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) :

  V_i ÷ T_i (K) = V_f ÷ T_f (K)

  or

  ^V_i/_Ti = ^V_f/_Tf

  where
  V_i is the initial (original) volume
  T_i is its initial (original) temperature (in Kelvin)
  V_f is its final volme
  T_f is its final tempeature (in Kelvin)

  V_i and V_f must be in the same units of measurement (eg, both in litres)
  T_i and T_f 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.

Please do not block ads on this website.
No ads = no money for us = no free stuff for you!

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 -23^oC 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.

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:

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:

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:

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:

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 -23^oC (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).

Calculations : ^V_i/_Ti = ^V_f/_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 = V_i

temperature = T_i

In the beginning of the experiment,

V_i ÷ T_i = 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 = V_f

temperature = T_f

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

V_f ÷ T_f = the same constant as at the beginning of the experiment = k

Therefore V_i ÷ T_i = k = V_f ÷ T_f

So V_i ÷ T_i = V_f ÷ T_f

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, V_i

Find the final volume, V_f

Find the initial temperature, T_i

Find the final temperature, T_f

Worked Example of Charles' Law to Calculate Volume of Gas

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

   V_i = 1.2 L

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

   T_f = 0°C
   Convert temperature in °C to K
   T_i + 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 T_f:

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

4. Substitute in the values and solve for V_f

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

5. Is your answer plausible?

   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":

   V_f = 0.88 L

Worked Example of Charles' Law to Calculate Temperature of Gas

Question : A helium filled balloon had a volume of 75 L at 25^oC.
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
   T_f = ? 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.

   V_i = 75 L

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

   V_f = 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 T_f:

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

   Multiply both sides of the equation by T_i:

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

   Divide both sides of the equation by V_i

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

4. Substitute in the values and solve for T_f

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

5. Is your answer plausible?

   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 ":

   T_f = 397 K