According to this experimental gas law, the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature. An important point to note here is that the amount of ideal gas (i.e., number of moles) is kept fixed and a constant temperature is maintained. Under this condition, the pressure of the gas will tend to increase if we decrease the volume of the container and vice-versa.
Mathematically, it can be expressed as:
Where āPā is the pressure exerted by the gas and āVā is the Volume of the container.
Graphically, we can observe the behaviour of the curve as follows:
You can see how the value of Pressure decreases when the Volume is increased or vice-versa.
In order to understand it better, let us go through a simple example:
Example: If 20$c{m^3}$ gas at 1 atm. is expanded to 50 $c{m^3}$ at constant T, then what is the final pressure Options: (a) $20 \times \frac{1}{{50}}$ (b) $50 \times \frac{1}{{20}}$ (c) $1 \times \frac{1}{{20}} \times 50$ (d) None of these Answer: (a) Solution: At constant $T,{P_1}{V_1} = {P_2}{V_2}$ $1 \times 20 = {P_2} \times 50$; ${P_2} = \frac{{20}}{{50}} \times 1$
We hope you have understood, how to apply the Boyleās Law Formula. Now in order to test your understanding, try and solve the Question by yourself.
Example: An ideal gas exerts a pressure ofĀ 3atmĀ in aĀ 3LĀ container. The container is at a temperature ofĀ 298K. What will be the final pressure if the volume of the container changes toĀ 2L? Options: (a) 2 atm (b) 3.5 atm (c) 4.5 atm (d) 4 atm Answer: (c) Solution: Boyle's law can be written as follows: \[{P_1}{V_1} = {P_2}{V_2}\] Use the given volumes and the initial pressure to solve for the final pressure. (3atm)(3L)=(2L)P_{2} P_{2}=4.5atm