Boyle's Law

1
Boyle's Law
2
Sir Robert Boyle
(1627-1691)

• The first chemist to perform a truly quantitative
  experiment
• Investigated the relationship between volume and
  pressure in gases
• Result Boyles Law

3
Boyles Law
The volume of a definite quantity of dry gas is
inversely proportional to its pressure, provided
the temperature remains constant.
4
Boyles Law An Explanation

• As the pressure exerted on a gas increases, its
  volume decreases, and vice versa

• In a flexible container at equilibrium, the
  pressure exerted on a gas always equals the
  pressure exerted by the gas

• Definite quantity constant number of moles

• Temperature must be constant

5
Boyles Law The Mathematics

• V ? 1/P

The graph of an inverse proportion is hyperbolic.
Volume is proportional to 1/P (meaning volume is
inversely proportional to P).

• V k(1/P)

Volume is equal to some constant, k, times 1/P
(the value of k depends only on number of moles
and temperature)

• VP k

Any volume times the corresponding pressure
(called the pressure-volume product) is equal to
the constant k.
The working form of the equation is
V1P1 V2P2
The graph of a direct proportion is a straight
line, the slope of which is equal to the constant
k.
Any pressure-volume product equals any other so
long as T and n are constant.
6
Boyles Law The Mathematics

• V ? 1/P

The graph of an inverse proportion is hyperbolic.
Volume is proportional to 1/P (meaning volume is
inversely proportional to P).

• V k(1/P)

Volume is equal to some constant, k, times 1/P
(the value of k depends only on number of moles
and temperature)

• VP k

Any volume times the corresponding pressure
(called the pressure-volume product) is equal to
the constant k.
The working form of the equation is
V1P1 V2P2
The graph of a direct proportion is a straight
line, the slope of which is equal to the constant
k.
Any pressure-volume product equals any other so
long as T and n are constant.
7
V1P1 V2P2This is the working form of the
equation, used to predict changes in V or P for a
trapped gas, such as in a balloon or engine
cylinder

• V1 initial volume

• P1 initial pressure

• V2 final volume

• P2 final pressure

• Three of the four must be known (or knowable)

• Any consistent units for V and P are acceptable

• T and n must be constant

• The identity of the gas is irrelevant.

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Important Point
The identity of the gas is irrelevant. At the
same temperature, equal numbers of moles of all
gases exhibit precisely the same behavior with
respect to volume and pressure. One mole of He, 1
mol of oxygen, 1 mol of steam, and 1 mol of
uranium hexafluoride would all occupy the same
volume provided they are at the same temperature
even though they are vastly different in terms of
molecular size and mass. This is a notable
observation and will be explained in detail
later. In the mean time, consider the
implications. Does this seem remarkable? If so,
why?
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Instructions for Example Problems
As you move through the examples, make sure you
understand each step before proceeding. In each
case, the first step is extracting the values
from the text of the problem. Next, units are
made compatible with appropriate conversions.
Finally, substitutions are made and the problems
is solved.
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2.5 liters
Example 1 A balloon filled with helium has a
volume of 2.5 liters at sea level (where the
pressure of 1 atm). It is released and rises to
an altitude where the pressure is 550 torr. What
would its volume be at that altitude.
1 atm
550 torr
What would its volume be?
1. V1 2.5 L
3. V2 ?
2. P1 1 atm
4. P2 550 torr
760 torr
760 torr
Remember, any unit for P works, but both must be
expressed in the same unit (equivalency 1 atm
760 torr). Here, we will convert atm to torr.
Substituting values into the equation
V1 P1 V2 P2
(2.50 L)
(760 torr)
V2 (550 torr)
V2 3.45 L
(2.50 L) (760 torr) V2 (550 torr)
(The final answer (V2) is in liters because the
initial volume (V1) is in liters.)
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Example 1 A balloon filled with helium has a
volume of 2.5 liters at sea level (where the
pressure of 1 atm). It is released and rises to
an altitude where the pressure is 550 torr. What
would its volume be at that altitude.
1. V1 2.5 L
3. V2 ?
2. P1 1 atm
4. P2 550 torr
760 torr
Remember, any unit for P works, but both must be
expressed in the same unit (equivalency 1 atm
760 torr). Here, we will convert atm to torr.
V1 P1 V2 P2
(2.50 L)
(760 torr)
V2 (550 torr)
V2 3.45 L
(2.50 L) (760 torr) V2 (550 torr)
(The final answer (V2) is in liters because the
initial volume (V1) is in liters.)
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10.0 liters
227 kPa
Example 2 At 227 kPa, a certain gas has a volume
of 10.0 L. If the gas is released into a 50.0
liter container, what pressure would it then
exert?
what pressure?
50.0 L
3. V2 50.0 L
1. V1 10.0 L
2. P1 227 kPa
4. P2 ?
Remember, any unit for V works as long as both
are expressed in the same unit. In this case
both are already in liters.
Substituting values into the equation
V1 P1 V2 P2
(10.0 L)
(227 kPa)
(50.0 L) P2
P2 45.4 kPa
(10.0 L)(227 kPa) P2 (50.0 L)
(The final answer (P2) is in kPa because the
initial pressure (P1) is in kPa.)
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Example 2 At 227 kPa, a certain gas has a volume
of 10.0 L. If the gas is released into a 50.0
liter container, what pressure would it then
exert?
3. V2 50.0 L
1. V1 10.0 L
2. P1 227 kPa
4. P2 ?
Remember, any unit for V works as long as both
are expressed in the same unit. In this case
both are already in liters.
V1 P1 V2 P2
(10.0 L)
(227 kPa)
(50.0 L) P2
P2 45.4 kPa
(10.0 L)(227 kPa) P2 (50.0 L)
(The final answer (P2) is in kPa because the
initial pressure (P1) is in kPa.)
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For homework, print out and complete Boyles Law
Worksheet Practice Problems for this
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