Custom Search

 Boyle's Law

In 1662, Robert Boyle made the first systematic study of the relationship between volume and pressure in gases. Boyle’s law simply states that the volume of a confined gas at a fixed temperature is inversely proportional to the pressure exerted on the gas. This can also be expressed as

PV = k

P is inversely related to V. This makes sense if you think of a balloon. When the pressure around a balloon increases, the volume of the balloon decreases, and likewise, when you decrease the pressure around a balloon, its volume will increase.

Boyle’s law to can also be expressed in the following way, and this is the form of the law that you should memorize:

P1V1 = k = P2V2

P1V1P2V2

 Increasing Pressure on a Gas Decreasing Pressure on a Gas
 We can see that the data fit into a pattern called a hyperbola.  If, however we plot pressure against 1/volume we get a linear (straight line) graph.   Boyle's Law works best at low pressures. Less intermolecular attractions. Gases act more ideally.

Sample Question
Sulfur dioxide (SO2) gas is a component of car exhaust and power plant discharge, and it plays a major role in the formation of acid rain. Consider a 3.0 L sample of gaseous SO2 at a pressure of 1.0 atm. If the pressure is changed to 1.5 atm at a constant temperature, what will be the new volume of the gas?

Explanation

P1V1 = P2V2

(1.0 atm) (3.0 L) = (1.5 atm) (V2)

V2 = 2.0 L.

Note**as the pressure of the system increases, the volume should decrease.

Note**Boyle's Law Questions are easy to spot. You will see the term "CONSTANT TEMPERATURE". This is how you know it is a Boyle's Law Question, temperature does not change.

Boyle's Law Quiz

AP  Try these

1. A 1.53L sample of gaseous CO2 at a pressure of 5.6 x 104Pa. If the pressure is changed to 1.00atm at constant temperature, what will be the new volume?

 Answer--> P1V1 = P2V2 (5.6 x 104Pa)(1.53L)=(1.00atm x 101,325Pa/1 atm)V2 (5.6 x 104Pa)(1.53L) =V2 (101,325Pa) 0.85L =V2
2. A balloon filled to a volume 7.00 x 102mL at a pressure of of 770 mmHg. The balloon then is released at constant temperature and the pressure changes to  temperature of 3.35 x 102torr. What is the final volume of this balloon?
 Answer--> P1V1 = P2V2                                P2= 3.35 x 102torr =335mmHg (770 mmHg    )(7.00 x 102mL)=(335mmHg)V2 (770 mmHg)(7.00 x 102mL) =V2 (335mmHg) 1609mL =V2