Section 3.7—Gas Laws
How can we calculate Pressure, Volume and Temperature of
our airbag?
Pressure Units
Several units are used when describing pressure
Unit
Symbol
atmospheres
atm
Pascals, kiloPascals
Pa, kPa
millimeters of mercury
mm Hg
pounds per square inch
psi
1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi
Definition
Kelvin (K)– temperature scale with an
absolute zero
Temperatures cannot fall below an absolute zero
A temperature scale with absolute zero is needed in Gas Law
calculations because you can’t have negative pressures or
volumes

C  273  K
Definition
Standard Temperature and Pressure
(STP) – 1 atm (or the equivalent in
another unit) and 0°C (273 K)
Problems often use “STP” to indicate quantities…don’t forget
this “hidden” information when making your list!
Gas Laws
KMT and Gas Laws
The Gas Laws are the experimental
observations of the gas behavior that
the Kinetic Molecular Theory explains.
“Before” and “After” in Gas Laws
This section has 4 gas laws which have
“before” and “after” conditions.
For example:
P1 P2

n1 n2
Where P1 and n1 are pressure and # of moles “before”
and P2 and n2 are pressure and # of moles “after”
Both sides of the equation are talking about the same sample of
gas—with the “1” variables before a change, and the “2” variables
after the change
Avogadro’s Law
Avogadro’s Law relates # of particles (moles) and
volume.
Where Temperature and Pressure are held constant
V1 V2

n1 n2
Example:
V = Volume
n = # of moles of gas
The two volume units must match!
A sample with 0.15 moles of gas has a volume of 2.5 L.
What is the volume if the sample is increased to 0.55 moles?
Avogadro’s Law
Avogadro’s Law relates # of particles (moles) and
volume.
Where Temperature and Pressure are held constant
V1 V2

n1 n2
Example:
V = Volume
n = # of moles of gas
The two volume units must match!
A sample with 0.15 moles of gas has a volume of 2.5 L.
What is the volume if the sample is increased to 0.55 moles?
n1 = 0.15 moles
V1 = 2.5 L
n2 = 0.55 moles
V2 = ? L
2.5 L
V2

0.15mole 0.55mole
0.55mole  2.5 L
 V2
0.15mole
V2 = 9.2 L
Boyles’ Law
Boyles’ Law relates pressure and volume
Where temperature and # of molecules are
held constant
P1V1  P2V2
P = pressure
V = volume
The two pressure units must match and
the two volume units must match!
Example:
A gas sample is 1.05 atm when 2.5 L. What volume is it if
the pressure is changed to 0.980 atm?
Boyles’ Law
Boyles’ Law relates pressure and volume
Where temperature and # of molecules are
held constant
P1V1  P2V2
P = pressure
V = volume
The two pressure units must match and
the two volume units must match!
Example:
A gas sample is 1.05 atm when 2.5 L. What volume is it if
the pressure is changed to 0.980 atm?
P1 = 1.05 atm
V1 = 2.5 L
P2 = 0.980 atm
V2 = ? L
1.05atm 2.5L  0.980atmV2
1.05atm  2.5L
 V2
0.980atm
V2 = 2.7 L
Charles’ Law
Charles’ Law relates temperature and pressure
V1 V2

T1 T2
Example:
V1 = 10.5 L
T1 = 25C
Where pressure and # of molecules are held
constant
V = Volume
T = Temperature
The two volume units must match and
temperature must be in Kelvin!
What is the final volume if a 10.5 L sample of gas is changed
from 25C to 50C?
Temperature needs to be in Kelvin!
25C + 273 = 298 K
V2 = ? L
T2 = 50C
50C + 273 = 323 K
Charles’ Law
Charles’ Law relates temperature and pressure
Where pressure and # of molecules are held
constant
V1 V2

T1 T2
Example:
V = Volume
T = Temperature
The two volume units must match and
temperature must be in Kelvin!
What is the final volume if a 10.5 L sample of gas is changed
from 25C to 50C?
V1 = 10.5 L
T1 = 25C = 298 K
V2 = ? L
T2 = 50C = 323 K
10.5 L
V2

298 K 323K
323K 10.5 L
 V2
298K
V2 = 11.4 L
Combined Gas Law
P1V1 P2V2

n1T1 n2T2
Example:
P = Pressure
V = Volume
n = # of moles
T = Temperature
Each “pair” of units must
match and
temperature must be in
Kelvin!
What is the final volume if a 0.125 mole sample of gas at 1.7
atm, 1.5 L and 298 K is changed to STP and particles are
added to 0.225 mole?
Combined Gas Law
P = Pressure
V = Volume
n = # of moles
T = Temperature
P1V1 P2V2

n1T1 n2T2
Example:
P1 = 1.7 atm
V1 = 1.5 L
What is the final volume if a 0.125 mole sample of gas at 1.7
atm, 1.5 L and 298 K is changed to STP and particles are
added to 0.225 mole?
n1 = 0.125 mole
T1 = 298 K
P2 = 1.0 atm
V2 = ? L
n2 = 0.225 mole
T2 = 273 K
Each “pair” of units must
match and
temperature must be in
Kelvin!
STP is standard temperature (273 K) and pressure (1 atm)
1.7 atm 1.5 L
1.0atm  V2

0.125mole  298K 0.225mole  273K
0.225mole  273K 1.7atm 1.5L
 V2
1.0atm  0.125mole  298K
V2 = 4.2 L
Why you really only need 1 of these
The combined gas law can be used for all “before”
and “after” gas law problems!
P1V1 P2V2

n1T1 n2T2
For example, if volume is held constant, then
V1  V2
and the combined gas law becomes: P1V1
P2V1

n1T1 n2T2
When two variables on opposites sides are the same, they cancel
out and the rest of the equation can be used.
P1
P2

n1T1 n2T2
Transforming the Combined Law
Watch as variables are held constant and the
combined gas law “becomes” the other 3 laws
Hold pressure and
temperature constant
P1V1 P2V2

n1T1 n2T2
Avogadro’s Law
Hold moles and
temperature constant
P1V1 P2V2

n1T1 n2T2
Boyles’ Law
Hold pressure and
moles constant
P1V1 P2V2

n1T1 n2T2
Charles’ Law
The Ideal Gas Law
The Ideal Gas Law does not compare situations—it
describes a gas in one situation.
PV  nRT
P = Pressure
V = Volume
n = moles
R = Gas Law Constant
T = Temperature
There are two possibilities for “R”:
L * atm
0.0821
mole * K
L * kPa
8.31
mole * K
Choose the one with
units that match your
pressure units!
Volume must be in Liters when using “R” to allow the unit to cancel!
The Ideal Gas Law Example
The Ideal Gas Law does not compare situations—it
describes a gas in one situation.
PV  nRT
Example:
P = Pressure
V = Volume (in L)
n = moles
R = Gas Law Constant
T = Temperature
A sample with 0.55 moles of gas is at 105.7 kPa and 27°C.
What volume does it occupy?
The Ideal Gas Law Example
The Ideal Gas Law does not compare situations—it
describes a gas in one situation.
P = Pressure
V = Volume (in L)
n = moles
R = Gas Law Constant
T = Temperature
PV  nRT
Example:
A sample with 0.55 moles of gas is at 105.7 kPa and 27°C.
What volume does it occupy?

n = 0.55 moles
P = 105.7 kPa
(105.7kPa) V  (0.55mole)  8.31 L * kPa
T = 27°C + 273 = 300 K
V=?
R = 8.31 L kPa / mole K
V

(0.55mole)  8.31 L * kPa
Chosen to match the kPa in the “P” above
mole * k
(105.7 kPa)
mole * k
 (300K )
 (300K )
V2 = 13 L
Let’s Practice
Example:
What is the final volume if a 15.5 L sample of gas at 755 mm
Hg and 298 K is changed to STP?
Let’s Practice
Example:
What is the final volume if a 15.5 L sample of gas at 755 mm
Hg and 298 K is changed to STP?
P1 = 755 mm Hg
V1 = 15.5 L
T1 = 298 K
P2 = 760 mm Hg
STP is standard temperature (273 K) and pressure (1 atm or
760 mm Hg)
“moles” is not mentioned in the problem—therefore it is
being held constant.
It is not needed in the combined law formula.
P1V1 P2V2

n1T1 n2T2
V2 = ? L
T2 = 273 K
755mm Hg 15.5L 760mm Hg  V2

298K
273K
273K  755mm Hg 15.5 L
 V2
760mm Hg  298 K
V2 = 14.1 L
What did you learn about
airbags?
Airbags
Use different
States
of
Matter
With different
Work because of changes
Changes
To produce
Which is a
Gas
Properties
Properties explained by
One of which is
Density
Kinetic
Molecular
Theory
Explanation for
Gas Laws