Introduction
Boyle's Law is one of many fundamental principles behind the way that gases in containers behave. Recall that the temperature, volume, and pressure of gases are very important to understand and identify when studying gases. If the temperature of a gas over time is kept constant, then there is a special relationship between the gas's pressure and volume: they are inversely proportional.
An Equation for Boyle's Law
The equation for Boyle's Law refers to the change of the gas's pressure and volume between points in time. The initial time is denoted with a subscript of 1 on the pressure and volume and the final time is denoted with a subscript of 2 on these same quantities. In general:
\(P_1\) is the initial pressure.
\(P_2\) is the final pressure.
\(V_1\) is the initial volume [of the container].
\(V_2\) is the final volume.
As the pressure increases the volume of the container decreases, and vice-versa. That is how this inverse relationship works. For a constant temperature:
$$P_1V_1 = P_2V_2$$
This equation can be manipulated in different ways, but before we get to that, it is important that the container in which the gas exists must be expandable and contractable; in other words, the volume must be able to change. If the container is rigid, that means the volume is constant, and this equation does not apply! When solving problems, watch the wordings of the problems carefully to avoid falling into this trap.
Manipulations of Boyle's Law
In general it is more efficient to solve for your given variable in terms of the other variables before substituting values. This can save a lot of grief and confusion with significant figures and units.
For instance, if you want to solve for the initial pressure of the gas, isolate \(P_1\) in the Boyle's Law equation by dividing both sides of the equation by \(V_1\):
$$P_1 = \frac{P_2V_2}{V_1}$$
It is not advised to memorize each such separate form of Boyle's Law. Rather, know how to derive these forms.
Example 1: For a gas at constant temperature, find the final volume of the gas in terms of the initial volume, the initial pressure, and the final pressure.
Solution: According to Boyle's Law:
$$P_1V_1 = P_2V_2$$
We are being asked to solve for the final volume,\(V_2\). Isolate \(V_2\) in the equation by dividing both sides of the equation by \(P_2\):
$$V_2 = \frac{P_1V_1}{P_2}$$
Note that the units on both sides of the equation will always be the same. In Example 1, the units on the left side are liters, while on the right side they are atmospheres (the default unit for pressure in these types of problems) times liters divided by atmospheres, which in turn is just liters.
Problem-Solving
At this point, we should see how Boyle's Law is applied to quantitative problems. We will show two different problems involving Boyle's Law to give you a good idea of how the calculations are followed through. Also note that if the container the gas is held in is nonrigid, then its volume is a variable and is not constant.
Example 2: A sample of gas is kept at a constant temperature and in a nonrigid container. The initial pressure is 2.00 atmospheres, the final pressure is 5.60 atmospheres, and the final volume is 12.54 liters. Find the initial volume of the container.
Solution: Our target variable is the initial volume, \(V_1\). According to Boyle's Law, with constant temperature:
$$P_1V_1 = P_2V_2$$
Solve for \(V_1\) by dividing both sides of the equation by \(P_1\):
$$V_1 = \frac{P_2V_2}{P_1}$$
Substitute the known values:
$$V_1 = \frac{5.60 \; atm \cdot 12.54 \; L}{2.00 \; atm} = 35.1 \; L$$
Here is another example with more real-world context.
Example 3: You are working in the chemical industry, observing how a sample of silicon tetrafluoride gas at a constant temperature behaves in a specially designed nonrigid container. The special property of this container is that regardless of how much it expands and contracts, it remains in the shape of a perfect cube. The initial side length of the cube is \(2.50\) centimeters, while the initial pressure of the gas is 15.60 atmospheres. Your team continually adds pressure to the gas by decreasing the pressure of the surrounding air over the course of 10 minutes. After 5 minutes, though, one team member trips on his shoelace onto the machine and knocks the barometer (which measures the pressure of the gas) off the machine. After 10 minutes, the side length of the cube is measured to be \(3.89\) centimeters. With a broken barometer, find the pressure of the gas after ten minutes, in liters.
Solution: