Boyles Law Worked Sample Chemistry Problem

Boyle's Law Worked Sample Chemistry Problem 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 Boyles Law, after Robert Boyle, who discovered it in 1660. Key Takeaways: Boyle's Law Chemistry Problems Simply put, Boyles 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.Boyles 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 Boyles Law problems. Problem A sample of helium gas at 25Â °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 Its always a good idea to write down the values of all known variables, indicating whether the values are for initial or final states. Boyles 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 Source Levine, Ira N. (1978). Physical Chemistry. University of Brooklyn: McGraw-Hill.