Lesson video

Hello, my name is Mrs. Niven, and today we're going to be talking about gas volumes as part of our unit on calculations involving masses.

Now, you may have some experience of what we talk about today from your previous learning, but today's lesson will help us to not only answer that big question of what are substances made of, but also help us to better appreciate how the properties of a gas can affect how we quantify the amounts of gases and how we talk about them or discuss them in terms of the reactions in which they're involved.

So by the end of today's lesson, you should hopefully feel more confident calculating the mass and the volume of a gas reactant or a gas product given a balanced symbol equation.

Now, throughout the lesson, I will be referring to some keywords, and these include: volume, mole, molar gas volume, stoichiometry, and density.

Now the definitions for these keywords are given in sentence form on the next slide, and you may wish to pause the video here so you can read through them and perhaps make a note of them so you can refer back to those definitions later in the lesson or later on in your learning.

So today's lesson will be broken into two parts.

Firstly, we'll look at the properties of gases and how they affect how we calculate mass and volume, and then we'll move on to look at the density of gases.

So let's get started by looking at that special case of gases.

You may recall that volume is the key word that's used to refer to how much 3D space matter occupies, and that volume tends to be measured in centimetres cubed.

You may also recall that gas particles, by their very nature, occupy the entire volume in which they exist.

Now these volumes or 3D spaces tend to be much larger than the volumes for a solid or a liquid.

And because of this, a gas volume tends to be more frequently measured in decimeters cubed or metres cubed.

So how does this compare to our standard centimetre cubed volume unit? Well, 1000 centimetres cubed is equal to one decimeter cubed is equal to 1000th of a metre cubed.

Now, the volume or 3D space that any gas occupies will depend on several factors, and these include: the number of gas particles that are present and the conditions of temperature and pressure.

Now, the volume occupied by one mole of any gas is referred to as the molar gas volume.

So that's the volume occupied by one mole or 6.

02 x 10 to the 23 particles of that gas.

Now, all gas particles are affected by temperature and pressure.

So it stands to reason then that if the temperature or pressure changes so does a gas' volume.

For instance, at higher temperatures, all particles have more kinetic energy and will occupy more space.

So at a higher temperature, the volume of a gas will also be higher.

Likewise, the property of a gas is that it can be compressed, meaning that its particles can be pushed closer together in this space, which means that the higher the pressure, the lower the volume of a gas.

So let's assume then that we can keep the temperature and pressure constant.

So they aren't changing, the volume isn't increasing or decreasing because of the temperature or pressure.

How might the volume then of a gas be affected if different gases are used? Well, what's important here is that the different sized particles doesn't matter.

The gas will still fill the volume in which it's found.

So really the overriding feature in terms of the the volume or 3D space that a gas is occupying is not the size of the particles but actually the space between those particles.

Let's stop here for a quick check.

Izzy and Alex are discussing the volume of two gases at the room temperature and pressure.

Who do you agree with? Izzy reckons that there's so much space between the gas particles that the particle size won't affect the gases volume.

Alex thinks that larger gas particles will take up more space, and so the volume of the neon will be larger than that for the fluorine.

Well done if you chose Izzy.

You are exactly right.

It's the space between the particles that has the larger effect on the volume of a gas.

So very well done if you got that correct, guys.

Great start.

In fact, Avogadro's law states that when temperature and pressure are the same, equal numbers of moles of gases, or the same number of particles of gases, have the same volume.

So 0.

5 moles of helium will take up the same space as 0.

5 moles of ammonia and 0.

5 moles of oxygen.

So the volume or 3D space of a gas is not affected by the size of those gas particles.

Now, we said earlier that a molar gas volume was the volume or 3D space that one mole of a gas will occupy, but we also said that gases and the volume of a gas is affected by a change in temperature or pressure.

Because of this, molar gas volumes are usually quoted at a particular temperature and pressure.

Now, RTP refers to room temperature and pressure.

That means that we're talking about gases that are at 20 degrees Celsius and under one atmosphere or 101 kilopascals of pressure.

So at RTP, that's room temperature and pressure, the molar gas volume of any gas will be 24 decimeters cubed per mole or 24,000 centimetres cubed per mole.

Which means, at RTP, one mole of helium, oxygen, or ammonia will all have the same volume.

24 decimeters cubed or 24,000 centimetres cubed.

Let's stop here for a quick check.

True or false, at 25 degrees Celsius and one atmosphere of pressure, 0.

5 moles of gaseous water, steam, has a larger volume than 0.

5 moles of hydrogen? Well done if you chose false, but which of these statements best justifies that answer? Well done if you chose B.

Whilst A is a correct statement on its own, we've been told that the temperature is at 25 degrees Celsius and therefore has been kept constant.

So these equal number of moles will have an equal volume.

So well done if you got that correct.

There exists then a mathematical relationship between a gas' total volume, the number of moles or particles of those gas that are present, and the molar gas volume, which is, if you remember, the volume that one mole of a gas occupies at a constant temperature and pressure.

So that relationship is the volume is equal to the moles times that molar gas volume.

So if I have two moles of oxygen at RTP, which is room temperature and pressure, and I want to know its volume, well I have the number of moles provided and it's molar gas volume then, I need to remember RTP.

I have two options.

The first one is decimeters cubes.

So taking the two moles of oxygen, timesing it by 24 decimeters cubed per mole and I get a volume of 48 decimeters cubed.

Alternatively, I could multiply my moles by the centimetres cubed version of the molar gas volume, which is 24,000, which gives me a volume of 48,000 centimetres cubed.

The thing here that we need to remember is that our total volume then, its units, will depend on the molar gas volume value and its units that we are using in this calculation.

So take care that we are always keeping track of those units and the value that we're using for that molar gas volume.

Let's go through another example.

I'd like to know the volume in centimetres cube of 1.

4 moles of gaseous water at room temperature and pressure to two significant figures.

So I'm going to need to use that equation of volume is equal to moles times the molar gas volume.

I know the number of moles, it was given to me in the question, but I don't know which value to use for my molar gas volume so I need to go back through my question for some clues.

Well, this was taking place at room temperature and pressure and I need the volume in centimetres cubed.

Centimetres cubed at RTP, the molar gas volume is 24,000.

So I use that value for my molar gas volume and I get an answer of 33,600 centimetres cubed, but to two significant figures then is 3.

4 x 10 to the 4 centimetres cubed.

So to ensure that I have those two significant figures, I'm giving my answer in standard form.

What I'd like you to do now then is to calculate the volume in decimeter cube of 0.

72 moles of oxygen at RTP to two significant figures.

Feel free to use the calculations that were given on the left to help guide you in your own calculations, but pause the video here and come back when you're ready to check your answer.

Okay, let's see how you got on.

If you've done your calculations correctly, you should have had a final answer of 17 decimeters cubed for your two significant figures.

So very, very well done if you've managed to get that correct.

If you've got a different answer, double check that you've used the correct value for the molar gas volume, but very well done if you got that correct.

Recall that earlier in the lesson we said that at the same temperature and pressure, equivalent moles of gases will occupy the same volume.

So because of that, that stoichiometry, or a molar ratio, of a balanced symbol equation can be used to help us deduce the volume of a reactant or product gas.

And we can do that because at room temperature and pressure, we know that one mole of a gas occupies 24 decimeters cubed or 24,000 centimetres cubed.

So let's look at an example.

I have here the reaction equation for the production of ammonia.

And the stoichiometry or molar ratios are given by those coefficients.

And I know that I have gases presence because of that state symbol that's provided for each of my substances.

So if I wanted to calculate the volume for each of these gases at room temperature and pressure, I simply need to multiply the coefficient for my gas times that value for the molar gas volume.

And for this one, I'm using decimeters cubed.

So for nitrogen, it would be 24 decimeters cubed, hydrogen is 72 decimeters cubed, and the ammonia would take up 48 decimeter cubed of volume.

Now the great thing about the stoichiometry or molar ratio of gases is that they don't change if different volumes are used.

Now, we said earlier that one mole of nitrogen at room temperature and pressure would take up 24 decimeters cubed volume.

But if I had six decimeters cubed of nitrogen instead, I can see that I needed to multiply that molar ratio of one times six to get that volume.

I can do the same then with the other molar ratios for the similar gases.

So for hydrogen, it would now be 18 'cause I multiplied three by six, and two times six then would give me a volume of 12 decimeters cubed now for my ammonia.

Similarly, if I had 200 centimetres cubed for my nitrogen, I would have to multiply that molar ratio by 200, which would give me 600 centimetres cubed of hydrogen and 400 centimetres cubed for the ammonia.

But a word of warning, this shortcut by calculating volumes only works for gases.

If you needed to try and find a mass using molar ratios, you have to use a different process.

So do take care that you are checking those state symbols out before you use this shortcut.

Let's stop for another quick check.

The reaction below is shown to take place at room temperature and pressure.

What volume of oxygen is required for this reaction to take place? Well done if you chose C, 120 decimeters cubed.

We can see here from our coefficients, if we multiply by that molar gas volume of 24 decimeters cubed for RTP, five times 24 gives us 120 for our volume.

So very well done if you got that correct.

Let's move on to the first task of today's lesson.

What I'd like you to do is to use words from the box to complete the summary about the molar gas volumes.

Take care, you will not need to use all the words, so do choose carefully.

Pause the video and come back when you're ready to check your answers.

Okay, let's see how you got on.

If you've chosen correctly, your summary should read like this.

Avogadro's law states that when the temperature and pressure of a gas remain constant, the volume of any gas is the same.

RTP stands for room temperature and pressure, which is 20 degrees Celsius and one atmosphere.

At RTP, molar gas volume is 24 decimeters cubed or 24,000 centimetres cubed.

The relationship between the volume of a gas, the number of particles of gas, and molar gas volume is volume equals moles times Vm or molar gas volume.

Very well done if you managed to get those correct.

If you did miss some of these out or got some incorrectly, please do pause the video here so you can make your corrections or add in any that you missed out so you have this lovely summary ready for you to use to revise later on.

For the next part of this task, I'd like you to calculate the volume of each gas that's described.

Now each of these gases is at room temperature and pressure and I'd like you to give your answers to two significant figures.

So please do pause the video and come back when you're ready to check your work.

Okay, let's see how you got on.

Now, because I've been asked to calculate volume, I need to use that mathematical relationship of volume is equal to the moles times the molar gas volume.

So using that, A should be 32 decimeters cubed, B is 150 decimeters cubed, C is 3.

0 x 10 to the 5 centimetres cubed, and D is 2.

2 x 10 to the 5 centimetres cubed.

If you didn't get those values, do take care that you have double checked you've used the correct molar gas volume for RTP for that particular question.

Remember, if it's decimeter cubed, you're using the value of 24, and if it's centimetres cubed, you should be using the value of 24,000.

Very well done if you managed to get those correct, guys.

Great job.

For the last part of this task, I'd like you to consider the reaction that takes place when zinc nitrate decomposes.

And I'd like you to calculate the volume of gas produced if the reaction takes place at RTP.

So room temperature and pressure.

You may wish to discuss some of your ideas with the people nearest to you, but definitely pause the video and come back when you're ready to check your answer.

Okay, let's see how you got on.

Well, you need to remember when we're looking at a reaction involving gases that one mole of a gas has a volume of 24 decimeters cubed at room temperature and pressure and that the coefficients for a gas in a balanced symbol equation can be read as 'moles of'.

So using that information, I calculate then that 96 decimeters cubed of NO2 was produced and 24 decimeters cubed of oxygen was produced.

But the total volume of gas produced in the reaction, I need to add those values together for a final answer of 120 decimeters cubed produced in the decomposition of zinc nitrate.

So very, very well done if you managed to get that correct, guys.

Great job.

Now that we're feeling a little more comfortable discussing gases, let's move on to talk about their densities.

We said earlier that one mole of a gas will occupy the same volume at room temperature and pressure, but we need to recall that each gas' particles actually have a different relative mass.

For instance, helium has a relative mass of four while chlorine has a relative mass of 71.

Now chemists can use that relationship of the mass in grammes is equal to the relative mass times the number of moles to calculate the mass of a gas sample.

So for the helium, we would take the number of moles times its relative mass, giving us a mass of four grammes, and the chlorine, using the same process, has a mass of 71 grammes.

Now density, if you recall, is simply a measure of how much matter, or the moles number of particles, that are located in a particular volume or 3D space.

So density is equal to the mass divided by the volume of a particular substance.

So if we look again at our balloons that we have here of helium and chlorine, we have the mass that was calculated earlier and the volume's been given to us is 24 decimeters cubed.

So the density for the helium then using this calculation comes out as 0.

167 grammes per decimeters cubed, and the density for chlorine then is calculated to be 2.

96 grammes per decimeter cubed.

So the density of any gas at room temperature and pressure can be calculated using two very simple calculation grids using that relationship of the density is equal to the mass divided by the volume.

Now mass in grammes, we already know, is equal to the moles or the number of particles times the relative mass of each particle.

And the volume then is equal to the moles or number of particles times the molar gas volume at that temperature and pressure.

So if we set up two calculation grids, one is going to be for the mass and the other will be for the volume.

And you'll notice here, I start with moles at the top of my grid and the mass or volume at the bottom of my grid.

So the last label for these grids then is the outstanding or unused variable thus far.

So if I want to find the density of my example here, two moles of carbon dioxide, I have two moles, so I'm going to put those into the mole row for each calculation grid, I can find the relative mass of my carbon dioxide using a periodic table, which is 44, and then when I multiply those values, I get a mass of 88 for this sample.

Going on to look at the volume then, the RTP or room temperature and pressure value that I need for the molar gas volume is 24, and the volume then when I multiply those together comes out at 48.

Using those values then for the mass and the volume, I simply pop those into my density equation and I come up with a value of 1.

833.

So to two significant figures, the density of two moles of carbon dioxide is 1.

8 grammes per decimeter cubed.

Let's go through another example.

At RTP, what is the density in grammes per centimetre cubed of 0.

50 of moles of oxygen gas? So I know I need three different equations.

I need to calculate the density and then I need the two equations for the mass and the volume that make up that density relationship.

I need then two calculation grids, one for the mass and one for the volume.

Now the moles was 0.

50 for the oxygen and the relative mass for oxygen then is found using a periodic table to be 32.

Now, when I multiply those together then, I get a mass of 16.

Moving on to the volume calculation grid, the molar gas volume value that I need is dictated by the units for volume for density.

Because that's centimetres cubed, I need to use the value of 24,000.

When I multiply that by the number of moles, I get a volume of 12,000.

Putting the values for mass and volume into my density relationship, I get a final answer to two significant figures of 0.

0013 grammes per centimetre cubed.

What I'd like you to do now then is to calculate at RTP what the density is in grammes per decimeter cubed of four moles of gaseous water.

Use the calculations on the left as a guide, but pause the video now and come back when you're ready to check your answer.

Okay.

If you've calculated out your mass in volumes correctly, you should have had a final answer of 0.

75 grammes per decimeter cube.

So very well done if you managed to get that correct.

If not, I would recommend you first look to make sure that you have used the correct value for the molar gas volume or Vm.

Because it's in decimeters cubed, we need to be using 24 here, but very well done if you got that correct.

Now you may be sat there wondering what is the point of calculating the density of these different gases.

And there is actually a bit of a use to it.

If you recall that under the same conditions, so that's the same temperature and pressure, one mole of any gas will occupy the same volume.

But those balloons that are filled with the same amounts, so one mole of different gases, would float or sink relative to the density of their surroundings.

So let's take a closer look at this.

At room temperature and pressure, the density of air is 1.

21 grammes per decimeter cubed.

Any gas with a density less than air would appear to float in that environment and any gas that has a density more than air would sink in that environment.

So if you had a balloon that was filled with one mole of different gases, if it's gas density was less than that of air, it would float to the ceiling, and if the density of that gas was more than air, it would sink to the floor.

So let's look at an example.

Now, we said a little bit ago that at room temperature and pressure, the density of air is 1.

21 grammes per decimeter cubed.

And earlier in the lesson, we looked at finding the density of helium was 0.

167 grammes per decimeter cubed.

And we compare those two densities, we can see that helium's density is less than that of air.

And because of that, the balloon filled with helium should float in that environment.

If we look at chlorine, its density was 2.

96 grammes per decimeter cubed.

And when we compare that to the density of air, it's actually greater than the density of air.

And because of that, a balloon filled with the same amount of chlorine, so one mole of chlorine, would sink in air.

So if we had these two balloons held, say at waist height, in a room and let go of them, the helium would appear to float, and then the chlorine balloon would appear to sink.

Let's stop here for a quick check.

I'd like you to decide if each of these balloons would float or sink in air.

Now each balloon is at room temperature and pressure, and as a reminder, the density of air at room temperature and pressure is 0.

00121 grammes per centimetre cubed.

You may wish to pause the video here so you can discuss your ideas with the people nearest you and come back when you're ready to check your answers.

Okay, let's see how you got on.

So the balloon filled with carbon dioxide would sink because its density is greater than that of air, but a balloon filled with carbon monoxide would float because its density is less than that of air.

And if we look at a balloon that is filled with steam, it would also float because its density is also less than air.

So very well done if you got those correct, guys.

Great job.

Time to move on to the last tasks for today's lesson.

In this first part, I'd like you to calculate the density for each gas described below to three significant figures.

And a reminder that all the gases described are at room temperature and pressure.

So this is going to take a little bit of time.

Definitely pause the video and come back when you're ready to check your answers.

Okay, let's see how you got on.

So for part A, I needed to find the density of 0.

50 moles of neon in grammes per decimeter cubed.

So I know I need to use my three equations, one for density, mass in grammes, and for volume.

And if I use my calculation grid as shown earlier in the lesson, my final answer should come out to be 0.

833 grammes per decimeter cubed.

If you didn't get that answer, definitely pause the video and double check your calculation grids, and I would always start with that molar gas volume value that you used.

Was it appropriate? Does it match the volume units? Because we needed our density in decimeter cubed, we needed to use that molar gas volume of 24 here.

So very well done if you started off getting this one correct, guys.

Great job.

And for question 1-B then, using that same process should give you an answer of 1.

52 grammes per decimeter cubed.

Very well done if you got that correct.

For letter C, 3.

75 moles of ammonia, I'm going to use the same process, but because I need to do this in centimetres cu, sorry, grammes per centimetre cubed, my molar gas volume is 24,000.

That gives me an answer of 7.

08 x 10 to the -4 grammes per centimetre cubed.

If you chose not to give your answer in standard form, that's absolutely fine.

I have given it there as well for you to double check you got that correct.

I would always recommend using standard form though because it's so easy to not put the correct number of zeros between your decimal point and that first value of the seven for instance.

So just be careful if you're not going to use standard form, but well done if you got that correct.

And finally, in part D, using that same process as before, you should have had a final answer of 2.

67 x 10 to the -3 grammes per centimetre cubed.

Very, very well done if you managed to carry out these calculations correctly, guys.

I'm so impressed.

For the next part of this task, I'd like us to consider four balloons that have been filled with 1.

0 moles of a different gas and left in that room.

So we have here then the four balloons that are labelled with their densities, and a reminder that at room temperature and pressure, air has a density of 1.

21 grammes per decimeter cubed.

In considering all of this, I'd like you to tell me which balloons would you expect to sink and which would you expect to float and to explain your answers.

So I'm looking for a because clause.

And finally, I'd like you also to tell me which balloon do you think might float the fastest and which might sink the fastest, and as well, explain your answers for me.

So you may wish to pause the video whilst you discuss your ideas with the people nearest you and come back when you're ready to check your work.

Okay, let's see how you got on.

So for this first part, I asked you to tell me which balloons you thought would sink and which would float and to explain why.

And as a reminder, we said that if the density of the gas was less than the density of the air, it should float, and if the density of the gas was more than the density of air, it should sink.

And using that information then, the balloons filled with methane and nitrogen would float because their densities are less than 1.

21 grammes per decimeter cubed.

And then the balloons filled with N2O and CO2 would sink because their densities are more than 1.

21 grammes per decimeter cubed.

So firstly, well done if you manage to choose the balloons that would sink or float and very well done if you managed to include that because clause to explain your answers.

Great job, guys.

And finally then I asked you to tell me which balloon you thought would float the fastest and which would sink the fastest, but also to explain your answers with that because clause.

So how did you get on? Well, I would've said that the balloon filled with methane would float the fastest because it has the lowest density of those four balloons and it's still lower in density than air.

The balloon filled with N2O should sink the fastest because it has the highest density that was more than the density of air.

So very well done if you managed to choose the correct balloon that would float or sink the fastest, and incredibly well done if you supported your answer with that because clause.

Great job, guys.

Wow, we've gone through a lot of new information in today's lesson, so let's just take moment to summarise what we've learned.

Well, we've learned that equal moles of gases occupy the same volume if the temperature and pressure remain constant.

And that room temperature and pressure is shortened to RTP and refers specifically to a temperature of 20 degrees Celsius and a pressure of one atmosphere.

And that the molar volume or the volume of one mole of any gas at room temperature and pressure is 24 decimeter cubed per mole or 24,000 centimetres per mole.

And that if the moles of gas are known, it's density can be calculated using a molar gas volume and that relationship: the mass in grammes is equal to the relative mass times the number of moles that are present.

I hope you've had a good time learning with me today.

I've certainly had a good time learning with you, and I hope to see you again soon.

Bye for now.