10/3/2016 Summary of Gas Laws
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
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, 10/3/2016 Summary of Gas Laws
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
file://pminas/users/sgarcia/Downloads/gasoutline.html 2/12
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
file://pminas/users/sgarcia/Downloads/gasoutline.html 1/12
, 10/3/2016 Summary of Gas Laws
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
file://pminas/users/sgarcia/Downloads/gasoutline.html 2/12
