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Chapter 9. The chemical potential and open systems. Chemical Potential. We give a brief introduction to this concept as we will need to refer to it later in the course. As the name suggests, it is very important for studying chemical reactions and your will encounter - PowerPoint PPT Presentation
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1
Chapter 9The chemical potential and open systems
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Closed system: System does not exchange matter with surroundings.
Open system: Quantity of matter not fixed.
Chemical Potential.We give a brief introduction to this concept as we will
need to refer to it later in the course. As the name suggests, it is very important for studying chemical reactions and your will encounter this concept in a first course in physical chemistry. Just as the temperature governs the flow of energy between two systems, the chemical potential governs the flow of particles. When the chemical potential of the two systems are equal, they are in diffusive equilibrium.
We are now going to consider the introduction of matter into a system. If we introduce, say, dn kilomoles of matter into a system, there will be a change in energy of the system and dU will be proportional to dn
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For a pure substance, with a constant number of particles, we had, for a quasi-static process: )1.......(PdVQdU
and for a reversible process, dU=TdS –PdV ………(2)
We now consider a more general case where a system may consist of several different constituents. Furthermore, the system is no longer necessarily closed so that the number of moles of the various constituents may vary.
in = number of moles of constituent i
We now generalize equation (2) to allow for the possibility of adding of removing particles from our system. Before we had U(S,V) and now
sonVSU i ),,( )3.....(dnn
UdV
V
UdS
S
UdU i
n,V,Siin,Sn,V j
In the first two derivatives, all the n are constant and in the third derivative all the n are constant except in
đ
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Comparing equations (2) and (3)
nSnVV
UP
S
UT
,,
and now we define the chemical potentials by
jn,V,Si
i n
U
Equation (3) can now be written as )4.....(ii
idnPdVTdSdU
Solving for dS: ii
i dnT
dVT
PdU
TdS
1
and sojnVUi
i
nUnVn
S
TV
S
T
P
U
S
T,,,,
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The Helmholtz function is F=U-TS
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SdTTdSdnPdVTdSSdTTdSdUdF ii
i )(
soanddnSdTPdVdF ii
i
jnTVii
nVnTn
F
T
FS
V
FP
,,,,
EXAMPLE: Mixture of two ideal monatomic gases We will take as given the Helmholtz Thermodynamic Potential.
),,,( 21 nnTVF From the viewpoint of thermodynamics this comes from
experiment. In statistical mechanics this is derived from a model in which two types of molecules are present.
22
222222
21
111111
2ln
23
)ln(ln23
ln
2ln
23
)ln(ln23
ln
h
kmnnNnnTnVn
h
kmnnNnnTnVn
RTF
A
A
(reciprocity relations)
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m=molecular mass =Avogadro’s number h=Planck’s constant AN
nRTPVRTnnPVV
n
V
nRTP
V
FP
nT
)( 2121
,
If we have 02 n
If we have01 n
V
nRTP 1
1
V
nRTP 2
2
so 21 PPP
Dalton’s Law The pressure of a mixture of ideal gases is equal to the sum of the partial pressures. The partial pressure of a gas is the pressure that it would exert if it alone occupied the volume V at temperature T.
)(23
21
,
nnRT
FS
T
FS
nV
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2,,11
nTVn
F
21
11
2ln
23
11)ln(ln23
lnh
kmNnTVRT A
21
11
2ln
23
)ln(ln23
lnh
kmNnTVRT A
)(
23
21 nnRT
FTFTSFU
2121 )(23
UUUnnRTU
From the fact sheet we obtain G=F+PV so RTnnFG )( 21
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22
22222
21
11111
2ln
23
)ln(ln23
ln
2ln
23
)ln(ln23
ln
h
kmnNnnTnVn
h
kmnNnnTnVn
RTG
A
A
21 GGG
Comparison with the expression for and a similar expression for gives
1 22211 nnG In general i
iinG
For a system consisting of just one constituent (one phase) we have
gn
GornG
We see that for such a simple system, the chemical potential is just the Gibb’s function.
{We mentioned earlier that the Gibbs Potential was particularly important in physical chemistry.}