Gases
Chapter 5
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Gases – Ch. 5
1. Draw the following:
a. A closed monometer attached to a flask filled with CO at 250 torr
b. An open monometer at sea level attached to a flask filled with
N2O at 600 torr
Gases – Ch. 5
Gases – Ch. 5
2. Determine if the following are directly or inversely proportional –
assume all other variables are constant
a. Pressure and volume
b. Pressure and temperature
Gases – Ch. 5
3. The valve between two tanks is opened. See below. Calculate the
ratio of partial pressures (O2:Ne) in the container after the valve is
opened.
a.
b.
c.
d.
e.
1.31
1.60
2.10
0.477
0.615
3.25-L
8.64 atm
O2
2.48-L
5.40 atm
Ne
Gases – Ch. 5
Combined Gas Law
If one variable (P, V, n or T) of a gas changes
at least one other variable must change
P1V1 = P2V2
n1T1 n2T2
Gases – Ch. 5
4. In an experiment 300 m3 of methane is collected over water at 785
torr and 65 °C. What is the volume of the dry gas (in m3) at STP?
The vapor pressure of water at 65 °C is 188 torr.
Gases – Ch. 5
5. Consider a sample of neon gas in a container fitted with a movable
piston (assume the piston is mass-less and frictionless). The
temperature of the gas is increased from 20.0°C to 40.0°C. The
density of neon
a.
b.
c.
d.
e.
increases less than 10%.
decreases less than 10%.
increases more than 10%.
decreases more than 10%.
does not change.
Gases – Ch. 5
6. A gaseous mixture containing 1.5 mol Ar, 6 mol He and 3.5 mol Ne
has a total pressure of 7.0 atm. What is the partial pressure of Ne?
a.
b.
c.
d.
e.
1.4 atm
2.2 atm
3.8 atm
4.6 atm
2.7 atm
Gases – Ch. 5
7. A mixture of oxygen and helium is 92.3% by mass oxygen. What is
the partial pressure of oxygen if atmospheric pressure is 745 Torr?
Gases – Ch. 5
8. A sample of oxygen gas has a volume of 4.50 L at 27°C and 800.0
torr. How many oxygen molecules are in the sample?
a.
b.
c.
d.
e.
1.16 × 1023
5.8 × 1022
2.32 × 1024
1.16 × 1022
none of these
Gases – Ch. 5
Ideal gas law
Considering one set of variables
for a gas under “ideal” conditions
PV = nRT
R ⇒ Universal gas constant
R = 0.08206 atmL/molK
R = 62.37 torrL/molK
R = 8.314 KPaL/molK or J/molK
Gases – Ch. 5
9. Consider the combustion of liquid hexane:
2 C6H14 (l) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (l)
1.52-g of hexane is combined with 2.95 L of oxygen at 312K and
890 torr. The carbon dioxide gas is collected and isolated at 297 K
and 0.930 atm. What volume of carbon dioxide gas will be
collected, assuming 100% yield?
a.
b.
c.
d.
e.
0.504 L
1.93 L
2.23 L
0.607 L
4.04 L
Gases – Ch. 5
10. A 3.54-g sample of lead(II) nitrate (molar mass = 331 g/mol) is
heated in an evacuated cylinder with a volume of 1.60 L. The salt
decomposes when heated, according to the following equation:
2 Pb(NO3)2 (s)  2 PbO (s) + 4 NO2 (g) + O2 (g)
Assuming complete decomposition, what is the pressure (in atm)
in the cylinder after decomposition and cooling to a temperature
of 300. K? Assume the PbO(s) takes up negligible volume.
Gases – Ch. 5
11. 2.5 mol of O2 gas and 3.0 mol of solid carbon, C (s) are put into a
3.50-liter container at 23°C. If the carbon and oxygen react
completely to form CO (g), what will be the final pressure (in atm)
in the container at 23°C?
Gases – Ch. 5
12. The density of an unknown gas at STP is 0.715 g/L. Identify the
gas.
a. NO
b. Ne
c. CH4
d. O2
Gases – Ch. 5
Molar mass (M) can be used to
identify an unknown substance
M = DRT
P
or
mRT
M=
PV
Gases – Ch. 5
13. Air is 79% N2 and 21% O2 by volume. Calculate the density of air
at 1.0 atm, 25°C.
a.
b.
c.
d.
e.
0.590 g/L
1.18 g/L
2.46 g/L
14.1 g/L
none of these
Gases – Ch. 5
14. These plots represent the speed distribution for 1.0 L of oxygen at
300 K and 1000 K. Identify which temperature goes with each
plot.
Gases – Ch. 5
Average Speed
𝑈𝑎𝑣𝑒 =
8𝑅𝑇
𝜋𝑀
Gases – Ch. 5
15. These plots represent the speed distribution for 1.0 L of He at 300 K
and 1.0 L of Ar at 300 K. Identify which gas goes with each plot.
Gases – Ch. 5
16. Calculate the temperature at which the average velocity of Ar (g)
equals the average velocity of Ne (g) at 25°C.
a.
b.
c.
d.
e.
317°C
151°C
49.5°C
25°C
none of these
Gases – Ch. 5
17. Order the following according to increasing rate of effusion if all
gases are at the same T and P.
F2, Cl2, NO, NO2, CH4
Gases – Ch. 5
18. It takes 12 seconds for 8 mL of hydrogen gas to effuse through a
porous barrier at STP. How long will it take for the same volume
of carbon dioxide to effuse at STP?
Gases – Ch. 5
Graham’s Law
If two or more gases are
effusing under the same conditions
𝑡𝑖𝑚𝑒1
𝑡𝑖𝑚𝑒2
=
𝑀1
𝑀2
𝑜𝑟
𝑟𝑎𝑡𝑒1
𝑟𝑎𝑡𝑒2
𝑀2
𝑀1
Gases – Ch. 5
19. The diffusion rate of H2 gas is 6.45 times faster than that of a
certain noble gas (both gases are at the same temperature). What is
the noble gas?
a.
b.
c.
d.
e.
Ne
He
Ar
Kr
Xe
Gases – Ch. 5
20. Consider two 5 L flasks filled with different gases. Flask A has
carbon monoxide at 250 torr and 0 °C while flask B has nitrogen
at 500 torr and 0 °C.
a. Which flask has the molecules with the greatest average kinetic
energy?
b. Which flask has the greatest collisions per second?
Gases – Ch. 5
Useful equations
KEavg = 3/2RT
KE = 1/2mu2
𝐶𝑜𝑙𝑙𝑖𝑠𝑖𝑜𝑛 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦⇒ 𝑍 =
4𝑁𝑑 2
𝑉
𝜋𝑅𝑇
𝑀
Gases – Ch. 5
21. Under what conditions will a gas behave the most like an ideal gas?
Gases – Ch. 5
Gases – Ch. 5
22. Which of the following gases will have the lowest molar volume at
STP?
a. He
b. CH2Cl2
c. CO2
Gases – Ch. 5
The molar volume
𝑉
𝑛
can be derived from the ideal gas law:
𝑉
𝑛
= 𝑅𝑇
𝑃
At STP the molar volume of an ideal gas is 22.41 L/mol
As a gas deviates from ideal behavior the molar volume decreases
Gases – Ch. 5
𝑣𝑎𝑛 𝑑𝑒𝑟 𝑊𝑎𝑎𝑙𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛
[𝑃𝑜𝑏𝑠 + a
𝑛 2
𝑉
](V – nb) = nRT
a ⇒ compensates for the attractive forces between gas particles
b ⇒ compensates for the volume of the gas particles
Gases – Ch. 5
Gas
a (atmL2/mol2)
b (L/mol)
He
0.034
0.0237
Ne
0.211
0.0171
Ar
1.35
0.0322
Kr
2.32
0.0398
Xe
4.19
0.0511
N2
1.39
0.0391
CO2
3.59
0.0427
CH4
2.25
0.0428
NH3
4.17
0.0371
H2O
5.46
0.0305
Gases – Ch. 5
You have completed ch. 5
Answer key – Ch. 5
1. Draw the following:
a. A closed monometer attached to a flask filled with CO at 250
torr
b. An open monometer at sea level attached to a flask filled with
N2O at 600 torr
a.
b.
Answer key – Ch. 5
2. Determine if the following are directly or inversely proportional –
assume all other variables are constant
a. Pressure and volume inversely
b. Pressure and temperature directly
Answer key – Ch. 5
3. The valve between two tanks is opened. See below. Calculate the
ratio of partial pressures (O2:Ne) in the container after the valve is
opened.
a. 1.31
Each gas is affected by the valve opening
b. 1.60
P1V1 = P2V2 where n and T are constant
c. 2.10
n1T1 n2T2
d. 0.477
P1V1
solving
for
P
⇒
P
=
2
2
e. 0.615
V2
for O2 ⇒ P2 = (8.64 atm)(3.25 L)
(3.25L+2.48L)
P2 = 4.9 atm
for Ne ⇒ P2 = (5.4 atm)(2.48 L) = 2.3 atm
(3.25L+2.48L)
𝑃
The ratio of partial pressures ⇒ 𝑃 𝑂2 = 4.9 atm= 2.1
𝑁𝑒 2.3 atm
Answer key – Ch. 5
4. In an experiment 300 m3 of methane is collected over water at 785 torr
and 65 °C. What is the volume of the dry gas (in m3) at STP? The
vapor pressure of water at 65 °C is 188 torr.
When a gas is collected over water there will be water vapor in the collection
chamber. A dry gas implies that the water vapor has been removed.
The PCH4 = Ptotal – PH2O = 785 torr – 188 torr = 597 torr
At STP the temperature and pressure are 273K and 760 torr respectively.
Using the combined gas law where n is constant
P1V1 = P2V2
n1T1 n2T2
3)(273 K)
(597
torr)(300
m
P
V
T
solving for V2 ⇒ V2 = 1 1 2 ⇒ V2 =
P2 T 1
(760 torr)(65 + 273 K)
V2 = 190 m3
Answer key – Ch. 5
5. Consider a sample of neon gas in a container fitted with a movable piston
(assume the piston is mass-less and frictionless). The temperature of the
gas is increased from 20.0°C to 40.0°C. The density of neon
a.
b.
c.
d.
e.
𝑚𝑎𝑠𝑠
increases less than 10%.
Density = 𝑣𝑜𝑙𝑢𝑚𝑒
D2 = (m2/V2) where mass is constant
decreases less than 10%.
D1 (m1/V1)
increases more than 10%.
D2 = V1
decreases more than 10%.
D1 V2
does not change.
Using P1V1 = P2V2 where P and n are constant
n1T1 n2T2
V
Solving for V1 ⇒ 1 = T1
V2 V2 T2
So D2 = T1 = (20+273K) ⇒ D2 = 0.936 or 93.6% ⇒
D1 T2 (40 + 273K) D1
the density decreased by 6.4 %
Answer key – Ch. 5
6. A gaseous mixture containing 1.5 mol Ar, 6 mol He and 3.5 mol Ne
has a total pressure of 7.0 atm. What is the partial pressure of Ne?
a. 0.95 atm
b. 2.2 atm
PNe = XNePtotal
c. 3.8 atm
XNe = nnNe
d. 2.7 atm
total
e. 4.8 atm
X = 3.5mol = 0.318
Ne
11mol
PNe = (0.318)(7.0 atm) = 2.2 atm
Answer key – Ch. 5
7. A mixture of oxygen and helium is 92.3% by mass oxygen. What is
the partial pressure of oxygen if atmospheric pressure is 745 Torr?
If you had a 100 g sample ⇒ 92.3 g of O2 and 7.7 g of He
92.3 g O2 = 2.88 mol O
2
31.998 g/mol
7.7 g of He = 1.92 mol He 4.80 total moles
4.0026 g/mol
PO2 = XO2Ptotal = 2.88 mol O2 745 torr = 447 torr
4.80 total mol
Answer key – Ch. 5
8. A sample of oxygen gas has a volume of 4.50 L at 27°C and 800.0 torr.
How many oxygen molecules does it contain?
a. 1.16 × 1023
# of molecules can be derived from moles
b. 5.8 × 1022
PV = nRT
c. 2.32 × 1024
𝑃𝑉
n
=
22
d. 1.16 × 10
𝑅𝑇
(800torr)(4.5L)
e. none of these
n=
(62.37torrL/molK)(300K)
n = 0.192 mol
0.192 mol x 6.022x1023 molecules/mol =
1.16x1023 molecules
Answer key – Ch. 5
9. Consider the combustion of liquid hexane:
2 C6H14 (l) + 19 O2 (g) → 12 CO2 (g) + 14 H2O (l)
1.52-g of hexane is combined with 2.95 L of oxygen at 312K and 890 torr.
The carbon dioxide gas is collected and isolated at 297 K and 0.930 atm.
What volume of carbon dioxide gas will be collected, assuming 100% yield?
a. 0.504 L
Need to determine the limiting reagent
b. 1.93 L
1.52 g = 0.0177 mol vs.
n
=
C
H
6 14
c. 2.23 L
86.07 g/𝑚𝑜𝑙
d. 0.607 L
nO2 = PV (890 torr/760 torr/atm)(2.95L) = 0.135 mol
RT (0.08206 atmL/molK)(312K)
e. 4.04 L
0.135/19 < 0.0177/2 so O2 is the LR
0.135 mol O2 (12 mol CO2) = 0.0853 mol CO2
(19 molO2)
V = nRT = (0.0853 mol)(0.08206 atmL/molK)(297K) = 2.23 L
P
(0.93atm)
Answer key – Ch. 5
10. A 3.54-g sample of lead(II) nitrate (molar mass = 331 g/mol) is heated in an
evacuated cylinder with a volume of 1.60 L. The salt decomposes when heated,
according to the following equation:
2 Pb(NO3)2 (s)  2 PbO (s) + 4 NO2 (g) + O2 (g)
Assuming complete decomposition, what is the pressure (in atm) in the cylinder
after decomposition and cooling to a temperature of 300. K? Assume the PbO(s)
takes up negligible volume.
The reaction produces 2 gases so the pressure in the container
is the total pressure ⇒ Ptotal = ntotalRT
V
3.54 g Pb(NO3)2 (1mol Pb(NO3)2 ) 𝑥(4 mol NO2+1mol O2) = 0.0276 mol gas
(331 g Pb(NO3)2)
2 mol Pb(NO3)2
Ptotal = (0.0267 mol)(0.08206atmL/molK)(300K) = 0.41 atm
(1.6L)
Answer key – Ch. 5
11. 2.5 mol of O2 gas and 3.0 mol of solid carbon, C (s) are put into a
3.50-liter container at 23°C. If the carbon and oxygen react
completely to form CO (g), what will be the final pressure (in atm)
in the container at 23°C?
2 C (s) + O2 (g)  2 CO (g)
Determine limiting reagent
C ⇒ 3 mol C/2 = 1.5
O2 ⇒ 2.5 mol/1 = 2.5 ⇒ C is the LR
Since C is the LR ⇒ in addition to the CO formed there will be excess
O2 in the container so the pressure will be the total pressure ⇒
Ptotal = ntotalRT/V
…continue to next slide
Answer key – Ch. 5
11. …continued

2 C (s)
O2 (g)
3 mol
2.5 mol
-3
-3(2)
+3(2)
0
1 mol
3 mol
1
2 CO (g)
0 mol
2
Since there is 4 mol of gas in the container
Ptotal =
(4 𝑚𝑜𝑙)(
0.08206𝑎𝑡𝑚𝐿
)(23+273𝐾)
𝑚𝑜𝑙𝐾
3.5 𝐿
Ptotal = 27.8 atm
Answer key – Ch. 5
12. The density of an unknown gas at STP is 0.715 g/L. Identify the
gas.
a. NO
Molar mass can be useful to identify a substance
b. Ne
M = DRT
P
c. CH4
(0.715 g/L)(0.08206atmL/molK)(273K)
M
=
d. O2
(1atm)
M = 16 g/mol
Unknown gas is CH4
Answer key – Ch. 5
13. Air is 79% N2 and 21% O2 by volume. Calculate the density of air
at 1.0 atm, 25°C.
a. 0.590 g/L
b. 1.18 g/L
Density is in the equation ⇒
c. 2.46 g/L
M = DRT
P
d. 14.1 g/L
MP
D
=
e. none of these
RT
Since we have 2 gases ⇒
D = (MN2 PN2 )+(MO2 PO2)
RT
D=
(28g/mol)(0.79atm)+(32g/mol)(0.21 atm)
(0.08206 atmL/molK)(298K)
D = 1.18 g/L
Answer key – Ch. 5
14. These plots represent the speed distribution for 1.0 L of oxygen at
300 K and 1000 K. Identify which temperature goes with each
plot.
Average
Speed of A
Average
Speed of B
According to the average speed
equation ⇒ uavg = (8RT/π M)1/2
we can see the relationship between
average speed and temperature
as T ↑ uavg ↑
since uavg B > uavg A ⇒ TB>TA
Plot A ⇒ 300K
Plot B ⇒ 1000K
Answer key – Ch. 5
15. These plots represent the speed distribution for 1.0 L of He at
300 K and 1.0 L of Ar at 300 K. Identify which gas goes with
each plot.
According to the average speed
equation ⇒ 𝑈𝑎𝑣𝑒 =
Average
Speed of A
Average
Speed of B
8𝑅𝑇
𝜋𝑀
we can see the relationship between
average speed and molar mass
as molar mass ↑ uavg ↓
since uavg B > uavg A ⇒ MB < MA
Plot A ⇒ Ar
Plot B ⇒ He
Answer key – Ch. 5
16. Calculate the temperature at which the average velocity of Ar (g)
equals the average velocity of Ne (g) at 25°C.
a. 317°C
b. 151°C
𝑢𝑎𝑣𝑒 = 8𝑅𝑇
𝜋𝑀
c. 49.5°C
uave Ar = uave Ne
d. 25°C
8𝑅𝑇𝐴𝑟
8𝑅𝑇𝑁𝑒
=
e. none of these
𝜋𝑀𝐴𝑟
𝜋𝑀𝑁𝑒
8,R, and π constant
TAr = TNe ⇒ T = MArTNe
Ar
MNe
MAr MNe
TAr = (298K)(39.95 g/mol)
(20.18 g/mol)
T = 590 K or 317°C
Answer key – Ch. 5
17. Order the following according to increasing rate of effusion:
F2, Cl2, NO, NO2, CH4
As molar mass ↑ average speed ↓ rate of effusion ↓
Since the relative molar masses are
Cl2 (70.9 g/mol) > NO2 (46.01 g/mol) > F2 (38 g/mol) >
NO (30.01 g/mol) > CH4 (16.042 g/mol)
Therefore the relative rates of effusion are
Cl2 < NO2 < F2 < NO < CH4
Answer key – Ch. 5
18. It takes 12 seconds for a given volume of hydrogen gas to effuse
through a porous barrier. How long will it take for the same
volume of carbon dioxide?
timeCO2 = MCO2
timeH2
MH2
timeCO2 = timeH2
MCO2
MH2
44.01g/mol
2.016g/mol
timeCO2 = 56 s
timeCO2 = (12 s)
Answer key – Ch. 5
19. The diffusion rate of H2 gas is 6.45 times as great as that of a
certain noble gas (both gases are at the same temperature). What is
the noble gas?
a. Ne
Molar mass can be used to identify
b. He
rateH2
Munk
c. Ar
=
rateunk
MH2
d. Kr
rateH2
e. Xe
Munk =MH2
rateunk
Munk = 2.016 g/mol 6.45
Munk = 83.87 g/mol
Unknown gas is Kr
Answer key – Ch. 5
20. Consider two 5 L flasks filled with different gases. Flask A has carbon
monoxide at 250 torr and 0 °C while flask B has nitrogen at 500 torr and 0
°C.
a. Which flask has the molecules with the greatest average kinetic
energy? According to KEavg = 3RT ⇒ we see the relationship ⇒ as T
2
↑ KEavg↑ ⇒ since both flasks are at the same T they will have the
same KEavg
b. Which flask has the greatest collisions per second? According to
𝑍=
4𝑁𝑑 2
𝑉
𝜋𝑅𝑇
𝑀
we see three relationships ⇒ as T↑ Z↑ or as molar mass↑ Z↓ or as N/V (or
P)↑ Z↑ ⇒ so since both flasks have the same T and molar mass but
the PB > PA ⇒ ZB > ZA
Answer key – Ch. 5
21. Under what conditions will a real gas behave like an ideal gas?
An “ideal” gas is one that in reality adheres to the ideal gas law ⇒ meaning
experimental values agree with calculated values using PV = nRT
Gases are more likely to behave “idealy” when the pressure is low and/or the
temperature is high
Deviations from the ideal gas law is due to the attractive forces between the
gas particles and the volume of the gas particles relative to the volume of
the container
Answer key – Ch. 5
22. Which of the following gases will have the lowest molar volume at STP?
a. He
b. CH2Cl2
The molar volume of an “ideal” gas is 22.4 L/mol
c. CO2
as the attractive forces of the gas particles ↑
the molar volume ↓ – later (in ch 16)
we will learn the specifics of attractive forces
however for now we can use the relationship that as molar
mass ↑ attractive forces ↑ (an exception is water – although water
is rather on the light side it has quite strong attractive
forces called H-Bonds which we’ll see further in ch 16)
Therefore since CH2Cl2 has the highest molar mass it has the
strongest attractive forces and the lowest molar volume