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Transcript
```Basic Review of
Thermodynami
cs
Byeong-Joo Lee
POSTECH - MSE
[email protected]
Byeong-Joo Lee
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
Byeong-Joo Lee
1-1. Understanding and Utilizing Thermodynamic
Laws
 Why define state function ?
 Why Thermodynamic Laws are important ?
 Statistical vs. classical thermodynamics
 Why use Gibbs energy ?
Byeong-Joo Lee
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.
UB – UA = -w
 Generally:
UB – UA = q - w
dU = δq – δw
dU  0 : as a state function

 Special processes
1.Constant-Volume Process: ΔU ＝ qv
2.Constant-Pressure Process:
ΔH ＝ qp
q＝0
4.Reversible Isothermal Process:
ΔU ＝ ΔH ＝ 0
Byeong-Joo Lee
Second Law of thermodynamics - Reversible vs.
Irreversible
△S = measurable quantity + un-measurable quantity
=
q/T
+
△Sirr
=
qrev/T
Byeong-Joo Lee
Second Law of thermodynamics - Maximum Work
UB U A  q  w
q  dU system  w
dS system 
dS system 
q
T
 dS irr
dU system  w
T
 dS irr
w  TdS system  dU system  TdS irr
w  wmax  TS system  U system
Byeong-Joo Lee
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.
w  TdS system  dU system  TdS irr
PdV  TdS system  dU system  TdS irr
0  dU system  TdS irr
Byeong-Joo Lee
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
Byeong-Joo Lee
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.
Byeong-Joo Lee
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
Byeong-Joo Lee
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.
Byeong-Joo Lee
Thermodynamic Relations - For a closed system
dU = TdS – PdV
dH = TdS + VdP
dF = – SdT – PdV
dG = – SdT + VdP
Byeong-Joo Lee
Thermodynamic Relations - For a multicomponent
system
▷ G  G (T , P, ni , n j , n k , )
 G
 G 
 G 
 G 

dG  
dT  
dP  
dni  


 n
 T  P ,ni ,n j ,
 P  T ,ni ,n j ,
 ni  T , P ,n j ,nk ,
 j
▷ Chemical Potential


dn j  

 T , P ,ni ,nk ,
 G 


 i

n
 i T , P ,n j ,nk ,
n
dG   SdT  VdP    i dni
w  PdV   i dni
  i dni
1
is the chemical work done by the system
Byeong-Joo Lee
Gibbs Energy
▷ Effect of Temperature on G
▷ Effect of Pressure on G
Byeong-Joo Lee
Gibbs Energy for a Unary System - from dG = –SdT
+ VdP
Gibbs
Energy as a function of T and P
dG   S dT  V dP
T
P
To
Po
G(T , P)  G(To , Po )   S dT   V dP
@ constant P
C P (T )
dT
T
T
S (T )  S o  
To
T
G(T )  G(To )  
To
T C (T )


P
S

dT
 o To
 dT
T


P
@ constant T
G( P)  G( Po )   V dP
Po
Byeong-Joo Lee
Gibbs Energy for a Unary System - from G = H - ST
Gibbs Energy as a function of T and P
G  H  ST
G(T , P)  ( H (T , P)  H (To ,1))  H o  (S (T , P)  S (To ,1)  So )T
 H  H o  (S  So )T  H o  SoT  H  TS
@ constant P
T
S (T )  S o  
T
H (T )  H o   C P (T ) dT
To
To
T
T
To
To
G(T )  H o  S oT   C P (T ) dT  T 
C P (T )
dT
T
C P (T )
dT
T
@ constant T
H  
P
Po
P
 H 

 dP  P
o
 P T
P
P 
  S 


 V 
  V  dP  P V (1  T ) dP
T    V  dP  Po  T 
o

P

T
T
  T

 

P  S 
P  V 
P
S     dP   
dP


V dP


Po
P
P
o
o
 P T
 T T
Byeong-Joo Lee
Gibbs Energy for a Unary System - Temperature
Dependency
T
G(T )  G(To )  
To
T C (T )


P
S

dT
dT
 o To

T


T
T
To
To
G(T )  H o  S oT   C P (T ) dT  T 
C P (T )
dT
T
Are both of above expressions the same?
Use Empirical Representation of Heat Capacities
cP  a  bT  cT 2
to verify that the above expressions are exactly the same.
Byeong-Joo Lee
Gibbs Energy for a Unary System - Effect of
Pressure
Molar
volume of Fe
Expansivity
= 7.1 cm3
= 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
Expansivity
= 10 cm3
= 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) =
-0.00052 e.u. for Fe
-0.00167 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
Byeong-Joo Lee
Gibbs Energy for a Unary System
T
T
To
To
G(T )  H o  S oT   C P (T ) dT  T 
P
C P (T )
dT   V dP
Po
T
• 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
Byeong-Joo Lee
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
Byeong-Joo Lee
Numerical Examples
1. 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.
H melting  4810 J / mole
C p (l )  32.4  3.1 103 T J / mol  K
C p ( s )  9.75  103 T J / mol  K
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
H mSn  7070 J at Tm = 505 K
C p,Sn(l )  34.7  9.2 103T J / K
C p,Sn( s )  18.5  26 103T J / K
Byeong-Joo Lee