Chapter 11
Gas Laws
The Gas Phase
• Gases have no distinct volume or shape.
• Gases expand to fill the volume of their
container.
• Gas particles are miscible with each other.
• Evidence for gas particles being far apart :
 We can see through gases
 We can walk through gases
 Gases are compressible
 Gases have low densities
The Air We Breathe
Composition of Earth’s Atmosphere
Compound
Nitrogen
Oxygen
%(Volume)
78.08
20.95
Mole Fractiona
0.7808
0.2095
Argon
Carbon dioxide
Methane
Hydrogen
0.934
0.033
2 x 10-4
5 x 10-5
0.00934
0.00033
2 x 10-6
5 x 10-7
a. mole fraction = mol component/total mol in mixture.
Kinetic Theory of Gases
Kinetic Theory Postulates:
• Gas particles are sizeless relative to the volume of the
gas
• Gas particles are in constant rapid motion
• Gas particles have elastic collisions; means no kinetic
energy is lost on impact.
• The absolute temperature is directly proportional to
the kinetic energy of a gas.
• Gas particles have no attraction to each other; i.e. no
inter particle froces.
Parameters Affecting Gases
• Pressure (P); atm, mmHg, torr, lbs/in2
• Volume (V); L, mL
• Temperature (T); K (only)
• Number of Moles (n)
Pressure
Pressure is equal to force/unit area (P =F/A) lbs/in2
Force is a push which comes from gas particles
striking a container wall
Pressure Units
•
•
SI units = Newton/meter2 = 1 Pascal (Pa)
1 standard atmosphere (atm) = 101,325 Pa
 1 atm =760 mm Hg
 1 atm = 760 torr (torr is abbreviation of mmHg)



1 atm = 14.7 lbs/in2
1 atm = 1.013 barr
Barr = 100 kPa
Measurement of Pressure
What is above mercury?
Measurement of Pressure
Elevation and Atmospheric Pressure
Units for Expressing Pressure
Unit
Value
Atmosphere
1 atm
Pascal (Pa)
1 atm = 1.01325 x 105 Pa
Kilopascal (kPa) 1 atm = 101.325 kPa
mmHg
1 atm = 760 mmHg
Torr
1 atm = 760 torr
Bar
1 atm = 1.01325 bar
mbar
1 atm = 1013.25 mbar
psi
1 atm = 14.7 psi
Pressure Measurement
Open Tube Manometer
= 15 mm
Is the atmosphere or the gas
in the canister pushing
harder?
Pressure Measurement
Open Tube Manometer
Is the atmosphere or the gas
in the canister pushing
harder? Gas in the canister
If the atmospheric pressure is
= 15 mm 766 mm, then what is the
pressure of the canister?
Pressure Measurement
Open Tube Manometer
Is the atmosphere or the gas in
the canister pushing harder? Gas
in the canister
If the atmospheric pressure is 766
= 15 mm mm, then what is the pressure of
the canister?
P = 766 + 15 = 781 mm (torr)
gas
Pressure Measurement
Open Tube Manometer
gas
Is the atmosphere or the gas
in the canister pushing
harder?
Pressure Measurement
Open Tube Manometer
= 13 mm
gas
Is the atmosphere or the gas
in the canister pushing
harder? The atmosphere
What is the pressure of the
gas if the atmosphere is 766
mm?
Pressure Measurement
Open Tube Manometer
= 13 mm
gas
Is the atmosphere or the gas
in the canister pushing
harder? The atmosphere
What is the pressure of the
gas if the atmosphere is 766
mm? 753 mm
Pressure Measurement
Open Tube Manometer
gas
Now what is pushing harder,
the gas or the atomosphere?
Pressure Measurement
Open Tube Manometer
gas
Now what is pushing harder,
the gas or the atmosphere?
Neither, both the same.
Pressure Measurement
Open Tube Manometer
Now what is pushing harder,
the gas or the atmosphere?
Neither, both the same.
Is the gas canister empty?
gas
Pressure Measurement
Open Tube Manometer
Now what is pushing harder,
the gas or the atmosphere?
Neither, both the same.
Is the gas canister empty?
No, completely full of gas!
gas
Dalton’s Law of Partial Pressures
• For a mixture of gases in a
container
• PTotal = P1 + P2 + P3 + . .
.
Boyles Law
Atm
●
●
●
●
Consider a gas in a closed system
containing a movable plunger. If the
plunger is not moving up or down,
what can be said about the pressure
of the gas relative to the
atmospheric pressure?
Boyles Law
Atm
●●
●
●
●
●
●
Suppose we add some red gas to the
container, what would happen to
the collisions of gas particles with
container walls. Would they
increase, decrease or stay the same?
Boyles Law
Atm
●●
●
●
●
●
●
Suppose we add some red gas to the
container, what would happen to
the collisions of gas particles with
container walls. Would they
increase, decrease or stay the same?
More particles, more collisions, and
more pressure.
What happens to the plunger?
Boyles Law
Atm
●●
●
●
●
●
●
Suppose we add some red gas to the
container, what would happen to
the collisions of gas particles with
container walls. Would they
increase, decrease or stay the same?
More particles, more collisions, and
more pressure.
What happens to the plunger?
Boyles Law
●
●●
●
●
●
●
●
●
The number of
particles remain the
same, but the surface
area they have to strike
increases, thus the
number of collisions
per square inch
decrease as the
plunger goes up
exposing more surface
area causing a
decrease in pressure.
Boyle’s Law
Pressure and volume are
inversely proportional.
• P  1/V (T and n
fixed)
• P  V = Constant
• P1V1 = P2V2
Charles’s Law
• The volume of a gas is
directly proportional to
Kelvin temperature,
and extrapolates to
zero at zero Kelvin.
V  T
(P & n are constant)
V1 = V2
T1
T2
Combined Gas Law
• Combining the gas laws the relationship
P T(n/V) can be obtained.
• If n (number of moles) is held constant,
then PV/T = constant.
P1V1
T1
=
P2 V 2
T2
Temperature, K (only)
Pressure: Atm, mmHg, Torr, PSI, KPa
Volume: L, mL, cm3, …
Example
A balloon is filled with hydrogen to a
pressure of 1.35 atm and has a volume of
2.54 L. If the temperature remains constant,
what will the volume be when the pressure is
increased to 2.50 atm?
P1V1
P2 V 2 (1.35 atm)(2.54 L)
(2.50atm)V2
=
=
T1
T1
T1
T2
(1.35 atm)(2.54 L)
V2 = (2.50atm)
Constant Temp. means T1=T2
V2 = 1.37 L
Example
A sample of oxygen gas is at 0.500 atm
and occupies a volume of 11.2 L at 00C,
what volume will the gas occupy at 6.00
atm at room temperature (250C)?
Ideal Gas Law
PV = nRT
R = universal gas constant
= 0.08206 L atm K-1 mol-1
P = pressure in atm
V = volume in liters
n = moles
T = temperature in Kelvin
STP
•
“STP” means standard temperature and standard
pressure
 P = 1 atmosphere
 T = 0C
 The molar volume of an ideal gas is 22.42 liters
at STP (put 1 mole, 1 atm, R, and 273 K in the
ideal gas law and calculate V)
Example
Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.
Example
Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.
0.0821 L-atm
Mole-K
Example
Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.
0.0821 L-atm
Mole-K
3.3 L
Example
Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.
0.0821 L-atm
298 K
Mole-K
3.3 L
Example
Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.
0.0821 L-atm
298 K 1.2 mole
Mole-K
3.3 L
Example
Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.
0.0821 L-atm
298 K 1.2 mole = 8.9 atm
Mole-K
3.3 L
Example
An experiment shows that a 0.495 g sample
of an unknown gas occupies 127 mL at
98°C and 754 torr pressure. Calculate the
molar mass of the gas.
Example
An experiment shows that a 0.495 g sample
of an unknown gas occupies 127 mL at
98°C and 754 torr pressure. Calculate the
molar mass of the gas.
0.0821 L-atm
mole-K
Example
An experiment shows that a 0.495 g sample
of an unknown gas occupies 127 mL at
98°C and 754 torr pressure. Calculate the
molar mass of the gas.
0.0821 L-atm
mole-K
0.495 g
Example
An experiment shows that a 0.495 g sample
of an unknown gas occupies 127 mL at
98°C and 754 torr pressure. Calculate the
molar mass of the gas.
0.0821 L-atm
mole-K
0.495 g mL
10-3 L
Example
An experiment shows that a 0.495 g sample
of an unknown gas occupies 127 mL at
98°C and 754 torr pressure. Calculate the
molar mass of the gas.
0.0821 L-atm
mole-K
0.495 g mL
10-3 L
127 mL
Example
An experiment shows that a 0.495 g sample
of an unknown gas occupies 127 mL at
98°C and 754 torr pressure. Calculate the
molar mass of the gas.
0.0821 L-atm
mole-K
0.495 g mL
10-3 L
760 torr
127 mL atm
Example
An experiment shows that a 0.495 g sample
of an unknown gas occupies 127 mL at
98°C and 754 torr pressure. Calculate the
molar mass of the gas.
0.0821 L-atm
mole-K
0.495 g mL
10-3 L
760 torr
127 mL atm
754 torr
Example
An experiment shows that a 0.495 g sample
of an unknown gas occupies 127 mL at
98°C and 754 torr pressure. Calculate the
molar mass of the gas.
0.0821 L-atm
mole-K
0.495 g mL
10-3 L
760 torr
371 K
127 mL atm
754 torr
Example
An experiment shows that a 0.495 g sample
of an unknown gas occupies 127 mL at
98°C and 754 torr pressure. Calculate the
molar mass of the gas.
0.0821 L-atm
mole-K
0.495 g mL
10-3 L
760 torr
371 K
127 mL atm
754 torr
= 120 g/mole
Collecting a Gas Over Water
Practice
A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 25°C. The total volume of the
collected gas is 229 mL at a pressure of 755 torr.
How many moles of oxygen formed?
Hint: The gas collected is a mixture so use
Dalton’s Law to calculate the pressure of oxygen
then the ideal gas law to find the number of
moles oxygen.
PT = PO + P
2
H2O
Vapor Pressure of Water
Practice
A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 25°C. The total volume of the
collected gas is 229 mL at a pressure of 755 torr.
How many moles of oxygen formed?
Hint: The gas collected is a mixture so use
Dalton’s Law to calculate the pressure of oxygen
then the ideal gas law to find the number of
moles oxygen.
PT = PO+
PH2O
2
755 torr = PO2+ 23.8 torr
P
O2
= 755 – 23.8 = 731 torr
Practice
• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 25°C. The total volume of the
collected gas is 229 mL at a pressure of 755 torr. How
many moles of oxygen formed?
mole-K
0.0821 L-atm
Practice
• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 25°C. The total volume of the
collected gas is 229 mL at a pressure of 755 torr. How
many moles of oxygen formed?
mole-K
atm
0.0821 L-atm 760 torr
Practice
• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 25°C. The total volume of the
collected gas is 229 mL at a pressure of 755 torr. How
many moles of oxygen formed?
mole-K
atm
731 torr
0.0821 L-atm 760 torr
Practice
• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 25°C. The total volume of the
collected gas is 229 mL at a pressure of 755 torr. How
many moles of oxygen formed?
mole-K
atm
731 torr
298 K
0.0821 L-atm 760 torr
Practice
• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 25°C. The total volume of the
collected gas is 229 mL at a pressure of 755 torr. How
many moles of oxygen formed?
mole-K
atm
731 torr
10-3 L
298 K mL
0.0821 L-atm 760 torr
Practice
• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 25°C. The total volume of the
collected gas is 229 mL at a pressure of 755 torr. How
many moles of oxygen formed?
mole-K
atm
731 torr
10-3 L 229 mL
298 K mL
0.0821 L-atm 760 torr
= 9.00 X 10-3 mole
Stoichiometry and Gases
Calculate the volume of hydrogen gas at 25°C and 766 torr from
13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq)
13.4 g Zn
ZnCl2 (aq) + H2 (g)
Stoichiometry and Gases
Calculate the volume of hydrogen gas at 25°C and 766 torr from
13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq)
13.4 g Zn mole Zn
65.39 g Zn
ZnCl2 (aq) + H2 (g)
Stoichiometry and Gases
Calculate the volume of hydrogen gas at 25°C and 766 torr from
13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq)
mole H2
13.4 g Zn mole Zn
65.39 g Zn mole Zn
ZnCl2 (aq) + H2 (g)
Stoichiometry and Gases
Calculate the volume of hydrogen gas at 25°C and 766 torr from
13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq)
mole H2 0.08206 L-atm
13.4 g Zn mole Zn
65.39 g Zn mole Zn mole H2-K
ZnCl2 (aq) + H2 (g)
Stoichiometry and Gases
Calculate the volume of hydrogen gas at 25°C and 766 torr from
13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
mole H2 0.08206 L-atm 298.15 K
13.4 g Zn mole Zn
65.39 g Zn mole Zn mole H2-K
Stoichiometry and Gases
Calculate the volume of hydrogen gas at 25°C and 766 torr from
13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
mole H2 0.08206 L-atm 298.15 K 760 torr
13.4 g Zn mole Zn
65.39 g Zn mole Zn mole H2-K
Stoichiometry and Gases
Calculate the volume of hydrogen gas at 25°C and 766 torr from
13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
mole H2 0.08206 L-atm 298.15 K 760 torr
13.4 g Zn mole Zn
atm
65.39 g Zn mole Zn mole H2-K
766 torr
Stoichiometry and Gases
Calculate the volume of hydrogen gas at 25°C and 766 torr from
13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq)
ZnCl2 (aq) + H2 (g)
mole H2 0.08206 L-atm 298.15 K 760 torr
13.4 g Zn mole Zn
atm
65.39 g Zn mole Zn mole H2-K
766 torr
= 4.97 L
The End