Jump to a key chapter

- We will firstly explore the definition of molar volume of gas
- We will then explore the molar volume of gas equation at room temperature and pressure.
- Then we will explore molar gas volume units.
Then we will explore another type of molar volume of gas equation, which occurs in non-room temperature and pressure.

Finally, we will go through some examples of molar volume of gas calculations.

## Molar Volume of Gas definition

So, what exactly is the molar volume of gas?

So if we break it down, it allows us to see that it is the measurement of the volume of one mole of gas, this can be in room temperature and pressure or non-room temperature and pressure. The room temperature is 298.15 K and the standard pressure is 1 atm.

**The molar volume of gas is the space one mole of gas takes up under specified conditions.**

Let us look at those gas cylinders, they can store many types of gases such as, oxygen, which you may see in hospitals or carbon dioxide cylinders which are used to make soft drinks fizzy. Do you know how to calculate how much one mole of gas may take up in one of these gas cylinders? Read on to find out!

To calculate a temperature from degrees Celsius to Kelvin, we add 273.15 to the degrees Celsius temperature

To explore this further, let us put it into some context. We previously mentioned a bottle of fizzy drink. By using the measurements of the fizzy drink and as long as we have two from the volume, moles or molar volume we can find the answer that we are looking for.

## Molar Volume of Gas formula

**Molar volume is the volume of one mole of gas** under specified conditions (temperature and pressure). Interestingly, all gasses have roughly the same volume under the same conditions. If you are working under so-called standard conditions ( 1 atm pressure and 298.15 K) all gases can be said to take up roughly 24.5 dm^{3}per mole of gas. For all non-standard conditions, you have to actually calculate it!

Memorizing 24.5 dm3 per mole of gas under standard conditions helps to skip a lot of calculations in the A-level exam to save time.

Volume is the amount of space a 3D shape takes up. Again, if we explore this through a fizzy drink bottle, we know that there are varying sizes. For example, 330 ml and 1 l are both examples of volume.

Finally, we have moles. This is the amount of substance. If this needs to be calculated there are a couple of ways to do it for example it can be done by dividing the weight of a substance by the molecular mass of the substance. This will result in the number of moles of the substance. For example, if we had 24g of oxygen, to calculate the moles we would do the mass of 24g divided by the molecular mass which is 32 g/mol for oxygen:

\[moles = mass \div \text{molecular mass}\]

\[\text{moles of oxygen} = 24 \div 32\]

\[\text{moles of oxygen} = 0.75 moles\]

The equation for volume of a gas based the amount of gas and the molar volume is the following:

\[volume = moles \times \text{molar volume}\]

Applying our rough estimation of 24.5 dm^{3} to the previous example of oxygen gas (0.75 mol) gives us the following:

\[V = 24.5 dm^3mol^{-1} \times 0.75 moles = 18.4 dm^3\]

### Molar gas volume unit

So as we explored earlier in the article, we know that at standard temperature and pressure the molar gas volume is around 24.5 dm^{3}. Nevertheless, it is key to remember that at non-standard temperatures this figure can be different depending on the gas.

In our last example, you could see that molar volume has a unit and it is dm^{3} divided by moles, It is pretty intuitive if you consider how we found this number, by measuring how much volume 1 mol of gas takes up. Unsurprisingly in the next section, you will see that by simply following the ideal gas law you get back to the same results for the unit.

### Molar Volume of Gas in non-standard conditions formula

**Ideal gases do not exist, but ****we like to use them for calculations. They ****behave very similarly to most gases under most conditions, and so we can use the handy equation for ideal gases to calculate things about real gases too.** So, what's this handy equation you say? Here it is:

\[ p \times V = n \times R \times T\]

Here you need to know a couple of things about the gas to use it, namely:

p = pressure [Pa].

V = volume [m^{3}]

n = moles [mol]

R = this is the gas constant, which is roughly 8.314 [JK^{-1} mol^{-1}]

T = temperature [K]

As with any equation, you can actually leave out one parameter (this will be the unknown) and still calculate the results. Molar volume is just how much space (V) some mole of gas ( n) takes up. Dividing both sides by n we get an equation for molar volume:

\[\text{molar volume} = V \div n = R \times T \div p\]

**This means that molar volume only depends on 2 conditions temperature and pressure. **It will increase with temperature and decrease with pressure. So if you would like to transport a lot of gas, you would compress it ( pressure up) and cool it down ( temperature down). This is actually how companies transport natural gas with ships, this is called liquified natural gas and is just one of the examples of using chemistry in the real world.

Let us now use this equation to calculate some properties of gases now, and after that, we will get right to measuring stuff!

## Molar Volume of Gas examples

Now that, we have explored how to calculate the molar volume of gas at room temperature and pressure, which are the standard conditions and what equation to use for all other conditions,** let us work through a couple of examples together!**

Let's say we have a flask with a volume of 1500 cm^{3} which contains 5.45 g of an unknown gas. The pressure in the flask was 250 kPa and the temperature was 21 °C.

Calculate the relative molecular mass of the gas.

1. First, we need to re-arrange the equation to calculate the moles:

\[ p \times V = n \times R \times T\]

2. Now we need to calculate what the values would be to input into the equation:

p = 250 kPa which equals to 250,000 Pa

V = 1500 cm^{3} which equals to 0.001500 m^{3}

R = 8.31441 JK^{-1}mol^{-1}

T = 21 + 273.15 = 294.15 K

3. Now let us plug these values into the equation:

\[ 250000Pa \times 0.0015 m^3 \div 8.314 \times 294.15 = n = 0.1533mol\]

4. Finally, to calculate the molecular mass, we use the following equation:

\[\text{molar mass} = M = m\div n = 5.44g \div 0.1533mol = 25.5 gmol^{-1}\]

While there are many possibilities as to what this gas can be and we can't be sure, however, this suspiciously looks like Cl_{2} gas. Let's try to not break it!

As a second example let's take a look at a much more human-friendly gas oxygen:

Calculate the volume occupied by 0.666 mols of oxygen at a pressure of 180 kPa and a temperature of 27 °C

1. We need to re-arrange the equation to calculate the volume, this would be:

\[V=\frac{{n}\times {R}\times {T}} {{p}}\]

2. Now we need to calculate what the values would be to input into the equation:

n = 0.666 moles

R = 8.31441 JK-1 mol-1

T = 273.15 + 27 = 300.15 K

p = 180 kPa which equals to 180,000 kPa

3. Finally, we need to plug all these values into an equation to get our answer:

\[v = 0.666mol \times 8.314 \times 300.15 \div 180000 Pa = 0.00923m^3 = 9.23dm^3\]

As we have mentioned previously most gases act very much like ideal gases and we can use these calculations to learn their volume, pressure or number of moles in them. However, not all gases are like ideal gases and sometimes "they act roughly like ideal gases" just does not cut it. **If you really want to know (as with almost everything) you have to measure it! So let's get measuring!**

## Measuring the molar volume of gases

So far, we have gone through the two equations that we can use to calculate the molar volume of gases, but how can we measure the molar volume of gases? This is what we are going to explore now, and don't worry, it's not too difficult.

We will be using the reaction between ethanoic acid aka. acetic acid and calcium carbonate for this experiment. The full equation for this reaction is:

\[ CaCO_3(s) + 2CH_3COOH(s) \rightarrow CO_2(g) + H_2O(l) + Ca(CH_3COO)_2(aq)\]

Do you see those little subscripts? They will tell you what state the reactant or product is. Solid components are represented with (s) while gases are represented with a (g). Similarly (l) means it is a liquid and (aq) states for aqueous which just means dissolved in water. You can see we are forming gas in this reaction and by controlling how much CaCO_{3} we are using in this reaction, we can control how much gas we are producing. However just producing gas in known quantities is not our goal, we want to measure the volume after the gas is produced. For this we will need a special setup and we are going to take a look at that next.

### Equipment

We need quite a few things, but rest assured your technician or teacher will ensure you have everything and will make sure it is safe. The set-up looks like this:

As you can see in the picture above you will need a lot to build this setup. Specifically you will need the following to measure the volume of a gas:

- Stand and clamp
- Boiling tube
- Water bath for gas collection
- Bung fitted with a delivery tube to fit boiling tube
- Test tube
- Boiling tube
- Balance
- 100cm
^{3}measuring cylinder - 50cm
^{3}measuring cylinder - 1 mol dm
^{-3}ethanoic acid - Powdered calcium carbonate

Now that you have all this let's get to measuring!

### Method

This is what you going to need to do in order to measure the volume:

- Measure 30cm
^{3}of ethanoic acid using a 50cm^{3}measuring cylinder and add this to a test tube. - Now set up the equipment as presented in the diagram above and ensure the bung can fit the boiling tube to ensure no gas escapes.
- Now measure 0.05g of calcium carbonate using the balance and add this to the test tube and weight the test tube (keep a note of this measurement).
- Remove the bung from the boiling tube and quickly add the calcium carbonate and put the bung back on.
- You will now notice bubbles travelling to the 100cm
^{3}measuring cylinder, when this has stopped the reaction is complete. - Looking at the 100cm
^{3}measuring cylinder, we can see how much gas was produced, note this down as well. - Finally, reweight the test tube that had the calcium carbonate.
- Now, repeat the experiment 5 times, increasing the mass of calcium carbonate by 0.05g each time. Ensure you do not use more than 0.4g in a single time.
- After we have got all the figures, we can now analyse our findings.

Starting to get tricky and work-intensive right? Remember how easy it was to use the equation, well that's the reason most people just use ideal gases, they are super convenient! But we are not yet done, we have to analyse the data you just collected!

### Analysis

We have now completed the experiment and have some figures to work with. First, what we are going to do is plot our results on the graph, with the mass of calcium carbonate on the x-axis and the volume of gas collected on the y-axis. From this, we can now explore the different gas volumes with different masses of calcium carbonate.

You can see that it is very close to a straight line, which is great since our ideal gas law equation predicted this to be a straight line. The slope of this line will tell you the volume of gas in dm^{3} produced from a gram of calcium carbonate. If you now multiply this slope by the molar mass of calcium carbonate you can get the molar volume of the gas produced in this reaction (CO2).

I hope you are now confident in your ability to measure the molar volume of gas but if you don't have the equipment at home you can at least estimate it with a handy equation!

## The Molar Volume of a Gas - Key takeaways

- The molar volume of gas is the measurement of the volume of one mole of gas.
- This can be in room temperature and pressure or non-room temperature and pressure.
- We can use the equation volume = moles x molar volume to calculate moles and volume using the molar volume of gas figure in room temperature and pressure.
- In instances where we have non-room temperature and pressure, we use the equation: PV = nRT

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##### Frequently Asked Questions about The Molar Volume of a Gas

What is the formula for calculating the molar volume of gas?

We use the equation:

volume = moles x molar volume

What is the molar volume of a gas?

The value is 24 dm^{3} which is equivalent to 24,000 cm^{3}.

How do you find molar volume without STP?

We use the equation:

PV = nRT

What is an example of molar volume of gas?

Calculate the volume occupied by 0.666 mol of oxygen at a pressure of 180 kPa and a temperature of 27 °C

1. We need to re-arrange the equation to calculate the volume, this would be:

V = nRT ÷ P

2. Now we need to calculate what the values would be to input into the equation:

n = 0.666 moles

R = 8.31441 JK-1 mol-1

T = 273.15 + 27 = 300.15 K

P = 180 kPa which equals to 180,000 kPa

3. Finally, we need to plug all these values into an equation to get our answer:

V = 0.666 x 8.31441 x 300.15 ÷ 180000

V = 0.0092336 m^{3}

How do you calculate the molar volume of a gas?

In non room temperature and atmosphere conditions:

P = nRT ÷ V

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