1. Thermodynamics Basic Review of Byeong-Joo Lee Microstructure Evolution
POSTECH - MSE

2. Understanding and Utilizing Thermodynamic Laws
Objective
Understanding and Utilizing Thermodynamic Laws
State function
Thermodynamic Laws
Statistical thermodynamics
Gibbs energy
Extension of Thermodynamics
Multi-Phase System
Multi-Component System
Partial Molar Quantities
Utilization of Thermodynamics
Phase Diagrams
Defect Thermodynamics

3. Why define state function ? Why Thermodynamic Laws are important ?
1-1. Understanding and Utilizing Thermodynamic Laws
Why define state function ?
Why Thermodynamic Laws are important ?
Statistical vs. classical thermodynamics
Why use Gibbs energy ?

4. State function vs. Process variable First Law of Thermodynamics
Fundamentals
State function vs. Process variable
First Law of Thermodynamics
“The change of a body inside an adiabatic enclosure from a given initial state to a given
final state involves the same amount of work by whatever means the process is carried out”
It was necessary to define some function which depends only on the internal state of a system
– Internal Energy.
For adiabatic process: UB – UA = -w
Generally: UB – UA = q - w
dU = δq – δw : as a state function
Special processes
Constant-Volume Process: ΔU ＝ qv
Constant-Pressure Process: ΔH ＝ qp
Reversible Adiabatic Process: q ＝ 0
Reversible Isothermal Process: ΔU ＝ ΔH ＝ 0

5. Second Law of thermodynamics - Reversible vs. Irreversible
△S = measurable quantity + un-measurable quantity
= q/T △Sirr
= qrev/T

6. Second Law of thermodynamics - Maximum Work

7. Second Law of thermodynamics - Entropy as a Criterion of Equilibrium
※ for an isolated system of constant U and constant V,
(adiabatically contained system of constant volume)
equilibrium is attained when the entropy of the system is maximum.
※ for a closed system which does no work other than work of
volume expansion,
dU = T dS – P dV (valid for reversible process)
U is thus the natural choice of dependent variable for S and V
as the independent variables.
※ for a system of constant entropy and volume, equilibrium is attained
when the internal energy is minimized.

8. New Thermodynamic Functions – Reason for the necessity
※ Further development of Classical Thermodynamics results from the fact
that S and V are an inconvenient pair of independent variables.
+ need to include composition variables in any equation of state and
in any criterion of equilibrium
+ need to deal with non P-V work
(e.g., electric work performed by a galvanic cell)
dU = TdS – PdV
S, V are not easy to control. Need to find new state functions
which are easy to control and can be used to estimate equilibrium
→ F, G

9. Helmholtz Free Energy - Work Function, F ≡ U – ST
dF ≡ dU – TdS – SdT
For a reversible process
dF = [TdS – PdV – δw’] – TdS – SdT = – SdT – PdV – δw’
dFT = – PdV – δw’ = – δwT.Total
▷ Maximum work that the system can do by changing its state at Constant T, V = -ΔF.
For a irreversible isothermal process
△FT = [q – w] – T△S
T△S = q + T△Sirr
w = P△V + w’ = w’ For constant V
△FT,V + w’ + T△Sirr = 0
▷ If cannot do a maximum work, it’s due to the creation of Δsirr.
Under a Constant T, V, an equilibrium is obtained when the system has maximum (w’+ Δsirr) or minimum F.

10. Helmholtz Free Energy - Example of Application
Equilibrium between condensed phase and gas phase.
Use Helmholtz Free Energy Criterion to determine equilibrium amount of gaseous phase at a given temperature, and how it changes with changing temperature

11. Gibbs Free Energy - Gibbs Function, G ≡ U + PV – ST
dG ≡ dU + PdV + VdP – TdS – SdT
For a reversible process
dG = [TdS – PdV – δw’] + PdV + VdP – TdS – SdT = – SdT + VdP – δw’
dGT,P = – δw’
▷ Maximum work that the system can do by changing its state at Constant T, P = -ΔG.
For a irreversible isothermal process
△GT,P = [q – w] + P△V – T△S
T△S = q + T△Sirr
w = P△V + w’
△GT,P + w’ + T△Sirr = 0
▷ If cannot do a maximum work, it’s due to the creation of Δsirr.
Under a Constant T, P, an equilibrium is obtained when the system has maximum (w’+ Δsirr) or minimum G.

12. Thermodynamic Relations - For a closed system
dU = TdS – PdV
dH = TdS + VdP
dF = – SdT – PdV
dG = – SdT + VdP

13. Thermodynamic Relations - For a multicomponent system▷
▷ Chemical Potential
is the chemical work done by the system

14. Gibbs Energy
▷ Effect of Temperature on G
▷ Effect of Pressure on G

15. Gibbs Energy for a Unary System - from dG = –SdT + VdP
Gibbs Energy as a function of T and P
@ constant P
@ constant T

16. Gibbs Energy for a Unary System - from G = H - ST
Gibbs Energy as a function of T and P
@ constant P
@ constant T

17. Gibbs Energy for a Unary System - Temperature Dependency
Are both of above expressions the same?
Use Empirical Representation of Heat Capacities
to verify that the above expressions are exactly the same.

18. Gibbs Energy for a Unary System - Effect of Pressure
Molar volume of Fe = 7.1 cm3
Expansivity = 0.3 × 10-4 K-1
∆H(1→100atm,298) = 17 cal
※ The same enthalpy increase is obtained by heating from 298 to 301 K at 1atm
Molar volume of Al = 10 cm3
Expansivity = 0.69 × 10-4 K-1
∆H(1→100atm,298) = 23.7 cal
※ The same enthalpy increase is obtained by heating from 298 to 302 K at 1atm
∆S(1→100atm,298) = e.u. for Fe
e.u. for Al
※ The same entropy decrease is obtained by lowering the temperature from 298
by 0.27 and 0.09 K at 1 atm.
※ The molar enthalpies and entropies of condensed phases are
relatively insensitive to pressure change

19. Gibbs Energy for a Unary System
V(T,P) based on expansivity and compressibility
Cp(T)
S298: by integrating Cp/T from 0 to 298 K and using 3rd law of thermodynamics
(the entropy of any homogeneous substance in complete internal
equilibrium may be taken as zero at 0 K)
H298: from first principles calculations, but generally unknown
※ H298 becomes a reference value for GT
※ Introduction of Standard State

20. Statistical Thermodynamics
Basic Concept of Statistical Thermodynamics
Application of Statistical Thermodynamics to Ideal Gas
Understanding Entropy through the Concept of the
Statistical Thermodynamics
Heat capacity
Heat capacity at low temperature

21. Numerical Examples
The melting point of Pb is 600K at 1 atm. Show that solidification of supercooled
liquid Pb at 590K and 1 atm is a natural process, based on
(1) maximum-entropy criterion
(2) minimum-Gibbs-Energy criterion.
2. Estimate what will eventually happen (equilibrium state) if the Pb in problem #1
is contained in an adiabatic container.
3. A quantity of supercooled liquid Tin is adiabatically
contained at 495 K. Calculate the fraction of the Tin
which spontaneously freezes. Given
J at Tm = 505 K