Welcome to a tutorial on the ideal gas law, PV = nRT, in which we will gain an understanding
of R, the ideal gas constant.
Then look at where PV = nRT comes from, including how the different values for R
are derived, and then several problems using PV = nRT, including the use of mass and molar mass.
The four simple gas laws are shown here, Boyle's, Charles', Gay-Lussac's and Avogadro's Laws.
Gases have four variables that define their behavior: pressure, volume, temperature and
number of particles.
Each of these laws looks at the mathematical relationship of two of those four variables,
while holding the other two constant.
Each equation represents an initial and final condition of a gas indicated by the subscripts 1 and 2.
We can look at the relationship of all four variables by combining these equations into
one equation, showing the relationship of all four variables in one equation with an
initial and final condition.
We could keep going with changing conditions, and what this really tells us is that in any
particular condition of pressure, volume, temperature and number of particles, the ratio
of PV to Tn will always be constant for an ideal gas.
We give that constant a symbol, which is R. PV/Tn of any gas under ideal conditions
will always be R.
An ideal gas is defined as a gas whose pressure is low enough and temperature is high enough
so that gas particles are very far apart.
Meaning, the size of any single particle is irrelevant, and particles are moving fast enough
so that any collision is perfectly elastic.
In other words, there are no apparent attractive forces influencing the collision.
The temperature and pressure of the earth's atmosphere are ideal gas conditions,
we live in an ideal gas.
So if we multiply both sides by Tn, then Tn cancels on the left and we are left with
PV = RTn, which by convention is stated as PV = nRT or PVNRT, for short.
The equation has four variables, and one constant, R. Because R has many values depending on
the units used, you have to determine what is the appropriate value of R for a given problem
Let's go back to this form of the equation so we can determine two common values of R used in textbooks.
The numerical value of R depends on the units used, and so using atmospheres, liters, Kelvin, and mole
for PVT and n, gives the unit of R. However, we need to determine the numerical
value of R, which will be guided by Avogadro's hypothesis: that any two gases at the same
pressure, volume, and temperature will have the same number of particles.
Measuring one mole of gas at one atmosphere and 273 K, will occupy 22.4 liters, and so this
is an R of 0.082 atmospheres liters per Kelvin mole.
Kilopascals is also a common unit for pressure, and under these conditions, the pressure would be
101.3 kilopascals.
This gives an R of 8.314 kilopascals liters per Kelvin mole.
Note that each of these values express the same magnitude of R but have different numbers
due to different units used.
There are many values of R depending on the units used, but we will stick to these two
for this presentation.
Now we will go through four example problems to see how we can use PV = nRT
given various units and variations such as using mass and molar mass.
So before we start, let's write down the two values of R that we're going to be using.
And we need to take note that we have to have Kelvin, absolute temperature, because the gas
laws all rely on the specific amount of kinetic energy that the gasses have.
Let's list our variables here, and in the problem, notice that the temperature is given in Celcius,
so we have to convert by adding 273, and that will give us Kelvin.
So now we can list values for each variable.
We don't know what the volume is, so that is what we will solve for.
The problem gives us atmospheres and so we will use R that has atmospheres in it.
And then the calculated Kelvin is 303 K.
Let's rearrange PV = nRT to solve for volume by dividing both sides by pressure.
The pressure cancels out on the left and we are left with volume = nRT / P.
So let's plug in all the values.
Now moles cancel out top and bottom, atmospheres cancel out, Kelvin cancels out, and we are
left with liters, which is what we want, the volume.
So the answer is 6.3 liters of hydrogen gas.
Okay.
Second problem.
12.2 moles of CH4 gas occupy 432 liters at 688 millimeters of mercury.
What is the temperature of the gas?
I hope you notice we don't have an ideal gas constant R calculated for millimeters of mercury,
you can do one of two things.
You can either calculate a new R for millimeters of mercury or you could convert millimeters
of mercury to atmospheres, and use the R for atmosphere.
So let's convert, setting up a simple proportion because we know that 760 millimeters
of mercury equals one atmosphere.
And so 688 millimeters of mercury over 760 equals X over one atmosphere, that gives us
X equals 0.905 atmospheres that is equal to 688 millimeters of mercury.
So our pressure, we will use, is 0.905 atmospheres.
Volume is given in the problem.
Moles is given in the problem.
Now that we have atmospheres, we are going to use the R for atmospheres.
Temperature is the unknown, and so that is what we will solve for.
So if we divide both sides of the equation by nR, nR cancels on the right, and so we
have PV / nR = T. So let's go over here so we have a little more room.
So T = PV/NR, and so we'll just fill in all the values
and we can see that atmospheres top and bottom cancel, liters top and bottom cancel,
moles top and bottom cancel and we are left with Kelvin, the unit wanted.
391 Kelvin is the answer.
Problem number 3), 66.8 grams of an unknown gas at 123.0 kilopascals occupies 57.1 liters
at a temperature of 255K.
The question is asking for the molar mass of the gas, and we are also given the mass,
so what can we do to find the molar mass.
Mass divided by moles is molar mass.
So what we need to find is the moles, since we're given the mass, therefore once we find moles,
we can plug it in and find the molar mass.
So let's list our values given in the problem.
So now we have kilopascals as our unit of pressure, so we're using the value of R that
has kilopascals in it, 8.314 kilopascals liters per Kelvin mole, and temperature is given as 255 K.
So rearranging PV = nRT, we get PV/RT = n. And so plugging in our values to find n,
we get 3.31 moles and so what we're going to do next is simply divide the amount
of mass by the amount of moles that we found, 3.31 moles, and that gives us grams/moles,
which is the molar mass, in this case,
20.18 grams/mole.
Which just happens to be,
if you look on the periodic table, yes, you're right.
It is neon.
Problem number 4), gas with a molar mass of 26.0 grams per mole is at 1.8 atmospheres
and 338 K and occupies 19.0 liters.
What is the mass of the gas?
Ok, so again, we're dealing with mass and molar mass.
Let's see what we can do with this.
We're given the molar mass and we're asked for the mass of the gas.
So molar mass is grams over moles, and so if we multiply by the amount of moles that
we find, we will get grams, which is the mass.
So we need to find moles, let's fill in our values.
We have atmospheres, so we're going to use the R for atmospheres and temperature is 338K.
So again, as before, PV/RT gives us the number of moles.
So we plug in our values of PV and RT.
Now we have the moles and so we also have our molar mass.
So when we multiply the molar mass
x the number of calculated moles, that will give
us the mass of the gas.
26 grams per mole x 1.23 moles gives us
32.0 grams.
This has been a Mad Scientist Production.
See ya!