Gases
Boyle’s Law
1
Pressure and Volume
2
3
Boyle’s Law
• If pressure increases, the
volume decreases
4
• the volume of a gas is inversely
related to its pressure when T and n
are constant.
5
How can we write Boyle’s Law as a
formula?
Volume is inversely
This is more usually
proportional to the
pressure and can be written as:
written as:
Pressure a 1/volume PV=k
P1V1=P2V2
P=pressure in N/m2
V=volume in dm3 (litres)
k=constant
Boyle’s Law apparatus
Below are some results of an
experiment
Pressure p Volume V PxV
1.1 40 44
1.7 26
2.2 20
2.6 17
Calculate pV (pressure x volume) for each set of
results. What do you notice?
Here are the results of the experiment
Pressure p Volume V PxV
1.1 40 44
1.7 26 44
2.2 20 44
2.6 17 44
Did you notice that if p is doubled, V is halved?
If p increases to 3 times as much, V decreases to a
1/3rd . This means:
Volume is inversely proportional to pressure, or
V1
p
If we plot volume directly against
pressure we would get a downwards
curve showing that volume gets smaller
as the pressure gets larger, and vice
versa.
Another way of plotting the data
we plot the volume against the reciprocal
of pressure (ie. 1/p)
This time the points lie close to a straight
line through the origin.
This means volume is directly proportional
to 1/pressure or
volume is inversely proportional to pressure
Boyle’s Law
For a fixed mass of gas kept at
constant temperature the
volume of the gas is inversely
proportional to its pressure.
PV Constant in Boyle’s Law
In Boyle’s Law, the product P x V is constant as long
as T and n do not change.
Boyle’s Law can be stated as
P1V1 = P2V2 (T, n constant)
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Solving for a Gas Law Factor
The equation for Boyle’s Law can be rearranged to
solve for any factor.
P1V1 = P2V2 Boyle’s Law
To solve for V2 , divide both sides by P2.
P1V1 = P2V2
P2 P2
V1 x P1 = V2
P2
14
Boyles’ Law and Breathing
During an inhalation,
• the lungs expand.
• the pressure in the
lungs decreases.
• air flows towards
the lower pressure
in the lungs.
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
15
Boyles’ Law and Breathing
During an exhalation,
• lung volume
decreases.
• pressure within the
lungs increases.
• air flows from the
higher pressure in
the lungs to the
outside.
Copyright © 2005 by Pearson Education, Inc.
Publishing as Benjamin Cummings
16
Calculation with Boyle’s Law
Freon-12, CCl2F2, is used in refrigeration systems.
What is the new volume (L) of a 8.0 L sample of Freon
gas initially at 550 mm Hg after its pressure is changed
to 2200 mm Hg at constant T?
1. Set up a data table:
Conditions 1 Conditions 2
P1 = 550 mm Hg P2 = 2200 mm Hg
V1 = 8.0 L V2 = ?
17
Calculation with Boyle’s Law
(Continued)
2. When pressure increases, volume decreases.
Solve Boyle’s Law for V2:
P1V1 = P2V2
V2 = V1 x P1
P2
V2 = 8.0 L x 550 mm Hg = 2.0 L
2200 mm Hg
pressure ratio
decreases volume
18
Learning Check
For a cylinder containing helium gas indicate if
cylinder A or cylinder B represents the new volume for
the following changes (n and T are constant).
1) pressure decreases
2) pressure increases
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Solution
For a cylinder containing helium gas indicate if
cylinder A or cylinder B represents the new volume for
the following changes (n and T are constant):
1) Pressure decreases B
2) Pressure increases A
20
Learning Check
If a sample of helium gas has a volume of 120 mL
and a pressure of 850 mm Hg, what is the new
volume if the pressure is changed to 425 mm Hg ?
1) 60 mL 2) 120 mL 3) 240 mL
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Solution
3) 240 mL
P1 = 850 mm Hg P2 = 425 mm Hg
V1 = 120 mL V2 = ??
V2 = V1 x P1 = 120 mL x 850 mm Hg = 240 mL
P2 425 mm Hg
Pressure ratio
increases volume
22
Learning Check
A sample of helium gas in a balloon has a volume of
6.4 L at a pressure of 0.70 atm. At 1.40 atm (T
constant), is the new volume represented by A, B, or
C?
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Solution
A sample of helium gas in a balloon has a volume of
6.4 L at a pressure of 0.70 atm. At a higher pressure
(T constant), the new volume is represented by the
smaller balloon A.
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Learning Check
If the sample of helium gas has a volume of 6.4 L
at a pressure of 0.70 atm, what is the new
volume when the pressure is increased to 1.40
atm (T constant)?
A) 3.2 L B) 6.4 L C) 12.8 L
25
Solution
If the sample of helium gas has a volume of 6.4 L
at a pressure of 0.70 atm, what is the new
volume when the pressure is increased to 1.40
atm (T constant)?
A) 3.2 L
V2 = V1 x P1
P2
V2 = 6.4 L x 0.70 atm = 3.2 L
1.40 atm
Volume decreases when there is an increase in
the pressure (temperature is constant.)
26
Learning Check
A sample of oxygen gas has a
volume of 12.0 L at 600. mm
Hg. What is the new pressure
when the volume changes to
36.0 L? (T and n constant).
1) 200. mm Hg
2) 400. mm Hg
3) 1200 mm Hg
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Solution
1) 200. mm Hg
Data table
Conditions 1 Conditions 2
P1 = 600. mm Hg P2 = ???
V1 = 12.0 L V2 = 36.0 L
P2 = P1 x V1
V2
600. mm Hg x 12.0 L = 200. mm Hg
36.0 L
28
Problem:
A deep sea diver is
working at a depth
where the pressure is
3.0 atmospheres. He is
breathing out air
bubbles. The volume of
each air bubble is 2
cm2. At the surface the
pressure is 1
atmosphere. What is
the volume of each
bubble when it reaches
the surface?
We assume that the temperature is constant,
so Boyle’s Law applies:
P1 x V1 = P2 x V2
1.0 x 2 = 3.0 x V2
Now rearrange the numbers so that you have
V2 on one side, and the rest of the numbers
on the other side of the ‘equals’ symbol.
V2 = 3.0 x 2
1.0
therefore volume of bubbles = 6 cm3
Note that P1 and P2 have the same unit, as well as V1
and V2
