[ KEY ] Review – Unit 6 Test (Chp 10): Gases
Section I Multiple Choice
NO CALCULATOR
___1. A sample of an ideal gas is heated from 100.0 °C to 150.0 °C in a sealed container of constant volume.
Which of the following values for the gas will increase?
I. The average density of the gas same mass in same volume
II. The average number of collisions with the walls of the container
III. The average kinetic energy of the molecules
↓T, ↓ KE avg
(A) I only (B) II only (C) III only (D) I and II
↓T, ↓ KE avg so fewer collisions
(E) II and III
___2. A sample of 5.0 grams of an ideal gas at 37°C and 2.0 atmosphere pressure has a volume of
3.0 liters. Which of the following expressions is correct for the molar mass of the gas?
(A) (0.08206)(310)
(5.0)(2.0)(3.0)
(B) (5.0)(0.08206)(3.0)
(2.0)(310)
(C) (5.0)(0.08206)(310)
(2.0)(3.0)
List what you know: m = 5.0 g
T = 310 K
P = 2.0 atm
V = 3.0 L
M = ? g/mol
PV = nRT
PV = mRT
M
M = mRT
PV n = m
M
(D) (0.08206)(2.0)(3.0)
(5.0)(310)
(E) (2.0)(3.0)_____
(5.0)(0.08206)(310)
___3. A flexible metal tank contains nitrogen gas. Which of the following applies to the gas in the tank when additional nitrogen is added at constant temperature?
(A) The volume of the gas increases.
(B) The pressure of the gas decreases.
(C) The average speed of the gas molecules increases.
(D) The total number of gas molecules remains the same.
(E) The average distance between the gas molecules increases.
___4. Equal numbers of moles of F
2
(g), Cl
2
(g), and Br
2
(g) are placed in a glass vessel at room temperature. If the glass vessel is complexly broken in pieces, which of the following will be true regarding the rates of diffusion throughout the room of the three gases?
(A) R
F
2
> R
Cl
2
> R
Br
2
(B) R
F
2
> R
Br
2
> R
Cl
2
(C) R
Cl
2
> R
Br
2
> R
F
2
(D) R
Br
2
> R
F
2
> R
Cl
2
(E) R
F
2
= R
Br
2
= R
Cl
2 smaller molar mass, faster speed
↓ M , ↑ u
1

___5. A sample of 92 grams of sodium metal is added to an excess of sulfuric acid. The volume of hydrogen gas produced at standard temperature and pressure is _____ liters.
(A) 5.05
(B) 11.2
(C) 22.4
(D) 44.8
(E) 67.2
2 Na + H
2
SO
4

H
2
+ Na
2
SO
4
92 g Na x 1 mol Na x 1 mol H
2
x 22.4 L H
2
=
23 g Na 2 mol Na 1 mol H
2
4 x 1/2 x 22.4 =
2 x 22.4 = 44.8 L
___6. A mixture of He and Ne at a total pressure of 0.50 atm is found to contain 1.2 mol of He and
0.80 mol of Ne. The partial pressure of Ne is __________ atm.
(A) 0.20
(B) 0.30
C) 0.330
(D) 0.40
(E) 0.50
P
Ne
= P
T
∙ X
Ne
P
Ne
= (0.50)(0.80
/ 2.0) = 0.20 atm
___7. A hot-air balloon, shown right, rises. Which of the following is the best explanation for this observation?
(A) The pressure on the walls of the balloon increases with increasing temperature.
(B) The difference in temperature between the air inside and
outside the balloon produces convection currents.
(C) The cooler air outside the balloon pushes in on the walls of the balloon.
(D) The rate of diffusion of cooler air is less than that of warmer air.
(E) The air density inside the balloon is less than that of the surrounding air.
As T↑, air in balloon expands (↑V) and some leaves the balloon.
There is less mass left in the same volume for ↓density.
___8.
The system shown above is at equilibrium at 28 °C. At this temperature, the vapor pressure of water is 28 millimeters of mercury. The partial pressure of O
2
(g) in the system is ____ mmHg.
(A) 28
(B) 56
161 = 28 + P
O
2
P
O
2
= 161 – 28 (C) 133
(D) 161
(E) 189
P
O
2
= 133 mmHg
2

___9. Under which of the following conditions of temperature and pressure will H
2
gas be expected to behave least like an ideal gas?
(A) 50 K and 0.10 atm
(B) 50 K and 5.0 atm
(C) 500 K and 0.10 atm
(D) 500 K and 50 atm
At high pressure, the volume of each gas particle is negligible compared to the total volume, so the volume of gas is less than the measured container volume.
At low temperatures, the particles are moving with less speed and energy that the attractive forces between gas particles are significant (no longer negligible) and will decrease the pressure significantly.
___10. The volume of a sample of air in a cylinder with a movable piston is 2.0 L at a pressure P
1
, as shown in the diagram above. The volume is increased to 5.0 L as the temperature is held constant. The pressure of the air in the cylinder is now P
2
. What effect do the volume and pressure changes have on the average kinetic energy of the molecules in the sample?
The average kinetic energy would ____________. (increase / decrease / stay the same )
The pressure would ____________. (increase / decrease / stay the same)
The density would ____________. (increase / decrease / stay the same)
The average molecular speed would ___________. (increase / decrease / stay the same )
Same KE avg
b/c Temperature is constant
↓Pressure b/c walls are farther away so fewer collisions
↓Density b/c Volume is greater while mass is the same (D = m/V)
Same u (speed) b/c Temp and Molar Mass are the same
V
(liters)
11.
A 1.0 mole sample of NO
2
(g) is pressurized at constant temperature.
On the graph, sketch the expected plot of volume versus pressure as the gas is pressurized (increased pressure).
↑P , ↓V b/c as walls get closer so more collisions occur
A 1.0 mole sample of NO
2
(g) is heated at constant pressure.
On the graph, sketch the expected plot of volume versus temperatre as the gas is heated (increased temperature).
↑T , ↑V b/c particles collide with walls with more energy and more frequently
A 1.0 mole sample of NO
2
(g) is heated at constant volume.
On the graph, sketch the expected plot of pressure versus temperature as the gas is heated (increased temperature).
↑T , ↑P b/c particles collide with walls with more energy and more frequently
V
(liters)
P
(atm)
P (atm)
T (K)
3

Questions 12-14 refer to the following.
The table below contains information about samples of four different gases at 273 K.
The samples are in four identical rigid containers numbered 1 through 4.
T (K)
___12. On the basis of the data provided above, the gas in container 3 could be
(A) CH
(B) O
2
(C) Ar
(D) CO
2
4 Container 4 has 64.1 g of SO
2
, and SO
2
has a Molar Mass of 64 g/mol . Therefore… there must be 1 mol of particles ( n =1) in container 4 at a pressure of 2 atm.
Container 3 has P = 2 atm (at same T & V), so must also have 1 mol of particles ( n =1).
1 mol of gas with a mass of 16 g gives a Molar Mass of 16 g/mol ( CH
4
).
___13. Under the conditions given, consider containers 1, 2, and 4 only.
The average speed of the gas particles is
(A) greatest in container 1 smaller molar mass, faster velocity
(B) greatest in container 2
↓ M , ↑ v
(C) greatest in container 4
(D) the same in containers 1, 2, and 4
___14. The best explanation for the lower pressure in container 4 is that SO
2
molecules
(A) have a larger average speed than the other three gases
(B) occupy a larger portion of the container volume than the other three gases
(C) have stronger intermolecular attractions than the other three gases
(D) contain pi bonds, while the other gases contain only sigma bonds
Ideal gases exhibit negligible IMAFs, but molecules with stronger IMAFs due large size & LDFs, dipole-dipole att’s, or H-bonding will exhibit attractive forces between gas particles are significant (not negligible) and will decrease the pressure noticeably.
15.
The graph below shows the speed distribution of molecules in a sample of a gas at a certain temperature. Sketch a graph on the same chart showing the speed distribution of the same molecules at a higher temperature (as a dashed curve).
4

___16. Suppose you had 1.0 mol samples of the following gases at STP. If the volume of each sample of gas was reduced to one tenth of its original size and the temperatures remained the same, which gas would have the highest pressure?
(A) He
(B) Xe
(C) O
3
(D) HCl
(E) CO
2
Reducing volume increases pressure in the system causing non-ideal gas behavior.
The gas that deviates least from ideal behavior in these conditions is He since it has the weakest IMAFs (small LDFs).
He would not stick together as much as the other gases and would continue to collide with the walls at a closer to ideal frequency.
___17. A chemist puts 1.0 mol of He (g) , 1.0 mol of Ar (g) , and 1.0 mol of Xe (g) in to a 1.0 L evacuated cylinder at STP. The volume of the cylinder is then reduced to one fifteenth of its original size.
Which if the following would be true?
(A)
(B)
(C)
P
P
P
Ar
He
Xe
< P
< P
< P
He
Ar
He
< P
< P
< P
Xe
Xe
Ar
Reducing volume increases pressure in the system causing non-ideal gas behavior.
The gas that deviates most from ideal behavior in these conditions has the strongest IMAFs (LDFs for He, Ar, & Xe).
(D)
(E)
P
P
Xe
He
< P
< P
Ar
Xe
< P
< P
He
Ar
Xe would stick together most and would collide with the walls at a lower than ideal frequency causing the lowest measured pressure, P
Xe
.
The opposite is true for P
He
.
___18. A hydrocarbon has a density of about 1.5 g/L at STP. What is the molecular formula of the hydrocarbon?
(A) CH
2
(B) CH
3
OH
(C) C
4
H
10
(D) CH
3
COOH
(E) C
3
H
8
D = 1.5 g/L
T = 273 K
P = 1.00 atm n = ? mol m = ? g
M = ?
g/mol
PV = nRT & n = m/M & D = m/V
PV = (m/M)RT
P = mRT
VM
P = DRT
M
M = DRT
P
…but NO CALCULATOR!
At STP , the molar volume of a gas, V m
= 22.4 L/mol
1.5g x 22.4 L = (1.5)(22.4) g = 22.4 + 11.2 = 33.6 g/mol CH
3
OH = 32 g/mol
1 L 1 mol 1 mol
M = m M = 11.0 M = 11.0 x 4 = 44 g/mol
(A) O
2
(B) Cl
2
(C) H
2
(D) Kr
(E) CO
2
V = 5.6 L
T = 273 K
P = 1.00 atm m = 11.0 g
M = ?
PV = nRT & n = m/M
PV = (m/M)RT
(1.00)(5.6) = (11.0)(0.8206)(273)
M
…but NO CALCULATOR!
At STP , the molar volume of a gas, V m
= 22.4 L/mol
5.6 L x 1 mol = 5.6 = ¼ mol
22.4 L 22.4
M = m
n
M = 11.0 g
¼ mol
M = 11.0 x 4 = 44 g/mol
5

[ KEY ] Review – Unit 6 Test (Chp 10): Gases
Section II Free Response
Calculator Allowed
CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS. It is to your advantage to do this, because you may earn partial credit if you do and little or no credit if you do not.
Attention should be paid to significant figures.
0.50 mol
O
3
(g)
1.0 mol
O
2
(g)
1.0 L 1.0 L
298 K 298 K
1.
Consider two containers of volume 1.0 L at 298 K, as shown above. One container holds 0.50 mol
O
3
(g) and the other holds 1.0 mol O
2
(g). Assume that the O
3
(g) and the O
2
(g) exhibit ideal behavior.
(a) Is the pressure in the container holding the O
2
(g) less than, greater than, or equal to the pressure in the container holding the O
3
(g) ? Justify your answer. (4)
The pressure is greater b/c there are more particles of gas and hence more collisions with the walls.
(b) The molecules of which gas, O
3
or O
2
, have the greater average speed? Justify your answer.
(1)
O
2
has a greater average speed b/c the speed is inversely proportional to molar mass and O
2
has a smaller (lighter) molar mass than O
3
.
(c) What measurement could be made that is directly proportional to the average kinetic energy of the O
3
(g) molecules in the container? (1) the temperature.
(d) If the volume of the container holding the O
3
(g) was increased to 2.0 L at 298 K, what would be the change in each of the following variables? In each case, justify your answer.
(i) The pressure within the container. (2)
( n and T are constant and are eliminated from the PV = nRT equation)
P
1
V
1
= P
2
V
2 The pressure would decrease by half because PV is a constant when the temperature and number of moles is held constant.
P
1
(1.0 L) = P
2
(2.0 L)
Therefore, if the volume is doubled, the pressure would be halved.
(1/2)P
1
= P
2
(ii) The average speed of the O
3
(g) molecules. (1)
The average speed is unchanged since the speed depends on temperature not on volume.
6

2.
Use appropriate chemical principles to explain the observation that methane gas behaves as an ideal gas at low pressure and high temperature. (2)
At low pressure, the volume of each gas particle is negligible compared to the total volume, so the volume of gas is not decreased significantly.
At high temperatures, the particles are moving with so much speed and energy that the attractive forces between gas particles are negligible and will not decrease the pressure significantly.
3.
A valve separates two flasks of gases as follows. A 2.0 L flask contains H
2
at a pressure of 5.0 atm, and a 6.0 L flask contains C
3
H
8
at a pressure of 2.0 atm. Calculate the total pressure of the system after the valve is opened. Assume the temperature remains constant. (2)
Calculate total pressure ( P
T
) of P
(H
2
)
+ P
(C
3
H
8
)
, but the V’s change, so P changes, too.
P total
= P
A
+ P
B
PV = nRT P
1
V
1
= R (constant) n
1
T
1
P
1
V
1
= P
2
V
2 n
1
T
1 n
2
T
2
List Variables:
P
1 (H
2
)
= 5.0 atm
V
1 (H
2
)
= 2.0 L
P
2
V
2
(H
2
)
= ? atm
(H
2
)
= 8.0 L
(total final volume after valve opened)
H
2
gas
P
1
V
1
= P
2
V
2 n
1
T
1 n
2
T
2
( n & T are constant)
P
1
V
1
= P
2
V
2
P
2
= P
1
V
1
V
2
P
2
= (5.0 atm)(2.0 L)
(8.0 L)
P
2
= 1.3 atm H
2
P
T
= P
2 (H
2
) +
P
2 (C
3
H
8
)
C
3
H
8
gas
P
1
V
1
= P
2
V
2 n
1
T
1 n
2
T
2
( n & T are constant)
P
1
V
1
= P
2
V
2
P
2
= P
1
V
1
V
2
P
2
= (2.0 atm)(6.0 L)
(8.0 L)
P
2
= 1.5 atm C
3
H
8
List Variables:
P
1 (C
3
H
8
)
= 2.0 atm
V
1 (C
3
H
8
)
= 6.0 L
P
2 (C
3
H
8
)
= ? atm
V
2 (C
3
H
8
)
= 8.0 L
(total final volume after valve opened)
P
T
= (1.3 atm) + (1.5 atm) = 2.8 atm
7

4.
Consider the hydrocarbon methane, CH
4
(molar mass 16.01 g/mol) a.
Write the balanced equation for the combustion of methane to yield carbon dioxide and water.
(1)
CH
4
( g ) + 2 O
2
( g )

2 H
2
O( l ) + CO
2
( g ) b.
What volume of dry carbon dioxide, measured at 20.0
o
C and 745 mmHg, will result from the complete combustion of 7.00 g of methane? (3)
P = 745 mmHg
V = ? L
T = 293 K n
CO2
= ? mol m
CH4
= 7.00 g
7.00 g CH
4
x 1 mol CH
4
x 1 mol CO
2
= 0.436 mol CO
2
16.04 g CH
4
1 mol CH
4
P V
CO2
= n
CO2
RT
(745/760) V
CO2
= (0.436)(0.0821)(293)
V
CO2
= 10.7 L
5.
c.
Which gas, CH
4
or He , will effuse at twice the rate of the other gas?
Use kinetic-molecular theory to justify your answer. (2)
KE = ½ mv 2 Molecular mass and molecular speed are inversely proportional.
At the same temperature, He will effuse at twice the rate of CH
4
because its molecular mass that is one fourth that of He.
↓ M , ↑ v
KE = ½ (16) v
CH
4
2
KE = ½ (4) v
He
2
At same T, KE
CH
4
= KE
He
, v
He
must be 2 times faster than v
CH
4
(then squared)
Lab question.
Study your Lab carefully . Be familiar with the equipment and procedure .
Be especially familiar with the calculations and the error analysis .
-be able to calculate molar mass using PV = nRT & n = m OR M = mRT
M PV
-know the measurements needed for this experiment
-be able to discuss sources of error for a molar mass too small or too large use M = mRT you can discuss if M will be too large or too small due to having a
PV too small or too large number in the numerator or denominator and having multiplied or divided by a it.
8