Energy F– Gibbs Free Energy G
• Entropy S
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-10
Energy Functions
Internal Energy U
Enthalpy H = U + PV
Helmholtz Free Energy F = U − TS
Gibbs Free Energy G = H − TS
6
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-11
Combined 1st & 2nd Laws
dU = TdS − PdV + δW/
dH = TdS + VdP + δW/
dF = − SdT − PdV + δW/
dG = − SdT + VdP + δW/
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-12
Coefficient Relations
ST
G
P
−=
∂∂
VP
G
T
=
∂∂
ST
F
V
−=
∂∂
PV
F
T
−=
∂∂
TS
H
P
=
∂∂
VP
H
S
=
∂∂
TS
U
V
=
∂∂
PV
U
S
−=
∂∂
7
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-13
Maxwell Relations
sV V
T
S
P
∂∂−=
∂∂
PS S
V
P
T
∂∂=
∂∂
VT T
P
V
S
∂∂=
∂∂
PT T
V
P
S
∂∂−=
∂∂
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-14
State Functions=f(T,P)(Table 4.5)
VdPSdTdG ++++−−−−====
8
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-15
Volume Relationsto Temperature & Pressure
( )PTVV ,=
VdPVdTdV βα −=
dPP
VdT
T
VdV
TP
∂∂+
∂∂=
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-16
State Functions=f(T,P)(Table 4.5)
dPVdTVdV ββββαααα −−−−====
VdPSdTdG ++++−−−−====
9
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-17
Entropy Relationsto Temperature & Pressure
( )PTSS ,=
( ) VdPdTTCdS P α−=
dPP
SdT
T
SdS
TP
∂∂+
∂∂=
TdSdTCQ Prev ========δδδδPT T
V
P
S
∂∂−=
∂∂
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-18
Thermodynamics State Functions - Table 4.5
dPVdTVdV ββββαααα −−−−====
dPVdTT
CdS P αααα−−−−====
(((( )))) (((( ))))dPTPVdTPVCdU P ααααββββαααα −−−−++++−−−−====
dPTVdTCdH P )1( αααα−−−−++++====
( ) dPPVdTPVSdF βα −+−=VdPSdTdG ++++−−−−====
10
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-19
Relations Between State VariablesIdentify the variables. Z=Z(X,Y)Write the differential form. dZ=MdX+NdYConvert dX & dY in terms of dT & dP.
dZ=M[XTdT+XPdP]+N[YTdT+YPdP]where dX=XTdT+XPdP; dY=YTdT+YPdPCollect terms. dZ=[MXT+NYT]dT+[MX P+NYP]dPObtain: Z=Z(T,P) & dZ=ZTdT+ZPdPSet: MXT+NYT=ZT MXP+NYP=ZP
Solve for M & N, integrate dZ=MdX+NdY between end points.
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-20
Example: Find T=T(S,V)
Identify the variables. T = T(S,V)
Rearrange. S = S(T,V)
Write the differential form. dS = MdT+NdV
Convert dT & dV in terms of dT & dP by substituting
dT = dT & dV = VαdT-VβdP
dS = MdT+N[VαdT-VβdP]
Collect terms.
dS = [M+NVα]dT-NVβdP
11
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-21
Example: Find T=T(S,V)dS = [M+NVα]dT-NVβdP
Obtain: S=S(T,P) & dS=(CP/T)dT-VαdPSet: [M+NVα]=(CP/T) & -NVβ= -VαSolve for M & N:
M = 1/T(CP-TVα2/β) & N = α/βInsert M & N in differential form: dS=MdT+NdV
dS = 1/T(CP-TVα2/β)dT+α/βdVNote the relation:CP − CV = TVα2/β
dS = (CV /T)dT+(α/β)dV
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-22
Example: Find T=T(S,V)dS= (CV /T)dT+(α/β)dV
Solve for dT: dT= (T/CV)dS-(αT/βCV)dV
For isentropic process: dTS= −(αT/βCV)dVS
Integrating:
( )121
2ln VVCT
T
V
−−=
βα
( )
−−= 1212 exp VV
CTT
Vβα
12
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
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UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-23
Ideal Gas
PV = nRT
α = 1/T β = 1/P
Monatomic: CP = 5/2 R CV = 3/2 R
Diatomic: CP = 7/2 R CV = 5/2 R
U & H depend only on temperature:
∆U = CV dT ∆H = CP dT
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-24
Ideal Gas
====
∂∂∂∂∂∂∂∂
====P
nRnRT
PTV
V PV
1αααα
TV
1====∴∴∴∴ αααα
−
−=
∂∂
−= 2
1
P
nRT
nRT
P
P
V
V TVβ
(((( )))) -1PnRT V ====
TP
nRV
====
====
∂∂∂∂∂∂∂∂
PnR
TV
P
(((( )))) 2−−−−−−−−====
∂∂∂∂∂∂∂∂
PnRTPV
T
PV
1====∴∴∴∴ ββββ
13
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-25
EXAMPLECompare the ∆S for the following processes.(a.) One gr-at of Ni is heated at 1 atm from 300 K to1300K.
Need S = S(T,P) evaluated for P = constant.
( ) VdPdTTCdS P α−=For constant P, dP=0.
(((( ))))dTTCdS P====
For this state function, integrate between limits.
(((( ))))∫∫∫∫====∆∆∆∆ 2
1
T
T P dTTCS
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-26
Compare the ∆S for the following processes.(a.) One gr-at of Ni is heated at 1 atm from 300 K to1300K.
From the Appendix.
KmolJbTaCNi
P −−−−++++====
Where a = 17.0 & b = 0.0295.
(((( ))))∫∫∫∫====∆∆∆∆ 2
1
T
T P dTTCS
[[[[ ]]]]∫∫∫∫ ++++====
++++====∆∆∆∆ 2
1
2
1ln
T
T
TTbTTadTb
Ta
S
14
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-27
Compare the ∆S for the following processes.(a.) One gr-at of Ni is heated at 1 atm from 300 K to1300K.
Substituting values for Ni and limits.
Where a = 17.0 & b = 0.0295.
[[[[ ]]]] 2
1ln T
TbTTaS ++++====∆∆∆∆
(((( ))))
−−−−++++====∆∆∆∆ 30013000295.03001300
ln17S
KatgrJ
S−−−−
====∆∆∆∆ 4.54
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-28
Compare the ∆S for the following processes.(b.) One gr-at of Ni is heated at 300 K is isothermallycompressed from 1 atm to 100 kbars.
Need S = S(T,P) evaluated for T = constant.
( ) VdPdTTCdS P α−=For constant T, dT=0.
VdPdS αααα−−−−====Assume α and V are independent of pressure.
(((( ))))21
2
1
PPVdPVSP
P−−−−====−−−−====∆∆∆∆ ∫∫∫∫ αααααααα
15
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA4-29
Compare the ∆S for the following processes.(b.) One gr-at of Ni at 300 K is isothermally compressed from 1atm to 100 kbars.
From Appendices:moleccV Ni
O /60.6====Substituting numerical values for α and V.
(((( ))))21 PPVS −−−−====∆∆∆∆ αααα
161040 −−−−−−−−==== KxNiVαααα
(((( ))))Kmol
atmccS
−−−−−−−−−−−−====−−−−⋅⋅⋅⋅⋅⋅⋅⋅====∆∆∆∆ −−−− 41.26101104060.6 56
KmoleJ
S−−−−
−−−−====⋅⋅⋅⋅−−−−====∆∆∆∆ 65.206.82
314.841.26
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-30
Compare the ∆S for the following processes.(c.) One mole of ZrO2 from 300 K to 1300 K at 1 atm.
Need S = S(T,P) evaluated for P = constant.
( ) VdPdTTCdS P α−=For constant P, dP=0.
(((( ))))dTTCdS P====For this state function, integrate between limits.
(((( ))))∫∫∫∫====∆∆∆∆ 2
1
T
T P dTTCS
16
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-31
Compare the∆S for the following processes.(c.) One mole of ZrO2 from 300 K to 1300 K at 1 atm.
From the Appendix.
KmolJcTbTaCZrO
P −−−−++++++++==== −−−−22
Where a = 69.6, b = 0.0075, c = -14.1x105.
(((( ))))∫∫∫∫====∆∆∆∆ 2
1
T
T P dTTCS
∫∫∫∫
−−−−++++====
++++++++====∆∆∆∆ 2
1
2
1
23 2ln
T
T
T
TTc
bTTadTTc
bTa
S
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-32
Compare the∆S for the following processes.(c.) One mole of ZrO2 from 300 K to 1300 K at 1 atm.
Substituting values for ZrO2 and limits.
Where a = 69.6, b = 0.0075, & c = -14.1x105.
(((( ))))[[[[ ]]]] 2
1
2
2lnT
TTcbTTaS −−−−++++++++====∆∆∆∆
(((( ))))
−−−−⋅⋅⋅⋅++++−−−−++++====∆∆∆∆ 225
3001
13001
101.1430013000075.0300
1300ln6.69S
KatgrJ
S−−−−
====∆∆∆∆ 177
17
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-33
Compare the∆S for the following processes.(d.) One mole of ZrO2 at 300 K is isothermallycompressed from 1 atm to 100 kbars.
Need S = S(T,P) evaluated for T = constant.
( ) VdPdTTCdS P α−=For constant T, dT=0.
VdPdS α−=Assume α and V are independent of pressure.
(((( ))))21
2
1
PPVdPVSP
P−−−−====−−−−====∆∆∆∆ ∫∫∫∫ αααααααα
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-34
Compare the∆S for the following processes.(d.) One mole of ZrO2 at 300 K is isothermallycompressed from 1 atm to 100 kbars.
From Appendices:
moleccV ZrOO /02.272 ====
Substituting numerical values for α and V.
(((( ))))21 PPVS −−−−====∆∆∆∆ αααα161072 −−−−−−−−==== KxZrO
Vαααα
(((( ))))Kmol
atmccS
−−−−−−−−====−−−−⋅⋅⋅⋅⋅⋅⋅⋅====∆∆∆∆ −−−− 9.1810110702.27 56
KmoleJ
S−−−−
−−−−====⋅⋅⋅⋅====∆∆∆∆ 92.106.82
314.89.18
18
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-35
Compare the∆S for the following processes.(e.) One mole of O2 from 300 K to 1300 K at 1 atm.
Need S = S(T,P) evaluated for P = constant.
( ) VdPdTTCdS P α−=For constant P, dP=0.
(((( ))))dTTCdS P====For this state function, integrate between limits.
(((( ))))∫∫∫∫====∆∆∆∆ 2
1
T
T P dTTCS
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-36
Compare the ∆S for the following processes.(e.) One mole of O2 from 300 K to 1300 K at 1 atm.
From the Appendix.
KmolJcTbTaCO
P −−−−++++++++==== −−−−22
Where a = 30.0, b = 0.0042, c = -1.7x105.
(((( ))))∫∫∫∫====∆∆∆∆ 2
1
T
T P dTTCS
∫∫∫∫
−−−−++++====
++++++++====∆∆∆∆ 2
1
2
1
23 2ln
T
T
T
TTc
bTTadTTc
bTa
S
19
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-37
Compare the∆S for the following processes.(e.) One mole of O2 from 300 K to 1300 K at 1 atm.
Substituting values for ZrO2 and limits.
Where a = 30.0, b = 0.0042, & c = -1.7x105.
(((( ))))[[[[ ]]]] 2
1
2
2lnT
TTcbTTaS −−−−++++++++====∆∆∆∆
(((( ))))
−−−−⋅⋅⋅⋅++++−−−−++++====∆∆∆∆ 225
3001
13001
107.130013000042.0300
1300ln0.30S
KatgrJ
S−−−−
====∆∆∆∆
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-38
Compare the∆S for the following processes.(f.) One mole of O2 at 300 K is isothermally compressedfrom 1 atm to 100 kbars.
Need S = S(T,P) evaluated for T = constant.
( ) VdPdTTCdS P α−=For constant T, dT=0.
dPp
RTT
dS1−−−−====
For 1 mole of an ideal gas.
PRT
V ====T1====αααα
VdPdS αααα−−−−====
20
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-39
Compare the∆S for the following processes.(f.) One mole of O2 at 300 K is isothermally compressedfrom 1 atm to 100 kbars.
∫∫∫∫∫∫∫∫ −−−−==== 2
1
2
1
P
P
S
SdP
PR
dS
====∆∆∆∆
2
1lnPP
RS
Substituting numerical values.
KmoleJ
S 72.95101
ln314.8 5 −−−−====
====∆∆∆∆
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-40
Compute the∆U when 12 liters of Ar at 273 K and 1 atm arecompressed to 6 liters with final pressure = 10 atm. (a.) FindU=U(P,V) & integrate.
),( VPUU ====
Compare to.
NdVMdPdU ++++====(((( ))))dPVdTVNMdPdU ββββαααα −−−−++++====(((( ))))dPNVMdTNVdU ββββαααα −−−−++++====
(((( )))) (((( ))))dPTPVdTPVCdU P ααααββββαααα −−−−++++−−−−====
αααααααα PVCNV P −−−−==== (((( ))))ααααββββββββ TPVNVM −−−−====−−−−
PVC
N P −−−−====αααα
ααααααααββββ
TVCM P −−−−====
EXAMPLE
21
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-41
Compute the∆U when 12 liters of Ar at 273 K and 1 atm arecompressed to 6 liters with final pressure = 10 atm. (a.) FindU=U(P,V) & integrate.
PVC
N P −−−−====αααα
Using:
T1====αααα
P1====ββββ
Using:
RCC VP ++++====
PdVR
CVdP
RC
dU VV ++++====
(((( ))))RTPV 1====
ααααααααββββ
TVCM P −−−−====
(((( ))))RCRV
M P −−−−==== (((( ))))RCRP
N P −−−−====
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-42
Compute the∆U when 12 liters of Ar at 273 K and 1 atm arecompressed to 6 liters with final pressure = 10 atm. (a.) FindU=U(P,V) & integrate.
∫∫∫∫∫∫∫∫ ++++====∆∆∆∆ 2
1
2
121
V
V
VP
P
V dVPR
CdPV
RC
U
Simulate 2 step process, constant pressure + constant volume.
∫∫∫∫∫∫∫∫ ++++====∆∆∆∆ 2
1
2
112
V
V
VP
P
V dVPR
CdPV
RC
U
P
V(P1V1)
(P2V2)
∫∫∫∫∫∫∫∫ ++++====∆∆∆∆ 22
11
22
11
VP
VP
VVP
VP
V PdVR
CVdP
RC
U
22
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-43
Compute the∆U when 12 liters of Ar at 273 K and 1 atm arecompressed to 6 liters with final pressure = 10 atm. (b.) Usethe temperature change.
2
22
1
11
nRTVP
nRTVP ====
111
222 T
VPVP
T ====
KT 365,1273121
6102 ====
⋅⋅⋅⋅⋅⋅⋅⋅====
TCU V ∆∆∆∆====∆∆∆∆
(((( )))) RRU 638,1273136523 ====−−−−====∆∆∆∆
CCCCCCCC
HHHHHHHH
AAAAAAAA
PPPPPPPP
TTTTTTTT
EEEEEEEE
R R R R R R R R
44444444
Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013
FKMFKM
UNIVERSITI TEKNOLOGI MALAYSIAUNIVERSITI TEKNOLOGI MALAYSIA09/19/20014-44
For one mole of nitrogen gas compute and plot thesurfaces that represent the variation with pressure andvolume over the range (1 atm, 22.4 l) to (10 atm, 8.2l)of (a) the internal energy.Use the result from 4.6b:
PdVR
CVdP
RC
dU VV ++++====
[[[[ ]]]])()(),(),(),( 111 RRV VVPPPV
RC
VPUVPUVPU −−−−++++−−−−====−−−−====∆∆∆∆
[[[[ ]]]]RRV VPPV
RC
VPU −−−−====∆∆∆∆ ),(
By analogy
[[[[ ]]]]RRP VPPV
RC
VPH −−−−====∆∆∆∆ ),(
EXAMPLE
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