Peter Atkins • Julio de Paula
Atkins’ Physical Chemistry
Eighth Edition
Chapter 5 – Lecture 1
Simple Mixtures
Copyright © 2006 by Peter Atkins and Julio de Paula
Homework Set #5
Atkins & de Paula, 8e
Chap 5
Exercises: all part (b) unless noted:
1, 5, 7, 8, 10, 11, 13, 14
Objectives:
• Introduce partial molar quantities
• Use chemical potential (μ) to describe physical
properties of a mixture (Gm = μ)
• Raoult’s and Henry’s Laws: Thermo properties μ(Xi)
• Effect of a solute on a solution: Colligative properties
Restrictions at this point:
• Binary mixture
• Unreactive species
• Nonelectrolytes
• e.g., Dalton’s law:
and:
P = PA + Pb + ...
XA + XB +... = 1
• Now apply partial molar concept to other extensive
state functions
Partial molar volume
Observe:
18
H2O
cm3
H2O
Vtotal = V + 18 cm3
So:
VH2O = 18 cm3/mol
V>> 18 cm3
Partial molar volume (continued)
Observe:
18 cm3 H2O
EtOH
Vtotal = V + 14 cm3
So:
VH2O (in EtOH) = 14 cm3/mol
V>> 18 cm3
Conclusion: VJ varies with
composition
Fig 5.1 Partial molar volumes of water and ethanol
 V 

VJ  
 n J  P ,T ,n
dV  VJdn J
Result when dnA
moles of A and dnB
moles of B are
added to the
mixture:
 V 
 V 



dV  
dn A  
dnB

 nA P,T,nB
 nB P,T,nA
Fig 5.2 Partial molar volume of a substance
Total volume decreases
Total volume increases
as A is added
 V
VJ  
 n J


 P ,T ,n
dV  VA dnA
as A is added
For a binary mixture of A and B:
 V
dV  
 n A

 V

dn A  
 P ,T ,nB
 nB


dnB
 P ,T ,nA
dV = VA dnA + VB dnB
Once VA and VB are known:
V = nA VA + nB VB
Molar volumes always > 0, but not necessarily
partial molar volumes: VMgSO4 = −1.4 cm3/mol
Fig 5.4 Chemical potential as function of amount
Partial molar Gibbs energies
 G 

 J  
 n J  P ,T ,n
dG   A dnA
For a binary mixture
of A and B at some specific
composition:
G = nAμA + nBμB
Recall that G = G(T,P,n). When all may change:
dG = VdP - SdT + μAdnA + μBdnB + ...
The fundamental equation of chemical thermodynamics
At constant temperature and pressure:
dG = μAdnA + μBdnB + ...
Since dG = dwadd,max,
dwadd,max = μAdnA + μBdnB + ...
Gives the maximum amount of additional (non-PV) work
that can be obtained by changing system composition.
Because G = G(n,μ), for infinitesimal changes:
dG = μAdnA + μBdnB + nAdμA + nBdμB
Recall that at constant temperature and pressure
the change in Gibbs energy:
dG = μAdnA + μBdnB
Therefore:
nAdμA + nBdμB = 0
Special form of the Gibbs-Duhem equation:
 n d
J
J
Significance: The chemical potential of one component
cannot change independently of the other components
J
0
Fig 5.1 Partial molar volumes of water and ethanol
 n d
J
J
J
0
nAdμA + nBdμB = 0
nAdμA = - nBdμB
d A  
nB
d B
nA
As partial molar
volume of H2O
decreases, that of
ethanol increases:
The thermodynamics of mixing
For a binary mixture
of A and B at some specific
composition:
G = nAμA + nBμB
Also, systems at constant P and T
tend towards lower Gibbs energy:
Fig. 3.18
Fig 5.6 Arrangement for calculating the thermodynamic
functions for the mixing of two perfect gases
μA  μo  RT ln PA
μB  μo  RT ln PB
From:
G = nAμA + nBμB
ΔG mix
PA
PB
 n ART ln
 nBRT ln
P
P
ΔGmix  nRT(x A ln x A  xB ln xB )
Fig 5.6 Arrangement for calculating the thermodynamic
functions for the mixing of two perfect gases
From:
 ΔGmix 
ΔS mix  

 T 
ΔS mix  nR(x A ln x A  xB ln xB )
Fig 5. 9 Entropy of mixing of two perfect gases
Since ΔG = ΔH – TΔS
it follows that:
ΔHmix = 0