The Gas Laws
Learning about the special behavior of
gases
Objective #3
The Ideal Gas Law, pg. 6
The Ideal Gas Law
• This equation considers a fourth
variable, the NUMBER of particles,
and the “ideal gas constant” (a.k.a.
Avogadro’s Hypothesis) into the
combined gas law.
• It would be good to familiarize
yourself with the content poster.
The Ideal Gas Law
• This equation considers a fourth
variable… the NUMBER of particles!
• This equation incorporates
Avogadro’s Hypothesis into the
combined gas law.
PV = nRT
Solving for “R”
Use the values of standard condition
• A: Using Standard Pressure in kPa
ONLY!
Solving for “R”
• A: Using Pressure in kPa
R = Pressure(P) x Volume(V)
# of mol(n)xTemp(K)
Solving for “R”
• A: Using Pressure in kPa
Volume at 1 mole
R = Pressure(P) x Volume(V) = (101.3kPa)x(22.4L)
# of mol(n)xTemp(K)
=
Standard temp.
(1 mol)x(273K)
Solving for “R”
• A: Using Pressure in kPa
R = Pressure(P) x Volume(V) = (101.3kPa)x(22.4L)
# of mol(n)xTemp(K)
=
kPa  L
R = 8.31 Mol  K
(1 mol)x(273K)
Solving for “R”
• B: Using Pressure in ATM
Solving for “R”
• B: Using Pressure in ATM
R = Pressure(P) x Volume(V)
# of mol(n)xTemp(K)
Solving for “R”
• B: Using Pressure in ATM
R = Pressure(P) x Volume(V) = (1ATM)x(22.4L)
# of mol(n)xTemp(K)
=
(1 mol)x(273K)
Solving for “R”
• B: Using Pressure in ATM
R = Pressure(P) x Volume(V) = (1ATM)x(22.4L)
# of mol(n)xTemp(K)
R = 0.0821
=
(1 mol)x(273K)
ATM  L
Mol  K
Solving for “R”
• C: Using Pressure in mm Hg
R = (P) x (V)
(n)x(K)
Solving for “R”
• C: Using Pressure in mm Hg
R = (P) x (V) = (760mmHg)x(22.4L)
(n)x(K)
=
(1 mol)x(273K)
Solving for “R”
• C: Using Pressure in mm Hg
R = (P) x (V) = (760mmHg)x(22.4L)
(n) x (K)
=
(1 mol)x(273K)
R = 62.4
mmHg  L
Mol  K
Example 1, pg. 7
• You fill a rigid steel cylinder that has a volume of 20
L with nitrogen gas to a final pressure of 20,000
kPa at 28o C. How many moles of nitrogen gas does
the cylinder contain?
• How many grams is this?
Example 1
• You fill a rigid steel cylinder that has a volume of 20 L
with nitrogen gas to a final pressure of 20,000 kPa at
28o C. How many moles of nitrogen gas does the cylinder
contain?
Rearrange the formula to find what we’re looking for:
• How many grams is this?
Example 1
• You fill a rigid steel cylinder that has a volume of 20 L
with nitrogen gas to a final pressure of 20,000 kPa at
28o C. How many moles of nitrogen gas does the cylinder
contain?
Rearrange the formula to find what we’re looking for:
n = PV
RT
• How many grams is this?
Example 1
• You fill a rigid steel cylinder that has a volume of 20 L
with nitrogen gas to a final pressure of 20,000 kPa at
28o C. How many moles of nitrogen gas does the cylinder
contain?
Rearrange the formula to find what we’re looking for:
n = PV = (20,000kPa)x(20L)
=
RT (8.31 kPaL/molK)x(301K)
• How many grams is this?
Example 1
• You fill a rigid steel cylinder that has a volume of 20 L
with nitrogen gas to a final pressure of 20,000 kPa at
28o C. How many moles of nitrogen gas does the cylinder
contain?
Rearrange the formula to find what we’re looking for:
/
n = PV = (20,000kPa)x(20L)
= _________ mol N2
/
RT (8.31 /kPaL/ /molK
/
/ )x(301K)
• How many grams is this?
Example 1
• You fill a rigid steel cylinder that has a volume of 20 L
with nitrogen gas to a final pressure of 20,000 kPa at
28o C. How many moles of nitrogen gas does the cylinder
contain?
Rearrange the formula to find what we’re looking for:
/
n = PV = (20,000kPa)x(20L)
= 159.92 mol N2
/
RT (8.31 /kPaL/ /molK
/
/ )x(301K)
• How many grams is this?
Example 1
• You fill a rigid steel cylinder that has a volume of 20 L
with nitrogen gas to a final pressure of 20,000 kPa at
28o C. How many moles of nitrogen gas does the cylinder
contain?
Rearrange the formula to find what we’re looking for:
/
n = PV = (20,000kPa)x(20L)
= 159.92 mol N2
/
RT (8.31 /kPaL/ /molK
/
/ )x(301K)
• How many grams is this?
159.92 mol N2
28 g N2
1
1 mol N2
=
Example 1
• You fill a rigid steel cylinder that has a volume of 20 L
with nitrogen gas to a final pressure of 20,000 kPa at
28o C. How many moles of nitrogen gas does the cylinder
contain?
Rearrange the formula to find what we’re looking for:
/
n = PV = (20,000kPa)x(20L)
= 159.92 mol N2
/
RT (8.31 /kPaL/ /molK
/
/ )x(301K)
• How many grams is this?
159.92 mol N2
28 g N2
1
1 mol N2
=
4,477.7 g N2
Example 2
• A deep underground cavern contains 2.24 x 106 L
of methane gas, CH4, at a pressure of 1.5 x 103
kPa and a temperature of 42o C. How many
grams of methane does this natural-gas deposit
contain?
Example 2
• A deep underground cavern contains 2.24 x 106 L
of methane gas, CH4, at a pressure of 1.5 x 103
kPa and a temperature of 42o C. How many
grams of methane does this natural-gas deposit
contain?
n = PV
RT
Example 2
• A deep underground cavern contains 2.24 x 106 L
of methane gas, CH4, at a pressure of 1.5 x 103
kPa and a temperature of 42o C. How many
grams of methane does this natural-gas deposit
contain?
n = PV = (1,500kPa)x(2.24 x 106L)
RT (8.31 kPaL/molK)x(315K)
Example 2
• A deep underground cavern contains 2.24 x 106 L
of methane gas, CH4, at a pressure of 1.5 x 103
kPa and a temperature of 42o C. How many
grams of methane does this natural-gas deposit
contain?
n = PV = (1,500kPa)x(2.24 x 106L) = 1.28X106 mol CH4
RT (8.31 kPaL/molK)x(315K)
Example 2
• A deep underground cavern contains 2.24 x 106 L
of methane gas, CH4, at a pressure of 1.5 x 103
kPa and a temperature of 42o C. How many
grams of methane does this natural-gas deposit
contain?
n = PV = (1,500kPa)x(2.24 x 106L) = 1.28X106 mol CH4
RT (8.31 kPaL/molK)x(315K)
1.28X106 mol CH4 x 16g CH4
1
1 mol CH4
Example 2
• A deep underground cavern contains 2.24 x 106 L
of methane gas, CH4, at a pressure of 1.5 x 103
kPa and a temperature of 42o C. How many
grams of methane does this natural-gas deposit
contain?
n = PV = (1,500kPa)x(2.24 x 106L) = 1.28X106 mol CH4
RT (8.31 kPaL/molK)x(315K)
1.28X106 mol CH4 x 16g CH4
1
1 mol CH4
=
2.05X107 g CH4
Example 3.
When the temperature of a rigid hollow sphere
containing 685 L of helium gas is held at 621 K, the
pressure of the gas is 1.89 x 103 kPa. How many
moles of helium does the sphere contain?
Example 3.
When the temperature of a rigid hollow sphere
containing 685 L of helium gas is held at 621 K, the
pressure of the gas is 1.89 x 103 kPa. How many
moles of helium does the sphere contain?
n = PV
RT
Example 3.
When the temperature of a rigid hollow sphere
containing 685 L of helium gas is held at 621 K, the
pressure of the gas is 1.89 x 103 kPa. How many
moles of helium does the sphere contain?
n = PV = (1.89x103 kPa)x(685L)
RT (8.31 kPa L/mol K)x(621K)
Example 3.
When the temperature of a rigid hollow sphere
containing 685 L of helium gas is held at 621 K, the
pressure of the gas is 1.89 x 103 kPa. How many
moles of helium does the sphere contain?
n = PV = (1.89x103 kPa)x(685L) = 250.88 moles He
RT (8.31 kPa L/mol K)x(621K)
Example 4
What pressure will be exerted by 0.45 mol of a gas at
25o C if it is contained in a 0.65 L vessel?
Example 4
What pressure will be exerted by 0.45 mol of a gas at
25o C if it is contained in a 0.65 L vessel?
P = nRT
V
Example 4
What pressure will be exerted by 0.45 mol of a gas at
25o C if it is contained in a 0.65 L vessel?
P = nRT = (0.45 mol)x(8.31 kPaL/
V
0.65 L
molK
)x(298K)
Example 4
What pressure will be exerted by 0.45 mol of a gas at
25o C if it is contained in a 0.65 L vessel?
P = nRT = (0.45 mol)x(8.31 kPaL/molK)x(298K) =1714.4
V
0.65 L
kPa
Example 5
A child has a lung capacity of 2.2 L. How many grams of air
do her lungs hold at a pressure of 102 kPa and a normal
body temperature of 37o C? Air is a mixture, but you may
assume it has a molar mass of 28 g/mole.
Example 5
A child has a lung capacity of 2.2 L. How many grams of air
do her lungs hold at a pressure of 102 kPa and a normal
body temperature of 37o C? Air is a mixture, but you may
assume it has a molar mass of 28 g/mole.
n = PV
RT
Example 5
A child has a lung capacity of 2.2 L. How many grams of air
do her lungs hold at a pressure of 102 kPa and a normal
body temperature of 37o C? Air is a mixture, but you may
assume it has a molar mass of 28 g/mole.
n = PV = __(102kPa)x(2.2L)__
RT (8.31kPaL/molK)x(310K)
Example 5
A child has a lung capacity of 2.2 L. How many grams of air
do her lungs hold at a pressure of 102 kPa and a normal
body temperature of 37o C? Air is a mixture, but you may
assume it has a molar mass of 28 g/mole.
n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air
RT (8.31kPaL/molK)x(310K)
Example 5
A child has a lung capacity of 2.2 L. How many grams of air
do her lungs hold at a pressure of 102 kPa and a normal
body temperature of 37o C? Air is a mixture, but you may
assume it has a molar mass of 28 g/mole.
n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air
RT (8.31kPaL/molK)x(310K)
0.87 mol air x 28g air
1
1 mol air
Example 5
A child has a lung capacity of 2.2 L. How many grams of air
do her lungs hold at a pressure of 102 kPa and a normal
body temperature of 37o C? Air is a mixture, but you may
assume it has a molar mass of 28 g/mole.
n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air
RT (8.31kPaL/molK)x(310K)
0.87 mol air x 28g air = 2.4g air
1
1 mol air
Example 6
What volume will 12 grams of oxygen gas occupy at
25o C and a pressure of 52.7 kPa?
Please do Example 6 with your neighbor.
Example 6
What volume will 12 grams of oxygen gas occupy at
25o C and a pressure of 52.7 kPa?
12g O2 x 1 mol O2 = 0.375 mol O2
1
32g O2
V = nRT = (0.375 molO2)x(8.31kPaL/molK)x(298K) =
P
(52.7 kPa)
17.62 L O2
It’s always a good idea to
regularly review the notes
we’ve take up to this point.
Avogardro’s Hypothesis
• Although gas molecules of different
gases are different sizes, equal
volumes of gases at the same
temperature and pressure contain
equal number of particles.
In other words…
• Even though Chlorine gas molecules
are 35 times bigger than hydrogen
gas molecules, equal numbers of the
two gases would occupy the same
volume at the same temperature and
pressure.
Example 1
Determine the volume occupied by .202
mole of a gas at STP
Example 1
Determine the volume occupied by .202
mole of a gas at STP
.202 mol x 22.4 L
1
1 mol
Example 1
Determine the volume occupied by .202
mole of a gas at STP
.202 mol x 22.4 L = 4.52 L
1
1 mol
Example 2
What is the volume occupied by .742
mol of Argon at STP?
Example 2
What is the volume occupied by .742
mol of Argon at STP?
.742 mol Ar x 22.4L Ar
1
1 mol Ar
Example 2
What is the volume occupied by .742
mol of Argon at STP?
.742 mol Ar x 22.4L Ar = 16.62 L Ar
1
1 mol Ar
Example 3
• Determine the volume occupied by 14
grams of nitrogen gas at STP?
Example 3
• Determine the volume occupied by 14
grams of nitrogen gas at STP?
14g N2 x 1 mol N2
1
28g N2
Example 3
• Determine the volume occupied by 14
grams of nitrogen gas at STP?
14g N2 x 1 mol N2 = 0.5 mol N2
1
28g N2
Example 3
• Determine the volume occupied by 14
grams of nitrogen gas at STP?
14g N2 x 1 mol N2 = 0.5 mol N2
1
28g N2
0.5 mol N2 x 22.4 L N2
1
1 mol N2
Example 3
• Determine the volume occupied by 14
grams of nitrogen gas at STP?
14g N2 x 1 mol N2 = 0.5 mol N2
1
28g N2
0.5 mol N2 x 22.4 L N2 = 11.2 L N2
1
1 mol N2