10/3/2016 Summary of Gas Laws

file://pmi­nas/users/sgarcia/Downloads/gasoutline.html 1/12

A Quick Introduction to Gases

These notes were written to help you quickly get to the point where you can work gas law problems on the

quizzes and exams in this course. They are not a detailed development of the topic.

The problems which will appear on exams will involve 4 "empirical" gas laws and one additional law known as

the ideal gas law.

The empirical gas laws are as follows:

Name of Gas Law Equation Parameters That Must Be Constant

Boyle's Law P1V1 = P2V2 Temperature, Number of moles

Charles' Law V1

/ T1 = V2

/ T2 Pressure, Number of moles

Amonton's Law P1

/ T1 = P2

/ T2 Volume, Number of moles

Combined Gas Law P1V1

/ T1 = P2V2

/ T2 Number of moles

The problems which you will encounter with these laws involve having a gas under an initial set of conditions

(state 1) and then having the gas undergo some process which changes its state of existence to another set of

conditions (state 2). All the gas parameters will be specified except one, and the one missing can be solved for

algebraically. Generally, I ask for one of the state 2 parameters.

Boyle's Law Sample Problems

Boyle's Law Example 1

A gas was confined in a cylinder fitted with a movable piston. At 22.6

oC, the gas occupied a volume of 4.573

L under a pressure of 1.152 atm. The gas was isothermally compressed, reducing its volume to 2.963 L. What

pressure was exerted by the comperssed gas?

Solution to Boyle's Law Example 1

We recognize that this is a Boyle's Law problem because the temperature remains constant (we are told the

process is "isothermal", which means constant temperature). Presumably, there is also no change in the number

of moles of gas, since no amount of gas was ever mentioned in the problem. Although the temperature is given,

it is not needed, since it does not appear in Boyle's Law.

Before the process (compression), the gas volume is 4.573 L, so this is V1

, and the gas pressure is 1.152 atm, so

this is P1

. After the process (compression), the gas volume is 2.963 L, so this is V2

. The variable left

unspecified is the pressure after the process, (P2

) so this is the variable we must solve for.

The equation P1V1 = P2V2 can be solved for P2

to get

P2 = P1

( V1

/ V2

) = 1.152 atm ( 4.573 L / 2.963 L ) = 1.778 atm <===== ANSWER

10/3/2016 Summary of Gas Laws

file://pmi­nas/users/sgarcia/Downloads/gasoutline.html 2/12

Notice that the "new" pressure is just the "old" pressure multiplied by a ratio of the two volumes. A pattern like

this exists in all problems involving the empirical gas laws. It is possible to solve such problems by inspection,

without actually having to set up the algebra, once you get used to it. Notice in this case, for example, it makes

sense that the larger volume is in the numerator. Since the gas is being compressed, the final pressure should be

larger than the initial one. Therefore, the original pressure has to be multiplied by a number greater than one, to

create a number larger than we started with. The only other possible ratio we could write for the two volumes is

( 2.963 L / 4.573 L ), but this ratio would be less than one, and when multiplied by the origianl pressure, would

yield a smaller pressure than we started with. This would not make sense, in light of the fact that the gas is

being compressed, thus we could have selected the proper ratio without ever setting up the algebra.

Boyle's Law Example 2

A gas was confined in a cylinder fitted with a movable piston. At 19.6

oC, the gas occupied a volume of 7.336

L under a pressure of 3.196 atm. The gas was isothermally compressed until its pressure reached 5.500 atm.

What volume was occupied by the compressed gas?

Solution to Boyle's Law Example 2

We recognize this as a Boyle's Law problem because the temperature remains constant ("isothermal") and there

is apparently no change in the number of moles of gas, since moles (or amount of gas in any units, for that

matter), is not mentioned in the problem. The temperature is specified in the problem, but not needed, since

temperature does not appear in Boyle's Law.

Reading through the problem, we can make the following variable assignments:

V1 = 7.336 L P1 = 3.196 atm

V2 = ? P2 = 5.500 atm

The Boyle's Law equation P1V1 = P2V2 can be solved for V2

to give

V2 = V1

( P1

/ P2

) = 7.336 L ( 3.196 atm / 5.500 atm ) = 4.263 L <===== ANSWER

Notice that the "new" volume is just the "old" volume multiplied by a ratio of the pressures. It makes sense that

the smaller pressure is on top in the ratio. This makes the ratio less than one, so when multiplied by the original

volume, is gives us a smaller volume than we started with. Since the gas is being compressed, it makes sense

that the final volume should be smaller than the orignial.

Boyle's Law Example 3

A gas was confined in a cylinder fitted with a movable piston. At 21.3

oC, the gas occupied a volume of 1.994

L under a pressure of 4.357 atm. The pressure on the gas was then reduced to 3.264 atm, and the gas was

allowed to expand isothermally to a new volume. What was the volume of the expanded gas?

Solution to Boyle's Law Example 3

We recognize this as a Boyle's Law problem because the temperature does not change ("isothermal") and the

number of moles apparently does not change. The temperature has been given, but is not needed in the

calculation. Reading through the problem, we can make the following variable assignments:

V1 = 1.994 L P1 = 4.357 atm

Powered by qwivy(www.qwivy.org)