In this screencast we are going to answer the question to what do you do when you cannot
recall a gas constant R or easily look it up in a table. Which is you still want to
use the ideal gas law. To do some kind of calculation. One way to do this is to convert
from standard conditions. What I mean by that is say you are using the ideal gas law. To
figure out the volume of gas x. At some certain pressure and temperature. But you cannot recall
the gas constant R. If we rewrite this using standard conditions and I will define those
in just a second. We will write Ps and some molar volume Vs. That is going to be equal
to some gas constant R, and the standard temperature. If we divide that first equation by the second
equation. We are left with something of the form that looks like this. What this allows
us to do is we know that for any ideal gas. Given a pressure of 1 atm, and a temperature
of 0 degree Celsius. We will call these standard conditions. It is going to have a molar volume,
of 22.4 L/mol. Now this might also be written as 22.4 meters cubed per kmol, and of course
depending on the units there are other numbers that could be memorized. These numbers might
be a lot easier to memorize then say a gas constant with the appropriate units. So given
this information and the memory of say 22.4 L/mol. We are going to use these numbers at
anytime without ever having to use the gas constant R. Let's provide an example of this.
Let's take a look at an example where we are flowing propane gas into say a grill, and
pardon my horrible drawings. Let's say that gas is at 45 degree Celsius, 2 atm, and we
know that we are flowing at a mass flow rate of 2.2 kg/hr. What we want to know is the
volumetric flow rate. The first place to start is to determine the molar flow rate. So that
we have all the variables, necessary, for our ideal gas law. In this case the molecular
weight of propane is 44 g/mol or 44 kg/kmol. Since we are working with kg an hour we can
go ahead and convert that to kg per kmol. Taking the mass flow rate and dividing by the molecular
weight gives us 0.05 kmol/hr. So now in this case we cannot look up the gas constant R,
but if you recall from above, we know that P over Ps, V over V-bar-s equals nT over Ts, and
we can rearranged this so that we can solve for our volumetric flow rate. So when we do
this it should look like the following, and we can start putting in our variables. We
have 0.05 kmol/hr and that is going to be times the teperature 273 plus 45 degree Celsius.
Is going to give us 318 kelvin. Over the standard temperature, which is 273 Kelvin. Multiply
this by our standard pressure of 1 atm over the pressure that we are interested in at
2 atm, and that number we can memorize is a little easier 22.4 meters cubed STP per
kmol. When we do this calculation we get a V-bar that is equal to 0.652 meters cubed
per hour. If those units are good enough we can stop there or we could convert this to liters,
which would give us 652 L per hour. So hopefully you can see that we never needed the gas constant
R, because we used the standard conditions.