Name_______________________
3-2 Boyle’s Law Solutions
3-2 Solutions
PV = k
or
P1V1 = P2V2
For the following problems calculate the value which the question asks for. Be sure to follow the proper steps and show
all your work (1) Convert values (temp) to correct units if necessary and list variables (2) write the law (3) solve the law
for desired variable (4) plug in the numbers with the correct units and (5) calculate the solution. (Answers are provided
for you to check your work) (Follow the appropriate steps as laid out for you in the first problem for full credit)
1. A sample of H2 has a volume of 10 liters at a pressure of 1 atmosphere (100 kPa). If the pressure is raised to 2
atmospheres what would the volume be? (V2 = 5 liters)
LIST VARIABLES
P1 = 100 kPa
V1 = 10 liters
P2 = 200 kPa
STATE THE LAW
Boyle’s Law
P1V1 = P2V2
V2 = ????
SOLVE THE LAW
÷ both sides by P2
P1V1 P2V2

P2
P2
CALCULATE SOLUTION
( 100kPa)(10liters)
( 200 kPa)
simplify
P1V1 P2V2

P2
P2
to get
P1
V1  V2
P2
2. A sample of gas has a volume of 4 liters at a pressure of 5 atmosphere (500 kPa). If the pressure is lowered to 2
atmospheres (200 kPa) what would the volume be? (V2 = 10 liters)
LIST VARIABLES
P1 = 500 kPa
STATE THE LAW
SOLVE THE LAW
Boyle’s Law
÷ both sides by P2
P1V1 = P2V2
P1V1 P2V2

P2
P2
V1 = 4 liters
P2 = 200 kPa
V2 = ????
simplify
P1V1 P2V2

P2
P2
to get
P1
V1  V2
P2
CALCULATE SOLUTION
V2 
P1
(500 _ kPa)
4 _ liters 
V1 
P2
(200 _ kPa)
V2 = 10 liters
3. A sample of Oxygen gas at 3 atmospheres has a volume of 100 liters. When compressed its volume is reduced
to 73 liters. What is the pressure on the gas when the volume is 73 liters? (P2 = 4.1 atm )
LIST VARIABLES
P1 = 3 atm.
STATE THE LAW
Boyle’s Law
V1 = 100 liters
P2 = ????
P1V1 = P2V2
V2 = 73 liters
SOLVE THE LAW
CALCULATE SOLUTION
÷ both sides by V2
P1V1 P2V2

V2
V2
simplify
P2 
P1V1 (3 _ atm.)(100 _ liters )

V2
(73 _ liters )
P2 = 4.1 atmospheres
P1V1 P2V2

V2
V2
to get
P1V1
 P2
V2
4. A sample of Oxygen gas at 25 atm is allowed to expand to a lower pressure. After expanding to the volume of
30 liters the pressure is 20 atm. What was the volume when the pressure was 25 atm? (V1 = 24 liters)
LIST VARIABLES
P1 = 25 atm.
STATE THE LAW
Boyle’s Law
V1 = ???? liters
P2 = 20 atm.
V2 = 30 liters
P1V1 = P2V2
SOLVE THE LAW
CALCULATE SOLUTION
÷ both sides by P1
P1V1 P2V2

P1
P1
simplify
P1V1 P2V2

P1
P1
to get
V1 
P2V2
P1
V1 
P2V2 (20 _ atm.)(30 _ liters )

P1
(25 _ atm)
V1 = 24 liters
5. Air in a scuba divers tank takes up a volume of about 10 liters and is compressed to about 17 atm (1700
kPascals). If the air were released to the atmosphere to create a large soap bubble, how large would the bubble
be? (V2 =170 liters)
LIST VARIABLES
P1 = 17 atm.
STATE THE LAW
Boyle’s Law
V1 = 10 liters
P1V1 = P2V2
P2 = 1 atm.
V2 = ???? liters
SOLVE THE LAW
CALCULATE SOLUTION
÷ both sides by P2
P1V1 P2V2

P2
P2
simplify
V2 
P1V1 (17 _ atm.)(10 _ liters )

P2
(1 _ atm)
V2 = 170 liters
P1V1 P2V2

P2
P2
to get
P1
V1  V2
P2
6. Calculate the constant k for a gas with a volume of 12.5 liters at a pressure of 100 atmospheres . What are the
units of the constant?
PV = k, (100 atm)(12.5 liters) = 1250 liters·atmospheres
a. What would the volume of the gas be if the pressure is decreased to 75 atmospheres?
V
k (1250 _ litersatm .)

 16.67 _ liters
P
(75 _ atm.)
b. What would the volume of the gas be if the pressure is decreased to 50 atmospheres?
V
k (1250 _ litersatm .)

 25 _ liters
P
(50 _ atm.)
c. Calculate the volume for the temperatures in the list below in the same way as you have done for part a)
and b)
d. When you have calculated all the volumes. Plot the points on the graph at right.
e. Describe the graph – Is there a mathematical equation that will give this graph? What is it?
Y = 1/X this is an inverse relation
f.
Is the graph Increasing or Decreasing from left to right? Decreasing
g. Is the graph Decreasing at a constant rate, increasing rate or a decreasing rate? Decreasing, it is
becoming less steep.