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CLEP 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. 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
(E) 293 K
A. Incorrect.
Be sure to use Kelvin in Gas Law problems.
B. Incorrect.
Be sure to use Kelvin in Gas Law problems.
C. Correct.
You correctly converted to Kelvin and used the Gas Law!
Feedback
D. Incorrect.
Check your algebra.
E. Incorrect.
Convert to Kelvin before performing the algebra calculations.
Use Kelvin for gas problems: °C + 273 = K
Combined gas law:
P1V1 P2V2

n1T1 n2T2
No mention of moles or pressure, so they’re held constant:
27 °C + 273 = 300 K
V1 V2

T1 T2
Solution
3.0 L 4.0 L

300 K
T2
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
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.
2. A 4.40 g piece of 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
(E) The problem cannot be solved with this information.
A. Incorrect.
Check your algebra.
B. Incorrect.
Check your algebra.
C. Incorrect.
Check your algebra.
Feedback
D. Correct.
You correctly used the ideal gas law with molar mass.
E. Incorrect.
Use the ideal gas law with mass divided by molar mass substituted for moles.
Ideal gas law with mass and 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
The correct answer is (D).
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Question No. 3 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.
3. For the following picture, the first flask is 1.00 L and is at 180 mm Hg. The second flask
is 1.00 L and is at 342 mm Hg. The third flask is 2.00 L and is at 188 mm Hg. What is the
final pressure of the total system when both the stopcocks are opened?
Question
(A) 177.5 mm Hg
(B) 224.5 mm Hg
(C) 710 mm Hg
(D) 616 mm Hg
(E) 74.8 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.
You correctly used the combined gas law and Dalton's law of partial pressure!
C. Incorrect.
You need to adjust their pressures to their new volume before adding them together.
Feedback
D. Incorrect.
You need to adjust their pressures to the new final volume with the combined gas law and
then add them together.
E. Incorrect!
Add the new pressures together instead of finding their average.
Dalton’s Law of partial pressure:
Combine gas law:
Ptotal   Pof each gas
P1V1 P2V2

n1T1 n2T2
Before using Dalton’s law, we need to find the pressure of each gas in the new, larger
volume.
There is no mention of temperature of mole changes, so we assume them to be constant.
P1V1  P2V2
Gas 1:
Solution
(180mm Hg)(1.00L)  P2 (4.00L)
P2 = 45.0 mm Hg
Gas 2:
(342mm Hg)(1.00L)  P2 (4.00L)
P2 = 85.5 mm Hg
Gas 3:
(188mm Hg)(2.00L)  P2 (4.00L)
P3 = 94.0 mm Hg
Ptotal = 45.0mm Hg + 85.5mm Hg + 94.0mm Hg
Ptotal = 224.5 mm Hg
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Question No. 4 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.
4. 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)
(E)
0.11
0.22
0.45
2.17
22.4
L
L
L
L
L
A. Correct.
You correctly used gas stoichiometry!
B. Incorrect.
There are 2 mole HgO for every 1 mole O2.
C. Incorrect.
Check your stoichiometry set-up.
Feedback
D. Incorrect.
Check your stoichiometry set-up.
E. Incorrect!
1 mole of gas is 22.4 L at STP, but you do not have 1 mole of O2 in this problem.
216.59 g HgO = 1 mole HgO (molar mass)
2 mole HgO = 1 mole O2 (balanced equation)
1 mole O2 = 22.4 L O2 at STP
2.17 g
HgO
Solution
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
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.
5. Which of the following is not valid for real gases?
(A)
(B)
(C)
(D)
Question
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.
(E) Gas particles are in rapid, random motion.
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.
Temperature is proportional to kinetic energy.
Feedback
D. Correct.
The volume of real gases is significant in the whole volume of the container.
E. Incorrect.
All gases, including real gases, are in constant, rapid, random motion.
Real gases have attractions/repulsions with other molecules
Real gases is one 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.
The correct answer is (D).
Solution
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Question No. 6 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.
6. Which of the following changes would DECREASE the volume of a 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.
(E) Putting it in a situation with lower atmospheric pressure.
A. Correct.
This would decrease the volume.
B. Incorrect.
This would increase the volume.
C. Incorrect.
This would increase the volume.
Feedback
D. Incorrect.
This would increase the volume.
E. Incorrect!
This would increase the volume.
(A) Would lower the volume.
(B) Would increase the volume.
(C) Would increase the volume if the pressure is held constant.
(D) Would have less interactions between the particles and would cause more
frequent collisions with the container and raise pressure, which would raise the
volume if constant pressure and temperature are held constant.
(E) Would increase the volume as the gas expanded to lower the internal pressure
to match the new lower external pressure.
The correct answer is (A).
Solution
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Question No. 7 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.
7. Which of the following gases would have the greatest deviation from the ideal
gas behavior?
Question
(A) H2
(B) He
(C) CO2
(D) H2O
(E) N2
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.
Molecules deviate when they are large or polar. This molecule is very polar.
E. Incorrect!
Molecules deviate when they are large or polar. This molecule is small and nonpolar.
Molecules deviate when they are large or polar. H2O is very polar.
The correct answer is (D).
Solution
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Question No. 8 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.
8. 2 moles each of H2 and CH4 are both placed in containers. Constant pressure
and temperature are maintained. Which container would have the smaller volume?
Question
(A) H2
(B) CH4
(C) They both are the same.
(D) It cannot be determined from this information.
(E) All gases are the same volume if they are at the same temperature and
pressure.
A. Correct.
A larger molecule would take up more room in the container, and therefore if they
are held at the same pressure and temperature, the larger molecule will have the
larger volume.
B. Incorrect.
A larger molecule would take up more room in the container, and therefore if they
are held at the same pressure and temperature, the larger molecule will have the
larger volume.
C. Incorrect.
Feedback
A larger molecule would take up more room in the container, and therefore if they
are held at the same pressure and temperature, the larger molecule will have the
larger volume.
D. Incorrect.
A larger molecule would take up more room in the container, and therefore if they
are held at the same pressure and temperature, the larger molecule will have the
larger volume.
E. Incorrect!
A larger molecule would take up more room in the container, and therefore if they
are held at the same pressure and temperature, the larger molecule will have the
larger volume.
A larger molecule would take up more room in the container, and therefore if they
are held at the same pressure and temperature, the larger molecule will have the
larger volume.
The smaller molecule will have the smaller volume.
The correct answer is (A).
Solution
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Question No. 9 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.
9. Real gases behave most like ideal gases at ____.
Question
(A)
(B)
(C)
(D)
(E)
High temperature and low concentration.
High temperature and high concentration.
Low temperature and high concentration.
Low temperature and low concentration.
Real gases never behave similar to ideal gases.
A. Correct.
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 space in the container.
C. Incorrect.
At high concentrations, the molecules take up space in the container.
Feedback
D. Incorrect.
At low temperatures, the molecules are more likely to have interactions with other
particles.
E. Incorrect!
Molecules moving at high speed (high temperature) are less likely to have
interactions and at low concentrations take up less space in the container.
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.
A low concentration would mean there aren’t many molecules in the container
“taking up space” with their own volumes.
The correct answer is (A).
Solution
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Question No. 10 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.
10. A lowering of temperature with constant pressure would have what affect on
volume?
Question
(A)
(B)
(C)
(D)
(E)
Decrease the volume.
Increase the volume.
Have no affect on volume.
It would depend on which gas was in the container.
It is impossible to tell from this information.
A. Correct.
Lower temperature means lower pressure. In order to maintain the original
pressure, the volume decreases.
B. Incorrect.
Lower temperature means lower pressure. Increasing the volume would lower it
even more rather than increasing back to the original.
C. Incorrect.
Feedback
Changing the temperature will have an effect on the volume.
D. Incorrect.
All gases will behave the same way to a decrease in temperature.
E. Incorrect!
Changing the temperature will have a predictable effect on the volume.
Lower temperature means the molecules move slower and have a lower pressure.
In order to maintain the same pressure, the volume decreases to increase the
pressure back up to its original value.
PV=nRT
The correct answer is (A).
Solution
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