Hi and welcome to gas density and the ideal gas law PV = nRT, in which we
will look at what gas density refers to, mathematically incorporating density
into PV = nRT, and then a couple of practice problems with density
and PV = nRT. So like any density gas density is the
ratio of its mass to volume, and gases occupy their entire container, so its
density is the mass of the gas divided by the containers volume.
Let's make this helium, and we can see how density
would change by comparison
to other gases such as oxygen.
Note that from the periodic table we see that the helium to oxygen
mass ratio is 4:32, but this ratio only holds if comparing the same number
of particles.Which is the case here, but having the same amount of particles can
be assumed from Avogadro's Hypothesis: any two gases at the same pressure
temperature and volume will have the same number of particles.
Both illustrations have the same number of particles in the same volume but the
periodic table mass of the oxygen molecule is 8x the mass of the
helium atom, and so the density of
oxygen will be 8 fold higher.
Let's go back to our 10 atoms of helium.
What would happen if we increase the amount of helium
atoms but keep the volume constant?
Our example here of 20 atoms versus 10 means
the 2nd mass is twice the 1st mass, and so the density is doubled as well,
assuming the same volume for each.
Ok now that we know at gas density is, how does
it fit mathematically into the ideal gas law PV=nRT? For reference let's
keep all of our variables and constants defined on the side here, and some of the
common units used in chemistry are shown as well. Let's take PV = nRT and
isolate volume since that is the denominator in density, which is
mass/volume. Dividing both sides by pressure allows us to cancel pressure on
the left, and we then have the equation rearranged in terms of volume V.
We know that density is mass divided by volume, in this case gas volume,
so if we divide mass by volume on the left, that gives density. But we would
then have to divide the right side into mass to keep things equal. Pressure is
now in the denominator of a denominator and so comes to the numerator.
Density, is mass times pressure divided by nRT.
This presents something very interesting
because mass over moles is molar mass.
And we can then substitute molar mass
for mass/n in the equation,
to get density equals moler mass times
pressure divided by RT. We now have two equations expressing the ideal gas law
in terms of density. Let's see how we can use information in a problem to
determine which equation to use. Let's do two problems. The first gives us pressure
and temperature, but neither mass nor molar mass. However we do have the
identity of the gas, carbon dioxide, and so from the periodic table, we find the
molar mass to be 44 grams per mole. Since we now have molar mass, we can see the
second equation would be appropriate. Plugging in the values from the problem
as well as using the gas constant with the appropriate units, in this case
liters since we want the gas density to be expressed with liters, and
kiloPascals is the unit in the problem, while Kelvin has to be used since the
gas constant relies on absolute temperature, and moles is the common way
to express the amount of gas particles in chemistry. We see that Kelvin cancels,
kiloPascals cancel and moles cancel, which leaves grams over liters, which is
mass over volume, which is density. If your units do not cancel correctly then
you have set up the problem incorrectly.
The density of carbon dioxide under the
given conditions is 1.32 grams per liter.
Last problem: Here we are given
moles, density, pressure, and Temperature. The question is asking for mass, so we
will use the first equation. Let's rearrange the equation to isolate mass,
and then plug in the given values. If we multiply both sides by nRT/P then
nRT and P cancel on the right, leaving mass by itself.
Let's plug everything in, and this time we use R with atm since that is
the unit used in the problem. Atmospheres, Kelvin, moles, and liters all cancel,
leaving grams, the unit for mass.
Plugging everything in gives a mass of 20.2 grams.
The second question asks What gas this is.
Well we have mass and we have moles.
If we divide the two, that gives us the molar mass of the gas.
Let's look on the periodic table and we see that argon matches are calculated molar mass,
and so it seems this is argon