AP Chemistry Ch 5 Gases
Ch 5/6 Quiz T 9/30
Ch 5/6 Exam Th 10/2
5.1 Pressure
 Pressure is a measure of the force that a gas exerts on its container. It can be described as force per unit area.
 Gas pressure is measured using barometers, which use mercury (hence the unit mmHg).
 A manometers is a device for measuring the pressure of a gas in a container. The pressure of the gas is given
by h (the difference in mercury levels).
 The most commonly used unit is atmospheres (atm), but others are used as well.
o 1 atm = 760.00 mmHg = 760.00 torr = 101.325 kPa + 1.013 x 103 Pa = standard pressure (sea level)
o 1 Pa = 1 N/m2
Exercise 5.1 Pressure Conversions
The pressure of a gas is measured as 49 torr. Convert this to both atm and Pa.
5.2 The Gas Laws of Boyle, Charles, and Avogadro
 Boyle’s law states that the volume of a confined gas is inversely proportional to the pressure exerted on the
gas. Another way to phrase this is “for a quantity of a gas at constant temperature, the product of pressure
and volume is a constant”. We call this product the “Boyle’s law constant”.
o All gases behave in this manner under ideal conditions—in this case, low pressure. Ideal gases
always follow Boyle’s law.
o Easiest form to memorize: P1V1 = P2V2
Exercise 5.2 Boyle’s Law
Sulfur dioxide is found in the exhaust of automobiles and power plants. Consider a 1.53 L sample of this gas at a
pressure of 5600 Pa. If the pressure is changed to 15000 Pa at a constant temperature, what will be the new volume
of the gas?
What is the Boyle’s law constant of this gas?
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Charles’s law states that if a given quantity of a gas is held at a constant pressure, then its volume is directly
proportional to the absolute temperature (meaning, temperature in Kelvin).
o Easiest form to memorize: V1/T1 = V2/T2; also V1T2 = V2T1
Exercise 5.3 Charles’s Law
A sample of gas at 15°C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38°C and 1 atm?
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Gay-Lussac’s law of combining volumes states that volumes of gases always combine with one another in
the ration of small whole numbers, as long as the volumes are measured at the same T and P.
o Easiest form to memorize: P1/P2 = T1/T2; also P1T2 = P2T1
Avogadro’s hypothesis states that equal volumes of gases under the same conditions of temperature and
pressure contain equal numbers of molecules.
o Avogadro’s law states that the volume of a gas, at a given T and P, is directly proportional to the
quantity of the gas. V1/n1 = V2/n2
Combined gas law: P1V1/T1 = P2V2/T2; also P1V1T2 = P2V2T1  memorize it
Exercise 5.4 Avogadro’s Law
Suppose we have a 12.2 L sample containing 0.50 moles of oxygen gas at a pressure of 1 atm and a temperature of
25°C. If all this oxygen is converted to ozone (O3) at the same temperature and pressure, what would be the volume
of the ozone? (answer = 8.1 L)
5.3 The Ideal Gas Law
 PV = nRT  R = 0.08206 Latm/molK, though it has other values and units as well; useful only at low
pressures and high temperatures
Exercise 5.5 Ideal Gas Law I
A sample of hydrogen gas has a volume of 8.56 L at a temperature of 0°C and a pressure of 1.5 atm. Calculate the
moles of hydrogen molecules present in this gas sample. (answer = 0.57 moles)
Exercise 5.6 Ideal Gas Law II
A sample containing 0.35 moles of argon gas at a temperature of 13°C and a pressure of 568 torr is heated to 56°C
and a pressure of 897 torr. Calculate the change in volume that occurs. (answer = decreases by 3 L)
5.4 Gas Stoichiometry
 Using PV = nRT to solve for the volume of one mole of gas at STP, we get V/n = RT/P. This is the molar
volume of a gas at STP.
 Use the ideal gas law to convert quantities that are not at STP.
Exercise 5.7 Gas Stoichiometry I
A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present? (answer = .0781 mol)
Exercise 5.8 Gas Stoichiometry II
Quicklime (calcium oxide) is produced by the thermal decomposition of calcium carbonate. Calculate the volume of
carbon dioxide at STP produced from the decomposition of 152 g of calcium carbonate. (answer: 34.1 L at STP)
Exercise 5.9 Gas Stoichiometry III
A 2.80 L sample of methane gas at 25°C and 1.65 atm was mixed with a 35.0 L sample of oxygen gas at 31°C and
1.25 atm. The mixture was then ignited to form carbon dioxide and water. Calculate the volume of carbon dioxide
formed at a pressure of 2.50 atm and a temperature of 125°C. (answer = 2.47 L)
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The Density of Gases
o “Molecular Mass kitty cat”—all good cats put dirt (dRT) over their pee (P). Ew, but it works.
o Remember that the densities of gases are reported in g/L not g/mL.
Exercise 5.10 Gas Density/Molar Mass
The density of a gas was measured at 1.50 atm and 27°C and foud to be 1.95 g/L. Calculate the molar mass of the
gas. (answer = 32.0 g/mole)
5.5 Dalton’s Law of Partial Pressures
 Dalton’s law states that the pressure of a mixture of gases is the sum of the pressures of the different
components of the mixture
o Ptotal = P1 + P2 + …. + Pn  no matter the identity of the gases in a mixture, unless it is stated that a
chemical reaction occurs…then you will need to use stoichiometry
o This law uses the concept of mole fractions. That’s not on the AP exam anymore, so we’ll leave this
as it is.
o The partial pressure of each gas in a mixture of gases in a container depends on the number of moles
of that gas.
Exercise 5.11 Dalton’s Law
Mixtures of helium and oxygen are used in scuba diving tanks to help prevent “the bends”. For a particular dive, 46
L He at 25°C and 1.0 atm, and 12 L oxygen gas at 25°C and 1.0 atm were pumped into a tank with a volume of 5.0 L.
Calculate the partial pressure of each gas and the total pressure in the tank at 25°C.
(answer: PHe =9.3 atm, P2 = 2.4 atm, Ptotal = 11.7 atm)
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Water displacement (application of Dalton’s law)
o It is common to collect a gas by water displacement
which means some of the pressure is due to water vapor
collected as the gas was passing through. You must
correct for this by looking up the partial pressure due to
water vapor, which depends on the temperature.
Exercise 5.12 Gas Collection over Water
A sample of solid potassium chlorate was heated in a test tube and decomposed according to the following equation:
2 KClO3(s)  2 KCl(s) + 3 O2(g)
The oxygen produced was collected by water displacement at 22°C at a total pressure of 754 torr. The volume of the
gas collected was 0.650 L. Calculate the partial pressure of oxygen in the gas collected and the mass of potassium
chlorate in the sample that was decomposed. (answer = 733 torr, 2.12 g KClO3)
5.6 The Kinetic Molecular Theory of Gases
 Assumptions
o All particles are in constant, random motion
o All collisions between particles are perfectly elastic
o The volume of the particles in a gas is negligible
o The average kinetic energy of the molecules in a gas it is its Kelvin temperature
 These assumptions ignore intermolecular forces.
 Reminder…gases expand to fill their containers; they are also compressible.
 Distribution of molecular speeds
5.7 Effusion and Diffusion
 The plot of gas molecules having various speeds vs. fraction of molecules with a particular speed is a curve.
This equation (Maxwell’s equation) is
o
o
o
Urms = root mean speed
T = temperature, Kelvin
MM = mass of a mole of gas particles in kg
(weird, I know—respect the math though)
o Use the “energy R” or 8.314510 J/molK for
this equation since kinetic energy is involved.
Exercise 5.13 Root Mean Square Velocity
Calculate the root mean square velocity for the atoms in a sample of helium gas at 25°C.
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The mean free path is the average distance a particle travels between collisions. It’s
on the order of a tenth of a micrometer, and very erratic.
The graph on the right shows the effect of temperature on the numbers of particles
with a given velocity. It’s no surprise that increasing the temperature increases the
mean velocity, but the shape of the curve changes as well. There will always be the
odd particle present with zero or near-zero velocity, so as mean temperature
increases, the curve gets pulled to the right and squashed. Keep in mind that all
three of these curves represent the same number of particles. You may see graphs
like this on the AP exam where you have to identify the highest temperature based
on the shape of the graph.
Now, for Graham’s Law of Diffusion and Effusion.
o Diffusion describes the mixing of gases; the rate of diffusion is the rate of
mixing.
o Effusion describes the passage of a gas through a tiny orifice into
an evacuated chamber. The rate of effusion measures the speed at
which the gas is transferred into the chamber.
o The rates of effusion of two gases are inversely proportional to the
square roots of their molecular masses at the same temperature and
pressure.
o
Remember that rate is a change in a quantity over time.
Exercise 5.14 Effusion Rates I
Calculate the ratio of the effusion rates of hydrogen gas and uranium hexafluoride gas. (answer = 13.2)
Exercise 5.15 Effusion Rates II
A pure sample of methane is found to effuse through a porous barrier in 1.50 minutes. Under the same
conditions, an equal number of molecules of an unknown gas effuse through the barrier in 4.73 minutes. What is
the molar mass of the unknown gas? (answer = 160 g/mole)
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Classic example of diffusion
5.8 Real Gases
 Most gases behave ideally until you reach high pressure and low
temperature (gases also tend to liquefy at these conditions).
 The van der Waals equation corrects for the negligible volume of
molecules and accounts for inelastic collisions leading to IMFs. a and b are van der Waals constants.
o The “a” term increases
the pressure to correct
for lower pressure due to
IMFs.
o The “b” term accounts
for decreased free
volume, since particles
take up space.
o It is more important to
understand the concepts
behind this equation
than to plug in values
and use it.