The Ideal Gas Law - Online Tutor, Practice Problems & Exam Prep

Ideal Gases and the Ideal Gas Law

Hey, guys. So throughout our discussion on thermodynamics, we'll often be talking about gases. And specifically, we'll be talking about these things called ideal gases. So in this video, I want to introduce to you what an ideal gas is and the equation that we use for them, which is called the ideal gas law. So let's get started here. What is an ideal gas? Well, the definition I like to use is it's kind of like a simplified perfect gas. The analogy is that when we talked about forces and drew free body diagrams, it didn't matter whether it was a plane or a train or a car or whatever. We drew everything as boxes because it was the simplest model for an object. And similarly, the simplest model for a gas is called an ideal gas. Now it basically just satisfies a couple of conditions here, and I want to go over them real quickly because you may need to know for a conceptual question on a homework or test. So let's get started. The first one is the gas has a low density. So what happens is that the particles in a container are going to be kind of spread out doing their own thing. They're not going to be super, super tightly crammed in together. So generally, what this means is that the pressure is pretty low, but the temperature is pretty high. You won't see any extreme values or anything like that. The second condition is that there are no forces between the gas particles. So in real life, in real gases, you have particles that exert some forces on each other. These are usually, like, electromagnetic forces. And for an ideal gas, we just consider that they don't exist. The third condition is that the particles have zero physical size. And what this means is that we're going to just treat ideal gas particles like points. So you basically just treat them as tiny little balls that are moving. They're very, very tiny. You can treat them as point particles, whereas real gas molecules actually do take up space. But for an ideal gas, we just assume that it's 0. The last one is that the particles are moving in straight lines, and they collide elastically. This is the most important part here. So, basically, these tiny little balls or these tiny little points are moving around. They move in perfectly straight lines until they collide with either the walls of the container or even each other. They can bounce off of each other. However, when they do collide, they collide elastically. And what that means is the energy in the container is conserved. There's no energy loss because these things are bumping around into each other. Alright. So what you really need to know about everyday conditions and ideal gases is that, you know, even under everyday conditions like in our atmosphere and things like that, most real gases already behave very much like ideal gases. So, you know, we can still use this model for an ideal gas even when we talk about real gases like oxygen and nitrogen in our atmosphere. Alright. So let's get to the equation then.

So the equations that govern ideal gases are called the ideal gas law, and it relates 4 related variables. We have pressure, volume, temperature, and moles, which is the amount of substance for an ideal gas. And it actually works for any ideal gas. It doesn't matter what it is. Now there are 2 different versions of this, and you may see both of them. So I want to go over them. The first one is PV=nRT. If you've ever taken an introductory chemistry course or something like that, you may be pretty familiar with this one. The other one is PV=nk_BT, where N is a big N. The difference between them is that they both work for ideal gases, is that you'll use this equation whenever you have the number of moles of substance, and you'll use this one whenever you're asked to find or you're given the number of particles, in a gas. And these things actually have a pretty simple relationship between them. n=N∕N_A , where N_A is Avogadro's number, 6.02×1023.

There are 2 constants that you need to know in both of these equations. The first one is big R, and it's called the universal gas constant. It's sometimes called the ideal gas constant, and it's just this number here, 8.314 joules∕ mol∕ Kelvin. The other one is k_B. This k_B is called Boltzmann's constant, and it has a value 1.38×10−23. Now the last thing here is that we're working with absolute temperatures, not changes in temperature. So the temperature actually must be in Kelvin. That's really, really important. You have to plug in all your temperatures in Kelvin. And the very last thing I want to mention here is that this is something you might see in your problems. So something called STP. It's an acronym that stands for standard temperature and pressure. Basically, it's just a common set of conditions that's very much like what we experience here at Earth's surface, which is that the temperature is 0 degrees Celsius or 273 Kelvin, and the pressure is 1.01×105. This is basically just whenever you see STP, these are your numbers for temperature and pressure. Alright? So let's get to our example here, which is we're going to calculate the volume of exactly one mole of an ideal gas that is at STP. So we're going to use those conditions here. So we have STP, which means the temperature is 0 Celsius or 273 Kelvin. We also have that the pressure is 1.01×105, Pascals. Now if we have one mole of substance, that's n=1, we want to figure out what is the volume. So we have these 4 variables. Remember, they're all related using the ideal gas law. So which version are we going to use? Well, remember, we're given the number of moles of a substance, so that means we're going to use PV=nRT. So what happens here is that there are 5 variables. There are 5 letters in this equation, but one of them is a constant. So there's really only 4 variables you need to know. If you ever have 3, you can always figure out the other one. So we know what the pressure is, we know what the number of moles is, and we have the temperature. Remember, this is just a constant, so we can figure out what this volume is by just rearranging the equation. So what happens is this p goes to the other side like this and your v is going to be nRT∕p. So I'm just going to start plugging in. The 1 mole, we have 8.314, that's the gas constant. 273, that's the temperature, and then 1.01×105 pascals. So when you work this out, what you're going to get is 0.0224, and this is going to be meters cubed. Now there are also a couple of other volume conversions that you may need to know for these types of problems. One of them is that 1 centimeter cube is 1 milliliter, and then sort of if you work this out and you play around with the zeros, you'll figure out that 1 meter cubed is 1,000 liters. So one way you might see this number represented, 0.0224, is when they multiply it by this conversion factor, you're going to multiply by a 1000 liters divided by 1 meter cubed. Meters cubed cancels. You're going to get 22.4 liters, and that's the answer. Either one of these numbers will be the answer. And that's actually a really sort of special number here. So the idea here is that 1 mole of any ideal gas at STP, sort of at these special conditions, has a volume of exactly 22.4 liters. Notice how we didn't talk about what type of gas it is. All we had were these conditions. So what this means here is that any gas at STP, whether you have 1 mole, occupies this amount of space, and this is sometimes called the molar volume of a gas at STP. Alright. So, guys, that's it for this one. Let me know if you have any questions.

3 moles of an ideal gas fill a cubical box with a side length of 30cm. If the temperature of the gas is 20°C, what is the pressure inside the container?

• A. 2.7 × 10⁵ Pa
• B. 0.27 Pa
• C. 1.85 × 10⁴ Pa
• D. 8.12 × 10³ Pa

Hydrogen gas behaves very much like an ideal gas. If you have a sample of Hydrogen gas with a volume of 1000 cm³ at 30°C with a pressure of 1 × 10⁵ Pa, calculate how many hydrogen atoms (particles) there are in the sample.

• A. 2.42 × 10²²
• B. 2.39 × 10²²
• C. 2.39 × 10²⁸

Solving Ideal Gas Problems With Changing States

Hey, guys. So now that we've been introduced to the ideal gas law, in some of the problems you're going to run across, you're going to have to compare an initial and final state of a gas. In the example we're going to work out down here, we have some amount of moles of gas in a container, and we're given what the initial pressure and volume are. Then what happens is we’re going to compress the gas using a piston or something like that, and what we’re trying to figure out is what's the final pressure. So, basically, what happens is we’re going to try to take a gas in its initial state and we’re going to change it. We’re going to figure out which one of the variables changes as a result. Now, in order to do that, we're going to use an approach similar to how we used energy conservation. We know the equation for ideal gases is PV=nRT. Now what happens is this is sort of like sometimes called an equation of state. It's sort of like a snapshot of these 4 variables at a specific moment in time. If you change the gas, however, from initial to final, this PV=nRT has to remain sort of conserved, if you will. So what we're going to do is we’re going to stick subscripts on each one of these letters here. So we have PinitialVinitial=ninitialRTinitial. Now remember, this R is the universal gas constant. Right? It doesn’t actually need a subscript. And then we have PfinalVfinal=nfinalRTfinal. So what happens is these two things are equal to each other. We’re going to set these two equations equal. And in doing so, what happens is that the R is going to cancel. Right? It's just a constant that cancels out from both sides. And this equation has a lot of equal signs. So what we’re going to do is we're going to divide by n's and T's on each side, and what you end up with is this equation over here. PinitialVinitialninitialTinitial=PfinalVfinalnfinalTfinal. So this is the one equation that you need to solve any kind of problem where an ideal gas is changed from one state to another. Now let me show you how to use this using a step-by-step approach that's going to get you the right answer every single time here. So the first step, if we’re just going to use this equation, is to actually write the equation out. We're just going to write this equation every time. It’s always going to work. So the first step here is PinitialVinitial/ninitialTinitial=PfinalVfinal/nfinalTfinal. Notice how, again, the R is missing because it actually gets canceled out when you set these things equal to each other. Alright. So the next thing we have to do is we’re going to have to cancel out the constant variables. So what ends up happening here is that in most of these problems, 2 out of the 4 variables are going to remain constant, which is actually just going to let you cancel even more things inside of this equation. So let's go through the our variables and figure out which ones we can cancel out. Now the first part of the problem tells us that we have 2 moles of an ideal gas. So we know that n equals 2, but we’re also told that no gas can leak out or in. So what that means here is that the change in number of moles is just 0. It's the same amount of gas. So that means that n can just be canceled out. So for instance, if we put 2 for 2 moles inside of the left and right sides of the equation, it’s just going to cancel out. Right? Those things are going to cancel each other out. Alright. So the other thing we can do is we can say that the container is then at constant temperature. So that's the other variable that has remained constant. So we know that T_i is equal to T_f. Whatever that number is, it doesn't really matter that we know it or not because it's just going to get canceled out. So what we're left with is really just these four variables over here. Now we know that the initial pressure is going to change to some final pressure, and we’re told that the initial volume goes from 0.05 to 0.01. So that means here that we have the initial pressure with the initial volume. We have the final volume and we want to calculate what is. So all these variables are going to change. We can't cancel them out. And now the last thing we do is we just go ahead and solve. So what happens here, is we just end up with PinitialVinitial/Vfinal=Pfinal. So we just move this down to the other side. And now we just start plugging in. So the initial pressure is 1 times 10 to the fifth. The initial volume is 0.05, and the final volume is 0.01. When you work this out, what you're going to get is 5 times 10 to the fifth Pascals, and that is the answer. Alright? So that's how to go through these steps here. Now if you think about what's happened here, what we've had is that the initial volume went from 0.05 to 0.01. So what ended up happening was we had the volume that decreased by a factor of 5. And as a result, our pressure went from 1 times 10 to the fifth to 5 times 10 to the fifth. So it increased by a factor of 5. So what we say here is that P and V are sort of indirectly or inversely proportional to each other. And that brings me to this point here, which is that when scientists were studying these gases a hundred years ago, hundreds of years ago, they actually sort of came up with these three relationships that are historically called the gas laws. And they're called Charles or Boyle's, Charles, and Gay Lussac's law. Now the thing is these are actually special cases of this ideal gas law which we now know how to use. So I want to go over them quickly just in case you need to know them. Basically, what they did is they held the number of moles and one of the other variables as fixed, and they found that these relationships, between the gas variables. Now what we actually just saw was Boyle's law. Boyle's law says that if the temperature is constant, basically if there is no change in the temperature of the gas, then what happens is that P is inversely proportional to V. That's what we just saw here. And basically how you get to these equations is you just cancel out the constant variables and what we end up with is this relationship which we just saw. Right? If one goes down by a factor of 5, the other one goes up by a factor of 5, so they’re inversely proportional. Now the other one's called Charles' law and basically if you held the pressure constant, V is directly proportional to T. So if you have n and P that get sort of canceled out like this, then what happens is you just end up with V/T on both sides. Now these are directly proportional even though it’s a ratio. And if you think about it, what's happening is that if this increases by a factor of 5, then this also has to increase by a factor of 5 in order to keep that number, whatever it is, constant. The same works by the way for Gay Lussac's law. If V is constant, basically cancel out these two things here, then you just get P/T and you have the exact same relationship. So you might need to know that, but basically those are just the gas laws. So now that we know how to solve the ideal gas law problems, let me know if you have any questions. That's it for this one.

A balloon contains 3900cm³ of a gas at a pressure of 101 kPa and a temperature of –9°C. If the balloon is warmed such that the temperature rises to 28°C, what volume will the gas occupy? Assume the pressure remains constant.

• A. 0.012 m³
• B. 0.0041 m³
• C. 0.39 m³
• D. 0.0044 m³

Doubling Pressure & Temperature

Hey, everybody. So let's get started with our problem here. We have an ideal gas in a sealed container. We're told what the initial volume, pressure, and temperature are, and we're told that the pressure and the volume are both going to double. Let me just go ahead and draw out a sketch of what's going on here. So imagine that I have this kind of sealed container like this, but I have a sort of plunger that I can move up and down. This thing can sort of go up and down like this. So what happens here is that this plunger, imagine it's kind of like halfway through this container, which means that the gas is sort of contained inside of just this area over here. This is the gas. We have an initial volume, pressure, and temperature. Now, what happens here is I'm going to move the plunger up so that it goes like this. So that's basically the top of the container. So now, all the gas has now effectively sort of doubled in size. Right? The container is now sort of bigger than it was before. This is what's going on in this problem here. We have a final volume, final pressure, and final temperature. What we're trying to find in this first part here is what the final temperature is.

We've got an ideal gas. We're going to change some of the characteristics or variables like pressure and volume. We want to figure out what's the final temperature. In order to do that, we're just going to stick to the steps here. The first step we're going to do is write out our changing ideal gas equation. So I'm going to do p_initialv_initialn_initialT_initial = p_finalv_finaln_finalT_final. What we're looking for is T_final.

Now, we want to cancel out any constant variables because they'll just cancel off from both sides of the equation. Pressure and volume clearly don't remain constant because we're both told that they both double. However, this is a sealed container. So if it's a sealed container, what happens is that nothing is allowed to go in or come out. So, the number of moles, the amount of gas inside the container does actually remain constant. Alright. So now that that step's done, we're going to go ahead and solve for our target variable. T_final equals p_finalv_finalT_initialp_initialv_initial.

Since the pressure and volume both double, I can rewrite the initial pressure of the final pressures as multiples of the initial. It means that p_final, if it's just double the initial pressure, I can write this as 2p_initial. I can do the same exact thing for the volume. The volume also doubles, so v_final is just 2v_initial. Then, T_final just becomes 4T_initial. Whatever you do to one side, you have to do to the other. As both p_initial and v_initial are multiplied by 2, T_final has to increase by 4 times.

The initial temperature, which is 40 degrees Celsius, is first converted to Kelvin (adding 273 to it) giving 313 Kelvin. So, T_final just becomes 4 × 313, yielding a final temperature of 1252 Kelvin.

Now with part b, we are supposed to figure out how many moles of gas there are. Remember the moles of gas is just n. There are different ways I can do this. If I'm looking for n, that was the only variable that I canceled out of my equation. I can just go ahead and use pv = nRT to solve for this. It doesn’t matter whether you use the initials or the finals, as n remains constant. So, your p_initialv_initial divided by RT_initial equals n. The initial pressure is 0.15 atmospheres, converting it with 1 atmosphere = 1.01 × 10⁵ Pascals gives 1.515 × 10⁴ Pascals. The initial volume of 2.8 liters, converted to cubic meters (1 liter = 0.001 cubic meters), gives 0.0028 cubic meters. Working this out gives approximately 0.016 moles, which is your final answer. Let me know if you guys have any questions.