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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


Increasing Pressure on a Gas Decreasing Pressure on a Gas

Even MoreVacuum Demonstration Videos


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?


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)



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)



1609mL =V2

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