Pressure and temperature of a gas
The Pressure Law
Heating a container filled with a mass of gas.
An experiment to investigate the relationship between pressure and temperature of a gas can be carried out using a container filled with a mass of gas. The temperature and pressure of the gas are recorded. During the experiment:
- temperature is varied – in \(10^{\circ}C \)steps between \(0^{\circ}C\) and \(100^{\circ}C \)
- volume is kept constant
Results from this experiment produce results leading to the following graph of pressure versus temperature in degrees celsius.
This graph tells us that as temperature increases, kinetic energy increases and this in turn increases the pressure.
But if the same data is used and the graph extended to the left, at what point will it cut the x–axis?
When the graph is extrapolated backwards, it can be seen from the graph that when the pressure is 0, then the temperature is \(-273^{\circ}C \).
Lord Unit of temperature on the absolute scale. Named after Lord Kelvin. For example, 0 degrees Celsius is 273 Kelvin. used this theory to create a new temperature scale, the Kelvin scale.
IMPORTANT
\(- 273^\circ C = 0K\)
\(\Rightarrow 0^\circ C = 273K\)
\(\Rightarrow 100^\circ C = 373K\)
A temperature increase of \(1K\) = a temperature increase of \(1^{\circ}C \).
This is a temperature scale where temperature and pressure are directly related to one another.
If you double the temperature in Kelvin – the pressure will double.
In this experiment, the gas is trapped inside the container which has a fixed size (its volume cannot change).
When the gas is heated the particles gain kinetic energy which makes them move faster. This means they collide with the walls more frequently and with greater force and so the pressure increases.
If the temperature of the gas is measured on the Scale of temperature directly relating temperature to pressure. we find that the pressure is A relationship between two variables, eg in a gas. As temperature increases, the pressure would also increase proportionally. (If the temperature doubled, the pressure would double). to the temperature.
Using these results leads to the following relationship between pressure and kelvin temperature:
\( \frac{p_{1}}{T_{1}}=\frac{p_{2}}{T_{2}}\) (Sometimes known as the Pressure Law)
Where:
- \(p_{1}\) is the starting pressure (measured in any relevant unit of pressure, eg Pascals)
- \(T_{1}\) is the starting temperature (must be in Kelvin)
- \(p_{2}\) is the finishing pressure (same units as p1)
- \(T_{2}\) is the finishing temperature (must be in Kelvin)
This equation is true as long as the volume and mass of the gas are constant.
Question
A car tyre contains air at \(1.25 \times 10^{5}Pa\)when at a temperature of \(27^{\circ}C\). Once the car has been running for a while the temperature of the air in the tyre rises to \(42^{\circ}C\).
If the volume of the tyre does not change what is the new pressure of the air in the tyre?
First convert the temperatures into Kelvin.
\(T_{1}=27\,+273=300K \)
\(T_{2}=42\,+273=315K \)
Now calculate the new pressure;
\(T_{1}=300K \)
\(T_{2}=315K \)
\(p_{1}=1\cdot25\,\times10^{5}Pa \)
\(\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}}\)
Therefore;
\(p_{2}=\frac{p_{1T_{2}}}{T_{1}} \)
\(p_{2}=\left (\frac{1\cdot25\,\times10^{5}\times315}{300} \right ) \)
\(p_{2}=1\cdot31\times10^{5}Pa \)
The new pressure is \(p_{2}=1\cdot31\times10^{5}Pa \)
