 # Boyle's Law (Mariotte's Law) Chemistry Tutorial

## Key Concepts

• Boyle's Law is also known as Mariotte's Law.
• Boyle's Law (Mariotte's Law) states that
At constant temperature, the volume of a given quantity of gas is inversely proportional to its pressure :

 V ∝ 1   P

where V = volume of the gas
and P = pressure of the gas

So at constant temperature, if the volume of a gas is doubled, its pressure is halved.

• This relationship can be written as an equation if we use a constant of proportionality, that is:
At constant temperature for a given quantity of gas, the product of its volume and its pressure is a constant :

PV = constant

OR

PV = k

where k is a constant
• At constant temperature for a given quantity of gas :

PiVi = PfVf

where
Pi is the initial (original) pressure
Vi is its initial (original) volume
Pf is its final pressure
Vf is its final volume

(a) Pi and Pf must be in the same units of measurement (eg, both in atmospheres or both in kPa)
(b) Vi and Vf must be in the same units of measurement (eg, both in litres or both in millilitres)

• All gases approximate Boyle's Law at high temperatures and low pressures.
(a) A hypothetical gas which obeys Boyle's Law at all temperatures and pressures is called an Ideal Gas.

(b) A Real Gas is one which approaches Boyle's Law behaviour as the temperature is raised or the pressure lowered.

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## Graphical Representations of Boyle's Law

Consider an experiment in which a known amount of hydrogen gas in a syringe has a volume of 23 mL at atmospheric pressure (760 mm Hg or 1 atm or 101.3 kPa).

You then apply an external pressure of 912 mm Hg (1.2 atmospheres or 121.6 kPa) by pressing down on the plunger in the syringe.

The volume of hydrogen gas is then recorded as 19.2 mL.

You continue to apply external pressure by pushing the plunger down further, recording the volume of hydrogen gas as shown in the table below:

Pressure
(mm Hg)*
Volume
(mL)
Trend
760 23 Increasing the pressure applied to the plunger causes a reduction in the gas volume.

Decreasing the applied pressure increases the volume of the gas.

912 19.2
1064 16.4
1216 14.4
1368 12.8
1520 11.5
* A pressure of 760 mm Hg is equal to 1 atmosphere (atm) or 101.3 kilopascals (kPa)

If we plot these points on a graph, the graph looks like the one below:

 volume (mL) Gas Volume versus PressurePressure (mm Hg)

Note that this is not a linear relationship, the line in the graph is curved, it is not a straight line.

But look what happens if we multiply volume and pressure (P × V):

Pressure
(mm Hg)
Volume
(mL)
P × V Trend
760 23 1.75 × 104 P × V is a constant!

For this amount of gas at this temperature:

P × V = 1.75 × 104

912 19.2 1.75 × 104
1064 16.4 1.75 × 104
1216 14.4 1.75 × 104
1368 12.8 1.75 × 104
1520 11.5 1.75 × 104

For a given amount of gas at constant temperature we now we can write the equation:

P × V = constant

If we divide both sides of the equation by P, we get:

 V = constant × 1   P

Recall that the equation for a straight line that runs through the point (0,0) is

y = mx

where m is the slope (or gradient) of the line

Then a graph of V against 1/P, should be a straight line with a slope (or gradient) equal to the value of the constant.

The table below shows what happens if we calculate 1/P for each volume, V, in the experiment above and then graph the results:

Volume
(mL)
Pressure
(mm Hg)
1/Pressure
(1/mm Hg)*
11.5 1520 6.6 × 10-4 As gas volume (V) increases, the value of 1/P increases.

As gas volume (V) decreases, the value of 1/P decreases.

12.8 1368 7.3 × 10-4
14.4 1216 8.2 × 10-4
16.4 1064 9.4 × 10-4
19.2 912 1.1 × 10-3
23 760 1.3 × 10-3

By plotting these points on a graph, we can see that the relationship is linear:

 volume (mL) Gas Volume versus 1/Pressure1/Pressure (1/mm Hg)

We now have a simple method for determining the value of the constant:

Recall that we can calculate the slope (gradient, m) of a straight line using two points on the line
 m = (y2 - y1) (x2 - x1)

Choosing the points (0.00094,16.4) and (0.0013,23)
 m = (23 - 16.4)   (0.0013 - 0.00094) = (6.6)   (0.00036) = 1.8 × 104

and the equation for this straight line is

 V = 1.8 × 104 × 1 P

This equation then allows us to calculate the volume of the gas at any pressure, as long as we use the same amount of gas and keep the temperature the same.

Let us say we have a specific amount of gas and keep the temperature constant, then initially at pressure Pi the gas has a volume of Vi and we know that:

PiVi = constant

If we maintain the same temperature and the same amount of gas, but change the pressure to Pf, then the new gas volume will be Vf, and

PfVf = the same constant

So, as we long as we use the same amount of gas at the same temperature:

PiVi = constant = PfVf

that is:

PiVi = PfVf

This means that if we know the initial conditions (Pi and Vi), and, we know the final pressure (Pf), we can calculate the final volume (Vf):

 Vf = Pi × ViPf

or we can calculate the final pressure (Pf) if we know the final volume (Vf):

 Pf = Pi × ViVf

Similarly, if we know the final conditions (Pf and Vf), and, we know the initial pressure (Pi), we can calculate the initial volume (Vi):

 Vi = Pf × VfPi

or we can calculate the initial pressure (Pi) if we know the initial volume (Vi):

 Pi = Pf × VfVi

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## Worked Example: Calculating the Final Pressure of a Gas Sample

Question : A certain mass of gas occupies a volume of 2.5 L at 90 kPa pressure.
What pressure would the gas exert if it were placed in a 10.0 L container at the same temperature?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate final gas pressure
Pf = ? kPa

2. What data (information) have you been given in the question?

Extract the data from the question:

Conditions of the experiment: constant amount of gas at constant tempertaure.

Vi = initial gas volume = 2.5 L
Pi = initial gas pressure = 90 kPa
Vf = final gas volume = 10.0 L

3. What is the relationship between what you know and what you need to find out?
Because the amount of gas and temperature are constant, we can use Boyle's Law:

PiVi = PfVf

Rearrange this equation by dividing both sides by Vf:

 PfVf Vf = PiVi Vf Pf = PiVi Vf

4. Substitute in the values and solve for Pf
 Pf = PiVi Vf = 90 × 2.5 10.0 = 22.5 kPa
Consider that the volume has increased from 2.5 to 10, an increase of 10/2.5 = 4.
If the volume increases 4 times, then the pressure must decrease and the new pressure will be 1/4 of the initial pressure, that is, 1/4 × 90 = 22.5 kPa
Since this is the same value as we calculated above, we are reasonably confident that our answer is correct.
6. State your solution to the problem "final gas pressure":

Pf = 22.5 kPa

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## Worked Example: Calculating the Final Volume of a Gas Sample

Question : 4.5 L of gas at 125 kPa is expanded at constant temperature until the pressure is 75 kPa.
What is the final volume of the gas?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate final volume
Vf = ? L

2. What data (information) have you been given in the question?

Extract the data from the question:

Conditions: constant amount of gas and temperature.

Pi = 125 kPa
Vi = 4.5 L
Pf = 75 kPa

3. What is the relationship between what you know and what you need to find out?
Because the amount of gas and temperature are constant, we can use Boyle's Law:

PiVi = PfVf

Rearrange this equation by dividing both sides by Pf:

 PfVf Pf = PiVi Pf Vf = PiVi Pf

4. Substitute in the values and solve for Vf

 Vf = PiVi Pf = 125 × 4.5 75 = 7.5 L