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The Second Law ofThermodynamics (II)
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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)
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In chapter 3 we discussed total integrals.
Properties of the internal energy
We can express U as a function of S and V, i.e. U = f ( S,V )
If z = f (x,y) then:
dU = TdS – pdV
c.f.
We have
discovered that
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Properties of the internal energy
Recall the test for exactness:
If the differential is exact then:
All state functions have exact differentials
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Properties of the internal energy
Therefore:
Where:
Because this is exact we may write:
We have obtained our first Maxwell relation!
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Relationships between state functions: Be prepared!
U and S are defined by the first and second laws of thermodynamics, but H, Aand G are defined using U and S.
The four relationships are:
We can write the fundamental thermodynamicequation in several forms with these equations
dU = TdS – PdV
dH = TdS + VdP
dA = -SdT - PdV
dG = -SdT + VdP
Gibbs Equations
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Properties of the internal energy
Also consider dH = TdS + Vdp, and writing H = f ( S,p )
Where:
Because this is exact we may write:
We have obtained our second Maxwell relation!
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The Maxwell Relations
(a) U = q + w
(b) S = qrev/T
(c) H = U + pV
(d) A = U – TS
(e) G = H - TS
1.
2.
3.
4.
1.dU = TdS – pdV
2.dH = TdS + Vdp
3.dA = -SdT - pdV
4.dG = -SdT + Vdp
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The Maxwell Relations: The Magic Square
V
A
T
G
P
H
S
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
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Example:
Calculate the change in enthalpy if the pressure on one mole ofliquid water at 298 K is increased from 1 atm to 11 atm,assuming that V and α are independent of pressure. At roomtemperature α for water is approximately 3.0 × 10-4 K-1.
(The expansion coefficient)
The volume of 1 mole of water is about 0.018 L.
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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
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Properties of the Gibbs energy
dG = Vdp - SdT
V is positive so G is
increasing with
increasing p
G
T (constant p)
Slope = -S
G
P (constant T)
Slope = V
S is positive (-S is negative)
so G is decreasing with
increasing T
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Dependence of G on T
Using the same procedure asfor the dependence of G on pwe get:
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
Let G/T = x
This is the Gibbs-Helmholtz
Equation
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Dependence of G on T
Two expressions:
Gibbs-Helmholtz Equation
Changes in entropy or, more commonly, changes in enthalpy can be usedto show how changes in the Gibbs energy vary with temperature.
For a spontaneous (G < 0) exothermic reaction (H < 0) the change inGibbs energy increases with increasing temperature.
G/T
T (constant p)
Slope = -H/T2 = positive for exothermic reaction
Very negative
Less negative
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Dependence of G on p
It would be useful to determine the Gibbs energy at onepressure knowing its value at a different pressure.
dG = Vdp - SdT
We set dT = 0 and integrate:
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Dependence of G on p
Liquids and Solids.
Only slight changes of volume with pressure mean that we can effectivelytreat V as a constant.
Often V p is very small and may be neglected i.e. G for solids and liquidsunder normal conditions is independent of p.
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Dependence of G on p
Ideal Gases.
For gases V cannot be considered a constant with respect topressure. For a perfect gas we may use:
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Dependence of G on p
Ideal Gases.
We can set pi to equal the standard pressure, p ( = 1 bar).
Then the Gibbs energy at a pressure p is related to itsstandard Gibbs energy, G, by:
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Dependence of G on p
Exercise 5.8(b) When 3 mol of a perfect gas at 230 K and 150 kPais subjected to isothermal compression, its entropy decreases by15.0 J K-1. Calculate (a) the final pressure of the gas and (b) Gfor the compression.
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Dependence of G on p
Real Gases.
For real gases we modify the expression for a perfect gasand replace the true pressure by a new parameter, f, whichwe call the fugacity.
The fugacity is a parameter we have simply invented toenable us to apply the perfect gas expression to real gases.
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Dependence of G on p
Real Gases.
We may show that the ratio of fugacity to pressure is called the fugacitycoefficient:
Where is the fugacity coefficient
Because we are expressing the behaviour of real gases in terms of perfectgases it is of little surprise that is related to the compression factor Z:
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 theGibbs-Helmholtz equation.
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Exercise 5.15 (a) (first bit)
Evaluate (S/ V)T for a van der Waals gas.
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Preparation for Chapter 6:
So far we have only considered G = f ( p, T ).
To be completely general we should consider Gas a functionof p, T and the amount of each component, ni.
G = f ( p,T, ni )
Then:
where
is the chemical potential.