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.

**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