Solution thermodynamics theory—Part I

Chapter 11

topics

• Fundamental equations for mixtures• Chemical potential• Properties of individual species in solution

(partial properties)• Mixtures of real gases • Mixtures of real liquids

A few equations

dTnSdPnVnGd

PVUH

from

dTnSnSTdnHdnGd

TSHG

)()()(

d(nH)obtain

)()()()(

For a closed system

Total differential form, what are (nV) and (nS)Which are the main variables for G??What are the main variables for G in an open system of k components?

G in a mixture (open system)

dTT

nGdP

P

nGnGd

nPnT ,,

)()()(

G in a mixture of k components at T and P

i

k

iidndTnSdPnVnGd

1

)()()(

How is this equation reduced if n =1?

2 phases (each at T and P) in a closed system

i

k

iidndTnSdPnVnGd

1

)()()(

)()()( nMnMnM

dTnSdPnVnGd )()()(

Apply this equation to each phase

Sum the equations of both phases, take into account that

In a closed system:

We end up with

0 ii

iii

i dndn

How are dnia and dni

b related at constant n?

For 2 phases, k components at equilibrium

ii

PP

TT

For all i = 1, 2,…k

Thermal equilibrium

Mechanical equilibrium

Chemical equilibrium

In order to solve the VLE problem

• Need models for i in each phase

• Examples of models of i in the vapor phase

• Examples of models of i in the liquid phase

Now we are going to learn:

• Partial molar properties

• Because the chemical potential is a partial molar property

• At the end of this section think about this– What is the chemical potential in physical terms– What are the units of the chemical potential– How do we use the chemical potential to solve a VLE

(vapor-liquid equilibrium) problem

Partial molar property

ijnTPii n

nMM

,,

)(

Solution property

Partial property

Pure-species property

example

ijnTPii n

nVV

,,

)(

0lim

~)(

)(

wn

ww

ww

nVnV

nVnV

Open beaker: ethanol + water, equimolarTotal volume nVT and P

Add a drop of pure water, Dnw

Mix, allow for heat exchange, until temp T

Change in volume ?

Total vs. partial properties

ii

i

ii

i

MnnM

MxM

See derivation page 384

Derivation of Gibbs-Duhem equation

ii

i

ii

ixPxT

MxM

dxMdTT

MdP

P

MdM

,,

Gibbs-Duhem at constant T&P

P& Tconstant 0 ii

i Mdx

Useful for thermodynamic consistency tests

Binary solutions

112

121

dx

dMxMM

dx

dMxMM

See derivation page 386

Obtain dM/dx1 from (a)

Example 11.3

• We need 2,000 cm3 of antifreeze solution: 30 mol% methanol in water.

• What volumes of methanol and water (at 25oC) need to be mixed to obtain 2,000 cm3 of antifreeze solution at 25oC

• Data:

watermolcmV molcmV

methanol molcmV molcmV

/07.18/77.17

/73.40/63.383

23

2

31

31

solution• Calculate total molar volume of the 30% mixture

• We know the total volume, calculate the number of moles required, n

• Calculate n1 and n2

• Calculate the total volume of each pure species needed to make that mixture

Note curves for partial molar volumes

From Gibbs-Duhem:

P& Tconstant 0 ii

i Mdx

02211 VdxVdx

Divide by dx1, what do you conclude respect to the slopes?

Example 11.4

• Given H=400x1+600x2+x1x2(40x1+20x2) determine partial molar enthalpies as functions of x1, numerical values for pure-species enthalpies, and numerical values for partial enthalpies at infinite dilution

• Also show that the expressions for the partial molar enthalpies satisfy Gibbs-Duhem equation, and they result in the same expression given for total H.