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Need help with Matlab: Adsorption isotherms are phase coexistence relationships that relate the partial pressure of a gas phase substance to the amount of gas

Need help with Matlab: Adsorption isotherms are phase coexistence relationships that relate the partial pressure of a gas phase substance to the amount of gas molecules adsorbed inside a porous material (i.e., loading). In thermodynamics, we have learned how to predict the properties of two coexisting phases, such as vapor-liquid equilibria (VLE). If we replace the liquid phase with an adsorbed phase and treat the adsorbed phase using a formalism similar to Raoults law,
we have the ideal adsorbed solution theory (IAST). This assignment and the next will walk you through a few important coding concepts. At its completion, you will be able to make your own IAST predictions.
Complete Part (1) Part (5) for this assignment.
Use water_data, ehtanol_data, and binary_data when referencing data used
1. Inspect the provided dataset by plotting the single-component adsorption isotherms,
Ni(pi), using the logarithmic scale for the horizontal axis. Use empty blue circles for
water (i =1) and empty red up triangles for ethanol (i =2).
Note: For this part, you need to search for the plotting command and read its docu-
mentation to learn how to use it.
Interpretation of the plot: If we flow water vapor at a given pressure, say p1=
11302.26 Pa, over silicalite, 1.367 water molecules will be adsorbed per unit cell of
the zeolite sorbent at equilibrium.
2. For each of the two unary isotherms, write a function to implement Ni(ln pi) by fitting
to the given data points. You can either use a file or the anonymous function syntax
@(x)(). A user needs to be able to call your function via, e.g., isotherm1(t) with
the independent variable t = ln pi. pi can be any pressure not necessarily from the
provided data set.
Note: You may want to read the documentation of F = griddedInterpolant(x, y,
linear). To use the returned object with fsolve in (3), you can create a true
function isotherm1(t)=@(t)F(t).
3. Find the roots ln p1,0 and ln p2,0 for the two isotherm functions using fsolve. In the
same figure of (1), plot your fitted functions from ln pi,0 to 2 ln pi,max, where pi,max is
the largest pressure from the provided dataset.
In a binary adsorption experiment, we put the water/ethanol mixture in contact with the
sorbent, and both molecules will adsorb. Of course, they will do so in different quantities,
resulting in preferential adsorption that can be used to separate the mixture. For a mixture
at given T and p, the specification of its composition (e.g., expressed in terms of the mole
fraction of ethanol, x2) then completely fixes the chemical potentials of water and ethanol,
which can be converted to equivalent partial pressures. The first two columns of the mixture
isotherm data give the partial pressures, p1 and p2. The last two columns give the measured
loadings for water (N1) and ethanol (N2).
4. In a new figure, plot both water and ethanol loadings as a function of p2, using the
same symbol conventions as in (1).
1
The IAST equations for a C-component mixture are as follows:
p0
i zi\gamma i = fi pi, i =1... C (1)
\Psi =\pi A
RT =
Z ln p0
1
ln p1,0
N1(ln p1) d ln p1=
Z ln p0
i
ln pi,0
Ni(ln pi) d ln pi, i =2... C (2)
where A is the (unkwown) surface area of the sorbent material, R is the gas constant, and
T is temperature. Eqn. 2 also defines spreading pressure \pi and adsorption potential \Psi .
When using IAST to predict binary adsorption (C =2) from single-component isotherms
Ni(ln pi), the partial pressures (pi) are specified, and adsorbed-phase activity coefficients (\gamma i)
are assumed to be unity. Using the three independent equations (two of Eqn. 1 and one of
Eqn. 2), one can solve for the adsorbed-phase composition, z1, and the two pressures, ln p0
i .
5. Using the two fitted isotherm functions, Ni(ln pi), write another two functions (one for
water and the other for ethanol) to compute the adsorption potential as a function of
ln p0
i : \Psi i(ln p0
i )= R ln p0
i
ln pi,0 Ni(ln pi) d ln pi, where the integration lower limit is what you
found in (3). That is, a user needs to be able to call ads pot(t), where t = ln p0
i , to
obtain the value of the integral.

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