2
The Fundamental Equation
We have shown that:
dU = dq + dw
plus
dwrev = -pdV and dqrev = TdS
We may write:
dU = TdS – pdV (for constant composition)
3
In chapter 3 we discussed total integrals.
Properties of the internal energy
dyy
zdx
x
zdz
xy
We can express U as a function of S and V, i.e. U = f ( S,V )
If z = f (x,y) then:
dVV
UdS
S
UdU
SV
dU = TdS – pdVc.f.
TS
U
V
pV
U
S
We have
discovered that
4
Properties of the internal energy
Recall the test for exactness:
dyy
zdx
x
zdzGdyFdx
xy
If the differential is exact then:
yxx
G
y
F
All state functions have exact differentials
5
dVV
UdS
S
UdUGdVFdS
SV
Properties of the internal energy
Therefore:
Where:T
S
UF
V
pV
UG
S
Because this is exact we may write:
VS S
G
V
F
VS S
p
V
T
We have obtained our first Maxwell relation!
6
Relationships between state functions: Be prepared!
U and S are defined by the first and second laws of thermodynamics, but H, A and G are defined using U and S.
The four relationships are:
pVAGTSUA
TSHGpVUH
We can write the fundamental thermodynamic equation in several forms with these equations
dU = TdS – PdV
dH = TdS + VdP
dA = -SdT - PdV
dG = -SdT + VdP
Gibbs Equations
7
dpp
HdS
S
HdHNdpMdS
Sp
Properties of the internal energy
Also consider dH = TdS + Vdp, and writing H = f ( S,p )
Where:T
S
HM
p
Vp
HN
S
Because this is exact we may write:
pSS
N
p
M
pSS
V
p
T
We have obtained our second Maxwell relation!
8
The Maxwell Relations
(a) U = q + w(b) S = qrev/T(c) H = U + pV(d) A = U – TS(e) G = H - TS
dyy
zdx
x
zdzNdyMdx
xy
yxx
N
y
M
pSS
V
p
T
VS S
p
V
T
VT T
p
V
S
pTT
V
p
S
1. 2.
3. 4.
1. dU = TdS – pdV
2. dH = TdS + Vdp
3. dA = -SdT - pdV
4. dG = -SdT + Vdp
(b)(a)
(c)(b)(a)
(d)(b)(a)
(e)(b)(a)
9
The Maxwell Relations: The Magic Square
V A T
G
PHS
U
“Vat Ug Ship”
Each side has an energy ( U, H, A, G )
Partial Derivatives from the sides
Thermodynamic Identities from the corners
Maxwell Relations from walking around the square
10
Example:
Calculate the change in enthalpy if the pressure on one mole of liquid water at 298 K is increased from 1 atm to 11 atm, assuming that V and α are independent of pressure. At room temperature α for water is approximately 3.0 × 10-4 K-1.
pT
V
V
1 (The expansion coefficient)
The volume of 1 mole of water is about 0.018 L.
11
Properties of the Gibbs energy
G = H - TS
dG = dH –TdS - SdT
dG = dU + pdV + Vdp –TdS - SdT
dU = TdS –pdV
dG = TdS – pdV + pdV + Vdp –TdS - SdT
dG = Vdp - SdT G = f ( p, T )
dH = dU +pdV + Vdp
H = U + pV
12
Properties of the Gibbs energy
dG = Vdp - SdT
ST
G
p
Vp
G
T
V is positive so G is increasing with increasing p
G
T (constant p)
Slope = -SG
P (constant T)
Slope = V
S is positive (-S is negative)so G is decreasing with increasing T
13
Dependence of G on TS
T
G
p
Using the same procedure as for the dependence of G on p we get:
TSdGd
To go any further we need S as a function of T ?
Instead we start with: G = H - TS
-S = (G – H)/T
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Dependence of G on T
T
HGS
ST
H
T
G
Let G/T = x ST
Hx
2T
H
T
x
p
2
)/(
T
H
T
TG
p
This is the Gibbs-Helmholtz Equation
2
)/(
T
H
T
TG
p
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Dependence of G on T
2
)/(
T
H
T
TG
p
ST
G
p
Two expressions:
Gibbs-Helmholtz Equation
Changes in entropy or, more commonly, changes in enthalpy can be used to show how changes in the Gibbs energy vary with temperature.
For a spontaneous (G < 0) exothermic reaction (H < 0) the change in Gibbs energy increases with increasing temperature.
G/T
T (constant p)
Slope = -H/T2 = positive for exothermic reaction
Very negative
Less negative
16
Dependence of G on p
It would be useful to determine the Gibbs energy at one pressure knowing its value at a different pressure.
dG = Vdp - SdT
We set dT = 0 and integrate:
f
i
)()( if
p
pVdppGpG
f
i
f
i
p
p
G
GVdpGd
f
i
p
pVdpG
17
Dependence of G on p
f
i
)()( if
p
pVdppGpG
Liquids and Solids.
Only slight changes of volume with pressure mean that we can effectively treat V as a constant.
)()()( ifif ppVpGpG
pVpGpG )()( if
Often V p is very small and may be neglected i.e. G for solids and liquids under normal conditions is independent of p.
18
Dependence of G on p
f
i
)()( if
p
pVdppGpG
Ideal Gases.
For gases V cannot be considered a constant with respect to pressure. For a perfect gas we may use:
i
fi
if
ln)(
)()(f
i
p
pnRTpG
p
dpnRTpGpG
p
p
19
Dependence of G on p
Ideal Gases.
i
fif ln)()(
p
pnRTpGpG
We can set pi to equal the standard pressure, p ( = 1 bar).Then the Gibbs energy at a pressure p is related to its standard Gibbs energy, G, by:
p
pnRTGpG f
f ln)(
20
Dependence of G on p
Exercise 5.8(b) When 3 mol of a perfect gas at 230 K and 150 kPa is subjected to isothermal compression, its entropy decreases by 15.0 J K-1. Calculate (a) the final pressure of the gas and (b) G for the compression.
21
Dependence of G on p
Real Gases.
p
fRTGpG ln)( mfm
For real gases we modify the expression for a perfect gas and replace the true pressure by a new parameter, f, which we call the fugacity.
The fugacity is a parameter we have simply invented to enable us to apply the perfect gas expression to real gases.
22
Dependence of G on p
Real Gases.
We may show that the ratio of fugacity to pressure is called the fugacity coefficient:
p
fWhere is the fugacity coefficient
Because we are expressing the behaviour of real gases in terms of perfect gases it is of little surprise that is related to the compression factor Z:
dpp
Zp
0
1ln
1
2lnf
fRTG We may then write
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Summary
1. The four Gibbs equations.
2. The four Maxwell relations. (The Magic Square!)
3. Properties of the Gibbs energy
• Variation of G with T• The Gibbs-Helmholtz equation.• Variation of G with p • Fugacity
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Exercise:
For the state function A, derive an expression similar to the Gibbs-Helmholtz equation.
26
Preparation for Chapter 6:
So far we have only considered G = f ( p, T ).
To be completely general we should consider Gas a function of p, T and the amount of each component, ni.
G = f ( p,T, ni )
Then:
...2
,,21
,,1,,
dnn
Gdn
n
GdT
T
Gdp
p
GdG
jjii npTnpTnpnT
...1
,,1
jnpTn
Gwhere is the chemical potential.
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