 # Charles' Law Chemistry Tutorial

## Key Concepts

• Charles' Law describes the relationship between the volume and temperature of a gas at constant pressure.
• Charles' Law states that for a gas at constant pressure,
the volume, V, of a given quantity of gas is directly proportional to the absolute temperature of the gas, T :

V ∝ T (in Kelvin)

So at constant pressure, if the temperature (in Kelvin) is doubled, the volume of gas is also doubled.

• Charles' Law can be represented as mathematical equation:
For a quantity of gas at constant pressure:

V ÷ T(K) = a constant

or

V/T(K) = a constant

• At constant pressure for a given quantity of gas that undergoes a change in temperature (or volume) :

Vi ÷ Ti (K) = Vf ÷ Tf (K)

or

Vi/Ti = Vf/Tf

where
Vi is the initial (original) volume
Ti is its initial (original) temperature (in Kelvin)
Vf is its final volme
Tf is its final tempeature (in Kelvin)

Vi and Vf must be in the same units of measurement (eg, both in litres)
Ti and Tf must be in Kelvin NOT celsius.

• All gases approximate Charles' Law at high temperatures and low pressures.1
A hypothetical gas which obeys Charles' Law at all temperatures and pressures is called an Ideal Gas.

A Real Gas is one which approaches Charles' Law as the temperature is raised or the pressure lowered.

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## Charles' Law Concepts

Consider the following experiment to measure the expansion of hydrogen gas:

• A known amount of hydrogen gas is drawn up into a syringe at -23oC and 100 kPa pressure.
• The volume of hydrogen gas is recorded as 25 mL.
• The hydrogen gas in the syringe is then heated while maintaining a constant pressure of 100 kPa.
• The volume of hydrogen gas is recorded at various temperatures.
• The results of the experiment are shown in the table.

Volume and Temperature of a Quantity of H2(g) at 100 kPa
Volume
(mL)
Temperature
(oC)
Trend
25 -23 As volume increases, temperature increases.

As volume decreases, temperature decreases.

As temperature increases, volume increases.

As temperature decreases, volume decreases.

30 27
35 77
40 127.5
45 177

Is there a simply relationship between the temperature of a gas in °C and its volume?
Let's try dividing volume by temperature and see, the results are in the table below:

Volume and Temperature of a Quantity of H2(g) at 100 kPa
Volume
(mL)
Temperature
(oC)
V ÷ T(°C)
25 -23 -1.1
30 27 1.1
35 77 0.45
40 127.5 0.31
45 177 0.25

No, there doesn't appear to be a simple relationship between the volume of the gas and its temperature in °C.

But, what happens if we convert all the temperatures in °C to temperatures in kelvin (K)?
The table below shows the temperature conversions:

Volume and Temperature of a Quantity of H2(g) at 100 kPa
Volume / mL Temperature / oC Temperature / K
T(K) = 273 + T(°C)
25 -23 250
30 27 300
35 77 350
40 127.5 400.5
45 177 450

Is there now a simple relationship between volume of gas and its temperature in Kelvin?
The table below shows the results of dividing the volume of a gas by its temperature in Kelvin:

Volume and Temperature of a Quantity of H2(g) at 100 kPa
Volume / mL Temperature / K
T(K) = 273 + T(°C)
V ÷ T (K)
25 250 0.1
30 300 0.1
35 350 0.1
40 400.5 0.1
45 450 0.1

Yes, we can now see a clear relationship between the volume of this gas (V) and its temperature in Kelvin (T) at a constant pressure of 100 kPa:

V ÷ T = 0.1

In general we could write:

V ÷ T = "a constant"

By rearranging this equation we can write:

V = "a constant" × T

Which is the equation for a straight line that goes through the origin (0,0) and has a slope (or gradient) equal to the value of "a constant".

The points are plotted and the line is extrapolated back to 0 (volume = 0 mL and temperature = 0 K) in the graph below:

 volume (mL) Expansion of Hydrogen Gas at Constant Pressuretemperature (K)

The graph of gas volume against temperature is a straight line.
We say that there is a linear relationship between the volume of a gas and its temperature at constant pressure.

Extrapolation of the line back to (0,0) assumes that at temperatures below -23oC (250 K), the linear relationship between volume and temperature will be maintained.

It is unlikely that this assumption will hold at very low temperatures for 100 kPa pressure as the hydrogen is likely to condense into a liquid first.

The extrapolation of the graph actually suggests that at 0 K an ideal gas has no volume (0 mL on our graph).

From the graph we see that:

• As the temperature increases, the volume of hydrogen gas increases in a linear fashion, that is

V ∝ T

Using a constant of proportionality we can write an equation:

V = T × "constant"

Dividing both sides of the equation by volume (V) we arrive at:

 V T = "constant"

• The slope (gradient) of the straight line gives us the constant of proportionality.

slope (gradient) = change in volume ÷ change in temperature

Using any two points on the line we can calculate the slope (gradient)

for example, take the points (250, 25) and (450, 45), then
(slope (gradient) = (45 - 25) ÷ (450 - 250) = 0.1

for example, take the points (300, 30) and (400.5, 40), then
slope (gradient) = (40 - 30) ÷ (400.5 - 300) = 0.1

• Charles' Law equation for this expansion of hydrogen gas can then be written as:

 V T = 0.1

As long as we assume that the hydrogen gas continues to behave like an ideal gas, and that the pressure and amount of gas does not change, then this equation can then be used to calculate the

(a) volume of hydrogen gas at any temperature:

V = 0.1 × T

(b) temperature of any volume of hydrogen gas :

T = 0.1 ÷ V

• This also means that for any 2 points on the line, (T1,V1) and (T2,V2):

 V1 T1 = 0.1

and

 V2 T2 = 0.1

So,

 V1 T1 = V2 T2

And this is an important and useful general description of Charles' Law because it allows us to calculate the volume of a gas after a temperature change (at constant pressure), or, the temperature of a gas after its volume is changed (at constant pressure).

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## Calculations : Vi/Ti = Vf/Tf

Consider an experiment in which we have a known quantity of gas in vessel such as a syringe in which the piston (plunger) can move freely up or down in order to change the volume occupied by the gas.

In the beginning of an experiment, a known amount of gas at a specified pressure has

volume = Vi

temperature = Ti

In the beginning of the experiment,

Vi ÷ Ti = constant = k

The temperature of the gaseous system is then changed (the system is heated or cooled) while constant pressure is maintained.

At the end of the experiment, the gas will have a different volume and a different temperature:

volume = Vf

temperature = Tf

As long as the amount of gas has not changed, and the pressure has not changed, then

Vf ÷ Tf = the same constant as at the beginning of the experiment = k

Therefore Vi ÷ Ti = k = Vf ÷ Tf

So Vi ÷ Ti = Vf ÷ Tf

This equation can then be rearranged to find the volume or temperature of a known amount of gas at specified pressure during the course of an experiment:

Find the initial volume, Vi

 Vi = Ti × Vf Tf

Find the final volume, Vf

 Vf = Vi × Tf Ti

Find the initial temperature, Ti

 Ti = Tf × Vi Vf

Find the final temperature, Tf

 Tf = Ti × Vf Vi

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## Worked Example of Charles' Law to Calculate Volume of Gas

Question : A sample of unknown gas had a volume of 1.2 L at 100oC and 100 kPa pressure.
What would its volume be at 0oC at the same pressure?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate final volume of gas
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 pressure (100 kPa).

Vi = 1.2 L

Ti = 100°C
Convert temperature in °C to K
Ti + 273 = 100 + 273 = 373 K

Tf = 0°C
Convert temperature in °C to K
Ti + 273 = 0 + 273 = 273 K

3. What is the relationship between what you know and what you need to find out?
Assume ideal gas behaviour.

Because the amount of gas and pressure are constant, we can use Charles' Law:

 Vi Ti = Vf Tf

Multiply both sides of the equation by Tf:

 Tf × Vi Ti = Tf × Vf Tf Tf × Vi Ti = Vf

4. Substitute in the values and solve for Vf

 Vf = Tf × Vi Ti = 273 ×1.2 373 = 0.88 L

Consider that the gas is made up of particles in constant motion.
If you cool the gas down from 100°C to 0°C the particles will have less kinetic energy, they move about less. If the volume that contained the gas was fixed (constant), the pressure exerted by the gas particles would be less.
But, in our example, the volume of the container is not fixed (since we are asked to calculate the new volume), so, in order to maintain the same gas pressure, the volume of the container must decrease.
Our calculated final volume of gas (0.88 L) is less than the initial volume of gas (1.2 L), so we are reasonably confident that our answer is plausible.
6. State your solution to the problem "calulate final volume of gas":

Vf = 0.88 L

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## Worked Example of Charles' Law to Calculate Temperature of Gas

Question : A helium filled balloon had a volume of 75 L at 25oC.
To what does the temperature, in Kelvin, need to be raised in order for the balloon to have a volume of 100 L at the same pressure?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate final temperature
Tf = ? K

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

Extract the data from the question:

Conditions: constant amount of gas and pressure.

Vi = 75 L

Ti = 25°C
Convert temperature in °C to K
Ti + 273 = 25 + 273 = 298 K

Vf = 100 L

3. What is the relationship between what you know and what you need to find out?
Assume ideal gas behaviour.

Because the amount of gas and pressure are constant, we can use Charles' Law:

 Vi Ti = Vf Tf

Multiply both sides of the equation by Tf:

 Tf × Vi Ti = Tf × Vf Tf Tf × Vi Ti = Vf

Multiply both sides of the equation by Ti:

 Ti × Tf × Vi Ti = Ti × Vf Tf × Vi = Ti × Vf

Divide both sides of the equation by Vi

 Tf × Vi Vi = Ti × Vf Vi Tf = Ti × Vf Vi

4. Substitute in the values and solve for Tf

 Tf = Ti × Vf Vi Tf = 298 × 100 75 = 397 K

Consider: V ∝ T (in K)
Increasing the volume of gas increases its temperature (in order to maintain constant pressure).
Double the the volume and the temperature (in K) doubles.
In the question the volume has increased by a factor of 100/75 = 1.333
So the temperature must have increased by a factor of 1.333
That is, final temperature = 1.333 × 298 = 397 K
Since this is the same value as we calculated above, we are reasonably confident that our answer is plausible.
6. State your solution to the problem "calculate final temperature in K ":

Tf = 397 K

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1. Deviations from Ideal Gas Behaviour:

As a Real Gas is cooled at constant pressure from a point well above its condensation point, its volume begins to increase linearly.
As the temperature approaches the gases condensation point, the line begins to curve (usually downward) so there is a marked deviation from Ideal Gas behaviour close to the condensation point.
Once the gas condenses to a liquid it is no longer a gas and so does not obey Charles' Law at all.
Absolute zero (0K, -273oC approximately) is the temperature at which the volume of a gas would become zero if it did not condense and if it behaved ideally down to that temperature.