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Boyle's Law - Worked Chemistry Problems

Concept and Example

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This is a graph of Boyle's original data, leading to the formulation of Boyle's Law.

This is a graph of Boyle's original data, leading to the formulation of Boyle's Law.

Marc Lagrange, Wikipedia Commons

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.

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 cm3 to 0.240 cm3. 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: P1 = ?; V1 = 200 cm3; n1 = n; T1 = T

Final: P2 = 3.00 cm Hg; V2 = 0.240 cm3; n2 = n; T2 = T

P1V1 = nRT (Ideal Gas Law)

P2V2 = nRT

so, P1V1 = P2V2

P1 = P2V2/V1

P1 = 3.00 cm Hg x 0.240 cm3/200 cm3

P1 = 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

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