The Second Law of Thermodynamics (II)
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The Second Law of Thermodynamics (II). The Fundamental Equation. We have shown that: dU = dq + dw plus dw rev = - pdV and dq rev = TdS We may write: dU = TdS – pdV (for constant composition). c.f. dU = TdS – pdV. Properties of the internal energy.

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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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c.f.

dU = TdS – pdV

Properties of the internal energy

In chapter 3 we discussed total integrals.

If z = f (x,y) then:

We can express U as a function of S and V, i.e. U = f ( S,V )

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, A and G are defined using U and S.

The four relationships are:

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


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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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2.

1.

(a) DU = q + w

(b) DS = qrev/T

(c) H = U + pV

(d) A = U – TS

(e) G = H - TS

3.

4.

  • dU = TdS – pdV

  • dH = TdS + Vdp

  • dA = -SdT - pdV

  • dG = -SdT + Vdp

The Maxwell Relations


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The Maxwell Relations: The Magic Square

V

A

T

“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

G

U

S

H

P


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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 αfor 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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H = U + pV

dH = dU +pdV + Vdp

dU = TdS –pdV

Properties of the Gibbs energy

G = H - TS

dG = dH –TdS - SdT

dG = dU + pdV + Vdp –TdS - SdT

dG = TdS – pdV + pdV + Vdp –TdS - SdT

G = f ( p, T )

dG = Vdp - SdT


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G

G

Slope = -S

Slope = V

T (constant p)

P (constant T)

Properties of the Gibbs energy

dG = Vdp - SdT

V is positive so G is

increasing with

increasing p

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 as for the dependence of G on p we 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

This is the Gibbs-Helmholtz

Equation

Let G/T = x


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Less negative

Slope = -DH/T2 = positive for exothermic reaction

DG/T

Very negative

T (constant p)

Dependence of G on T

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 (DG < 0) exothermic reaction (DH < 0) the change in Gibbs energy increases with increasing temperature.


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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:


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Dependence of G on p

Liquids and Solids.

Only slight changes of volume with pressure mean that we can effectively treat V as a constant.

Often V Dp is very small and may be neglected i.e. G for solids and liquids under 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 to pressure. 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 its standard 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 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) DG for the compression.


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Dependence of G on p

Real Gases.

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.


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Dependence of G on p

Real Gases.

We may then write

We may show that the ratio of fugacity to pressure is called the fugacity coefficient:

Where  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:


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  • Summary

  • The four Gibbs equations.

  • The four Maxwell relations. (The Magic Square!)

  • 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.


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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 function of p, Tand the amount of each component, ni.

G = f ( p,T, ni )

Then:

m is the chemical potential.

where