Pressure and volume of a gas
Boyle's Law
Imagine a gas is trapped in a cylinder by a piston.
If the piston is pushed in, the gas particles will have less room to move as the volume the gas occupies has been decreased.
Watch this video for a practical demonstration proving Boyle's Law, the relationship between pressure and volume.
Process of equilibrium
Piston prior to being pushed in and piston pushed into cylinder.
Because there has been a decrease in volume the particles will collide more frequently with the walls of the container. Each time they collide with the walls, they exert a force on them. More collisions mean more force, so the pressure will increase.
When the volume decreases the pressure increases. This shows that the pressure of a gas is A relationship between two variables where as one variable increases, the other variable decreases, eg as the volume doubled, the pressure decreased by half. to its volume.
From this we can derive the following equation:
\({p_1}{V_1} = {p_2}{V_2} \) (Sometimes known as Boyle's Law)
where
- \(p_{1}\) is the starting pressure (measured in any relevant unit of pressure, eg pascals)
- \(V_{1}\) is the starting volume (measured in any relevant unit of volume, eg litres)
- \(p_{2}\) is the finishing pressure (must be measured in the same units as \(p_{1}\)
- \(V_{2}\) is the finishing volume (must be measured in the same units as \(V_{1}\))
Question
A sealed syringe contains \(10 \times 10^{-6}m^{3}\) of air at \(1 \times 10^{5} Pa\). The plunger is pushed until the volume of trapped air is \(4 \times 10^{-6}m^{3}\). If there is no change in temperature what is the new pressure of the gas?
\({p_1} = 1 \times {10^5}Pa\)
\({V_1} = 10 \times {10^{ - 6}}{m^3}\)
\({V_2} = 4 \times {10^{ - 6}}{m^3}\)
\({p_1}{V_1} = {p_2}{V_2}\)
Therefore;
\({p_2} = \frac{{{p_1}{V_1}}}{{{V_2}}}\)
\({p_2} = \frac{{1 \times {{10}^5} \times 10 \times {{10}^{ - 6}}}}{{4 \times {{10}^{ - 6}}}}\)
\({p_2} = 2.5 \times {10^5}Pa\)
The new pressure in the syringe is \(2.5 \times 10^{5}Pa\)
