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LECTURE 6
Maxwell's relations are a set of equations in thermodynamics which are derivable from the
definitions of the thermodynamic potentials. These relations are named for the nineteenthcentury physicist James Clerk Maxwell.
The Maxwell relations are statements of equality among the second derivatives of the
thermodynamic potentials. They follow directly from the fact that the order of differentiation of
an analytic function of two variables is irrelevant. If Φ is a thermodynamic potential and xi and xj
are two different natural variables for that potential, then the Maxwell relation for that potential
and those variables is:
Maxwell relations(general)
where the partial derivatives are taken with all other natural variables held constant. It is seen that for
every thermodynamic potential there are n(n − 1)/2 possible Maxwell relations where n is the number of
natural variables for that potential
The four most common Maxwell relations are the equalities of the second derivatives of each of
the four thermodynamic potentials, with respect to their thermal natural variable (temperatureT;
or entropyS) and their mechanical natural variable (pressureP; or volumeV):
Maxwell's relations(common)
where the potentials as functions of their natural thermal and mechanical variables are the
internal energyU(S, V), EnthalpyH(S, P), Helmholtz free energyA(T, V) and Gibbs free
energyG(T, P). The thermodynamic square can be used as a mnemonic to recall and derive these
relations
Derivation
Derivation of the Maxwell relation can be deduced from the differential forms of
the thermodynamic potentials:
The differential form of internal energy U is
This equation resemble total differentials of the form
It can be shown that for any equation of the form
that
Consider, the equation
that
. We can now immediately see
Since we also know that for functions with continuous second derivatives, the
mixed partial derivatives are identical (Symmetry of second derivatives), that is,
that
we therefore can see that
and therefore that
Derivation of Maxwell Relation from Helmholtz Free energy
The differential form of Helmholtz free energy is
From symmetry of second derivatives
and therefore that
The other two Maxwell relation can be derived from differential form of enthalpy
and the differential from of Gibbs free energy
in a similar way. So all Maxwell Relationship above
follows from one of the Gibbs equations.
Extended derivation
Combined form first and second law of thermodynamics,
(Eq.1)
U, S, and V are state functions. Let,
Substitute them in Eq.1 and one gets,
And also written as,
comparing the coefficient of dx and dy, one gets
Differentiating above equations by y, x respectively
(Eq.2)
and
(Eq.3)
U, S, and V are exact differentials, therefore,
Subtract eqn(2) and (3) and one gets
Note: The above is called the general expression for Maxwell's thermodynamical
relation.
Maxwell's first relation
Allow x = S and y = V and one gets
Maxwell's second relation
Allow x = T and y = V and one gets
Maxwell's third relation
Allow x = S and y = P and one gets
Maxwell's fourth relation
Allow x = T and y = P and one gets
Maxwell's fifth relation
Allow x = P and y = V
=1
Maxwell's sixth relation
Allow x = T and y = S and one gets
=1
The Clausius-Clapeyron equation
Phase changes, such as the conversion of liquid water to steam, provide an important example of
a system in which there is a large change in internal energy with volume at constant temperature.
Suppose that the cylinder contains both water and steam in equilibrium with each other at
pressure P, and the cylinder is held at constant temperature T. The pressure remains equal to the
vapour pressurePvap as the piston moves up, as long as both phases remain present. All that
happens is that more water turns to steam, and the heat reservoir must supply the latent heat of
vaporization, λ = 40.65 kilojoules per mole, in order to keep the temperature constant.
The results of the preceding section can be applied now to find the variation of the boiling point
of water with pressure. Suppose that as the piston moves up, 1 mole of water turns to steam. The
change in volume inside the cylinder is then ΔV = Vgas − Vliquid, where Vgas = 30.143 litres is the
volume of 1 mole of steam at 100 °C, and Vliquid = 0.0188 litre is the volume of 1 mole of water.
By the first law of thermodynamics, the change in internal energy ΔU for the finite process at
constant P and T is ΔU = λ − PΔV.
The variation of U with volume at constant T for the complete system of water plus steam is thus
(48)
A comparison with equation (46) then yields the equation
(49) However, for
the present problem, P is the vapour pressure Pvapour, which depends only on T and is
independent of V. The partial derivative is then identical to the total derivative
giving the Clausius-Clapeyron equation
(50)
(51)
This equation is very useful because it gives the variation with temperature of the pressure at
which water and steam are in equilibrium—i.e., the boiling temperature. An approximate but
even more useful version of it can be obtained by neglecting Vliquid in comparison with Vgas and
using
(52)
from the ideal gas law. The resulting differential equation can be integrated to give
(53)
For example, at the top of Mount Everest, atmospheric pressure is about 30 percent of its value at
sea level. Using the values R = 8.3145 joules per K and λ = 40.65 kilojoules per mole, the above
equation gives T = 342 K (69 °C) for the boiling temperature of water, which is barely enough to
make tea.