If you trap a sample of air and measure its volume at different pressures (constant temperature), then you can determine a relation between volume and pressure. If you do this experiment, you will find that as the pressure of a gas sample increases, its volume decreases. In other words, the volume of a gas sample at constant temperature is inversely proportional to its pressure. The product of the pressure multiplied by the volume is a constant:

PV = k or V = k/P or P = k/V

where P is pressure, V is volume, k is a constant, and the temperature and quantity of gas are held constant. This relationship is called **Boyle's Law**, after Robert Boyle, who discovered it in 1660.

### Key Takeaways: Boyle's Law Chemistry Problems

- Simply put, Boyle's states that for a gas at constant temperature, pressure multiplied by volume is a constant value. The equation for this is PV = k, where k is a constant.
- At a constant temperature, if you increase the pressure of a gas, its volume decreases. If you increase its volume, the pressure decreases.
- The volume of a gas is inversely proportional to its pressure.
- Boyle's law is a form of the Ideal Gas Law. At normal temperatures and pressures, it works well for real gases. However, at high temperature or pressure, it is not a valid approximation.

### Worked Example Problem

The sections on the General Properties of Gases and Ideal Gas Law Problems may also be helpful when attempting to work Boyle's Law problems.

**Problem**

A sample of helium gas at 25°C is compressed from 200 cm^{3} to 0.240 cm^{3}. Its pressure is now 3.00 cm Hg. What was the original pressure of the helium?

**Solution**

It's always a good idea to write down the values of all known variables, indicating whether the values are for initial or final states. Boyle's Law problems are essentially special cases of the Ideal Gas Law:

Initial: P_{1} = ?; V_{1} = 200 cm^{3}; n_{1} = n; T_{1} = T

Final: P_{2} = 3.00 cm Hg; V_{2} = 0.240 cm^{3}; n_{2} = n; T_{2} = T

P_{1}V_{1} = nRT (Ideal Gas Law)

P_{2}V_{2} = nRT

so, P_{1}V_{1} = P_{2}V_{2}

P_{1} = P_{2}V_{2}/V_{1}

P_{1} = 3.00 cm Hg x 0.240 cm^{3}/200 cm^{3}

P_{1} = 3.60 x 10^{-3} cm Hg

Did you notice that the units for the pressure are in cm Hg? You may wish to convert this to a more common unit, such as millimeters of mercury, atmospheres, or pascals.

3.60 x 10^{-3} Hg x 10mm/1 cm = 3.60 x 10^{-2} mm Hg

3.60 x 10^{-3} Hg x 1 atm/76.0 cm Hg = 4.74 x 10^{-5} atm

### Source

- Levine, Ira N. (1978).
*Physical Chemistry*. University of Brooklyn: McGraw-Hill.