In this screen cast we are going to look at an example of mass transfer through diffusion. A specific
case of diffusion, that of equimolar countercurrent diffusion between two species. So we are going
to do this through an example problem. Here we have two tanks that are connected by a
tube, and one tank contains nitrogen and another tank contains carbon dioxide. The pressure
and temperature are maintained at 200 kPa and 25 degrees Celsius. Now at some initial
time, t equals zero, diffusion occurs, a separator is removed between the two gases. And we want
to know how long it will take for the mole fraction of the carbon dioxide to reach 0.7
on one end and 0.25 on the other end. So at this time determine the rates and directions
of mass transfer for the carbon dioxide and the nitrogen, the velocities of each species
and the molar average velocity of the mixture.
Now even though I drew this separator here
in the middle we are going to pretend that we are using this entire distance. So we are going to
assume that the only mass transfer resistance is in the tube. We are going to assume that
we have perfectly well mixed, ideal gases based on these properties. Because of these
we are going to get equimolar counterdiffusion. So we want to know when the mole fraction
of nitrogen at point 1 is going to be 0.7, and when it will also be at point 2, equal
to 0.25. So for equimolar counterdiffusion, or EMD, we know that the equation for mass
transfer can be defined as the following. Where we have the net molar flux of nitrogen
or CO2, is equal to a concentration times our diffusivity, times our gradient- in this case
written in terms of our mole fractions, I am going to use y to indicate a gas mole fraction
and we are looking at nitrogen. So when we do this there is no bulk flow term, that has
been crossed out. So this is the equation we use. So basically we want to solve for
these variables so we can determine where we are going to have our mole fractions equal
to 0.7 and 0.25. So first let's calculate the area at which this diffuses to. So remember
our flux is through some certain area so we are looking at a cross section of this pipe
and so our area is going to just be equal to pi D squared over 4 and in this case we are told we have
a 5 mm diameter pipe. I am going to work with cm since that is what we are given for our
diffusivity. And this gives me 0.1964 centimeters squared. We also need our total concentration of our
gas. So using our assumption of ideal gas our concentration C, we can figure out from
the ideal gas law, PV=nRT in this case we are looking for n over V. this is going to
be equal to P/RT. This is going to give me 80.71 moles per meter cubed. So when we convert into centimeters
cubed I get that our concentration is 8.0714 times ten to the minus five mole per centimeter cubed. So now we
have our concentration, we have part of our flux, the area, the rest of that part is going to
be our molar flow rate. We are given the diffusivity, we know a length between our two points so
really, at this point we are only looking for two more variables. We are given variables
for our differential mole fraction. So we should be able to solve for our flux at this
point. So I start plugging in values at this point and here we are looking for the mole
fraction to be 0.7 on one side and 0.25 on the other. And this is all over our distance
which was half a meter, or in this case 50 cm. To make sure our units work out. This
gives us a flux equal to 5.81 times ten to the minus eight mole per centimeter squared second. So the first part
of the question asks us to determine the rates and direction of mass transport. We have a net
flux for nitrogen, for equimolar counterdiffusion we know that net flux must also be the same
for the CO2 in the opposite direction. So we want to determine the molar flow rate which
we are going to write as N for the molar flow rate for N2 and this is just going to be our
flux times our area. This gives me 1.14 times ten to the minus eight moles per second. So the molar flow rate of
nitrogen is to the left and therefore, because of the equimolar counterdiffusion we'll also have 1.14 times
ten to the minus 8 moles per second, for our
molar flow rate of CO2 and this is to the right based again on our schematic. So the
second part of this was to determine the individual species' velocities. So we know, because we
know there is no bulk flow, that the species velocity, Vi, is just due to the diffusive
velocity of that species. So in this case if we wanted to determine the velocity of
CO2 we could set this equal to our flux of CO2 divided by our concentration of CO2, because
we know our velocity times our concentration of a species will give us our flux. We could
now re-write this in terms of things that we already have, our flux of CO2, which we calculated,
and then we are going to use the mole fraction of CO2 times the total concentration. So both
of these values we already have numbers for- so now we could get our species velocity as
a function of our mole fraction, yCO2. This is going to give us, finally a value of 7.2 times ten to the negative 4
over our mole fraction of carbon dioxide. So, again, because no bulk flow, nothing else
changes, we also know that the velocity of nitrogen is going to have the same equation
except over the mole fraction of nitrogen. So you see that the velocity does depend on
the mole fraction of the species and that answers the second part of the question- the
third part was to determine the average molar velocity of the mixture. So you could pick
a point in our system, let's just do this for yCO2 equals 0.7, so that means that our yN2
will be 0.3. So if we plug these two values in and we know that the velocities are in
opposite directions then we could use these values with our two equations to determine
that the velocity here for CO2 would be 1.03 times ten to the minus three centimeters per second.
The velocity of Nitrogen at this point is going to be
2.4 times ten to the minus three centimeters per second, and we are going to indicate since CO2 is going to the
right this is positive and for our system coordinates the velocity of nitrogen is to
the left and we will say is negative. So now our molar average is just going to be yCO2
for calculating this for the mixture, times the velocity of CO2 plus yNitrogen times the
velocity of nitrogen. Which you should see that this does in fact equal zero. And as we would
expect for equimolar counterdiffusion, that since we have no bulk flow, our average molar velocity of
the mixture is equal to zero.