
3.5 OPEN SYSTEMS: ENTHALPY FLUX 53
the ideal gas law underestimates the density of air by 1 g m
−3
compared to
van der Waals’ equation. We can safely use the ideal gas law for all practical
applications.
The above derivation uses an interaction potential based on a constant
average density throughout the gas, the so-called mean field method. This
ignores correlations between the positions of individual molecules (the at-
tractive force makes molecules preferentially cluster together on average).
Such correlations can be taken into account in a systematic way to find more
accurate approximations to the free energy. Typically, the pressure or the
free energy is written as a power series in the density. Such an expansion is
called a virial expansion. The virial expansion for the pressure has the form
p = RT
1 + B
2
(T) +B
3
(T)
2
+ ...
(3.52)
where the B
i
for i = 2, 3 ...are virial coefficients, which are functions of the
temperature. For small densities the equation of state reduces to that of an
ideal gas. For a van der Waals gas, the second and third virial coefficients
are given by
B
2
=
b
−
a
2
RT
,B
3
=
b
2
2
, (3.53)
with the molar mass of the substance, R its specific gas constant, and a and
b its van der Waals coefficients, see Problem 3.9.
At high densities the free energy cannot be determined as a modification of
the free energy of the ideal gas. For extremely high densities the free energy is
dominated by entropy variations due to the possible configurations of densely
packed molecules. Here an ordered state provides most freedom of motion
to the molecules and will therefore correspond to the largest entropy: the
molecules undergo an entropy driven phase transition to a crystalline state.
3.5 OPEN SYSTEMS: ENTHALPY FLUX
The first law expresses conservation of energy for closed systems, systems
that interact with their environment through heat and work exchange but not
through mass exchange. An open system allows mass exchange. Figure 3.3
shows a schematic of an open system. For simplicity the open system has one
inlet and one outlet. We allow an amount of mass ıM
1
to flow in through
the inlet and an amount ıM
2
to flow out of the outlet. The system gains an
amount of energy equal to U
1
= ıM
1
u
1
, with u
1
the specific internal energy
of the matter flowing into the inlet port. But to make this flow enter the
system an amount of work equal to p
1
dV
1
= p
1
ıM
1
v
1
has to be performed
on the system, with p
1
the pressure at the inlet. The total energy gain of the
system is therefore ıM
1
(u
1
+ p
1
v
1
) = ıM
1
h
1
with h
1
the specific enthalpy
at the inlet. Similarly, the total energy loss is ıM
2
u
2
plus the work p
2
dV
2
done by the system to expel the fluid. The total energy loss of the system is