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Maxwell Relations
Starting with the first law
dU = TdS - PdV,
we can express the enthalpy as
dH = dU + PdV + VdP = TdS + VdP,
the Helmholtz free energy as
dA = dU - TdS - SdT = -SdT - PdV,
and the Gibb's free energy as
dG = dH - TdS - SdT = -SdT + VdP.
These four relations are exact differentials.
dU = TdS - PdV
dH = TdS + VdP
dA = -SdT - PdV
dG = -SdT + VdP
For each of these we can derive an expression based on the
equality of the second cross derivative. These relations, also
known as Maxwell relations are fundamentally useful for
thermodynamics.
dU = U dS + U dV
S V
V S
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Let us begin the with the internal energy. We can write the
internal energy as a total derivative
Note that this implies that
U = T and
S V
U
V
=–P
S
Now take the second derivative (i.e. the cross derivative)
 2U = T
SV S V
and
S
 2U
VS
= – P
S
V
V
And since the second cross derivatives are equal (for an exact
differential)
 2U =  2U
SV S VS
so
V
T
V
= – P
S
S
V
This symmetry leads us to state that the natural variables of the
internal energy are S and V. You might ask why I do not consider
V and T. After all we can write
dU = U dT + U dV
T V
V T
And we can see immediately that
dCv = U
T
So
V
dU = CvdT + U dV
V T
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But then we are left with the second term. It is important to note
that
U
V
–P
T
But rather
U
V
=–P
S
The difference between these two partial derivatives is in the
variable that is held fixed. How do we find the value of (U/V)T?
We will show a Maxwell relation can be used to find an expression
for (U/V)T. However, it is clear that the variables V and T are
going to lead to a more complicated expression for U than the
variables V and S did.
Let us consider the Maxwell relations for A.
dA = A dT + A dV
T V
V T
This implies that
A = –S and
T V
A
V
= –P
T
Now take the second cross derivative of each of these.
 2 A = – S
TV T
V
and
T
2A
VT
= – P
T
V
V
Because of the equality of the second cross derivatives we have
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S
V
= P
T
T
Maxwell Relation for A
V
This is perhaps the most useful of the Maxwell relations. Let us
use it first to find (U/V)T.
Start with A = U - TS. Take the derivative with respect to V at
constant temperature.
A
V
= U
V
T
– T S
V
T
T
Use the Maxwell relation for A above to substitute for (S/V)T.
Note that (A/V)T = -P so we have
– P = U
V
– T P
T
T
V
Which rearranges to
U
V
= – P + T P
T
T
V
While this would be inconvenient to use in the first law, it is a
useful equation and is sometimes known as a thermodynamic
equation of state. For example, for an ideal gas P = nRT/V and
therefore (P/T)V = nR/V. If we substitute into the
thermodynamic equation of state we find
U
V
= – P + T P
T
T
= – P + T nR = – P + P = 0
V
V
This is not a surprise. We have shown that for an ideal gas U =
3/2nRT and therefore the internal energy depends only on the
temperature and not on the volume.
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McQuarrie and Simon show that the same is true for a hard
sphere gas P(V-nb) = nRT (page 891), but not for van der Waal's
or other equations of state that include "a" parmeters that depend
on the intermolecular potential energy.
We now use the Maxwell relation for A to derive a general
relationship between Cp and Cv. Recall that we have stated that for
an ideal gas Cp - Cv = nR. Let us further examine the relationship.
Even though V and T are not the natural variables for U we can
start with
dU = U dT + U dV
T V
V T
dU = U dT + U dP
T P
P T
Likewise, we can express U in terms of the variables P and T
dU = U dT + U
T V
V
T
V dT + V dP
T P
P T
We can find a natural relationship between these two using the
differential expression for the volume.
dV = V dT + V dP
T P
P T
Rearranging we have
dU =
U
T
We see that
+ U
V
V
T
V
T
P
dT + U
V
T
V dP
P T
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U
T
=
P
U
T
+ U
V
V
T
V
T
P
Using the fact that U = H - PV we have
H
T
– P V
T
P
= U
T
P
+ U
V
V
T
V
T
P
And noting that
H
T
= C p and U = Cv
T V
P
we obtain
C p – Cv = P + U
V
T
V
T
P
The Maxwell relation allows us to write this as
C p – Cv = T P
T
V
T
V
P
Noting that
P
T
V
T
V
P
P
V
=–1
The chain relation
T
We can substitute in for (P/T)V to obtain
C p – Cv = –T P
V
T
V
T
2
P
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We define the isothermal compressibility
 = – 1 V
V P
T
And the coefficient of thermal expansion
 = 1 V
V T
P
Using these definition the difference in the heat capacities becomes
2

C p – Cv = TV
Although we have focused on the Maxwell relations for U and A,
the same procedure leads to Maxwell relations for H and G. These
are described in Chapter 22.
The Maxwell relations and associated machinery are NOT
something you memorize. Everything can be derived from dU =
TdS – PdV which is a combined statement of the First and Second
Laws.
Since you can get dH, dG, and dA from this expression, you can
obtain the Maxwell relations from the expression of each of these
as a total derivative.
State function
dU
dH
dA
dG
Total derivative
TdS – PdV
TdS + VdP
–SdT – PdV
–SdT + VdP
Maxwell relation
(T/V)S = -(P/S)V
(T/P)S = (V/S)P
(S/V)T = (P/T)V
-(S/P)T = (V/T)P
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The Standard State for a Gas
– S
P
= V
T
T
Maxwell Relation for G
P
The standard state for a gas at any temperature is the hypothetical
ideal gas at one bar. To calculate corrections to the entropy due to
deviations from ideality we begin with the Maxwell relation for
dG.
This expression allows us to calculate the pressure dependence of
the entropy.
S = –
P2
P1
V dP
T
(constant T)
Suppose we wish to determine the deviation from ideality at 1 bar
of pressure (i.e. at the standard state). At some very low pressure
all gases behave ideally so we consider an ideal gas at pressure Pid.
P id
S P
id
– S 1 bar = –
1 bar
V dP =
T P
1 bar
P id
V dP
T P
The quantity (V/T)P can be determined from the equation of
state for the gas. For a hypothetical ideal gas this same expression
is
1 bar
S 1 bar – S P = –
0
id
P id
V dP =
T P
1 bar
–
P id
R dP
P
Adding these last two equations together gives the difference from
ideality at 1 bar of pressure
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1 bar
S 0 1 bar – S 1bar =
P id
V
T
– R dP
P
P
Now we need only find an equation of state that we can use to
determine (V/T)P. We can use the virial equation of state for
example
PV = 1 + B2V(T) + ....
RT
RT
From the virial expansion we have
V
T
dB2V
=R+
+ ....
P
dT
P
Which gives a particularly simple expression for the deviation
from the ideal entropy
S 0 1 bar = S 1 bar +
dB2V
P = 1bar + ....
dT
The temperature and pressure dependencies of the Gibb's free
energy
G =
P2
VdP
P1
We shown that dG = -SdT + VdP. This differential can be used to
determine both the pressure and temperature dependence of the
free energy. At constant temperature SdT = 0 and dG = VdP. The
integrated form of this equation is
For one mole of an ideal gas we have
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G = RT
P2
P1
dP = RT ln P2
P
P1
The same formula can be obtained from
G = H – TS
We can use the above expression to indicate the free energy at
some pressure P relative to the pressure of the standard state P = 1
bar.
G T = G 0 T + RT ln
P
1 bar
G0(T) is the standard molar Gibb’s free energy for a gas. As
discussed above the standard molar Gibb’s free energy is the free
energy of one mole of the gas at 1 bar of pressure. The Gibb’s free
energy increases logarithmically with pressure. This is entirely an
entropic effect.
If we are dealing with a liquid or a solid the molar volume is more
or less a constant as a function of pressure. Actually, it depends on
the isothermal compressibility,  = -1/V(V/dP)T, but since k is
typically a number of the order 10-4 atm-1 for liquids and 10-6 atm-1
for solids. For our purposes we can treat the volume as a constant
and we obtain
G T = G0 T + V P – 1
The Gibb’s free energy also depends on temperature. The easiest
way to determine the temperature dependence is to begin with G =
H – TS and divide through by T.
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G=H –S
T T
We differentiate this expression with respect to T keeping P fixed.
G/T = – H + 1 H
T
T 2 T T
– S
T
P
P
The last two terms cancel since (S/T)P = CP/T.
G/T = – H
T
T2
Which can be applied to any process in which case it is written as
G/T = – H
T
T2
You might think that we could calculate the temperature
dependence of the free energy directly from G = H – TS and,
in fact, we can. We can calculate the enthalpy referenced to the
enthalpy at absolute zero and the absolute entropy. However, as
discussed on page 903 and 904 of McQuarrie and Simon, this is a
cumbersome procedure and involves some duplication of effort
since fusS = fusH/Tfus and vapS = vapH/Tvap so the calculations
done for entropies and enthalpies of phase transitions will cancel.
As we have already seen this is consistent with the fact fusG = 0
andvapG = 0. A plot of G(T) vs. T has a constant negative slope
with discontinuities at the phase transitions. The reason for the
negative slope (see Figure 22.7) is that
G
T
=–S
P
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Of course, the entropies of solid, liquid, and vapor are different
giving rise to the changes in slope at the phase transitions.