Chapter 4 Thermodynamic Variables and Relations

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Chapter 4Thermodynamic Variablesand Relations

• Notes on
• Thermodynamics in Materials Science
• by
• Robert T. DeHoff
• (McGraw-Hill, 1993).

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Thermal Expansion Coefficient

• Volumetric Thermal Expansion Coefficient

(SI Units K-1)
Linear Thermal Expansion Coefficient
(SI Units K-1)
Isotropic Material
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Thermal Expansion CoefficientIsotropic Material
If isotropic

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Coefficient of Compressibility

• Volumetric Coefficient of Compressibility

(SI Units atm-1)
Note negative sign in definition.
Approximately
E Elastic Modulus
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Heat Capacity

• At Constant Pressure
• dQrev,P CPdTP (SI Units J/mole-K)
• CP f(T,P,X,)
• Empirical Fit CP(T) a bT c/T2
• At Constant Volume
• dQrev,V CVdTV (SI Units J/mole-K)
• CV f(T,P,X,)
• In General CP gt CV CP - CV TVa2/b

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Internal Energy

• dU dQ dW dW/
• dQrev TdS
• dWrev -PdV
• 1st 2nd Laws dU TdS - PdV dW/
• Coefficient relations

Maxwell relation
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Enthalpy

• Defining an energy state function H U PV
• Differentiating dH dU PdV VdP
• Substituting for dU dH TdS-PdVdW/PdVVdP
• 1st 2nd Laws dH TdS VdP dW/
• Good for isobaric processes dP 0
• Coefficient relations

Maxwell relation
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Helmholtz Free EnergyF (DeHoff) or A (Arbeiten)

• Defining an energy state function F U - TS
• Differentiating dF dU - TdS - SdT
• Substituting for dU dF TdS-PdVdW/-TdS-SdT
• 1st 2nd Laws dF - SdT - PdV dW/
• Good for isothermal processes dT 0
• Coefficient relations

Maxwell relation
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Gibbs Free Energy G (DeHoff) or F (Others)

• Defining an energy state function G H -
  TS
• Differentiating dG dUPdVVdP-TdS-SdT
• Substituting for dU dG TdS-PdVdW/PdVVdP-Td
  S-SdT
• 1st 2nd Laws dG - SdT VdP dW/
• Good for isothermal/isobaric processes dT0,
  dP0
• Coefficient relations

Maxwell relation
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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

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Combined 1st 2nd Laws

• dU TdS - PdV dW/
• dH TdS VdP dW/
• dF - SdT - PdV dW/
• dG - SdT VdP dW/

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Coefficient Relations
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Maxwell Relations
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State Functionsf(T,P)(Table 4.5)
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Volume Relationsto Temperature Pressure
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State Functionsf(T,P)(Table 4.5)
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Entropy Relationsto Temperature Pressure
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State Functionsf(T,P)(Table 4.5)
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Relations Between State Variables

• Identify the variables. ZZ(X,Y)
• Write the differential form. dZMdXNdY
• Convert dX dY in terms of dT dP.
• dZMXTdTXPdPNYTdTYPdP
• where dXXTdTXPdP dYYTdTYPdP
• Collect terms. dZMXTNYTdTMXPNYPdP
• Obtain ZZ(T,P) dZZTdTZPdP
• Set MXTNYTZT
  MXPNYPZP

Solve for M N, integrate dZMdXNdY between
end points.
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Example Find TT(S,V)
Identify the variables. T T(S,V)
Rearrange. S S(T,V) Write the
differential form. dS MdTNdV Convert dT
dV in terms of dT dP by substituting dT dT
dV VadT-VbdP dS MdTNVadT-VbdP Collect
terms. dS MNVadT-NVbdP
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Example Find TT(S,V)
dS MNVadT-NVbdP Obtain SS(T,P)
dS(CP/T)dT-VadP Set MNVa(CP/T)
-NVb -Va Solve for M N M
1/T(CP-TVa2/b) N a/b Insert M N in
differential form dSMdTNdV dS
1/T(CP-TVa2/b)dTa/bdV Note the relation CP - CV
TVa2/b dS (CV /T)dT(a/b)dV
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Example Find TT(S,V)
dS (CV /T)dT(a/b)dV Solve for dT dT
(T/CV)dS-(aT/bCV)dV For isentropic process
dTS -(aT/bCV)dVS Integrating
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Ideal Gas

• PV nRT
• a 1/T b 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
• DU CV dT DH CP dT

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Ideal Gas
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4.4 Compare the DS for the following
processes.(a.) One gr-at of Ni is heated at 1
atm from 300 K to 1300K.

• Need S S(T,P) evaluated for P constant.

For constant P, dP0.
For this state function, integrate between limits.
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4.4 Compare the DS for the following
processes.(a.) One gr-at of Ni is heated at 1
atm from 300 K to 1300K.

• From the Appendix.

Where a 17.0 b 0.0295.
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4.4 Compare the DS for the following
processes.(a.) One gr-at of Ni is heated at 1
atm from 300 K to 1300K.

• Substituting values for Ni and limits.

Where a 17.0 b 0.0295.
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4.4 Compare the DS for the following
processes.(b.) One gr-at of Ni is heated at 300
K is isothermally compressed from 1 atm to 100
kbars.

• Need S S(T,P) evaluated for T constant.

For constant T, dT0.
Assume a and V are independent of pressure.
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4.4 Compare the DS for the following
processes.(b.) One gr-at of Ni at 300 K is
isothermally compressed from 1 atm to 100 kbars.

• From Appendices

Substituting numerical values for a and V.
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4.4 Compare the DS 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.

For constant P, dP0.
For this state function, integrate between limits.
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4.4 Compare the DS for the following
processes.(c.) One mole of ZrO2 from 300 K to
1300 K at 1 atm.

• From the Appendix.

Where a 69.6, b 0.0075, c -14.1x105.
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4.4 Compare the DS 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.
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4.4 Compare the DS for the following
processes.(d.) One mole of ZrO2 at 300 K is
isothermally compressed from 1 atm to 100 kbars.

• Need S S(T,P) evaluated for T constant.

For constant T, dT0.
Assume a and V are independent of pressure.
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4.4 Compare the DS for the following
processes.(d.) One mole of ZrO2 at 300 K is
isothermally compressed from 1 atm to 100 kbars.

• From Appendices

Substituting numerical values for a and V.
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4.4 Compare the DS 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.

For constant P, dP0.
For this state function, integrate between limits.
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4.4 Compare the DS for the following
processes.(e.) One mole of O2 from 300 K to 1300
K at 1 atm.

• From the Appendix.

Where a 30.0, b 0.0042, c -1.7x105.
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4.4 Compare the DS 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.
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4.4 Compare the DS for the following
processes.(f.) One mole of O2 at 300 K is
isothermally compressed from 1 atm to 100 kbars.

• Need S S(T,P) evaluated for T constant.

For constant T, dT0.
For 1 mole of an ideal gas.
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4.4 Compare the DS for the following
processes.(f.) One mole of O2 at 300 K is
isothermally compressed from 1 atm to 100 kbars.
Substituting numerical values.
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4.6 Compute the DU when 12 liters of Ar at 273 K
and 1 atm are compressed to 6 liters with final
pressure 10 atm. (a.) Find UU(P,V)
integrate.
Compare to.
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4.6 Compute the DU when 12 liters of Ar at 273 K
and 1 atm are compressed to 6 liters with final
pressure 10 atm. (a.) Find UU(P,V)
integrate.
Using
Using
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4.6 Compute the DU when 12 liters of Ar at 273 K
and 1 atm are compressed to 6 liters with final
pressure 10 atm. (a.) Find UU(P,V)
integrate.
Simulate 2 step process, constant pressure
constant volume.
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4.6 Compute the DU when 12 liters of Ar at 273 K
and 1 atm are compressed to 6 liters with final
pressure 10 atm. (b.) Use the temperature
change.
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4.8 For one mole of nitrogen gas compute and plot
the surfaces that represent the variation with
pressure and volume over the range (1 atm, 22.4
l) to (10 atm, 8.2 l)of (a) the internal energy.
Use the result from 4.6b
By analogy