Tip:
Highlight text to annotate it
X
This screencast is related to certain physical properties of gas mixtures specifically that
of an ideal gas mixture. So the problem states that we have an ideal gas mixture at 2 atm
and 35 degree Celsius, and it has the follow composition by volume percent. We are asked
to determine the partial pressure of each of the 5 species. The mass fraction of just
oxygen and carbon monoxide in this mixture. The average molecular weight of the gas mixture,
and then finally the density of this mixture. So lets go ahead and start with that of the
partial pressure. If we assume that oxygen has a certain amount of moles to it then we
use the ideal gas law RT is related to PV. These gas mixture share a common volume, but
each component is going to have its own partial pressure. So that is how we relate the two.
What we can see that if we related it to the total pressure. If we divide the first equation
by the second equation what we see is that the moles of oxygen over the total moles is
going to be related to the partial pressures of oxygen over the total pressure. We can
rewrite this as the mole fraction of oxygen times the pressure is going to be equal to
the partial pressure. So what this means is for an ideal gas mixture we can take the mole
percents or mole fraction and multiply it by the total pressure to get out the partial
pressure of each species or what pressure those components contributing to that total
pressure. Also for an ideal gas mixture the mole percent is going to be the same as the
volume percent. So in this case for each of these 5 species we can just multiply the 0.15
say for oxygen times the 2 atm and we are going to get 0.3 atm as the partial pressure
of oxygen. So going through the rest of these, and when we go through the rest of these hopefully
you get the values that I have gotten here, and we can add up all the partial pressures.
So the sum of each of the partial pressures is therefore going to be the total pressure,
which we know as 2 atm. So that should answer part a. We are going to go to part b now.
Part b asks us to find the mass fractions of oxygen and carbon monoxide in the mixture.
As we are given mole percents or volume percents we need to convert these to a mass percent
or therefore mass fraction. Now to do this we are going to pick a basis. We are going
to say that this intire gas mixture is going to be 1 mole total. What that allows us to
do is use an average molecular weight formula that we learned before, and we multiply the
mole fractions. In this case 0.15 moles of oxygen per mole total, and we are going to
multiply that by our basis of 1 mole. This is going to give us the amount of moles of
oxygen, which then we can multiply by the molecular weight of oxygen. In this case 32
grams of oxygen per mole of oxygen, and we are going to do this for each species. So
that we can get a total mass for our gas mixture that we assumed a basis of 1 mole. So as you
can see up here I multiplied each of the mole fractions times the molecular weight. So that
I can get the amount of mass that each component contributes to our gas mixture. When we do
this you get each of the masses of each component that contributes to a total mass of the mixture
of 30.42 grams. So if we want the mass fraction of oxygen. We are going to take the contribution
of oxygen 4.8 grams divide that by the total mass. This results in a mass fraction of oxygen
at 0.16, and the mass fraction of carbon monoxide. Is equal to 1.96 grams CO, 30.42 grams of
the mixture. This results in a mass fraction of carbon monoxide of 0.06 grams carbon monoxide
per gram mixture. So that answers our question for part b. Part c asks for the average molecular
weight of the gas mixture. Load and behold we have already done that. Choosing the basis
of 1 mole, we have calculated what each component contributes to the total mixture. So we have
30.42 grams mass per 1 mole. So the average molecular weight is therefore 30.42 grams
per mole. The last part of the problem asks us to determine the density of the gas mixture.
Now when we look at the density of the mixture we know the density is some mass per volume.
Now in the case of an ideal gas mixture. The best way to do this is to rewrite the mass
component in terms if the actual molecular weight average of the mixture. Which as the
units of grams per mole. So we are going to choose the molar volume V hat, which as units
of Liters per mole. That way the moles cancel, and we will get our grams per liter. This
allows us to use the ideal gas relationship, which relates the molar volume of the gas.
In this case a mixture to gas law constant R times the temperature of the pressure. So
if we substitute in this to our density equation what we see is that the density of the mixture
is going to be equal to the pressure times the average molecular weight over the gas
constant times the temperature. We know all of these values. So we can just start plugging
them in, and it is important to put all of your units here. So you can see what our density
ends up actually being. In this case you should get a value around 2.4 grams per liter. or
if we write those in some units that might look more familiar 2.4 kilograms per meter
cubed. So does does our final answer of 2.4 kg/m^3 make sense? Well of we look at the
density of air at 1 atm, 25 degree Celsius it is around 1.2 kilograms per meter cubed.
So we have a density of about twice of that, but we are at an elevated pressure and slightly
temperature, and we have a different composition of gas. It is not just nitrogen and oxygen.