In this screencast we are going to work through an example of mass transfer via unimolecular
diffusion. So the problem states that a student leaves a 5 liter beaker that is 1 foot high,
containing 3 L of ethyl ether on their lab bench open to the atmosphere and we are in
Boulder, Colorado. So the lab temperature is 25 degrees Celcius and we will assume that the ether
evaporates though a stagnant layer of air in the beaker and is carried away once it
reaches the top of the beaker. We want to determine the initial mass flow rate of ether from the
beaker and the time for the ether to evaporate from the beaker.
So the first place to start
with any problem especially mass transfer problems is to draw a picture of our system.
We are going to assume a couple things to make this problem easier. One, that we have
a steady state mass transfer, that the mole fraction of ether in the air, which we
will say is yE, adjacent to the liquid surface here, can be determined using Raoult's Law,
and at the surface of our container we have some air carrying away this ether so we are
going to say that the mole fraction of ether at the surface of the container is just equal
to zero. So we do have a cylinder and we can get an idea of what this stagnant layer thickness
is based on the dimensions of the beaker. We know that 5,000 cm^3 or 5 L is equal to
our height of 1 foot, which, converted to centimeters, gives me 30.48 cm. This height times the
cross sectional area, pi r squared, allows us to determine that the radius is approximately
7.23 cm. So one thing we could also do is since the volume is 3 L of the A combined
5 L total, we could say that our height must be 60% of our beaker height which gives us
18.29 cm. This also allows us to calculate the film thickness which is the remaining
height to the top of the surface which I am just going to say is delta Z and I get 12.19
cm. What is important is that since we are diffusing ether through a stagnant layer of
air we are going to go ahead and use unimolecular diffusion and the appropriate flux equation
which I am going to write down here. So we have our flux of ether in the z direction,
the vertical direction, is equal to the concentration of our system times the diffusivity of ether
in air. This is over delta z, our film thickness. We have our mole fraction of ether
at the surface of the fluid minus that at the top of the container. This is all over
the log mean of 1 minus the mole fraction of ether. We can calculate the concentration
using ideal gas law. We are given the diffusivity in the problem statement. We can calculate
the mole fraction using Raoult's Law. We are told that this term is zero. We could calculate
the log mean of 1 minus the mole fraction. And we just calculated what delta z is. So
we now know that we can calculate our flux. So let's go through that. So we are going
to start with calculating our concentration of our gas. This can be calculated using the
ideal gas law. So concentration is the number of moles over our gas volume. Which, using
the ideal gas law is equal to P over RT. So now plugging in values for this we get the
concentration is equal to 33.57 moles per meter cubed. We are going to want to work with centimeters,
so by converting this we get 3.357 times ten to the minus 5 moles per centimeter cubed.
We have the diffusivity and our distance already
determined. So we want to determine the mole fraction of our ether at the surface. And to do that we are going
to use Raoult's Law. Now Raoult's Law says that the mole fraction of ether in the vapor
phase times the pressure is equal to the mole fraction in the liquid phase times its saturation
pressure. Since our solution is all ether then the mole fraction is 1 and we could rewrite
this as the following ratio. Plugging in the values that we know for this system the mole
fraction of ether at the surface is 0.855. So the last thing to calculate in our flux
equation is this 1 minus yE or the log mean of this. We write this out as 1 minus the
concentration at the top of our film which is zero, minus 1 minus our surface concentration
and then this is all over the natural log of this ratio. And we should get the following
value. So now we can plug in all of these values into our flux equation. And when I
plug in all these values into our flux equation we should get that the flux of ether in the
z direction is equal to 5.045 times ten to the minus seven moles per centimeter squared second.
Now that we have calculated the flux of ether
we can use this to determine what we were asked for in the problem statement, which
was the mass flow rate of ether. So to go from a flux to a molar flow rate and then
a molar flow rate to a mass flow rate, we are going to start with our molar flow rate
calculation. So this is our flux of ether in the z direction times our cross sectional
area in the z direction. And we are working with a cylindrical beaker so our molar flow
rate of ether is equal to 8.286 times ten to the minus five moles per second.
To go from a molar flow rate to a mass flow rate we need to multiply it by the
molecular weight of ether which comes out to 6.14 times ten to the minus three grams per second
Now the problem asked us to report it in kilograms per hour, so we need to multiply by 3600 seconds per hour
and divide by 1000 grams per kilogram. This works out to 2.21 times ten to the minus two kilograms per hour
of ether evaporating away from this beaker. Just to give you an idea in terms of volume that is about 32 mL per
hour that would be leaving our beaker. So this answers the first part of our question. The second
part of our question deals with how long it takes for the ether to evaporate. So you could
imagine that as this fluid is evaporating from the beaker this stagnant layer thickness
is going to increase. Now we neglect the accumulation of ether and air in this new film space so
really this becomes something that we term a pseudo steady state case. So for the second
part I am using a pseudo steady state case, because even though we know something is changing with
time, we are assuming that we have no accumulation and so therefore our flux for
our ether is just what is leaving our system. So we can also say that this flux must therefore
equal a negative accumulation in the beaker itself. So not in the stagnant layer but in
the beaker if we are looking at the ether that is there. So really this is a disappearance.
Going back to our flux, it is in a moles per area per time so we want to know how many
moles we have to start with. So our volume of ether, we can multiply that by the density
of the ether to get a mass. And if we take the mass and divide it by the molecular weight
we could get moles. So we are looking at a change in volume with time, and really the
area is not changing so we could pull that out and just say the change in height or in
this case the change in z with time. And so, given this, if we multiply this by our density
and divide by our molecular weight of ether then this gives us a molar flow rate. That
divided by our area gives us our flux. The nice thing about this flux is that we have
a relationship for it already that we used before. That looks like the following. So
the only thing that changed here is that instead of z2 minus z1 we have z, since that is changing,
and we have our mole fraction that we have constant at the surface of the fluid and constant
at the top of the container. so as you see now, if I just get rid of this term, set these
two equal to each other, it is just a matter of plugging in the density, the molecular
weight, our concentration, our diffusivity, we know delta y, we can calculate the log
mean as we did before. So it is just matter of evaluating the integral that we are going
to look at. So let me write that out. Bringing our time component to its own side. Set everything
else on the other side, we get the density over the molecular weight times 1 minus y of ether,
the log mean of that over delta y. This is multiplied by 1 over the concentration times
the diffusivity of ether in air. And then we are going to evaluate the integral from
z0 to zt, for zdz. Now z0 in this case is where we started which was a delta z or a distance
from the top of the surface of 12.192 cm. And the total height is 30.48 cm. So at some
time that will be our other z. So it is just a matter of plugging these into the integral
for our boundary conditions and solving for t. When I plug all of this into our time equation
to get a time t equal to 6.1X10^5 seconds which equates to 169 hours which is roughly 1 week.
For the three liters of ether in this beaker to evaporate at 25 degrees Celsius using the
assumptions and techniques that we used in this video.