FFKKMM Chapter 4 Thermodynamic Variables and …mazlan/?download=Adv Thermo Chapter 4.pdf1 CCCC HHHH AAAA PPPP TTTT EEEE R RR R 4444 Advanced Thermodynamics Advanced Thermodynamics

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Advanced Thermodynamics Advanced Thermodynamics -- Mazlan 2013Mazlan 2013

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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

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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

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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)

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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

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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

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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


TS

H

P

=

∂∂

VP

H

S

=

∂∂

PS S

V

P

T

∂∂=

∂∂Maxwell relation:

4

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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

∂∂=


CCCCCCCC

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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


ST

G

P

−=

∂∂ V

P

G

T

=

∂∂

PT T

V

P

S

∂∂−=


5

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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

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Energy Functions

Internal Energy U

Enthalpy H = U + PV

Helmholtz Free Energy F = U − TS

Gibbs Free Energy G = H − TS

6

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Combined 1st & 2nd Laws

dU = TdS − PdV + δW/

dH = TdS + VdP + δW/

dF = − SdT − PdV + δW/

dG = − SdT + VdP + δW/

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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

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Maxwell Relations

sV V

T

S

P

∂∂−=

∂∂

PS S

V

P

T

∂∂=

∂∂

VT T

P

V

S

∂∂=

∂∂

PT T

V

P

S

∂∂−=

∂∂

CCCCCCCC

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State Functions=f(T,P)(Table 4.5)

VdPSdTdG ++++−−−−====

8

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Volume Relationsto Temperature & Pressure

( )PTVV ,=

VdPVdTdV βα −=

dPP

VdT

T

VdV

TP

∂∂+

∂∂=

CCCCCCCC

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State Functions=f(T,P)(Table 4.5)

dPVdTVdV ββββαααα −−−−====

VdPSdTdG ++++−−−−====

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Entropy Relationsto Temperature & Pressure

( )PTSS ,=

( ) VdPdTTCdS P α−=

dPP

SdT

T

SdS

TP

∂∂+

∂∂=

TdSdTCQ Prev ========δδδδPT T

V

P

S

∂∂−=

∂∂

CCCCCCCC

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Thermodynamics State Functions - Table 4.5

dPVdTVdV ββββαααα −−−−====

dPVdTT

CdS P αααα−−−−====

(((( )))) (((( ))))dPTPVdTPVCdU P ααααββββαααα −−−−++++−−−−====

dPTVdTCdH P )1( αααα−−−−++++====

( ) dPPVdTPVSdF βα −+−=VdPSdTdG ++++−−−−====

10

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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.

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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.



(((( ))))dTTCdS P====For this state function, integrate between limits.

(((( ))))∫∫∫∫====∆∆∆∆ 2

1

T

T P dTTCS

16

CCCCCCCC

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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

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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

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Compare the∆S for the following processes.(d.) One mole of ZrO2 at 300 K is isothermallycompressed from 1 atm to 100 kbars.



VdPdS α−=Assume α and V are independent of pressure.

(((( ))))21

2

1

PPVdPVSP

P−−−−====−−−−====∆∆∆∆ ∫∫∫∫ αααααααα

CCCCCCCC

HHHHHHHH

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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

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Compare the∆S for the following processes.(e.) One mole of O2 from 300 K to 1300 K at 1 atm.



(((( ))))dTTCdS P====For this state function, integrate between limits.

(((( ))))∫∫∫∫====∆∆∆∆ 2

1

T

T P dTTCS

CCCCCCCC

HHHHHHHH

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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

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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

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Compare the∆S for the following processes.(f.) One mole of O2 at 300 K is isothermally compressedfrom 1 atm to 100 kbars.



dPp

RTT

dS1−−−−====

For 1 mole of an ideal gas.

PRT

V ====T1====αααα

VdPdS αααα−−−−====

20

CCCCCCCC

HHHHHHHH

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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

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TTTTTTTT

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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

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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

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∫∫∫∫∫∫∫∫ ++++====∆∆∆∆ 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


FKMFKM


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


FKMFKM


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

Documents

FFKKMM Chapter 4 Thermodynamic Variables and …mazlan/?download=Adv Thermo Chapter 4.pdf1 CCCC HHHH AAAA PPPP TTTT EEEE R RR R 4444 Advanced Thermodynamics Advanced Thermodynamics