Gases
Chapter 11
Tro, 2nd ed.
WHY WE STUDY GASES
Our atmosphere: thin layer surrounds us
and is critical to life on earth
78% N2 & 21% O2 at sea level, plus CO2
& water vapor & noble gases (next
slide)
POLLUTANTS: (see page 378)
SOx & NOx cause acid rain
CFC's destroy O3 uv protective layer
CO2 may increase global warming
(some is necessary or we would cool
too much at night)
Average Composition of
Dry Air
Gas Volume Percent
N2
O2
78.08%
20.95%
Ar
0.93%
CO2
0.033%
Ne
0.0018%
Gas Volume Percent
He
CH4
0.0005%
0.0002%
Kr
0.0001%
Xe, H2, and N2O Trace
BEHAVIOR & PROPERTIES
OF GASES:
- can be compressed greatly
- can expand to fill container uniformly
- have low density compared to liquids
& solids
- may be mixed - always homogeneous
mixtures because always in motion
- a confined gas exerts constant
pressure on walls of its container
uniformly in all directions
The Kinetic-Molecular Theory
KMT is based on the motions of gas
particles.
A gas that behaves exactly as outlined
by KMT is known as an ideal gas.
While no ideal gases are found in nature,
real gases can approximate ideal gas
behavior under certain conditions of
temperature and pressure.
Principle Assumptions of the KMT
(similar to pages 343-344)
1. Gases consist of tiny atomic
particles.
2. The distance between particles is
large compared with the size of
the particles themselves.
3. Gas particles have no attraction
for one another.
Principle Assumptions of the
KMT
4. Gas particles move in straight
lines in all directions, colliding
frequently with one another and
with the walls of the container.
5. No energy is lost by the collision
of a gas particle with another
gas particle or with the walls of
the container. All collisions are
perfectly elastic.
Principle Assumptions of the KMT
6. The average kinetic
energy for particles is the
same for all gases at the
same temperature, and its
value is directly
proportional to the Kelvin
temperature.
1 2
KE = mv
2
Although each gas particle (atom or molecule)
moves with its own velocity, the average velocity
of all the particles gives an average kinetic energy
to a container of gases.
GAS PROPERTIES: four
quantities define state of a gas
1.
2.
3.
4.
Quantity in moles or mass
Temperature in Kelvin
Volume in Liters
Pressure in Atmospheres (usually)
We already know about mass/moles,
temperature & volume.
Learn about pressure!
PRESSURE
Pressure = Force/unit area
Force = mass*acceleration
Force causes something to move a
distance D in work.
Gravity is a weak force, g
F = m*g, units = Newtons or lbs
Common English unit of pressure is
lbs/in2 (psi)
The pressure
resulting from
the collisions of
gas molecules
with the walls of
the balloon keeps
the balloon
inflated.
GAS PRESSURE
The pressure exerted by a gas depends
on:
- the number of gas molecules present
- the temperature of the gas
- the volume in which the gas is
confined
V = 22.414 L
T = 0.000oC
The pressure exerted by a gas is
directly proportional to the number
of molecules present.
Dependence of Pressure on
Temperature
The pressure of a gas in a fixed volume
increases with increasing temperature.
When the pressure of a gas increases, its
kinetic energy increases.
The increased kinetic energy of the gas
results in more frequent and energetic
collisions of the molecules with the
walls of the container.
Mercury Barometer
A tube of mercury
is inverted and
placed in a dish of
mercury.
The barometer is
used to measure
atmospheric
pressure.
The atmosphere above us exerts a pressure, called atmospheric pressure,
which is measured by a barometer as shown above.
Memorize:
1 torr = 1 mm Hg
1 atm = 760 torr exactly
1 atm = 14.7 lb/in2 (psi)
1 atm = 33.9 ft water
PRESSURE CONVERSION
PRACTICE
A storm is heralded by falling atmospheric
pressure. The weather report says the
pressure is down to 28.5 inches of Hg.
Convert this to torr and atmospheres.
28.5 inches * 25.4 torr = 723.9 torr
1 inch
723.9 torr * 1 atm = 0.953 atm
760 torr
Now convert 684 torr to mm Hg, atm and psi.
Boyle’s Law
At constant temperature (T), the
volume (V) of a fixed mass of gas
is inversely proportional to the
Pressure (P).
V = cb * 1/P, or V*P =cb
P1V1 = cb = P2V2
P1V1 = P2V2
Memorize this!!!
Graph of pressure versus volume. This shows the
inverse PV relationship of an ideal gas.
The effect of pressure on the volume of a gas.
Boyle’s Law Problem
An 8.00 L sample of N2 is at a pressure of 500.0
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).
First ask yourself if P will increase or decrease:
V decreased, therefore P should increase.
Second, rearrange Boyle’s Law: P2 = P1V1/V2
Third, plug in the data given:
P2 = 500.0 torr * 8.00 L/ 3.00 L
= 1333 torr
(or 1.33 x 103 torr)
Charles’ Law
At constant pressure the volume of a
fixed mass of gas is directly
proportional to the absolute
temperature.
V = cc * T, or V/T = cc
V1/T1 = cc = V2/T2
V1/T1 = V2/T2
Memorize this!
Volume-temperature relationship of methane (CH4).
Absolute Zero
on the Kelvin Scale
-273oC (more precisely –273.15oC) is
the zero point on the Kelvin scale. It
is the temperature at which an ideal
gas would have zero volume.
Effect of temperature on the volume of a gas. Pressure
is constant at 1 atm. When temperature increases at
constant pressure, the volume of the gas increases.
Charles’ Law Problem
A 255 mL sample of nitrogen at 75oC is confined at
a pressure of 3.0 atmospheres. If the pressure
remains constant, what will be the volume of the
nitrogen if its temperature is raised to 250.oC?
First, convert temperature to Kelvin ALWAYS!
T1 = 75oC = 348 K
T2 = 250.oC = 523 K
Second, rearrange Charles’ Law V2 = V1T2/T1
Third, plug in data:
V2 = 255 mL*523K/348K
= 383 mL
Gay-Lussac’s Law (added to chp)
The pressure of a fixed mass of gas, at
constant volume, is directly
proportional to the Kelvin
temperature.
P = cg*T, or P/T = cc
P1/T1 = cc = P2/T2
P1/T1 = P2/T2
Memorize this!
Gay-Lussac’s Law Problem
At a temperature of 40.oC an oxygen container is at
a pressure of 2.15 atmospheres. If the
temperature of the container is raised to 100.oC
what will be the pressure of the oxygen?
First convert to Kelvin:
T1 = 40.oC = 313 K
T2 = 100.oC = 373 K
Second, write and solve the equation for the
unknown: P2 = P1T2/T1 = 2.15 atm*373K/313K
= 2.56 atm
Standard Temperature and
Pressure
273.15 K or 0.00oC
Exactly 1 atm or 760 torr
or 760 mm Hg or 14.7 psi or…
Combined Gas Law
A combination of Boyle’s, GayLussac’s and Charles’ Law.
P1V1 = P2V2
T1
T2
Used when pressure and
temperature change at the same
time while moles/mass is constant.
Solve the equation for any one of the
6 variables
Combined Gas Law Problem
A sample of hydrogen occupies 465 ml at STP.
If the pressure is increased to 950 torr and
the temperature is decreased to –15oC, what
would be the new volume?
First convert to Kelvin: T1 = 273 K, T2 = 258 K
Second, rearrange for V2 = P1V1T2/P2T1
Third, plug in data:
V2 = 760 torr * 465 mL * 258 K = 352 mL
950 torr * 273 K
Dalton’s Law of
Partial Pressures
Each gas in a mixture exerts a pressure that
is independent of the other gases present.
The total pressure of a mixture of gases is the
sum of the partial pressures exerted by
each of the gases in the mixture.
Ptotal = Pa + Pb + Pc + Pd + ….
Dalton’s Law Problem
A container contains He at a pressure of 0.50
atm, Ne at a pressure of 0.60 atm, and Ar at
a pressure of 1.30 atm. What is the total
pressure in the container?
Ptotal = PHe + PNe+ PAr
Ptotal = 0.5 atm + 0.6 atm + 1.30 atm = 2.40
atm
(Yes, it’s that easy.)
Collecting a Gas Sample Over
Water
The pressure in the collection
container is equal to the
atmospheric pressure.
The pressure of the gas collected plus
the pressure of water vapor at the
collection temperature is equal to
the atmospheric pressure.
Ptotal = Patm = Pgas + PH O
2
Oxygen collected over
water.
Dalton’s Law Problem
(again)
A sample of O2 was collected in a bottle over
water at a temperature of 25oC when the
atmospheric pressure was 760.0 torr. The
vapor pressure of water at 25oC is 23.8 torr.
What is the pressure of the O2 gas?
Pt = 760.0 torr = PO2 + PH2O
PO2 = 760.0 torr – 23.8 torr = 736.2 torr
Avogadro’s Law
Equal volumes of different gases at the
same temperature and pressure contain
the same number of molecules. (Section
11.7)
V is directly proportional to n, where n is
moles (if at same T & P)
V1/n1 = Ca = V2/n2
V1/n1 = V2/n2
Memorize this!
Avogadro’s Law Problem
If 0.500 moles of CO2 occupies 11.2
Liters, what volume will 0.670 moles
of CO2 occupy at the same T & P?
V2 = V1n2/n1
= 11.2 L * 0.670 mol/0.500 mol
= 15.0 L
IDEAL GAS LAW
PV = nRT
R = Ideal Gas Constant
= 0.082057 L-atm/mol-K
MEMORIZE THIS R VALUE AND UNITS!
Ideal Gas Law Problem
A balloon filled with 5.00 moles of helium gas is at a
temperature of 25oC. The atmospheric pressure is
750. torr. What is the balloon’s volume?
Convert T to K = 25oC + 273 = 298K
Convert pressure to atmospheres:
750. torr (1 atm/760 torr) = 0.987 atm
Rearrange:
V = nRT/P = 5.00 mol(0.082057 L-atm/mol-K)298K
0.987 atm
= 124 L
Another Ideal Gas Law
Problem
A big balloon of H2 has a volume of 3.20 x 104 L, T
is 20.0°C, P = 750.0 torr. How many moles of
gas are in the balloon?
n = PV
RT
= (750.0 torr/760 torr/atm) * 3.20 x 104 L
0.082057 L.atm/mol.K * 293.15 K
= 1.31 x 103 moles of H2 gas
PRACTICE WITH IDEAL GAS LAW:
1. What pressure will 1.00 mol H2 exert in a
250.0 mL container at 27oC?
(98.5 atm)
2. What pressure will 1.00 mol H2 exert in a
250.0 mL container at 327oC?
(197 atm)
3. What volume will 2.00 mol H2 exert if P
is 98.5 atm and T is 27oC?
(0.500 L)
SUPER-COMBINED
GAS LAW
Rearrange Ideal Gas Law to
R = PV/nT
At first set of conditions R = P1V1/n1T1
At second set R = P2V2/n2T2
Set them equal to each other:
P 1 V1 = P 2 V2
n1T1
n2T2
This can be used to find any of the four simple
gas laws.
GROUP WORK:
Derive Boyle’s Law, Charles’ Law, GayLussac’s Law, and Avogadro’s Law
from the super-combined gas law.
Mole-Mass-Volume Relationships
Volume of one mole of any gas at
STP = 22.414 L.
22.414 L at STP is known as the
molar volume of any gas.
22.414L
Standard Molar Volume
Problem
The density of neon at STP is 0.900
g/L. What is the molar mass of
neon?
(0.900g/L)(22.414L/mol) = 20.2 g/mol
Density of Gases
grams
m
d=
v
liters
Density of Gases
m
d=
v
depends
on T and P
Gas Density Problem
The molar mass of SO2 is 64.07 g/mol.
Determine the density of SO2 at STP.
D = (64.07 g/mol)(1mol/22.414 L)
= 2.858 g/L at STP
VARIATIONS OF THE IDEAL GAS LAW
ARE USEFUL!
Relationship to molar mass: moles =
mass/molar mass
n=m
PV = mRT
M
M
Rearrange to calculate molar mass or even
density (D = m/V):
M = mRT = DRT/P
m = PM = Density
PV
V
RT
Molar Mass & Density
Problem
What is molar mass of a gas if 0.681
grams occupies 442 mL (think
Density) at 49oC and 0.629 atm?
D = 0.681 g/0.442 L = 1.5407 g/L
M = DRT/P
=1.5407g/L(0.082057L.atm/mol.K)322 K
(0.629 atm)
= 64.7 g/mol
Molar Mass & Density Problem
(again)
An unknown gas A has a density of 1.429 g/L at
0.00oC and 1.00 atm. Find its molar mass.
MA = DRT/P
= 1.429 g/L* 0.082057 L.atm/mol.K * 273.15 K
1.00 atm
= 32.03g/mol
Or:
MA = (1.429 g/L)*(22.414 L/mol) = 32.03 g/mol
Molar Mass & Density Problem
(again)
What is density of CO2 at STP and at
25.00oC? (Two ways to solve this.)
a. D = 44.01 g/mol = 1.96 g/L
22.414 L/mol
b. D = PM = 1.00 atm*44.01 g/mol
RT
0.082057 * 298.15 K
= __________g/L
CHEMICAL REACTIONS
STOICHIOMETRY OF GASES:
Stoichiometry problems that you will love to
do!
At STP one mole of any gas occupies 22.4 L
This means that equal volumes of gases at STP
have the same number of moles!!!!
11.2 L of N2 = 0.500 moles = 11.2 L of O2 =
0.500 moles = 11.2 L of He, etc.
The volumes will all change equally with
changes in T and P, if all are at same T & P.
We can relate volumes of gases just the same
as relating moles of gases, using mol/mol
ratio as a vol/vol ratio.
CHEMICAL REACTIONS
STOICHIOMETRY OF GASES:
Example using vol/vol ratio from
mol/mol ratio:
N2(g) + 3 H2(g)  2 NH3(g)
Given 355 L of hydrogen, how many L of
nitrogen required? How many L of
ammonia produced?
355 L H2 * 1 vol N2 = 118 L
3 vol H2
355 L H2 * 2 vol NH3 = 237 L
3 vol H2
CHEMICAL REACTIONS
STOICHIOMETRY OF GASES:
If 12.0 g Zn react with excess sulfuric acid, how many
moles of gas will you have at STP?
Zn(s) + H2SO4(aq)  ZnSO4(aq) + H2(g)
12.0 g/(65.38 g/mol) = 0.1835 mol Zn
0.1835 mol Zn * 1 mol H2 = 0.1835 mol H2
1 mol Zn
0.1835 mol H2 * 22.414 L/mol = 4.11 L
The same problem but at other conditions like 25oC and
1.00 atm, use the Ideal Gas Law for step 4.
V = nRT/P = 0.1835 mol * 0.082057* 298 K = 4.49 L
1.00 atm
Gas Stoichiometry Map
Primary conversions involved in
stoichiometry.
Gas Stoichiometry Problem
What volume of hydrogen, collected at 30.0oC and
700. torr, will be formed by reacting 50.0 g of
aluminum with hydrochloric acid?
2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g)
50.0gAl(1mol/26.98g)(3molH2/2molAl)=2.78 molH2
Rearrange Ideal Gas Law to V = nRT/P, convert T to
Kelvin and P to atm.
V = 2.78 mol(0.082057)303.15 K/0.9211 atm
= 75.1 L of H2
Ideal Gas
An ideal gas obeys the gas laws.
The volume the molecules of an ideal
gas occupy is negligible compared to
the volume of the gas. This is true
at all temperatures and pressures.
The intermolecular attractions between
the molecules of an ideal gas are
negligible at all temperatures and
pressures.
Real Gases
Deviations from the gas laws occur at high
pressures and low temperatures.
At high pressures the volumes of the real
gas molecules are not negligible compared
to the volume of the gas
At low temperatures the kinetic energy of
the gas molecules cannot completely
overcome the intermolecular attractive
forces between the molecules.
Diffusion
The ability of two or more gases
to mix spontaneously until they
form a uniform mixture.
Stopcock closed
No diffusion occurs
Stopcock open
Diffusion occurs
Graham’s Law of Effusion
Effusion: a process by which gas molecules
pass through a very small orifice from a
container at higher pressure to one at lower
pressure.
The rates of effusion of two gases at the
same temperature and pressure are
inversely proportional to the square roots of
their densities, or molar masses.
rate of effusion of gas A
=
rate of effusion of gas B
dB
=
dA
molar mass B
molar mass A
What is the ratio of the rate of effusion of
CO to CO2?
effusion rate CO
=
effusion rate CO2
=
molar mass CO2
molar mass CO
44.0 g
 1.25
28.0 g