Volume and temperature of a gas
Charles's Law
Heating a container filled with a mass of gas.
To investigate the relationship between volume and temperature, at constant pressure, an experiment can be carried out where a fixed mass of gas is in a container, which is free to expand, without any gas escaping. This means that the gas is always at atmospheric pressure, which is assumed constant during the experiment. When the gas is heated, its volume increases. The particles gain kinetic energy and move faster. They collide more often with greater force on the container walls. Since the container can expand, the gas volume increases until the pressure of the gas is back to the constant atmospheric pressure.
Watch this video for a practical demonstration proving Charles Law, the relationship between temperature and volume.
Using results from this experiment leads to the following relationship between pressure and kelvin temperature:
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) (Sometimes known as Charles's Law)
where:
- \(V_{1}\) is the starting volume (measured in any relevant unit of volume, eg \(m^{3}\))
- \(T^{1}\) is the starting temperature (must be in Kelvin)
- \(V^{2}\) is the finishing pressure (same units as \(V_{1}\))
- \(T^{2}\) is the finishing temperature (must be in Kelvin)
This equation is true as long as the pressure and mass of the gas are constant.
Question
A balloon is blown up at night when the temperature is \(8^{\circ}C\) to a volume \(0.018m^{3}\). Next afternoon, the temperature is \(28^{\circ}C\).
If there is no change in air pressure, what is the new volume of the gas?
First convert the temperatures into kelvin.
\(T_{1}=8\,+273=281K,\,T_{2}=28\,+273=301K\)
\(V^{1}=0\cdot018\,m^{3} \)
\(\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}}\)
\(V^{2}=\frac{0\cdot018\,\times301}{281} \)
\(V^{2}=0\cdot019\,m^{3} \)