Fundamental Equations of Thermodynamics
한양대학교
4.1 Fundamental Equation for the Internal Energy
The entropy is a criterion of weather a change in an isolated system is spontaneous.
What would be the criterion at constant T and V or constant T and P?
The first law of thermodynamics dU =dq+dw
The second law of thermodynamics d
d q
S ≥ T
For a closed system with only reversible PV work dw = −P Vd and dq =T Sd dU =T Sd −P Vd
In 1876, Gibbs added the chemical potential µiterms.
1 1 2 2
1
d d d d d d d d
Ns
i i i
U T S P V µ n µ n T S P V µ n
=
= − + + += − +
∑
The chemical potential µi is a measure of the potential a species has to move from one phase to another or undergo a chemical reaction.
The total differential of U
한양대학교
1
d d d d
Ns
i i i
U T S P V µ n
=
= − +
∑
Conjugate variables: T and S, P and V, µi and ni Natural variables of U: S, V, {ni}U(S, V, {ni})
,{ } ,{ } 1 , ,
d d d d
s
i i j i
N
i
V n S n i i S V n n
U U U
U S V n
S V = n ≠
∂ ∂ ∂
= ∂ + ∂ +
∑
∂ The equation of state
,{ } ,{ } , ,
i i j i
i
V n S n i S V n n
U U U
T P
S V µ n
≠
∂ ∂ ∂
= ∂ = −∂ = ∂
The Maxwell relations
,{ } ,{ } , , ,{ }
, , ,{ } , , , ,
i i j i i
j i i i j j i
i
S n V n i S V n n V n
i i j
i S V n n S n j S V n n i S V n n
T P T
V S n S
P
n V n n
µ
µ µ µ
≠
≠ ≠ ≠
∂
∂ ∂ ∂
= − =
∂ ∂ ∂ ∂
∂
∂ ∂ ∂
− = =
∂ ∂ ∂ ∂
Spontaneity Criterion with the Internal Energy
ext
1
d d and d d d
Ns
i i i
q T S w P V µ n
=
≤ = − +
∑
ext
1
d d d d d d
Ns
i i i
U q w T S P V µ n
=
= + ≤ − +
∑
At constant S, V, and {ni}, U must be at a minimum. (d ) , ,{ } 0
S V ni
U ≤
1
d d d d
Ns
i i i
U T S P V µ n
=
= − +
∑
At constant T, P, and µi
1
Ns
i i i
U TS PV µ n
=
= − +
∑
Euler’s theorem
k=1 ( ,1 2, , ) 1
j i
N
N i
i i x x
k f x x x x f
= x ≠
∂
=
∑
∂
4.2 Legendre Transformation
한양대학교
A linear change in variables to define a new function from a mathematical function by subtracting one or more products of conjugate variables.
1
1
Define
d d d d
d d d d d d d
s
s
N
i i i
N
i i i
U T S P V n
H U P V V P T S V P n H U PV
µ
µ
=
=
= − +
= + +
=
+ +
+
=
∑
∑
The equation of state
,{ } ,{ } , ,
i i j i
i
P n T n i P T n n
H H H
T V
S P µ n
≠
∂ ∂ ∂
= ∂ = ∂ =∂
At constant S, P, and {ni}, H must be at a minimum. (d ) , ,{ } 0
S P ni
H ≤
At constant T, P, and µi
1
Ns
i i i
H TS µ n
=
= +
∑
Legendre Transformation U → A
1
1
Define
d d d d
d d d d d d d
s
s
N
i i i
N
i i i
U T S P V n
A U T S S T S T P V n A U TS
µ
µ
=
=
= − +
= − −
=
− +
−
= −
∑
∑
The equation of state
,{ } ,{ } , ,
i i j i
i
V n T n i T V n n
A A A
S P
T V µ n
≠
∂ ∂ ∂
= −∂ = −∂ =∂
At constant T, V, and {ni}, A must be at a minimum. (d ) , ,{ } 0
T V ni
A ≤
At constant T, P, and µi
1
Ns
i i i
A PV µ n
=
= − +
∑
Legendre Transformation U → G
한양대학교
1
1
Define
d d d d
d d d d d d d d d
s
s
N
i i i
N
i i i
U T S P V n
G U P V V P T S S T G U PV TS H T
S T V P S
n µ
µ
=
=
= − +
= + + − − = −
=
+ +
+ − = −
∑
∑
The equation of state
,{ } ,{ } , ,
i i j i
i
P n T n i T P n n
G G G
S V
T P µ n
≠
∂ ∂ ∂
= −∂ = ∂ =∂
At constant T, P, and {ni}, A must be at a minimum. (d ) , ,{ } 0
T P ni
G ≤
At constant T, P, and µi
1
Ns
i i i
G µ n
=
=
∑
i Gi
µ =
Thermodynamic relations based on the chemical potential
G(T, P, {ni}) If a thermodynamic potential is known as a function of its natural variables, all of the thermodynamic functions can be calculated.
At constant T and P, µi =Gi
and
P T
T P
P
T
G G
S V
T P
G G
U G PV TS G P T
P T
H G TS G T G T A G PV G P G
P
∂ ∂
= − ∂ = ∂
∂ ∂
= − + = − ∂ − ∂
∂
= + = − ∂
∂
= − = − ∂
Gibbs Energy
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A criterion for spontaneous change at constant T and P without considering the surroundings
G = H – TS
The Gibbs Free Energy ∆G < 0
∆G = 0
∆G > 0
∆S > 0
∆S = 0
∆S < 0
Spontaneous Equilibrium Impossible
universe system surroundings
system surroundings
system universe system
universe system system system
S S S
S H
T
S S H
T
T S T S H G
∆ = ∆ + ∆
∆ = −∆
∆ = ∆ − ∆
∆ = ∆ − ∆ ≡ −∆
Works other than the PV work
Surface work
Elongation work
dw=γdAS
dw= f Ld
s
S 1
S 1
, ,{ },
s , ,{ },
d d d d d d
d d d d d d
s
s
i
i
N
i i i
N
i i i
T P n A
T P n L
U T S P V n f L A
G S T V P n f L A
f G
L G A
µ γ
µ γ
γ
=
=
= − + + +
= − + + + +
∂
= ∂
∂
= ∂
∑
∑
Maximum Work
한양대학교
d d d
d d d
d( ) d
(d ) d (d ) d
T
T
U q w
U T S w
U TS w
A w
A w
= +
− + ≥ −
− − ≥ −
− ≥ −
≤
The decrease in A is an upper bound on the total work done in the surroundings
ext nonpv
ext nonpv
nonpv
, nonpv
, nonpv
d d d d d d
d d d d
d( ) d
(d ) d
(d ) d
T P
T P
U q w q P V w
U P V T S w
U PV TS w
G w
G w
= + = − +
− − + ≥ −
− + − ≥ −
− ≥ −
≤
The decrease in G is an upper bound on the non-PV work done on the surroundings
4.3 Temperature Effect on the Gibbs Energy
,{ }
= < 0
P ni
G S
T
∂ −
∂
,{ }i P n
G H TS H T G T
∂
= − = + ∂
2
,{ } ,{ }
/ 1
i i
P n P n
G T G G
T T T T
∂ ∂
= − +
∂ ∂
2
,{ }
/
P ni
H T G T
T
∂
= − ∂
2
,{ }
( / )
P ni
H T G T
T
∂ ∆
∆ = − ∂
the Gibbs–Helmholtz equation
4.4 Pressure Effect on the Gibbs Energy
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,{ }
0
T ni
G V
P
∂ =
∂
>
2 2
1 1
2 1
2 1
d d
d
G P
G P
P P
G V P
G G V P
=
= +
∫ ∫
∫
If the volume is nearly independent of pressure, as it is for a liquid or solid, G2 =G1 +V P( 2 −P1) For an ideal gas, nRT
V = P
2
2 1
1
d d ln
ln
ln
G P
G P
G nRT P
G G nRT P P G G G nRT P
P
=
= +
∆ = − =
∫
∫
Dependence of the Gibbs energy of formation of an ideal gas on the pressure of the gas
Pressure Effect on the Gibbs Energy
Example 4.1 Derive the molar thermodynamic properties for an ideal gas.
ln P G G RT
= + P
P
T
T P
P
S G
T V G
P
G G
U G PV TS G P T
P T
H G TS G T G T A G PV G P G
P
∂
= −∂
∂
= ∂
∂ ∂
= − + = − ∂ − ∂
∂
= + = − ∂
∂
= − = − ∂
ln
ln
P
P G
S S nR S
P T V RT
P
U U G T S RT H RT H H G T S
A A RT P P
∂
= − = − ∂
=
= = + − = −
= = +
= +
Pressure Effect on the Gibbs Energy
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Example 4.2 An ideal gas at 27℃ expands isothermally and reversibly from 10 to 1 bar.
Calculate q per mole and w per mole and ∆U, , , , .∆H ∆G ∆A and ∆S
1 1 1
2 1
1 2
(8.314 J K mol )(300.15 K) ln10 5746 J mol
lnV ln P 1
w RT RT
V P
− − −
− = −
= − = − =
5746 J mol 1
A w − −
∆ = = 0
∆ =U
0 5746 5746 J mol 1
q = ∆ − =U w + = −
( ) 0 0 0
H U PV
∆ = ∆ + ∆ = + =
1 1 1
1 10
(8.314 J K mol )(300.15 K) ln 1 5746 J mol 10
d ln 1
G V P RT 10 − − = − −
∆ =
∫
= =1
1 1
rev 5746 J mol
19.14 J K mol 300.15 K
S q T
− = − −
∆ = =
1
1 1
0 (5746 J mol )
19.14 J K mol 300.15 K
H G
S T
− − −
+ =
∆ − ∆
∆ = =
Pressure Effect on the Gibbs Energy
Example 4.3 An ideal gas at 27℃ expands isothermally into an evacuated vessel from 10 to 1 bar. Calculate q per mole and w per mole and ∆U, , , , .∆H ∆G ∆A and ∆S
0 w =
1 1 1
2 1
24.63 L
(8.314 J K mol )(300.15 K) ln 5746 J mol 2.463 L
lnV A RT
V
− − = − −
∆ = − = −
0
∆ =U
0 0 0
q = ∆ − =U w + =
( ) 0 0 0
H U PV
∆ = ∆ + ∆ = + =
1 1 1
1 10
(8.314 J K mol )(300.15 K) ln 1 5746 J mol 10
d ln 1
G V P RT 10 − − = − −
∆ =
∫
= =1
1 1
0 (5746 J mol )
H G + − − −
∆ − ∆ =
∆ = =
한양대학교
Pressure Effect on the Gibbs Energy
Example 4.4∆fG◦(CH3OH, g) = –161.96 kJ mol–1, ∆fG◦(CH3OH, l) = –166.27 kJ mol–1and ρ(CH3OH, l) = 0.7914 g/cm3at 298.15 K (a) Calculate ∆fG(CH3OH, g) at 10 bar at 298.15 K assuming the methanol vapor as an ideal gas (b) Calculate ∆fG(CH3OH, l) at 10 bar at 298.15 K.
1 1 1 1
f f
161.96 kJ mol + (8.314 10 kJ K mol )(298.15 K) ln 10 = 156.25 kJ mol
(a) ln
G G RT P P
− − − −
− × −
∆ = ∆ +
=
1 3 3
6 3 1
32.04 g mol 1 m 100 cm 0.7914 g cm
40.49 10 m mol
(b)
V
−
−
− −
×
= ×
=
f f
1
6 3 1 5
3 1
1
166.27 kJ mol
(40.49 10 m mol )(9 10 Pa) +
10 J kJ 166.23 kJ mol
( )
G G V P P
−
− −
−
−
−
× ×
−
∆ = ∆ + −
=
=
The Gibbs energy of formation of gaseous and liquid methanol at 298.15K
4.5 Fugacity and Activity
ln P G G RT
= + P for an ideal gas, G. N. Lewis devised fugacityf(T,P) for real gases.
0
ln lim 1
P
f f
G G RT
P → P
= + =
id id
At constant temperature, dG =V Pd for a real gas and dG =V d P for an ideal gas.
The difference in Gibbs energy between a real gas and an ideal gas can be integrated from some low pressure, P* to the pressure of interest, P.
id id
*
id
*
id * *id id
* *
* *id id id
*
0 id
As
d( ) ( )d
( ) ( ) ( )d
0, , ( ) ( )d
ln 1 ( ) 1
exp ( )d
d or
P P
P P
P P
P
P P
P P
P
G G V V P
G G G G V V P
P G G G G
f
V V P
f V V P
P V V P
P RT
RT φ
− = −
− − − = −
→ → − = −
= −
= = −
∫
∫
∫
∫
∫
∫
The Fugacity Coefficient
한양대학교
Using the compressibility factor, PV Z = RT
id
0 0 0
1 1 1
exp ( )d exp ( )d exp d
P P P
f ZRT RT Z
V V P P P
P RT RT P P P
φ = =
∫
− = ∫
− = ∫
− With the virial equation, PV 1 ' ' 2
Z B P C P
= RT = + + +
( )
20 0
1 '
ln d ' ' d '
2
P P
f Z C P
P B C P P B P
P P
=
∫
− =∫
+ + = + +For a van der Waals gas, 1
1 a
Z b P
RT RT
= + − +
0 0
1 1
ln d d
P P
f Z a a P
P b P b
P P RT RT RT RT
−
=
∫
=∫
− = − exp a P
f P b
RT RT
= −
Fugacity
Example 4.7 The van der Waals coefficients of nitrogen are a = 1.408 L2bar mol–2and b = 0.03913 L mol–1. Estimate the fugacity of nitrogen at 50 bar and 298K.
1.408 50
(50 bar) exp 0.03913
(0.083145)(298) (0.083145)(298) 48.2 bar
exp a P
f P b
RT RT
−
= −
=
=
Activity
한양대학교
ln i
i i RT a
µ = µ + to express the chemical potential of a species in a mixture in the reference state for which
for an ideal gas, for a real gas
for solutions where is for a pure solid or liquid if
the activity coefficie the pres
nt
1
1
i i i
i i
i i
i i
i i i
i
i
a
P f
a a
P P
a m
a
µ µ
γ γ
= =
= =
=
=
sure is close enough to the standard state pressure
If the effect of pressure is not negligible, assuming the molar volume to be constant at all reasonable pressures, for a pure solid or liquid
( , ) ( ) ( )
ln ( )
( )
exp
T P T V P P RT a V P P
V P P
a RT
µ = µ + −
= −
−
=
Activity
Example 4.8 What is the activity of liquid water at 1, 10, and 100 bar at 298K, assuming that the molar volume is constant?
1
1 1
( )
exp
At 1 bar, 1
(0.018 kg mol )(10 bar 1 bar)
At 10 bar, exp 1.007
(0.083145 L bar K mol )(298 K) At 100 bar, 1.075
V P P
a RT
P a
P a
P a
−
− −
−
=
= =
−
= = =
= =
4.6 The Chemical Potential
한양대학교
, ,{ j i} , ,{ j i} , ,{ j i} , ,{ j i}
i
i S V n i S P n i T V n i T P n
U H A G
n n n n
µ
≠ ≠ ≠ ≠
∂ ∂ ∂ ∂
= ∂ =∂ =∂ =∂
the partial molar Gibbs e At constant and T P, µi =Gi nergy
a species diffuses spontaneously from the phase with higher chemical potential to the phase with lower chemical potential
µi(β) µi(α)
( )
d , ( )d ( )d d[
( ) ( )]
d < 0, ( ) > ( ) d = 0, ( ) = ( )
i i i i i i i
T P
i i
i i
G n n n
G G
µ α µ β µ β µ α
µ α µ β µ α µ β
= − + = −
at equilibrium
The Chemical Potential
1
d d d d
Ns
i i
i
G S T V P µ n
=
= − + +
∑
For a mixture of ideal gases
,{ } ,{ }
i i
i i
i i
P n T n
S V
T P
µ µ
∂ ∂
− = ∂ = ∂
,{ }i ,{ }i ,{ }i
i i i i
i i
i i
T n T n T n
P
V V RT x
P P P P P
µ µ µ
∂ ∂ ∂ ∂
= = = ∂ = ∂ ∂ = ∂
,{ }i i
i T n i
RT
P P
µ
∂ =
∂
d d
i i
i
P i
i P i
RT P
P
µ
µ µ =
∫
∫
µi = µi + RTln PPiln i
i i
S S R P
= − P
4.7 Partial Molar Properties
한양대학교
At constant T, P, and µi
1 1
=
s s
N N
i i i i
i i
G n µ n G
= =
=
∑ ∑
µi =GiAll extensive properties of a one-phase system are additive.
,{ }
1 1
,{ }
1 1
the partial molar entropy
the partial molar volume
s s
i
s s
i
N N
i i i i
P n
i i
N N
i i i i
T n
i i
i
i
S n S n
T
V n V n
P
S
V µ
µ
= =
= =
∂
= − ∂ =
∂
= ∂ =
∑ ∑
∑ ∑
G(T, P, {ni})
1
d d d d
Ns
i i i
G S T V P µ n
=
= − + +
∑
2
,{ }
/
P ni
H T G T
T
∂
= − ∂
the Gibbs–Helmholtz equation
2
,{ }
1 1
the partial molar enthalpy
( / )
s s
i
N N
i i i i
P n
i i
i
H T T n H n
T H
µ
= =
∂
= −
∑
∂ =∑
Partial Molar Properties
, ,{ } , ,{ } , ,{ }
,{ }
,{ }
, ,{ } , ,{ } , ,{ }
j i j i j i
j i
j i j i j i
j i
i T P n i T P n i T P n
i i i
i
P n
P n
i T P n i T P n i T P n
G H TS
G H S
n n T n
G H T S
S G
T
S G G
n n T T n
µ
≠ ≠ ≠
≠
≠ ≠ ≠ ≠
= −
∂ =∂ − ∂
∂ ∂ ∂
≡ = −
∂
− = ∂
∂ ∂ ∂ ∂ ∂
−∂ = ∂ ∂ = ∂ ∂ ,{ }
,{ }
P nj i
i i
i
P n
S G
T T
µ
≠
≠
∂ ∂
− = ∂ = ∂
Partial Molar Properties
한양대학교
Ideal mixture at low pressures, mixtures of real gases are assumed to behave as ideal gases.
The chemical potential of a species in a ideal gas mixture
ln i
i i i i
RT y P P y P µ = µ + P ≡
and
ln ln ln ln
i i t i
t
i i i i i t i i i i t
n n X X
n
P P
G n RT n y n RT n y RT y y R n G
P
G n
T P
µ µ
µ = =
= + + = + + =
=
∑
∑ ∑ ∑ ∑
∑
∑
( )
,{ }
,{ }
ln ln ln ln
i
i
i i
i i i i t i i i t
i i
i i i i t
P
t n
n
i
i i i
T
P P
n S R n y n R n y S R y y R n
S G
T H G
P P S
n T S n H n y H n H n RT n V
P TS
V G
P
µ
= − − = − − =
∂
= −∂
= +
∂
= + = = =
=
= ∂ =
∑ ∑ ∑ ∑ ∑
∑ ∑ ∑
∑ ∑
Every gas is as vacuum to every other gas.
Ideal Gas Mixture
At constant pressure P1= P2= P
( ) ( )
0 1 1 1 2 2 2
mix 1 1 1 1 1 2 2 2 2 2
mix mix 0 1 1 2 2 1 1 2 2
ln ln
ln ln ln ln
ln ln t ln ln
P P
G n n RT n n RT
P P
P P
G n n RT y n RT n n RT y n RT
P P
G G G RT n y n y n RT y y y y
µ µ
µ µ
= + + +
= + + + + +
∆ = − = + = +
( ) ( )
1 2
0 1 1 2 2
1 2
mix 1 1 1 1 2 2 2 2
mix mix 0 1 1 2 2 1ln 1 2 2
ln ln
ln ln ln ln
ln ln t ln
P P
S n S n R n S n R
P P
P P
S n S n R y n R n S n R y n R
P P
S S S R n y n y n R y y y y
= − + −
= − − + − −
∆ = − = − + =− +
mixH 0
∆ =
4.8 Gibbs-Duhem Equation
한양대학교
1
1
1
1 1
d d d d
' 0
d ' d d d d d d d 0
s
s
s
s s
N i i i
N
i i i
N i i i
N N
i i i i
i i
U PV TS n
U P V T S n
U U PV TS n
U U P V V P T S S T n n
µ
µ
µ
µ µ
=
=
=
= =
= − + +
= − + +
= + − − =
= + + − − − − =
∑
∑
∑
∑ ∑
the complete Legendre transformation
1
d d d 0
Ns
i i
i
V P S T n µ
=
− −
∑
= The Gibbs-Duhem equation mandates that intensive variables for a system are not independent.4.9 Maxwell Relations
( , ) d d d
( , ) d d d
( , ) d d d
( , ) d
S V
S P
T S
T P
U S V U T S P V
V S
T V
H S P H T S V P
P S
S P
A T V A S T P V
V T
G T P
∂ ∂
= − ∂ = −∂
∂ ∂
= + ∂ =∂
∂ ∂
= − − ∂ = ∂
d d
T P
S V
G S T V P
P T
∂ ∂
= − + −∂ =∂
For a mole of a substance
Internal Pressure of a Real Gas
한양대학교
2
2 2
a van der Waals gas,
T T V
T
U S P
T P T P
V V T
RT a P V b V
U R R RT a a
V V b P V b V b V V
∂ = ∂ − = ∂ −
∂ ∂ ∂
= −
−
∂ = − = − − =
∂ − − −
Example 4.11 What is the molar internal energy change of propane isothermally expanding from 10 to 30 L?
2 2
1 1 2
1 2
2 2 5 1 3 3 1 2 6 2
6 2 1
3 3 1 3 3 1
(8.779 L bar mol )(10 Pa bar )(10 m L ) = 0.8779 Pa m mol
1 1
0.8779 Pa m mol 58.5 J mol
10 10 m mol 30 10 m mol
1 1
d d
( )
U V
U V
a
U U a V a
V V V
U
− − − − −
− −
− − − −
=
− =
× ×
∆ = = = −
∆ =
∫ ∫
Isothermal Expansion of a van der Waals Gas
2 2
1 1
2
2 1
d 1 d ln
T V
S V
S V
RT a P V b V
S P R
V T V b
V b
S S R V R
V b V b
= −
−
∂ = ∂ =
∂ ∂ −
∆ = = = −
− −
∫ ∫
The molar entropy of isothermal expansion of a van der Waals gas
Cubic Expansion Coefficient and Isothermal Compressibility
한양대학교 Using the cyclic rule,
1 1 1
P V
T
T T V
P
P P T T
V
V P T
T V
P
U S P T P
T P T P
V V V
V V T
C C P U
V
V T V T P V P
V
α κ
α κ
α κ
κ
∂
∂
∂ = − =
∂ ∂
∂ ∂ ∂
∂
∂ = ∂ − = ∂ −
∂
= = = − = −
− =
∂ ∂ ∂ ∂
∂ ∂ ∂
− = + ∂ ∂
2
T P
V TV T
α κ
∂ =
∂