Lesson video

Hello, my name's Mrs. Navin, 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.

Now, today's lesson is broken into two parts.

Firstly, we'll look at how the properties of gases affect how we calculate its mass and volume, and then we'll look at how we can use that information to discuss the density of gases.

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

You may recall that volume is the keyword 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, 1,000 centimetres cubed is equal to one decimeter cubed, is equal to 1,000th 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 times 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's 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 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 gas's 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.

A 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 kilo pascals 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 that 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's 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 cubed.

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 cubed 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 a answer of 33,600 centimetres cubed, but to two significant figures then is 3.

4 times 10 to the four 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 decimeters cubed 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 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 centimetres 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 in 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 times 10 to the five centimetres cubed.

And D is 2.

2 times 10 to the five 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 decimeters 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 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 centimetres 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 talking about gases and their volumes, let's move on to discuss their densities.

We said earlier that at a constant temperature and pressure, one mole of any gas will occupy the same volume.

But we need to remember that each gas's particles will have a different relative mass.

For instance, helium has a relative mass of 4.

0, and chlorine has a relative mass of 71.

0.

But chemists can use this 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, helium would have a mass of 4.

0 grammes, and the chlorine would have a mass of 71.

0 grammes.

And you may recall that the density of a substance is simply a measure of how much matter or how many particles or moles of particles are located in a particular volume or 3D space.

So, we could represent density mathematically as being the mass divided by its volume.

So, if we go back to those balloons we had of helium and chlorine, we already have the mass calculated for each of these gases, and the volume has been provided as being 24 decimeters cubed for each.

So, using that mathematical relationship of density equals mass divided by volume, the helium then has a density of 0.

167 grammes per decimeters cubed, while the chlorine has a density of 2.

96 grammes per decimeter cubed.

That means then that we can easily calculate the density of any gas at room temperature and pressure by using two very simple calculation grids.

Because density equals mass divided by volume, for the mass, we could use that relationship of the mass and grammes is equal to the number of moles times the relative mass of that substance.

And the volume then has the mathematical relationship of the volume is equal to the number of particles times the molar gas volume.

So, if we set out a calculation grid here where we have one for the mass and one for the volume, and you'll see that for the top of each of these grids, we have the number of moles, and the bottom of each of these grids, we have either the mass or the volume, which is what we need in order to calculate density.

That middle row then is occupied by the label for the missing variable, either the relative mass or the molar gas volume.

So, if I want to try and find the density for two moles of carbon dioxide gas, I'm going to put 2.

0 in for the moles of my gas into each calculation grid.

The relative mass for carbon dioxide then is calculated using a periodic table and found to be 44.

0.

If I multiply down then, my calculation grid for the mass, it comes out as 88 grammes.

The molar gas volume number or value that I use is related to RTP, so the room temperature and pressure.

And if you recall, it can be either 24 decimeters cubed or 24,000 centimetres cubed.

I'm going to use 24 here simply 'cause it's a smaller number and I've not been instructed to use a different value.

As before, with my mass calculation grid, I'm going to multiply down to find the volume of this particular sample, and that is 48.

So I now have a value for the mass and a value for the volume.

If I put those numbers then into my density relationship, I get an answer of 1.

833.

But to two significant figures then, my final density for two moles of carbon dioxide is 1.

8 grammes per decimeter cubed.

Let's go through another example.

I'd like to know the density of 0.

5 moles of oxygen at RTP and in grammes per centimetre cubed.

So, I'm going to need to use those three mathematical relationships.

Firstly, the density, 'cause that's what I've been asked to use.

And then a mass relationship and a volume relationship so that I can use those values to calculate the density.

So I have my two calculation grids, one for mass and one for volume.

The moles has been provided in the question as 0.

50.

Then I'm going to use my periodic table to calculate the relative mass of my oxygen, which is 32.

0.

Multiplying down that mass grid, I get a mass of 16 grammes.

For the molar gas volume that I need in order to complete the volume calculation grid, I need to look firstly at clues from the question.

It's at room temperature and pressure, and the density has been instructed to give in grammes per centimetre cubed.

And because of that, I need to use 24,000 for my molar gas volume value.

Multiplying then the number of moles by that molar gas volume, I get a volume of 12,000.

So I now have a mass of 16, a volume of 12,000.

And when I put those into my density relationship, I get a final answer of 0.

0013 grammes per centimetre cubed is the density of 0.

50 moles of oxygen.

What I'd like you to do now then is to calculate at RTP, what the density is in grammes per decimeter cubed of 4.

0 moles of gaseous water.

So, use the calculations on the left as a bit of a guide, but definitely work with people nearest you.

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

Okay, let's see how you got on.

If you've carried out your calculations correctly, you should have got a final answer of 0.

75 grammes per decimeter cubed.

If you didn't get that answer, the first thing I would do is double check you've used the correct value for your molar gas volume.

Remember, because they've asked for the density in grammes per decimeter cubed, you need to be using that value of 24 for the decimeter cube.

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

Great job.

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 or 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 its 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 decimeters 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've 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.

For part A, you were asked to find the density for 0.

50 moles of neon in grammes per decimeter cubed.

Now, because I need to calculate density, I need at least that equation which is equal to mass divided by volume, but I also need those equations to calculate the mass and to calculate the volume.

Using then those calculation grids that were outlined earlier in the lesson, I have a final answer of 0.

842 grammes per decimeter cubed as my density of the neon.

Now, if you didn't get that answer, it's worth pausing the video to double check your calculations.

And I would start by double checking you have the correct value for the molar gas volume, that VM.

Because the density is in grammes per decimeter cubed, that VM or molar gas volume value needs to match the decimeters cubed, which for here is 24.

But if you did get these correct, incredibly well done.

What a great start to this last task, guys.

Great job.

Moving on to part B then, and using the same process as we did for part A, you should have had a final answer of 1.

52 grammes per decimeter cubed.

Very well done if you got that one correct.

Moving on to part C then, in 3.

75 moles of ammonia in grammes per centimetre cubed, I'm using the exact same process as previously, but because I have to give my density in grammes centimetre cubed, my molar gas volume changes from 24 to 24,000, and that gives me a final answer then of 7.

08 times 10 to the minus four grammes per centimetre cubed.

Now, I've given my answer in standard form, but I've also shown it to you in nonstandard form.

Either is okay, as long as you're still showing it to three significant figures, but I would definitely recommend standard form for this because it's far easier to keep track of the number of significant figures, and it is very easy to accidentally miss out or put in extra zeros between that decimal place in your first significant figure.

So, do take care in providing your answers in this way.

But well done if you've got the calculations correct, guys.

Great job.

And for D then, using that same process as we used for part C, you should have a final answer of 2.

67 times 10 to the minus three grammes per centimetre cubed.

Very, very well done in these calculations, guys.

You've just been supremely well done.

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 decimeters 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 claws.

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 a moment to summarise what we've learned.

Well, we've learned that equal moles of gas will 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 20 degrees Celsius and a pressure of one atmosphere.

And that the molar gas volume or the mole of one mole of any gas at RTP is 24 decimeters cubed per mole or 24,000 centimetres cubed per mole.

And finally, we've learned that if the moles of the gas are known, its density can be calculated using a molar gas volume and that relationship of the massing grammes is equal to the relative mass times the number of moles in that sample.

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

I certainly had a good time learning with you.

And I hope to see you again soon.

Bye for now.