7.2 Ideal gas laws | Ideal gases

Boyle's law: Pressure and volume of an enclosed gas (ESBNW)

If you have ever tried to force in the plunger of a syringe or a bicycle pump while sealing the opening with your finger, you will have seen Boyle's Law in action! The following experiment will allow you to see this law in action.

Boyle's Law: The pressure of a fixed quantity of gas is inversely proportional to the volume it occupies so long as the temperature remains constant.

An informal experiment is included here. This is to verify Boyle's law and to verify Charles' law. This experiment is split into two parts and each part is given at the relevant section of the book. For the experiment on Boyle's law you will need a pressure gauge, $\text{10}$ $\text{ml}$ syringe, $\text{3}$ $\text{cm}$ silicon tubing and water bowl or Boyle's law apparatus. Learners will plot a graph of their results and use these to determine if Boyle's law has been verified. They should get a curved line for plotting pressure ($x$-axis) vs. volume ($y$-axis) and a straight line when they plot pressure against the inverse of volume ($\frac{\text{1}}{V}$).

Boyle's law

Aim

To verify Boyle's law.

Apparatus

Boyle's law apparatus (a syringe or bicycle pump attached to a pressure gauge.)

Method

Use the pump to fill the syringe (or glass tube) until the pressure gauge reads the maximum value. Note the volume and the pressure reading.
Slowly release some of the air until the pressure has dropped by about $\text{20}$ units (the units will depend on what your pressure gauge measures, e.g. $\text{kPa}$).
Let the system stabilise for about $\text{2}$ $\text{min}$ and then read the volume.
Repeat the above two steps until you have six pairs of pressure-volume readings.

Results

Record your results in the following table. Remember that your pressure and volume units will be determined by the apparatus you are using.

Pressure	Volume

What happens to the volume as the pressure decreases?

Plot your results as a graph of pressure versus volume (in other words plot pressure on the x-axis and volume on the y-axis). Pressure is the independent variable, which we are changing to see what happens to volume.

Plot your results as a graph of pressure versus the inverse of volume (in other words $\frac{\text{1}}{V}$, so for each volume reading you will work out the value of $\text{1}$ divided by the volume reading).

What do you notice about each graph?

Conclusion

If the volume of the gas decreases, the pressure of the gas increases. If the volume of the gas increases, the pressure decreases. These results support Boyle's law.

In the above experiment, the volume of the gas decreased when the pressure increased, and the volume increased when the pressure decreased. This is called an inverse relationship (or more-less relationship). The inverse relationship between pressure and volume is shown in Figure 7.3.

Figure 7.3: Graph showing the inverse relationship between pressure and volume.

Can you use the kinetic theory of gases to explain this inverse relationship between the pressure and volume of a gas? Let's think about it. If you decrease the volume of a gas, this means that the same number of gas particles are now going to come into contact with each other and with the sides of the container much more often. But we said that pressure is a measure of the number of collisions of gas particles with each other and with the sides of the container they are in. So, if the volume decreases, the number of collisions increases and so the pressure will naturally increase. The opposite is true if the volume of the gas is increased. Now, the gas particles collide less frequently and the pressure will decrease.

Robert Boyle is the scientist who is credited with discovering that the pressure and volume of a sample of gas are inversely proportional. This can be seen when a graph of pressure against the inverse of volume is plotted. When the values are plotted, the graph is a straight line. This relationship is shown in Figure 7.4.

Figure 7.4: The graph of pressure plotted against the inverse of volume, produces a straight line. This shows that pressure and volume are inversely proportional.

We have just seen that the pressure of a gas is inversely proportional to the volume of the gas, provided the temperature stays the same. We can write this relationship symbolically as:

\[p ∝ \frac{\text{1}}{V}\]

where $∝$ means proportional and we write $\frac{\text{1}}{V}$ to show that the proportionality is inverse .

This equation can also be written as follows:

\[p = \frac{k}{V}\]

where $k$ is a proportionality constant. If we rearrange this equation, we can say that:

\[pV = k\]

This equation means that, assuming the temperature and amount of gas is constant, multiplying any pressure and volume values for a fixed amount of gas will always give the same value ($k$). For example: \begin{align*} p_{1}V_{1} &= k \\ p_{2}V_{2} &= k \end{align*}

where the subscripts $\text{1}$ and $\text{2}$ refer to two pairs of pressure and volume readings for the same mass of any gas at the same temperature.

From this, we can then say that:

\[\boxed{p_{1}V_{1} = p_{2}V_{2}}\]

And we can also generalise this to say that: \begin{align*} p_{1}V_{1} &= p_{2}V_{2} \\ &= p_{3}V_{3} \\ &= p_{n}V_{n} \end{align*}

In other words we can use any two pairs of readings, it does not have to be the first and second readings, it can be the first and third readings or the second and fifth readings.

If you work out the value of $k$ for any pair of pressure readings from the experiment above and then work out $k$ for any other pair of pressure readings you should find they are the same.

For example if you have: \begin{align*} p_{1} & = \text{1}\text{ atm}\\ p_{2} & = \text{2}\text{ atm}\\ V_{1} & = \text{4}\text{ cm$^{3}$}\\ V_{2} & = \text{2}\text{ cm$^{3}$} \end{align*} then using the first pressure and volume gives: \begin{align*} k &= p_{1}V_{1} \\ k & = (\text{1}\text{ atm})(\text{4}\text{ cm$^{3}$}) \\ & = \text{4}\text{ atm·cm$^{3}$} \end{align*} and using the second pressure and volume gives: \begin{align*} k &= p_{2}V_{2} \\ & = (\text{2}\text{ atm})(\text{2}\text{ cm$^{3}$}) \\ & = \text{4}\text{ atm·cm$^{3}$} \end{align*}

Remember that Boyle's Law requires two conditions. First, the amount of gas must stay constant. If you let a little of the air escape from the container in which it is enclosed, the pressure of the gas will decrease along with the volume, and the inverse proportion relationship is broken. Second, the temperature must stay constant. Cooling or heating matter generally causes it to contract or expand, or the pressure to decrease or increase. In the experiment, if we were to heat up the gas, it would expand and require you to apply a greater force to keep the plunger at a given position. Again, the proportionality would be broken.

Did you know that the mechanisms involved in breathing also relate to Boyle's Law? Just below the lungs is a muscle called the diaphragm. When a person breathes in, the diaphragm moves down and becomes more “flattened” so that the volume of the lungs can increase. When the lung volume increases, the pressure in the lungs decreases (Boyle's law). Since air always moves from areas of high pressure to areas of lower pressure, air will now be drawn into the lungs because the air pressure outside the body is higher than the pressure in the lungs. The opposite process happens when a person breathes out. Now, the diaphragm moves upwards and causes the volume of the lungs to decrease. The pressure in the lungs will increase, and the air that was in the lungs will be forced out towards the lower air pressure outside the body.

Simulation: 23VW

Before we look at some calculations using Boyle's law we first need to know some different units for volume and pressure. Volume units should be familiar to you from earlier grades and will usually be $\text{cm$^{3}$}$ or $\text{dm$^{3}$}$ or $\text{m$^{3}$}$ or $\text{L}$. The SI unit for volume is $\text{m$^{3}$}$.

Pressure is measured in several different units. We can measure pressure in millimetres of mercury ($\text{mm Hg}$) or pascals ($\text{Pa}$) or atmospheres ($\text{atm}$). The SI unit for pressure is $\text{Pa}$. See Table 7.1 for converting between units.

Worked example 1: Boyle's law

A sample of helium occupies a volume of $\text{160}$ $\text{cm$^{3}$}$ at $\text{100}$ $\text{kPa}$ and $\text{25}$ $\text{℃}$. What volume will it occupy if the pressure is adjusted to $\text{80}$ $\text{kPa}$ and the temperature remains unchanged?

Write down all the information that you know about the gas.

\begin{align*} p_{1} &= \text{100}\text{ kPa}\\ p_{2} &= \text{80}\text{ kPa} \\ V_{1} &= \text{160}\text{ cm$^{3}$} \\ V_{2} &= ? \end{align*}

Use an appropriate gas law equation to calculate the unknown variable.

Because the temperature of the gas stays the same, the following equation can be used:

\[p_{2}V_{2} = p_{1}V_{1}\]

Substitute the known values into the equation, making sure that the units for each variable are the same. Calculate the unknown variable.

\begin{align*} (\text{80})V_{2} &= (\text{100})(\text{160}) \\ (\text{80})V_{2} &= \text{16 000} \\ V_{2} &= \text{200}\text{ cm$^{3}$} \end{align*}

The volume occupied by the gas at a pressure of $\text{80}$ $\text{kPa}$ is $\text{200}$ $\text{cm$^{3}$}$

Check your answer

Boyle's law states that the pressure is inversely proportional to the volume. Since the pressure decreased the volume must increase. Our answer for the volume is greater than the initial volume and so our answer is reasonable.

It is not necessary to convert to SI units for Boyle's law. Changing pressure and volume into different units involves multiplication. If you were to change the units in the above equation, this would involve multiplication on both sides of the equation, and so the conversions cancel each other out. However, although SI units don't have to be used, you must make sure that for each variable you use the same units throughout the equation. This is not true for some of the calculations we will do at a later stage, where SI units must be used.

Worked example 2: Boyle's law

The volume of a sample of gas is increased from $\text{2,5}$ $\text{L}$ to $\text{2,8}$ $\text{L}$ while a constant temperature is maintained. What is the final pressure of the gas under these volume conditions, if the initial pressure is $\text{695}$ $\text{Pa}$?

Write down all the information that you know about the gas.

\begin{align*} V_{1} &= \text{2,5}\text{ L} \\ V_{2} &= \text{2,8}\text{ L} \\ p_{1} &= \text{695}\text{ Pa} \\ p_{2} &= ? \end{align*}

Choose a relevant gas law equation to calculate the unknown variable.

The sample of gas is at constant temperature and so we can use Boyle's law: \[p_{2}V_{2} = p_{1}V_{1}\]

Substitute the known values into the equation, making sure that the units for each variable are the same. Calculate the unknown variable.

\begin{align*} (\text{2,8})p_{2} &= (\text{695})(\text{2,5}) \\ (\text{2,8})p_{2} &= \text{1 737,5} \\ p_{2} &= \text{620,5}\text{ kPa} \end{align*}

The pressure of the gas at a volume of $\text{2,8}$ $\text{L}$ is $\text{620,5}$ $\text{kPa}$

Check your answer

Boyle's law states that the pressure is inversely proportional to the volume. Since the volume increased the pressure must decrease. Our answer for the pressure is less than the initial pressure and so our answer is reasonable.

Boyle's law

Textbook Exercise 7.2

An unknown gas has an initial pressure of $\text{150}$ $\text{kPa}$ and a volume of $\text{1}$ $\text{L}$. If the volume is increased to $\text{1,5}$ $\text{L}$, what will the pressure be now?

\begin{align*} p_{2}V_{2} &= p_{1}V_{1}\\ \text{1,5}\text{ L}p_{2} & = (\text{150}\text{ kPa})(\text{1}\text{ L}) \\ p_{2} & = \frac{\text{150}}{\text{1,5}}\\ & = \text{100}\text{ kPa} \end{align*}

A bicycle pump contains $\text{250}$ $\text{cm$^{3}$}$ of air at a pressure of $\text{90}$ $\text{kPa}$. If the air is compressed, the volume is reduced to $\text{200}$ $\text{cm$^{3}$}$. What is the pressure of the air inside the pump?

\begin{align*} p_{2}V_{2} & = p_{1}V_{1}\\ \text{200}\text{ cm$^{3}$}p_{2} & = (\text{90}\text{ kPa})(\text{250}\text{ cm$^{3}$})\\ \text{200}\text{ cm$^{3}$}p_{2} & = \text{22 500}\text{ kPa}\\ p_{2} & = \frac{\text{22 500}}{\text{200}}\\ & = \text{112,5}\text{ kPa} \end{align*}

The air inside a syringe occupies a volume of $\text{10}$ $\text{cm$^{3}$}$ and exerts a pressure of $\text{100}$ $\text{kPa}$. If the end of the syringe is sealed and the plunger is pushed down, the pressure increases to $\text{120}$ $\text{kPa}$. What is the volume of the air in the syringe?

\begin{align*} p_{2}V_{2} & = p_{1}V_{1}\\ \text{120}\text{ kPa}V_{2} & = (\text{100}\text{ kPa})(\text{10}\text{ cm$^{3}$})\\ V_{2} & = \frac{\text{1 000}}{\text{120}}\\ & = \text{8,33}\text{ cm$^{3}$} \end{align*}

During an investigation to find the relationship between the pressure and volume of an enclosed gas at constant temperature, the following results were obtained.

Volume ($\text{dm$^{3}$}$)	$\text{12}$	$\text{16}$	$\text{20}$	$\text{24}$	$\text{28}$	$\text{32}$	$\text{36}$	$\text{40}$
Pressure ($\text{kPa}$)	$\text{400}$	$\text{300}$	$\text{240}$	$\text{200}$	$\text{171}$	$\text{150}$	$\text{133}$	$\text{120}$

Plot a graph of pressure (p) against volume (V). Volume will be on the x-axis and pressure on the y-axis. Describe the relationship that you see.

Plot a graph of p against $\dfrac{1}{V}$. Describe the relationship that you see.

The graph is a straight line. This straight line shows the inverse relation between pressure and volume.

Do your results support Boyle's Law? Explain your answer.

The graph of p against $\frac{1}{V}$ gives a straight line which shows that pressure is inversely proportional to volume. This is the expected result from Boyle's law.

Masoabi and Justine are experimenting with Boyle's law. They both used the same amount of gas. Their data is given in the table below:

	Masoabi		Justine
	Initial	Final	Initial	Final
Temperature ($\text{K}$)	$\text{325}$	$\text{350}$	$\text{325}$	$\text{325}$
Volume $\text{(dm$^{3}$)}$	$\text{1}$	$\text{3}$	$\text{1}$	$\text{3}$
Pressure ($\text{Pa}$)	$\text{650}$	$\text{233}$	$\text{650}$	$\text{217}$

Masoabi and Justine argue about who is correct.

Calculate the final pressure that would be expected using the initial pressure and volume and the final volume.

\begin{align*} p_{2}V_{2} & = p_{1}V_{1} \\ (\text{3})p_{2} & = (\text{650})(\text{1}) \\ \text{3}p_{2} & = \text{650}\\ p_{2} & = \text{217}\text{ Pa} \end{align*}

Who correctly followed Boyle's law and why?

From the above calculation we see that Justine has the correct pressure as predicted by Boyle's law. Masoabi did not get the expected reading.

Justine kept the temperature constant and so has correctly repeated Boyle's law. Masoabi changed the temperature for the second part of the experiment and so did not repeat Boyle's law (she actually repeated another gas law).

Charles' law: Volume and temperature of an enclosed gas (ESBNX)

Charles' law describes the relationship between the volume and temperature of a gas. The law was first published by Joseph Louis Gay-Lussac in 1802, but he referenced unpublished work by Jacques Charles from around 1787. This law states that at constant pressure, the volume of a given mass of an ideal gas increases or decreases by the same factor as its temperature (in Kelvin) increases or decreases. Another way of saying this is that temperature and volume are directly proportional.

Charles' Law: The volume of an enclosed sample of gas is directly proportional to its Kelvin temperature provided the pressure and amount of gas is kept constant.

An informal experiment is included here. This is to verify Boyle's law and to verify Charles' law (see notes under Boyle's law). In this second part of the experiment learners will verify Charles' law. You will need glass bottles, balloons, beakers or pots, water, hot plates. Learners place a balloon over the mouth of the bottle and then place it in a beaker or pot of water and see what happens. They can also place a bottle with a balloon into a freezer and see what happens.

Charles's law