Section 3.7—Gas Laws
Objectives:
•Compare different pressure units
•Relate the Kelvin and Celsius scales for
temperature
•Define standard temperature and pressure
•Define and perform calculations for 5 gas
laws
How can we calculate Pressure, Volume and
Temperature of our airbag?
Pressure Units
Pressure can be measured using several different
units.
Unit
Symbol
atmospheres
atm
Pascals, kiloPascals
Pa, kPa
millimeters of mercury or torr
mm Hg, torr
pounds per square inch
psi
Pressure Units
It is fairly easy to convert from one unit to
another. Use these as conversion factors.
1 atm =
101300 Pa =
101.3 kPa* =
760 mmHg =
760 torr =
14.7 psi
* This is the SI unit for pressure.
Practice
Convert the following pressure units.
1. 750 mm Hg to atm
2. 145 kPa to mm Hg
3. 1.5 atm to kPa
Temperature
The SI scale for temperature is the Kelvin
scale.
Kelvin (K)– temperature scale whose
lowest temperature is “0”, called 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
Temperature
It is very easy to convert temperatures in
oC to K through this expression:

C  273  K
Practice:
25 oC = ___ K
350 K = ___ OC
0 OC = ___ K
Standard Temperature and Pressure
To ensure that scientists worldwide discuss
gas behavior using the same temperature
and pressure units, the international scientific
community uses something called STP.
Standard Temperature and Pressure
(STP) = 1 atm (or the equivalent in
another unit) and 0°C (or 273 K)
Gas Laws
The gas laws summarize observations
regarding how gases behave under different
conditions.
Recall that the KMT explains why gases
behave as they do. The laws summarize how
they behave.
“Before” and “After” in Gas Laws
4 of the gas laws we will learn have “before” and
“after” conditions.
In the one of the laws we will learn, the expression is:
P1V1 = P2V2
P1 and V1 represent the starting conditions (or conditions
“before” they are changed) of pressure and volume for a
sample of gas.
P2 and V2 represent the resulting conditions (or conditions
“after” they are changed) of pressure and volume for the
same gas.
Avogadro’s Law
Avogadro’s Law relates # of moles and volume when
temperature and pressure are held constant.
V1 V2

n1 n2
V = Volume
n = # of moles of gas
Hint: The two volume units must be the same
on both sides of the equation!
Avogadro’s Law
V1 V2

n1 n2
n1 = 0.15 mol
V1 = 2.5 L
n2 = 0.55 mol
V2 = ?
Problem: 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?
2.5 L = V2
0. 15 mol
0.55 mol
V2 = (0.55 mol)(2.5L)
(0.15 mol)
V2 = 9.2 L
Practice Problems
 If a container holds 3.6 moles of helium in 4.7 L,
how many moles could it hold if the container is
increased to 5.8 L? (Hint: Use crossmultiplication to solve!)
 A sample of 9.5 mol of oxygen occupies 8.2 L.
What volume of the gas would be occupied by
13.0 mol?
 2.8 mol of a gas are put in a container that has a
volume of 15.0 L. If the size of the container is
increased to 25 L, how many moles of gas could
it then hold?
Boyles’ Law
Boyles’ Law relates pressure and volume when the
temperature and number of moles are held constant.
P1V1  P2V2
P = pressure
V = volume
The two pressure units must match and
the two volume units must match!
Boyle’s Law
P1V1  P2V2
V1 = 2.5 L
P1 = 1.05 atm
V2 = ?
P2 = 0.980 atm
Problem: A 2.5 L gas sample is held at
1.05 atm. What volume will it occupy if
the pressure is changed to 0.980 atm.
(1.05 atm)(2.5 L) = (0.980 atm)V2
V2 = (1.05 atm)(2.5 L)
(0.980 atm)
V2 = 2.7 L
Practice Problems
 If 98 mL of a gas at 50.0 kPa has its volume
reduced to 65 mL, what will the new pressure
be?
 A gas cylinder contains 25.0 L of gas under 5.0
atm of pressure. If the gas is transferred to a
30.0 L cylinder, what will the new pressure of
the gas be?
 A sample of gas has a volume of 12 L at 0.78
atm. The sample is put into a new tank under
1.4 atm of pressure. What is the volume of the
tank?
Charles’ Law
Charles’ Law relates temperature and Volume when
pressure and number of moles are held constant.
V1 V2

T1 T2
V = Volume
T = Temperature
The two volume units must match and
temperature must be in Kelvin!
Charles’ Law
V1 V2

T1 T2
Problem: What is the final volume if a
10.5 L sample of gas is heated
from 25C to 50C?
V2 = ?
V1 = 10.5 L
T1 = 25 oC = 298 K
T2 = 50 oC = 323 K
10.5 L = V2
298 K
323 K
V2 = (323K)(10.5 L)
298 K
V2 = 11.4 L
Practice Problems
 A gas sample at 40 0C occupies a volume of
2.32 L. If the temperature is raised to 75 0C,
what will the volume be, assuming the pressure
remains constant?
 A gas at 89 0C occupies a volume of 0.67 L. At
what temperature will the volume be increased
to 1.12 L?
 What is the volume of air in a balloon that
occupies 0.620 L at 25 0C if the temperature is
lowered to 0 0C?
Combined Gas Law
P1V1 P2V2

n1T1 n2T2
P = Pressure
V = Volume
n = # of moles
T = Temperature
Each “pair” of units must
match and
temperature must be in
Kelvin!
P1V1 P2V2

n1T1 n2T2
Problem
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
when enough gas is added to increase the number
of moles to 0.225 mole?
V1 = 1.5 L
n1 = 0.125 mol
P1 = 1.7 atm
T1 = 298 K
V2 = ?
n2 = 0.225 mol
P2 = 1.0 atm
T2 = 273 K
(1.7 atm)(1.5 L) =
(1.0 atm)V2_
(0.125 mol)(298 K)
(0.225 mol)(273 K)
V2 = (0.225 mol)(273 K)(1.7 atm)(1.5 L) = 4.2 L
(0.125 mol)(298 K)(1.0 atm)
Practice Problems
1. 3.0 mol of gas at 110 kPa and 30 0C fills a
flexible container with an initial volume of 2.0 L.
If the temperature is raised to 80 0C, the
pressure is increased to 440 kPa, and the
number of moles increases to 4.5 mol, what is
the new volume?
2. At STP, a 6.5 mol gas sample occupies 30 mL.
If the temperature is increased to 30 0C, the
entire sample is transferred to a 20 mL
container, and 3.5 additional moles of gas are
added, what will be the gas pressure inside the
container?
The Combined Gas Law Is All You Need
The combined gas law can be used for all of the
gas law problems we have learned so far!
P1V1 P2V2

n1T1 n2T2
For example, if the temperature and number of moles are
held constant, then T1 = T2 and n1 = n2. When variables
are the same, you can cancel them on both sides.
P1V1 P2V1

n1T1 n 2T2
The combined gas law becomes Boyle’s Law!

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
The value for “R” depends on the
unit that is used for pressure.
There are two possibilities:
L * atm
0.0821
mole * K
L * kPa
8.31
mole * K
Choose the one with
units that match your
pressure units!
The Ideal Gas Law Example
PV  nRT
Problem: A sample with 0.55 mol of gas is at
105.7 kPa and 27 oC. What volume does it
occupy?
n = 0.55 mol
P = 105.7 kPa
T = 27 oC = 300 K
V=?
L * kPa
R = 8.31
mole * K
(105.7 kPa)V =
(0.55 mol)(8.31 L * kPa )(300 K)
mole * K
V = (0.55)(8.31 L-kPa)(300)
(105.7 kPa)
V = 13 L
Practice Problems
1. Calculate the number of moles of gas contained in a 3.0 L
vessel at 3000 K with a pressure of 1.50 atm.
2. Determine the temperature of 2.49 moles of gas
contained in a 1.00 L vessel at a pressure of 143 kPa.
3. Calculate the pressure that a 0.323 mol sample of a gas
have at 265 K and a volume of 7.8 L.
R=
0.0821
L * atm
mole * K
or
8.31
L * kPa
mole * K
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