AP Chemistry - Problem Drill 14: The Gas Laws
Question No. 1 of 10
Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as
needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
1. The volume of a gas is directly proportional to its temperature. This relationship
between the temperature and volume of gas, which is known as Charles’ Law,
offers an explanation of how hot-air balloons work. A gas is 3.0 L when at 27°C. If
the gas is changed to 4.0 L, what is the final temperature?
Question
(A) 20.3 °C
(B) 36.0 °C
(C) 400 K
(D) 225 K
A. Incorrect.
Assume the pressure and moles are constant. Apply Charles’ Law. Be sure to use
Kelvin for temperature in any gas law problems.
B. Incorrect.
Assume the pressure and moles are constant. Apply Charles’ Law. Be sure to use
Kelvin for temperature in any gas law problems.
C. Correct.
Feedback
Good job! Assume the pressure and moles are constant. Apply Charles’ Law.
Convert the given temperature to Kelvin in any gas law problem.
D. Incorrect.
Assume the pressure and moles are constant. Apply Charles’ Law. Convert the
given temperature to Kelvin in any gas law problem. Check your algebra. You
inverted the volume ratio incorrectly.
Use Kelvin for gas problems: °C + 273 = K
The Combined Gas Law:
P1V1 P2V2

n1T1 n2T2
No mention of moles (n) or pressure (p), so they’re held constant:
27 °C + 273 = 300 K
Solution
V1 V2

T1 T2
T2 
…..
3.0 L 4.0 L

300 K
T2
4.0 L  300 K
 400 K
3.0 L
The correct answer is (C)
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Question No. 2 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
2. Most substances have three states – solid, liquid and gas. Going from solid state
to gaseous state, they usually turn into a liquid first (e.g. ice). Sublimation is when
a substance goes straight from a solid to a gas without turning into a liquid. Dry ice
(solid carbon dioxide) can sublime into a gas with the expansion of volume and
pressure. It can be used to blow up a balloon. A piece of 4.40 g solid CO2 sublimes
in a balloon. What is the pressure (in kPa) if the final volume is 1.00 L at 300 K?
Question
(A) 108 kPa
(B) 2.46 kPa
(C) 10969 kPa
(D) 249 kPa
A. Incorrect.
Use the ideal gas law with mass divided by molar mass substituted for moles (n)
and the gas constant of 8.31.
B. Incorrect.
Use the ideal gas law with mass divided by molar mass substituted for moles (n)
and the gas constant of 8.31.
C. Incorrect.
Feedback
Use the ideal gas law with mass divided by molar mass substituted for moles (n)
and the gas constant of 8.31.
D. Correct.
Good job! Use the ideal gas law with mass divided by molar mass substituted for
moles (n) and the gas constant of 8.31.
Ideal Gas Law (PV = nRT) where n = m/MM, m = mass and MM = molar mass:
PV 
m
RT
MM
Molar mass of CO2:
1 C  12.01 =
12.01
2 O  16.00 =
+ 32.00
44.01 g/mole
P(1.00 L) 
Solution


4.40 g
8.31 L  kPa
300 K
mole  K
44.01g / mole


4.40 g
8.31 L  kPa
300 K
mole  K
44.01g / mole
P
 249kPa
1.00 L
Note: The standard pressure is 101 kPa. It is then increased to 249 kPa due to the
dry ice to gas expansion. The balloon is blown up.
The correct answer is (D).
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Question No. 3 of 10
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(3) Pick the answer (4) Go back to review the core concept tutorial as needed.
3. According to Dalton’s Law of partial pressures, in a mixture of two or more gases, the
total pressure is the sum of the partial pressures of all the components. The partial pressure
of a gas is the pressure that gas would exert if it occupied the container by itself. For the
following picture, the first flask is 1.00 L at 180 mmHg. The second flask is 1.00 L at 342
mmHg. The third flask is 2.00 L at 188 mmHg. What is the final pressure of the total
system when both stopcocks are opened?
Question
(A) 177.5 mm Hg
(B) 224.5 mm Hg
(C) 710 mm Hg
(D) 616 mm Hg
A. Incorrect.
You need to adjust their pressures to the new final volume with the combined gas law and
then add them together.
B. Correct.
Good job! Apply the combined gas law and Dalton's law of partial pressures to calculate the
final pressure of the larger volume in the whole system.
Feedback
C. Incorrect.
You need to adjust their pressures to their new volume before adding them together.
D. Incorrect.
You need to adjust their pressures to the new final volume with the combined gas law and
then add them together.
Dalton’s Law of partial pressure:
Combine gas law:
Ptotal   Pof each gas
P1V1 P2V2

n1T1 n2T2
The strategy is to apply the Dalton’s law for the total pressures by calculating the partial
pressure of each gas. Using the gas law, we can find the pressure of each gas in the new,
larger volume. There is no mention of temperature (T) or mole (n) changes, so we assume
them to be constant. Apply Boyle’s Law: P1V1 = P2V2
Gas 1: (180 mmHg)(1.00L) = P2G1 (4.00L)
P2 (gas#1) = 45.0 mmHg
Gas 2: (342 mmHg)(1.00L) = P2G2 (4.00L)
Solution
P2 (gas #2) = 85.5 mmHg
Gas 3: (188 mmHg)(1.00L) = P2G3 (4.00L)
P3 (gas #3) = 94.0 mmHg
Ptotal = 45.0 mmHg + 85.5 mmHg + 94.0 mmHg
Ptotal = 224.5 mmHg
The Correct Answer is (B).
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Question No. 4 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
4. Gas stoichiometry is to calculate the amounts of reactants or products in a
reaction involving one or more gas(es), based on the ideal gas law and balanced
equation (molar relationship). For the reaction: 2 HgO (s)  2 Hg (l) + O2 (g), what
volume would be produced if 2.17 g HgO reacted at STP?
Question
(A)
(B)
(C)
(D)
0.11
0.22
0.45
2.17
L
L
L
L
A. Correct.
Good job! Apply the stoichiometric ratio and molar volume. Set up the dimensional
analysis to solve the final volume of O2 produced.
B. Incorrect.
Apply the stoichiometric ratio and molar volume. Set up the dimensional analysis to
solve the final volume of O2 produced.
C. Incorrect.
Feedback
Apply the stoichiometric ratio and molar volume. Set up the dimensional analysis to
solve the final volume of O2 produced.
D. Incorrect.
Apply the stoichiometric ratio and molar volume. Set up the dimensional analysis to
solve the final volume of O2 produced.
Gas stoichiometry is the study of relative amounts of reactants and products in
gaseous reactions. There are two key relationships used (1) The ideal gas law PV =
nRT; (2) The balanced equations – their stoichiometric coefficients (ratio).
216.59 g HgO = 1 mole HgO (molar mass)
2 mole HgO = 1 mole O2 (based on the balanced equation’s stoichiometric ratio)
1 mole O2 = 22.4 L O2 at STP (Vm = nRT/P = 1x0.082x273/1 = 22.4 L).
Set up the dimensional analysis based on the reaction coefficients and molar
volume of the gas O2.
Solution
2.17 g
HgO
1 mole
HgO
1 mole
O2
22.4 L
O2
216.59
g HgO
2 mole
HgO
1 mole
O2
= 0.11 L O2
The correct answer is (A).
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Question No. 5 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
5. Ideal gases follow the assumptions made for the Kinetic Molecular Theory of
Gases. The van der Waals equation for real gases makes the corrections on the
interaction (a-constant) and volume (b-constant). Which of the following is not
valid for real gases?
Question
(A)
(B)
(C)
(D)
Gases are made of atoms or molecules.
Gas particles are in constant motion.
Temperature is proportional to kinetic energy.
The volume of the particles is so small compared to the volume between
particles that the particle volume is insignificant.
A. Incorrect.
All gases, including real gases, are made of atoms or molecules.
B. Incorrect.
All gases, including real gases, are in constant, rapid, random motion.
C. Incorrect.
Feedback
Temperature is proportional to kinetic energy, for ideal gases or real gases.
D. Correct.
Good job! The volume of real gases is significant in the whole volume of the
container.
There are two important distinctions between ideal and real gases, their molecular
volumes and their interactions. Each of these terms is corrected in the real gases’
van der Waals equation with the constants of a and b.
Real gases have attractions/repulsions with other molecules.
Real gases are the ones in which the particle volume is important.
There is no option that discusses molecular attractions/repulsions
Option (D) discusses the volume of the particles being insignificant—this is valid for
ideal gases, but not for real gases.
Solution
The correct answer is (D).
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Question No. 6 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
6. Gases can be described in terms of four variables: pressure (P), volume (V),
temperature (T) and the moles of gas (n). There are three common gas laws –
Boyle’s Law, Charles’s Law and Avogadro’s Law). Each of these relationships is a
special case of a more general relationship known as the ideal gas equation PV =
nRT. Which of the following changes would DECREASE the volume of an ideal gas if
all other variables are held constant?
Question
(A) Decrease temperature.
(B) Decrease pressure.
(C) Increase number of molecules.
(D) Change from a more polar gas to a less polar gas.
A. Correct.
Good job! Apply the ideal gas law: PV = nRT. V is directly proportional to T if all
other variables are held constant. Therefore, the temperature drop would decrease
the volume.
B. Incorrect.
Apply the ideal gas law: PV = nRT. V is inversely proportional to P if all other
variables are held constant. The pressure drop would increase the volume.
C. Incorrect.
Feedback
Apply the ideal gas law: PV = nRT. V is directly proportional to n if all other
variables are held constant. The increase in the number of molecules would
increase the volume.
D. Incorrect.
Assuming that this is an ideal gas, the polarity is not one of the terms to consider.
Since this is an ideal gas, apply the ideal gas law PV = nRT.
With P and T constant, the volume V is directly proportional to T. The decrease in
temperature will cause the decrease in volume.
This is the basis of Charles’ Law: V1/T1 = V2/T2
As shown above, to decrease the volume, just decrease the temperature.
The correct answer is (A).
Solution
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Question No. 7 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
7. Real gases deviate from ideal behavior because they do have real volume and
attract each other. Which of the following gases would have the greatest deviation
from the ideal gas behavior?
Question
(A) H2
(B) He
(C) CO2
(D) H2O
A. Incorrect.
Molecules deviate when they are large or polar. This molecule is small and nonpolar.
B. Incorrect.
Molecules deviate when they are large or polar. This molecule is small and nonpolar.
C. Incorrect.
Feedback
Molecules deviate when they are large or polar. This molecule is small and nonpolar.
D. Correct.
Good job! Molecules deviate when they are large or polar. This molecule is very
polar.
Basic assumptions of Kinetic Molecular Theory apply to ideal gases. Real gases
deviate to a certain extent from the ideal gas behaviors.
Intermolecular forces of attractions contribute the deviation of real gas from ideal
gas behaviors. Polar molecules have stronger forces of attraction among molecules
and tend to deviate much more from ideality than non-polar molecules.
The second deviation comes from the real gases having real volume. Their molecule
sizes are not to be ignored. Bulkier molecules will deviate more from ideal
behaviors.
Solution
Molecules deviate when they are large or polar. H2O is more polar than other
molecules listed. H2O has two H-O polar bonds and cluster of H2O can be bonded
together via intermolecular forces (hydrogen bonding).
The correct answer is (D).
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Question No. 8 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
8. One of the most amazing things about gases is that, despite vast differences in
chemical properties, all the gases more or less obey the gas laws with the
assumptions of Kinetic Molecular Theory. The gas laws deal with how gases behave
with respect to pressure, volume, temperature, and amount. The relationships
among these variables are then described by Boyle’s Law, Charles’ Law and
Avogardro’s Law. The three laws can then be combined together into an ideal gas
law (PV=nRT), related by a gas constant R. Basically the P is inversely proportional
to V; while P & V are directly proportional to T. If both the volume and pressure of
an ideal gas are doubled, how will the absolute temperature change?
Question
(A) It will increase by four times of its original value.
(B) It will increase by two times of its original value.
(C) It will stay the same.
(D) It will decrease to one fourth of its original value.
A. Correct.
Good job! Apply the ideal gas law PV = nRT. Both P and V are directly proportional
to T. If both are doubled, the T will be quadrupled.
B. Incorrect.
Apply the ideal gas law PV = nRT. Both P and V are directly proportional to T. If
both are doubled, the T will be quadrupled.
C. Incorrect.
Feedback
Apply the ideal gas law PV = nRT. Both P and V are directly proportional to T. If
both are doubled, the T will be quadrupled.
D. Incorrect.
Apply the ideal gas law PV = nRT. Both P and V are directly proportional to T. If
both are doubled, the T will be quadrupled.
There are at least two ways to solve this.
(1) Use the ideal gas law PV = nRT. Notice that the PV are directly proportional
to T. If P and V are each doubled, the T will increase 4 times.
(2) Use the combined gas law P1V1/T1 = P2V2/T2. Since P2 = 2P1 and V2 = 2P1,
we can set it up to solve T2 in terms of T1.
P1V1/T1 = 2P1x2V1/T2. Therefore T2 = 4T1.
Solution
The absolute temperature will increase by four times of its original value.
The correct answer is (A).
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Question No. 9 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
9. In reality, the real gases don’t typically obey the ideal gas law because the
assumptions made for ideal gases are not completely true. The two main flaws of
the idea gas assumptions in Kinetic Molecular Theory are the omission of the gases’
own volume and the attractions with other gas molecules. Real gases behave most
like ideal gases at ____.
Question
(A)
(B)
(C)
(D)
High temperature and low concentration.
High temperature and high concentration.
Low temperature and high concentration.
Low temperature and low concentration.
A. Correct.
Good job! Molecules moving at high speed (high temperature) are less likely to
have interactions and at low concentrations take up less space in the container.
B. Incorrect.
At high concentrations, the molecules take up more space in the container.
C. Incorrect.
Feedback
At high concentrations, the molecules take up more space in the container.
D. Incorrect.
At low temperatures, the molecules are more likely to have interactions with other
particles.
Real gases have molecular volumes that matter and have interactions with other
particles in the sample.
Molecules that are moving at a high speed (high temperature) have less noticeable
interactions with other molecules. This is because the higher kinetic energy (at
higher temperature) of a molecule itself makes weak inter-molecular attraction
energy much smaller (ignorable).
A low concentration would mean there aren’t many molecules in the container
“taking up space” with their own volumes. In such a lower concentration, the
distance between molecules is much longer. Therefore, their interaction is weaker.
Solution
The correct answer is (A).
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Question No. 10 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
10. Determine the volume of O2, collected at 22.0 C and 728 mmHg in the
decomposition of 8.15 g of KClO3 at the reaction below.
2KClO3 (s)  2KCl (s) + 3O2 (g)
Question
(A)
(B)
(C)
(D)
2.51
1.23
22.4
2.13
L
L
L
L
A. Correct.
Good job! Use the reaction ratio from the balanced equation to compute the
amount of gas generated. Apply the ideal gas law to calculate the volume from the
moles. Be aware of the unit conversion and significant figures.
B. Incorrect.
Use the reaction ratio from the balanced equation to compute the amount of gas
generated. Apply the ideal gas law to calculate the volume from the moles. Be
aware of the unit conversion and significant figures.
C. Incorrect.
Feedback
Use the reaction ratio from the balanced equation to compute the amount of gas
generated. Apply the ideal gas law to calculate the volume from the moles. Be
aware of the unit conversion and significant figures.
D. Incorrect.
Use the reaction ratio from the balanced equation to compute the amount of gas
generated. Apply the ideal gas law to calculate the volume from the moles. Be
aware of the unit conversion and significant figures.
2KClO3 (s)  2KCl (s) + 3O2 (g)
To solve any gas stoichiometry problem, there are two steps to approach it.
(1) Use the stoichiometric ratio in the balanced reaction and calculate the gas
amount (the moles of a reactant or product) from any given data.
(2) Once the amount (moles) of the gas is obtained, use the ideal gas law to
calculate the other variables (P, V or T) from the given.
Solution
Step #1: Calculate the amount of O2 produced. The molar mass of KClO3 is 123.5
g/mol.
Therefore the amount of KClO3 (moles) = 8.15 g / 123.5 (g/mol) = 0.0660 mol.
The gas product O2 generated = (3O2/2KClO3) x 0.0660 mol KClO3 = 0.0990 mol.
Step #2: Apply ideal gas law to calculate the volume from the moles (step 1).
If the pressure unit is mmHg, the gas constant R = 62.4 L•mmHg/mol•K
The ideal gas law: PV = nRT
V = nRT/P = 0.0990 mol x 62.4 L•mmHg/mol•K x296 K / 728 mmHg = 2.51 L.
The correct answer is (A).
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