Gas Laws
Kinetic Theory
• A physical model that explains the
properties of gases, liquids, and
solids in terms of:
(A) forces between particles of
matter
(B) the en these particles possess.
3 Basic Assumptions of the Kinetic
Theory (K.T.):
1) Matter is composed of tiny
particles
• Chemical properties of matter
depend on their composition.
• Physical properties depend on the
forces they exert on each other and
the distance separating them.
2) Particles of Matter are in
continual motion
• Their average kinetic en depends
on temperature.
3) Total kinetic en of colliding
particles remains constant
• When particles collide, some lose
en while others gain en, but there is
no overall en loss. Collisions of this
type are said to be elastic.
• According to the K.T. a gas consists
of very small independent particles
which move randomly in space and
experience elastic collisions.
• The size of gas molecules is
insignificant when compared to the
distance between molecules. Thus, we
assume that the particles of a gas have
no attraction for each other. They are
called point masses since they are
considered to have no real volume or
diameter.
• A gas composed of point masses does
not actually exist. This imaginary gas
composed of molecules with mass, but
not volume or mutual attraction is
called an ideal gas.
Gases have 4 main characteristics:
1) Expansion - no definite shape or
volume. They expand to the shape
and volume of their container.
2) Pressure - their pressure increases
with an increase in temperature.
3) Low density - the density of a gas
is approximately 1/1000 of the same
density in the solid form.
4) Diffusion - the spontaneous
(without help) spreading out of a
material from areas of high
concentration to areas of low
concentration.
Graham’s Law of Diffusion - the
lighter gas diffuses faster
• At the same temperature two gases
have the same ave K.E.
K.E. = 1/2 mv2
(ex) Calculate the rate of diffusion of
hydrogen gas to that of oxygen gas.
(ex) Calculate the relative rate of
diffusion of carbon monoxide gas to
that of carbon dioxide gas.
• Calculate the velocity of sulfur
dioxide gas if the velocity of
bromine gas is .0892 m/s.
• Calculate the mass of an
unknown gas if the ratio of
diffusion of oxygen gas to the
unknown gas is 0.764.
• Variables we use to Measure a
Gas:
• (A) Volume (V) - the amount
of space a material occupies.
Units and conversions:
1 l = 1 dm3 = 1000 cm3
1 ml = 1cm3
Avogadro’s Principle - at equal
temps and pressures, equal volumes
of gases contain the same number of
molecules.
(B) Pressure (P) - the force on a given
amount of area caused by gas particles
colliding with the walls of its container.
Units of pressure:
1b/in2, mm Hg, torrs, pascals (pa),
kilopascals (kpa), atmospheres (atm).
Conversions for pressure:
1 atm = 760 mm Hg
1 mm Hg = 1 torr
1 mm Hg = 13.6 mm H2O
1 in Hg = 25.4 mm Hg
1 atm = 101.3 kpa
1 kpa = 7.5 mm Hg
Standard pressure
1 atm = 101.3 kpa
(C) moles (n) - the number of moles
in a gas depends upon the pressure
and the temp of the gas.
(D) Temperature (T) - a measure of
the ave K.E.
Units: F, C,K
Formulas (know these):
F = (1.8XC) + 32
C = (F -32)/1.8
K = C + 273
• We need an equation that relates all 4
variables together (V, P, n, and T) =
Ideal Gas Law!
• 3 of the variables will be measured
experimentally and we will solve for
the 4th.
Ideal Gas Law
• Suppose we take a sample of a gas
and measure its
P, V, n, and T.
• If you calculate
(Pa)(Va)/(na)(Ta) = #
If you do the same for a different gas
(Pb)(Vb)/(nb)(Tb) = #
• The numbers are the same = constant
called the Universal Gas Constant =
R
• The value of R depends on the units
of the other 4 variables:
• If P = atm, V = l, n = moles, and T =
K
R = 0.082 l atm/K mol
• If P = kpa, V = dm3, n = moles, and
T=K
3
R = 8.31 kpa dm /mol K
Ideal Gas Law = PV/nT = R
or
PV = nRT
T must be in K!
(ex) What volume will 92 moles of
oxygen gas occupy at 5.00 atm of
pressure and 81.0 C?
(534 = 530 l)
( ex) A sample of gas is collected in a
78.02 l flask at 752 torrs and 22.0C.
How many moles of gas are present?
(0.287 moles)
(ex) A sample of gas is collected at a
pressure of 12.5 mm Hg. The gas has a
volume of 17.2 ml and contains2.5 mol.
What is the temp in F, C, and K?
(0.0014 K, 1273 C, -459F)
(ex) A sample of gas has a volume of
91.5 l and was collected at 72.4 F.
What pressure will 36.2 mole of this
gas exert on its container? (9.58 atm)
Using the other value of R
(ex) What pressure would be exerted
by 0.622 mol of gas contained in a 9.22
dm3 vessel at 16.0 C?
(162 kpa)
(ex) At what temp is a gas if 0.0851
mol of it are found in a 604 cm3 vessel
at 100.4 kpa?
(85.8 K)
Variations of the Ideal Gas Law:
• The ideal gas law can easily be
altered to include mass.
(g) (1 mole)/(fm or mw) = moles
So if
PV = nRT
and n = mass/mw
then
(ex) What is the mass of 21.5 l of
hydrogen gas at 760 torrs and 0.0 C?
(1.94 g)
(ex) Find the molecular mass of a gas
whose volume is 91.7 ml. It is
collected at 37.2 atm and 71.4 C and
has a mass of 0.20 g.
(1.7 g)
(ex) What is the temperature of
107.5 g of chlorine gas which
occupies a volume of 95.2 l at a
pressure of 205 torrs?
(207 K)
(ex) Suppose we measure the mass
of the vapor of an unknown gas
3
contained in a 273 cm bulb. We find
that the bulb contains 0.750 g of gas
at 97.2 kpa of pressure and 61.0 C.
What is the molecular mass of the
gas?
(78.4 g/mol)
(ex) What is the molecular weight of
a gas if 150.0 cm3 has a mass of 0.922
g at 99.0 C and 107.0 kpa?
(ex) What will be the density of
oxygen gas at
100.5 kpa and 23.0 C?
(1.31 g/cm3)
Other Gas Laws:
What happens if we change the
conditions?
Squeeze it?
Heat it?
Add more moles to it?
What will happen to the other
variables?
Conditions 1
Conditions 2
n and T stay the same!
For state 1
For state 2
P1V1/n1T1 = R
P2V2/n2T2 =R
• Found that if n and T are constant,
then as P increases, V decreases inversely related.
(ex) Squeezing a ball
Scuba diver
Exploding teeth
To get Boyle’s Law from
P1V1/n1T1 = R
P2V2/n2T2 =R
P1V1/n1T1 = P2V2/n2T2
Let n1 = n2 and T1 = T2 since they are
constant.
Then
P1V1= P2V2 Boyle’s Law
• Note: R is gone, so the only
requirements for units is that all the
pressure units must be the same and all
the volume units must be the same.
(ex) 12.5 l of a gas at 5.43 atm of
pressure contract to a new volume as
the pressure increases to 9.21 atm.
What is the new volume?
(7.37 l)
(ex) If a gas at 920. mm Hg occupies a
volume of 21. Ml, what will be the new
pressure if the volume is cut in half?
(2.41 atm)
(ex) A gas is collected in a 242 cm3
container. The pressure of the gas in
the container is measured and
determined to be 87.6 kpa. What is the
new of this gas at standard pressure?
3
(ex)
Correct the following volume of
a gas from the indicated pressures to
the standard pressures:
3
1) 273 cm at 59.4 kpa
3
(160. cm )
2) 50.0 m3 at 55.1 kpa
(27.2 m3)
Demo: soda bottle and soda can.
Pressure is constant as well as the
number of moles!
T increases, and volume increases
T decreases, V decreases
T and V are directly related.
(ex) tires (winter vs. summer)
To get Charles’ Law:
P1V1/n1T1 = R
P2V2/n2T2 =R
P1V1/n1T1 = P2V2/n2T2
Let P1 = P2 and n1 = n2 since they are
constant.
Then
V1/T1= V2/T2 Charles' Law
Note: Volume can be in any units, but
temperature must be in Kelvin!
(ex) 19.21 l of a gas is at 27.2 C. Find
the volume if the temperature rises to
39.2 C.
(20.0 l)
(ex) A gas expands to 25.7 ml and is at
27.5 F. Its original volume was 10.2
ml. What was its original temperature?
(117 K)
(ex) 225 cm3 of a gas is collected at
58.0 C. What volume would this
sample of gas occupy at standard
temp? Assume P is constant.
(186 cm3)
(ex) Correct the following volumes of
gases for a change from the
temperature indicated to standard
temp. Assume P is constant.
3
2) 617 cm3 at 282 K
(597 cm3)
Combined Gas Law - This equation
allows one to calculate a change in P, V,
or T, assuming only n stays the same.
P1V1/n1T1 = R
P2V2/n2T2 =R
P1V1/n1T1 = P2V2/n2T2
Let n1 = n2 since they are constant.
Then
P1V1/T1= P2V2/T2 Combined Gas
Law
Combination of Charles’ and Boyle’s
Law.
Note: Volume and Pressure can be in
any units, but they must match.
Temperature must be in Kelvin!
(ex) 200.0 ml of a gas at 299 torrs and
25.2 C. Find the new volume if the
pressure increases to 760. torrs and the
temp decreases to -5.02 C
(70.7 ml)
(ex) 905 ml of a gas are at 25.0 atm
and 44.04 C. “Correct to standard
conditions” means find the new volume
at standard temp and pressure.
(19.5 l)
(ex) The volume of a gas measured at
75.6 kpa of pressure and 60.0 C is
corrected to correspond to the volume it
would occupy at STP. The measured
3
volume of the gas is 10.0 cm .
(6.12 cm3)
John Dalton’s Law of Partial
Pressures - the total pressure in a
container is the sum of the partial
pressures of the gases in the container.
• In other words, each gas would exert
the same pressure it would if it were
present alone.
PT = P1 + P2 + P3 + ...
• Therefore, if you collect a gas over
water the water vapor pressure exerts a
pressure and must be subtracted to find
the pressure of a dry gas.
Pg = PT - Pa
Collecting Gases
• To measure the pressure on a gas we
use a eudiometer.
Over Mercury (3 cases)
1) levels are equal
Pg = Pa
2) level inside is higher than the level
outside
P >P
Pa = Pg + Pc
Pg = Pa - Pc
3) level inside is lower than the level
outside
Pg > Pa
Pg = Pa + Pc
Examples
(1) Calculate the pressure of a gas
inside a eudiometer collected over Hg
if the barometric pressure is 750. mm
Hg and
(a) the levels are equal.
(750. mm Hg)
(B) the inside level is 60. mm higher
than the outside level.
(690 mm Hg)
(C) the inside level is 40. mm Hg
lower than the outside level.
(790 mm Hg)
Collecting Gases Over Water
• Hg is expensive and toxic!
• Water is 1/13.6 times as dense as
Hg, therefore, a given gas pressure
will support a column of water 13.6
X as great as an equivalent column of
Hg.
(ex) Oxygen gas is collected in a
eudiometer by water displacement. The
water level inside the tube is 27.2 mm
higher than the outside. The temp is
25.0 C. The barometric pressure is
741.0 mm Hg. What is the partial
pressure of the dry O2 in the
eudiometer?
(715.2 mm Hg)
(ex) Calculate the pressure of a dry gas
inside a eudiometer collected over
water. The temp is at 15.0 C and at an
atmospheric pressure of 750. mm Hg if
(a) the inside level and outside level
are equal
(732.2 mm Hg)
(b) the inside level is 68 mm higher
than the outside level.
(731.5 mm Hg)
(3) the inside level is 204 mm lower
than the outside level
(74.2. mm Hg)
(ex) A eudiometer contains 38.4 ml of
air, collected by water displacement at a
temp of 20.0 C. The water level inside
the eudiometer is 140. mm higher than
the outside. The barometric
pressure is 740.0 mm Hg. Calculate the
volume of dry air at STP.
(33.5 ml)
(ex) A quantity of a gas is collected
over water at 8.0 C is a 353 cm3 vessel.
The barometric pressure is 84.5 kpa.
What volume would the dry gas occupy
at standard pressure and 8.0 C?
3
(291 cm )
Deviations of Real Gases
• So far we have made 2
assumptions:
1) gas molecules have no volume
2) gas molecules have no attractive
forces.
• Only true for IDEAL GASES!
• For many real gases at low P, the
molecules closely approach the
behavior of ideal gas molecules.
• At low P, the molecules of both ideal
and real gases are far apart.
• As the P increases, the gas
molecules are forced closer together.
• All gases deviate from ideal gas
behavior at high P and low T.
• In real gases, atoms or molecules do
not move independently of one another
- they exert an attraction on one
another.
• Forces of attraction increase with a
decrease in distance and will decrease
with an increase in distance.
• Generally, the lower the critical temp
(the temp above which no amount of P
will liquefy a gas) of a gas the more
closely the gas obeys the ideal gas
laws.
Review of Stoichiometry
1) mass-mass
2) mass-volume
3) volume-volume
Path:
info known ----> moles known ----->
moles sought -----> info sought
(ex) What mass of sodium chloride is
produced when 55.4 g of sodium react
with chlorine?
(ex) What mass of aluminum oxide
will be produced when 75.0 liters of
oxygen gas reacts with aluminum at
STP?
(ex) What volume of nitrogen gas is
needed to react with 760. Liters of
hydrogen gas at STP to produce
ammonia?
Limiting Reactant Problems
• Convert all info to moles of what you
are seeking and finish the problem
• The problem with the least number of
moles. Whichever substance gave you
the least number of moles is said to be
the limiting reactant because that
substance limits how much product can
be produced.
• The limiting reactant will be totally
consumed in the reaction.
• The other substance is said to be in
excess because some of it will be left
over after the reaction is complete.
(ex) How many grams of carbon
dioxide are formed if 10.0 g of carbon
3
are burned in 20.0 dm of oxygen gas?
Assume STP.
(ex) How many grams of aluminum
sulfide are formed if 9.00 g of
aluminum react with 8.00 g of sulfur?