Gases:
·
Can be compressed
into smaller volumes
·
Exert pressure on
their surroundings
·
Pressure must be
exerted on gases to confine them
·
Diffuse into each
other, mixing completely
·
Properties of
gases are described by volume, pressure, temperature, and number of molecules
present
Pressure – force per unit
area
·
Measured by
barometer – inverted tube of mercury in a pool of mercury
·
Air exerts
pressure on the pool of mercury causing it to rise up in the tube
·
The height of the
column is related to the intensity of the air pressure
·
Units: 1 atm =
760 Torr = 760 mm Hg = 101.3 kPa
(at 0° C and sea-level elevation)
·
STP = standard
temperature and pressure
(0° C, and 1 atm).
·
Particles are
assumed to have no effect on each other -- no attractive forces.
Boyle's Law: If the
temperature of a gas remains constant, the pressure exerted by the gas varies
inversely with the volume.
• Pressure x
Volume = constant (PV = k)
• If P
increases, V must decrease
Boyle's Law allows us to predict the volume of a gas
at a new pressure.
P1(V1) = k and k
= P2(V2)
Therefore
... P1(V1) = P2(V2)
A gas with a volume of 952 cm3
at 0.86 atm would have what volume at standard pressure?
P1
= 0.860 atm V1 = 952 cm3
P2
= 1 atm V2 = ?
P1(V1) = P2(V2)
or (0.860)(952) = 1(V2)
V2 = (0.860) (952) = 819 cm3
1
A gas with a volume of 273 cm3
at 0.594 atm would have what volume at standard pressure?
Charles' Law
• The volume
of a quantity of gas, held at fixed pressure, varies directly with the Kelvin
temperature. (K = °C + 273)
• V = k
(T) ....... V/T = k
• Charles'
Law allows us to predict what volume a gas would fill when we change the
temperature.
• V1 = V2
T1 T2
What volume would a 225 cm3 sample of gas
at 57 °C occupy at standard temperature?
V1 = 225 T1
= 330 K
V2 = ? T2 =
273K
225 = V2
330 273 V2 = 186 cm3
What volume would a 2.90 m3 sample of gas
at 226 K occupy at 23 °C?
Combined Gas Law
From
Boyle's: P1V1 = P2V2
From
Charles': V1/T1 = V2/T2
Combined: P1V1 = P2V2
T1 T2
1)
Determine the
volume of a gas at STP if it occupies 7.51 m3 at 5° C and 59.9 kPa.
P1
= 59.9 kPa V1 = 7.51 m3
T1 = 278 K
P2
= 101.3 kPa V2 = ? T2 = 273 K
59.9(7.51) = V2(101.3)
278 273 V2 = 4.36 m3
2)
Determine the volume of a gas at 68 °C and 82.4 kPa
if it occupies 149 cm3 at 18 °C and 94.7 kPa.
P1
= 94.7 kPa V1 = 149 cm3 T1
= 291 K
P2
= 82.4 kPa V2 = ? cm3 T2 = 341 K
Equal
volumes of gases at the same temperature and pressure contain the same numbers
of particles.
·
At STP, 22.414
liters of any gas contains one mole of gas particles.
What is the molecular
weight of a gas which has a density of 1.41 g/L and 1.00 mol
of which occupies 27.0 Liters?
1.41 g/L ´ 27.0 L/mol = 38.1 g/mol
What
would its density be at STP?
38.1 g/mol x 1mol/22.4 L = 1.70 g/L
Ideal
gas law equation: Relates pressure,
volume, moles of particles, and Kelvin temperature for a gas.
PV = nRT
n
= moles of gas particles
R
= gas constant = 0.08206 (L.atm)/(mole.K)
Question 1: How many moles
of a gas will occupy 2.00 L at 91 °C and 0.500 atm?
PV
= nRT
(0.500)(2.00)
= n(0.08206)(364)
n
= (0.500)(2.00) = 0.0335
moles
(0.08206)(364)
Question
2: If this gas has a mass of 1.34 grams, what is its molar mass?
Molar
mass = grams/mole
= 1.34 g/0.0335 moles
= 40.0 grams/mole
Question
3: A small cylinder of helium for use in chemistry lectures has a volume of 334
mL. How many moles of helium are
contained in the cylinder at a pressure of 150 atm at 25° C?
PV = nRT
150
atm (0.334 L) =
n (0.08206)(298 K)
n
= 150(0.334) = 2.05 moles
0.08206(298)
Question
4: How many moles would need to be released to reduce the pressure to 100 atm?
PV = nRT
100
atm (0.334 L) =
n (0.08206)(298 K)
n
= 100(0.334) = 1.37 moles
0.08206(298)
2.05
moles - 1.37 moles = 0.68 moles released
Question
5: Calculate the density of C2H6 (ethane) at 0° C and
2.00 atm.
PV = nRT
For
1 mole:
(2.00
atm)(V) = 1(0.08206)(273K)
V=
1(0.08206)(273) = 11.2 Liters
2.00
1
mole of C2H6 weighs 30.0 grams
Density
= grams/liter
= 30.0
grams/11.2 liters
= 2.68 g/L
Question
6: Chemical analysis of a gaseous compound reveals that it has a percent
composition of 23.5% C, 1.98% H, and 74.5% F.
A 0.100 g sample of the gas exerts a pressure of 70.5 mm Hg in a 256 mL
container at 22.3°
C. Find the molar mass and the molecular
formula of the compound.
·
The total
pressure in a container is the sum of the partial pressures of the gases in the
container
·
Total P = PA + PB + PC
+…
A mixture of helium and oxygen being prepared for use
in a scuba diving tank is made by mixing 46 L of O2 with 12 L
He. Both gases are initially at 25° C and 1.0 atm.
The scuba tank has a volume of 5.0 L.
Calculate the total pressure in the tank.
·
Calculate moles
of each gas
·
Calculate total
pressure based on total moles
Many gases are collected over
water, so they are mixed with water vapor -- a measurement of the total
pressure exerted by the sample will include the pressure from water vapor as
well as from the gas being examined.
A gas sample collected over water at 20° C has a
pressure of 427 Torr. What pressure is exerted by the gas
alone?
Using
Total P =
P(H20) + P(gas)
427 Torr = 17.54
Torr + x
x
= 427 – 17.54
= 409 Torr
If this gas occupies 275 mL and weighs 0.197 g, how
many moles of gas are present? What is
the molar mass of the gas?
Kinetic Molecular Theory of
Gases
Assumptions:
·
Gas molecules are
very small and very far apart
·
Gas molecules are
in continuous straight-line motion
·
Collisions are
elastic
·
Gas molecules
exert no attractive forces on each other
KE = ½ mu2
m = mass in grams
u = velocity in meter/second
·
Kinetic energy is
directly proportional to temperature
·
Different gases
will have an equal average kinetic energy at the same temperature.
·
How would the
velocity of O2 molecules at 25° C compare to the velocity of H2 molecules at 25° C?
Consider the following laws
based on kinetic molecular theory:
Boyle’s Law: P1V1 = P2V2
·
Pressure comes
from collisions of molecules with the walls of the containers
·
In a smaller
volume, more collisions are likely
·
Pressure comes
from collisions of molecules with the walls of the containers
·
Having more gas
molecules present in the same area creates more collisions
Charles’ Law: V1/T1 = V2/T2
·
Pressure comes from
collisions of molecules with the walls of the containers
·
As temperature
decreases, average velocity decreases
·
At a slower
velocity, less collisions occur
·
If constant
pressure is maintained, volume must decrease.
Diffusion and effusion of
gases
·
Diffusion refers
to the movement of gas molecules within an enclosed environment
·
Effusion refers
to the escape of gas molecules through small openings (pores) in a container
The
rate at which gas molecules move is dependent on the kinetic energy of the
sample.
·
Heavy gas
molecules move more slowly at a given temperature than lighter gas molecules.
·
At a higher
temperature, a gas will move more quickly than at a lower temperature.
·
Gases deviate
from ideal behavior most significantly at high pressure and/or at low
temperature – when they are close to becoming liquids.
·
Under these
conditions actual values will differ somewhat from calculated values.
Stoichiometery and gas law calculations
·
Reactions which
generate gases can be examined using stoichiometry information combined with
gas law information
Example 1: CaCO3 (s) à CaO
(s) + CO2 (g)
Calculate the volume of CO2
at STP produced from the decomposition of 152 g CaCO3 by the
reaction.
Example 2: Air bags generate
N2 gas based on the following reaction:
2
NaN3 (s) à 2 Na (s) + 3 N2 (g)
How much sodium azide (grams) should be used to generate a pressure of 829
mm Hg at a temperature of 22.0° C in a 45.5 L air bag?