Answer Key
ALE 24. Ideal Gases, Real Gases and the Kinetic Molecular Theory
(Reference: Chapter 5 - Silberberg)
Instructions…
Use the following answers to check your responses, but use your own words to make corrections—
do not copy the answers below as little learning will take place!
For answers that involve a calculation you must show your work neatly using dimensional analysis
with correct significant figures and units to receive full credit. No work, no credit. Report numerical
answers to the correct number of significant figures. CIRCLE ALL NUMERICAL RESPONSES.
Section 5.5
The Ideal Gas Law and Reaction Stoichiometry
1. How many grams of potassium chlorate decompose to potassium chloride and 638 mL of O2 at 128 oC
and 752 torr?
2 KClO3(s) + heat  2 KCl(s) + 3 O2(g)
1.57 g KClO3
2. When 35.6 L of ammonia and 40.5 L of oxygen gas at STP burn, nitrogen monoxide and water are
produced. After the products return to STP, how many grams of nitrogen monoxide are present? Don’t
forget to balance the equation below!
Balanced Chemical Equation:
4 NH3 ( g ) + 5 O2 ( g )
4 NO ( g ) + 6 H2O (l )
43.4 g NO
3. How many liters of hydrogen gas are collected over water at 18.0 oC and 725 mmHg when 0.84 g
lithium metal reacts with water to produce aqueous lithium hydroxide and hydrogen gas?
Balanced Chemical Equation: 2 Li (s) + 2 H2O (l )
2 LiOH (aq) + H2 ( g )
1.52 L H2
Section 5.6 The Kinetic-Molecular Theory: A Model for Gas Behavior
4. Use the kinetic molecular theory to explain the change in gas pressure that results from warming a
sample of gas.
As the temperature of the gas sample increases, the most probable speed
increases. This will increase both the number of collisions per unit time and
the force of each collision with the sample walls. Thus, the gas pressure
increases.
5. How does the kinetic molecular theory explain why 1 mol of krypton and 1 mol of helium have the
same volume at STP?
At STP (or any identical temperature and pressure) the volume occupied by a
mole of any gas will be identical. This is due to the fact that at the same
temperature, all gases have the same average kinetic energy, resulting in the
same pressure.
Page 1 of 4
ALE 24 – Chem 140 – K. Marr
(Last Revised Fall 2008)
6. The graph below shows the distribution of molecular velocities of a gas at two different temperatures.
a.) Does curve 1 or 2 better represent the behavior of the gas at the lower temperature? Explain your
reasoning.
Curve 1
T Kinetic Energy molecular velocity (since Kinetic Energy = 1/2 mv2);
Curve 1 has the lower average molecular velocity and hence the lower average
kinetic energy, it must be at the lower temperature
b.) Which curve represents the sample with the higher average kinetic energy, Ek(ave.)? Explain your
reasoning.
Curve 2
T Kinetic Energy molecular velocity; Curve 2 has the higher average
molecular velocity, therefore it must have the higher average kinetic energy
c.) Which curve represents the sample that diffuses more quickly? Explain your reasoning.
Curve 2
T Kinetic Energy molecular velocity; Since curve 2 has the higher average
molecular velocity it will diffuse faster.
d.) Suppose that curves 1 and 2 represent two different gases at the same temperature. If the gases are
helium and neon, match the curves with each gas. Explain your reasoning.
Hints: i.) Substances at the same temperature have the same average kinetic energy; ii.) Kinetic
Energy of a particle is related to the mass of the particle and its velocity: Ek = ½ mv2
Curve 1 (lower average molecular velocity) = neon;
Curve 2 (higher average molecular velocity) = helium
Justification:
Since the gases are at same temperature: K.E. neon = K.E. helium,
Therefore: 1/2 mNevNe2 = 1/2 mHevHe2
Since molecular mass Ne > molecular mass He, the average molecular velocity of He must
be greater than the average molecular velocity of neon
ALE 24 - Page 2 of 4
ALE 24 – Chem 140 – K. Marr
(Last Revised Fall 2008)
Section 5.7 Real Gases: Deviations from Ideal Behavior
7.
a.) Explain why large gaseous molecules deviate from ideal behavior. That is, explain why gaseous
molecules of high molecular size have positive deviations from the PV/RT ratio of an ideal gas:
which variable in this ratio is affected? How is it affected? Why is it affected?
The ideal gas law assumes that molecular size (i.e. the amount of space the
molecules themselves occupy) is negligible when compared to the size of the
container. Consequently, for ideal gases: Vgas = V container However, if the
molecules are large, this assumption is incorrect. Hence, for real gases the
measured volume of a gas (i.e. the volume of the container) is an overstatement
of the volume of a gas since the actual volume occupied by the gas molecules is
significant relative to the size of the container they are in. This means the
measured volume for real gases is too large, which makes the ratio PV/RT too
large, resulting in positive deviation from ideal behavior.
b.) Use the data from Table 5.5 to rank Cl2, H2 and O2 in order of increasing magnitude of these
deviations.
Molecular size causes a positive deviation from ideal behavior.
Thus, VReal Gases > VIdeal Gases. Therefore, ranked in order of increasing deviations
from ideal behavior: H2 < O2 < Cl2.
8. Does sulfur hexafluoride, SF6, (boiling point 16 oC at 1 atm) behave more ideally at 150 oC or 20 oC?
Explain your reasoning.
SF6 behaves ideally at 150oC, a temperature well above its BP. Molecular
velocities are greater at higher temperatures. Higher molecular velocities help to
overcome intermolecular forces of attraction. Intermolecular forces of attraction
cause negative deviations from ideal behavior they result in lower than expected
pressure measurements, thus resulting in negative deviations in the PV/RT ratio.
20 oC is only four degrees above the boiling point of SF6, the temperature at
which it condenses from gas to liquid. Hence, intermolecular attractions would
cause severe negative deviations from ideal behavior at 20 oC.
Comprehensive Problems
9. Will the volume of a gas increase, decrease, or remain unchanged for each of the following sets of
changes. Justify your response in each case.
a.) The pressure is decreased from 2 atm to 1 atm, while the temperature is decreased from 200 oC to
100 oC.
increase: V2 = 1.6 V1
b.) The pressure is increased from 1 atm to 3 atm, while the temperature is increased from 100 oC to
300 oC.
decrease: V2= 0.5 V1
c.) The pressure is increased from 3 atm to 6 atm, while the temperature is increased from -73 oC to
127 oC.
unchanged: V2 = V1
ALE 24 - Page 3 of 4
ALE 24 – Chem 140 – K. Marr
(Last Revised Fall 2008)
10. Automobile air bags respond to a collision of a preset strength by electrically triggering the explosive
decomposition of sodium azide (NaN3) to its elements, sodium metal and nitrogen gas. In an industrial
lab simulation, 15.3 mL of nitrogen gas was collected over water at 25.0 oC and 755 torr. How many
grams of sodium azide decomposed?
Balanced Chemical Equation:
2 NaN3(s)
2 Na(s) + 3 N2 (g)
0.0261 g NaN3
11. Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was
the material used in the devastating and tragic explosion of the Oklahoma City federal building in
1995. How many liters of gas at 307 oC and 1 atm are formed by the explosive decomposition of 15.0
kg of ammonium nitrate to nitrogen gas, oxygen gas and water vapor?
Balanced Chemical Equation: 2 NH4 NO3 (s)
2 N2 (g ) + O2 (g ) + 4 H2 O ( g )
3.12 x 104 L gas
12. Cylinder A, below, contains 0.1 mol of an ideal gas. Choose the cylinder (B, C, or D) that correctly
represents the volume of the gas after each of the following changes. If none of the cylinders is correct,
state “none.”
a.) P is doubled at fixed n and T .
Cylinder: B
b.) T is reduced from 400 K to 200 K at fixed n and P.
Cylinder: B
c.) T is increased from 100oC to 200oC at fixed n and P. Cylinder: None
d.) 0.1 mol of gas is added at fixed P and T.
Cylinder: C
e.) 0.1 mol of gas is added and P is doubled at fixed T.
Cylinder: D
ALE 24 - Page 4 of 4