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An equation of state is a model for the relationship between properties that define the thermo - dynamic state of a substance. For example, equations

An equation of state is a model for the relationship between properties that define the thermo-
dynamic state of a substance. For example, equations of state for pure substances are commonly
expressed for the pressure P as a function of the molar volume V and temperature T. There are
many types of equations of state, but cubic equations of state are the simplest models that can si-
multaneously predict the properties of a substance when it coexists as a vapor and liquid. Models
for vapor-liquid equilibrium are important to design processes such as distillation.
The van der Waals cubic equation of state is
P=RTV-b-aV??(2),
where a and b are model parameters that can be expressed as functions of the critical pressure Pc
and critical temperature Tc, and R is the gas constant.
Our goal is to apply the van der Waals equation of state to predict the coexistence molar
volumes of vapor and liquid water at its normal boiling point.
Tasks
Derive expressions for a(Pc,Tc) and b(Pc,Tc) using the fact that at the critical point
The partial derivatives are also evaluated using the critical molar volume V??(c) and critical
temperature Tc. Calculate a and b for water using NIST reference data [1].
To use the van der Waals equation of state to find V? at a given T and P, it is numerically
convenient to rewrite it as a cubic polynomial in the compressibility factor Z :
Z3-(B+1)Z2+AZ-AB=0,
where
Determine ZL, the compressibility factor for liquid water and smallest root of this polyno-
mial, and ZV, the compressibility factor for water vapor and largest root of this polynomial,
at water's normal boiling point (T=100C,P=1(atm)) using:
(a) Fixed-point iteration. For only this part, it is recommended to rearrange the polynomial
to solve for the liquid as
ZL=B+ZL2B+1-ZLA,
and the equation to solve for the vapor as
ZV=B+1-AZV-BZV2.
You can implement fixed-point iteration in MATLAB using a loop. For example, this
loop iterates 100 times to solve our example problem from class starting from
{:x0=0) :
x=0;
for {:[i=1:100]x=exp(-x-2);
end
(b) The built-in fzero function in MATLAB. It is recommended to define an anonymous
function for the equation you will solve. For example, to solve e-x2-x=0 with initial
guess x0=0, you can do:
eqn =@(x)exp(-x-2)-x;
fzero(eqn,0.0)
(c) The built-in roots function in MATLAB. Read the MATLAB documentation for an
example of how to use this function.
Use your results from #2(c) to compute the liquid and vapor mass densities, then compare
to the properties of saturated water from NIST [1]. If there are discrepancies, comment on
what you think the most likely reason is.
A vapor often has a large molar volume, so its molar density =1V? is usually small.
The pressure can then be reasonably approximated by the virial expansion, which is a series
expansion in . Rewrite the van der Waals equation of state to use the molar density instead
of molar volume, i.e., as P(,T). Series expand P to second order in about =0, then
solve for for water vapor at the normal boiling point. How does your answer compare to
what you got in #3 by solving the cubic polynomial exactly?
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