1

The Second Law of Thermodynamics (II)

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

14

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

15

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

23

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

24

Exercise:

For the state function A, derive an expression similar to the Gibbs-Helmholtz equation.

25

Exercise 5.15 (a) (first bit)

Evaluate (S/ V)T for a van der Waals gas.

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.