# What is the Ideal Gas Law?

The Ideal Gas Law is P times V equals n times R times T. P stands for pressure, V stands for volume, N stands for number of moles, in other words, the amount. **Moles are used to measure chemical substances.**

T is the absolute temperature, always in Kelvin, and R is a universal gas constant. R takes different forms depending on the units that are needed. R can look like this, or R can look like this.

**Notice the only difference here is the unit for pressure.** Here, kilopascals are used, and here atmospheres are used. I want to go through an example problem so we can better understand the practical application of this gas law.

The example problem we have is: You have 30.0 liters of nitrogen gas at 373 Kelvin and 203 kilopascals. How many moles of nitrogen gas do you have? We’re looking for moles here. Remember, “moles” is N.

I’m going to put “N?”, because that’s what we’re looking for here. Let’s go ahead and write the equation. PV equals nRT. We’re looking for pressure. We know that pressure is expressed in those units.

Pressure is 203kPa times volume, which is 30.0 liters. We’re looking for moles. We don’t know what that is yet, so we’ll just leave n there. Then, we’re looking for R. We need to know which one to use.

The only difference here is the units. Since we see kilopascals right here, that means we must be looking for that constant right there. We have 8.31 times temperature (373 Kelvin). From here, we just need to divide by this right here.

Because we’re using algebra here, what we do to one side of the equation we also have to do the other. We’ll divide this side by the same thing. All of this right here crosses out, because we’re dividing one thing by the same thing.

The only thing left here is n, which is what we’re looking for. I’ll save you having to go through all the math here, but what you would do is just multiply these two numbers and then divide it by these two numbers that are multiplied by each other.

N equals 1.96. Now, we need to know what unit to use. In this case, we’re using moles, so I’ll just abbreviate it mol. That’s the answer we were looking for. Now, if you’re wondering exactly how we got the units, everything has to cross out for something like this to work.

We see kilopascals here and there, so that can cross out. We see liters here and here, so that can cross out. Right here, we have moles and Kelvin (times 373 Kelvin), so those right there cross out.

If you look at it like this, kPa is over moles-K, and then we’re multiplying like this. Actually, we don’t need that number. We’re just looking at units now. This is what the units actually look like.

That got crossed out from up there. We crossed that out like that, and then we’re just left with moles. That’s where we got the unit here. That’s the answer right there. That’s the practical application.

Because you have this equation now, if any of these are missing (if any of these variables are missing), you can find the missing one as long as you have all the other information.

Now, obviously, the missing information is never going to be this constant, because you already know it, but if you’re wondering what the pressure, the volume, and the number of moles or the temperature is, as long as you know the other three variables, you can figure out that missing variable.