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Ideal Gas Law - Concept
M.Ed., Columbia Teachers College
Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.
The Ideal Gas Law mathematically relates the pressure, volume, amount and temperature of a gas with the equation pressure x volume = moles x ideal gas constant x temperature; PV=nRT. The Ideal Gas Law is ideal because it ignores interactions between the gas particles in order to simplify the equation. There is also a Real Gas Law which is much more complicated and produces a result which, under most circumstances, is almost identical to that predicted by the Ideal Gas Law.
Alright. So we're going to talk about the ideal gas law and one word that might stick out to you is the word ideal. So we're going to talk about ideal gases versus real gases and gases in real life.
So ideally, when thinking about this and the kinetic molcular theory, there are two things postulates within that kinetic molcular theory that were kind of iffy or untrue. One of them was that gas particles have virtually no volume and we're going to basically have calculations which they do volume of the gas particles does not count. They're negligible. We know that's not true. Gas particles do have some sort of volume. We know it's really small and we know that a ga- real gas particles actually do contain some sort of volume. And also ideally in an ideal world, we're saying that gas particles have no intermolcular forces meaning that they don't attract or repel from each other and we know that's not true. All gas particles and all molcules actually have some sort of way that they are attracted or repelling from each other called IMFs. So really that's not really true. But in most cases, real gases actually behave extremely similarly to ideal gases and so we can actually use this ideal gas conditions in when we're making our calculations, they're pretty accurate. However, the only time that they are not accurate is when we're dealing with high pressure situations or low temperature situations. The reason high pressure situations are different is because most when you have high pressure, those gas particles they're being pushed together. And these intermolcular forces are going to start playing a major role. Also when you're dealing with low temperatures, those gas particles will start slowing down and they will start containing some sort of volume that is negligible and these IMFs will start playing a part too. So in these two scenarios, we can't use the ideal conditions. Otherwise, we can use them all the time which we are going to start doing.
Alright. So let's compare, let's say you have a gas. Using our combined gas law, let's say you have a gas and you don't have anything to compare it to. We can compare, we can always compare the gas law, any gas to its conditions at STP or standard temperature and pressure. And don't forget these conditions are one atmosphere or 101.3 kilo pascals or 760 milimetres of mercury for your pressure. Our molar volume will always be no matter what gas we're talking about 22.4 litres and our temperature will always be 273 kelvin or 0 degrees celsius. But we like things in kelvin because it's always positive that way.
So if we were to replace one of these guys, one of these pressure times volume over temperature with our conditions at STP, we get a certain number. And depending on our pressure, whatever unit of pressure we're talking about, we get different numbers. So we get, if we're using atmospheres we get 0.0821, if we're dealing with kilo pascals we get 8.314, if we're dealing with milimetres of mercury we get 62.4. And we're going to, this is always, will always be the case. We're just going ot make it a constant. And we're going to use the letter r to denote that constant. So, when we're dealing with, I'm going to grab a pen. When we're dealing with the combined gas law we want to actually compare something to a situation in a at STP, I can just replace this pv over t with the letter r. So in this case I'm going to say p1 times v1 over t1 equals r. Okay. Great. But let's say we're talking about one mol of gas. We're talking about one mol of gas, this all works fine because at one mol of gas, our volume is 22.4 litres.
Great. But what if we're talking about two mols of gas? We're going to have to multiply this number by two because we're multiplying that 22.4 litres by 2. Let's say we're talking about 1000 litres. 1000 mols. We have to multiply this whole thing by 1000. If we're talking about 0.55 mols, multiply this whole thing by 0.55. So multiply by the number of mols that we have in our sample.
So we're going to say the letter n denotes the number of mols we have in our sample. Okay, so if I rearrange this to make it much more easy to, easy to write down or remember, I'm going to rearrange this and bring the t over. So I'm going to say pv=nrt. Some people call it pivnert to remember the ideal gas law. So this combination of things is the ideal gas law. It's basically just the combined gas law we've rearranged using conditions at STP for r. Okay. So this actually uses, the ideal gas law uses all four variables. We have volume, pressure, number of mols and temperature. This number of mols, it's the first time it's been introduced. The other gas laws don't have the number of mols within them. So this actually is very very useful. Let's go over here and do a problem with them.
Alright, so our problem is what is the pressure in atmospheres of a 0.108 mol sample of helium gas at a temperature of 20 degrees celsius if its volume is 0.505 litres. Okay. I know right away this is an ideal gas law problem. how do I know that? Well, my problem had a number of mols in it. Now, remember I told you ideal gas laws, the only gas laws that actually can contain the number of mols within it, within it. So, I know when I'm given a mol sample or I'm asked about the number of mols, I know I'm going to always be using the ideal gas law.
Okay. So let's pull everything out. We have ideal gas law just to rewrite is pv=nrt. So our pressure in this case, we're looking for. We're looking for the pressure. So I'm going to say pressure is our variable. Our volume in this case is 0.505 litres. The number of mols we found is 0.108 mols and the r that we're going to use three r's to choose from. We're going to use r dealing with atmospheres. So pressure is wanted in the unit of atmospheres. So where r unit of atmosphere is this point 0821. And then our temperature in this case is 20 degrees celsius. Don't forget we always want things in kelvin. So we're going to add 273 to make it 293 kelvin. So if we multiplied all these together and divide by 5, sorry, 0.505 to isolate our pressure value, we're actually going to get 5.14 atmospheres. So we've just found out using ideal gas law that our new pressure or that our pressure in this scenario is 5.14 atmospheres. Awesome. Great. So this actually can be used, all you have to do is that make, you have to make sure that you get all your variables out and just plug this within this gas, within this ideal gas law.
But there's also lots of fun things you can do with the ideal gas law and one of them is to find the density of the gas. So we know density is mass over volume or grams over litres. And actually, if you rearranged the ideal gas law you're going to get this to find density. Let's actually do that together. Let's derive this. Alright. So we know what density is. So we know our mol grams per mol is equal to our molar mss. And I'm going to just substitute m m as molar mass because it's not milimetres. I'm just going to say, shorten it instead of having to write molar mass. Okay.
I'm going to isolate mols because in my ideal gas law, I have mols. So I'm going to isolate mols to just figure out what it solves for. So I'm going to multiply both sides by mol. So we know our grams equals molar mass times mol. But I want to isolate that mol so I'm going to divide by molar mass on both sides to cancel. So when I, our mol equals grams per molar mass. So in our ideal gas law, I'm going to substitute this n for grams per molar mass. And I have pressure times volume equals grams per molar mass times rt. Okay.
I want to get it so it's grams per litre, because that's our density value. So I'm going to mul- I'm going to divide both sides by rt to isolate the grams. So those cross out and I'll have pv over rt equals grams per molar mass. Okay. So now I want to get grams by itself so I'm going to multiply both sides by molar mass. So now my new formula is molar mass times pressure times volume over rt is equal to grams. And then I want to bring out my volume. Please don't forget my volume is measured in litres. So I want to divide both sides by volume, divide both sides by volume and I get molar mass. I can cross out my volume. Molar mass times pressure over rt equals grams over volume which is what I have written here which in other words is grams over litres. The new volume's in litres, which is our density. Yay. We found density using pv=nrt or ideal gas law.
So, there's lots of fun things you can do to find the ideal gas with the ideal gas law. Lots of fun different things you can do density being one of them and the ideal gas law is used in a lot of different ways.
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