CHE141
Chapter 10
Chapter 10
Gases
1. A sample of gas (24.2g) initially at 4.00 atm was compressed from 8.00 L to 2.00 L at
constant temperature. After the compression, the gas pressure was __________ atm.
(a). 4.00
(b). 2.00
(c). 1.00
(d).16.0
Explanation: This is an example of using Boyle’s law. The temperature and the
number of moles of the gas are constant. According to Boyle’s law P1V1 = P2V2
where 1 stands for the initial conditions and 2 for final. Need to find P2
P2 =
P1V1 4.00 atm ! 8.00 L
=
= 16.0 atm
V2
2.00 L
2. A sample of a gas (5.0 mol) at 1.0 atm is expanded at constant temperature from 10.0
L to 15.0 L. The final pressure is __________ atm.
(a). 1.5
(b). 7.5
(c). 0.67
(d). 3.3
Explanation: This is an example of using Boyle’s law. The temperature and the
number of moles of the gas are constant. According to Boyle’s law P1V1 = P2V2
where 1 stands for the initial conditions and 2 for final. Need to find P2 by
rearranging the above equation as follows:
P1V1
1.0 atm ! 10.0 L
= P2 ,
= 0.67 atm
V2
15.0 L
3. A balloon occupies 4.39 L at 44.0 ºC and a pressure of 729.0 torr. What temperature
must the balloon be cooled to, to reduce its volume to 3.78 L (at constant pressure)?
(a). 38.7 ºC
(b).0.00 ºC
(c). 2.6 ºC
(d). 30.9 ºC
Explanation: This is an example of using Charles’s law. The pressure and the
number of moles are held constant while the temperature is lowered. The initial
volume and temperature conditions can be labeled as V1 and T1 while the final
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 10
conditions can be labeled as V2 and T2. The temperature T2 (in Kelvin) can be found
by rearranging Charles’s law as follows:
V1 V2
V ! T 3.78 L ! 317.15 K
=
or T2 = 2 1 =
= 273.1 K
T1 T2
V1
4.39 L
This is approximately equal to 0.00 ºC.
4. A gas originally at 27.0 ºC and 1.00 atm pressure in a 3.90 L flask is cooled at
constant pressure until the temperature is 11.0 ºC. The new volume of the gas is
__________ L.
(a). 1.27
(b).3.69
(c). 3.96
(d). 4.12
Explanation: This is an example of using Charles’s law. The initial volume and
temperature conditions can be labeled as V1 and T1 while the final conditions can be
labeled as V2 and T2. The final volume V2 (in Liters) can be found by rearranging
Charles’s law as follows:
V1 T1
V ! T 3.90 L ! 284.15 K
=
or V2 = 1 2 =
= 3.69 L
V2 T2
T1
300.15 K
5. If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy
__________ L at STP.
(a). 3.92
(b). 50.8
(c). 12.9
(d).25.5
Explanation: Since the pressure and temperature are being held constant the volume
occupied by the gas should be directly proportional to the amount of the gas. The
initial conditions (of mass and volume) can be labeled as 1 and the final conditions as
2. The final volume can be now found by:
m1 m 2
m ! V1 129.3 g ! 10.0 L
=
or V2 = 2
=
= 25.5 L
V1 V2
m1
50.75 g
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CHE141
Chapter 10
6. A sample of He gas (2.35 mol) occupies 57.9 L at 300.0 K and 1.00 atm. What is the
volume of this sample at 423.0 K and 1.00 atm?
(a). 0.709
(b). 41.1
(c). 81.6
(d). 1.41
Explanation: This is an example of using Charles’s law. The initial volume and
temperature conditions can be labeled as V1 and T1 while the final conditions can be
labeled as V2 and T2. The final volume V2 (in Liters) can be found by rearranging
Charles’s law:
V1 V2
V ! T 57.9 L ! 423.0 K
=
or V2 = 1 2 =
= 81.6 L
T1 T2
T1
300.0 K
7. A sample of H2 gas (12.28g) occupies 100.0 L at 400.0 K and 2.00 atm. A sample
weighing 9.49 g occupies __________ L at 353.0 K and 2.00 atm.
(a). 109
(b).68.2
(c). 54.7
(d). 147
Explanation: Convert both the grams of H2 to moles (divide g by the molar mass of
H2) and then rearrange the ideal gas law to calculate the volume.
V1
V
Vn T
100.0 L ! (9.49 g/2.106 g) ! 353.0 K
= 2 or 1 2 2 = V2 =
= 68.2 L
n1T1 n 2T2
n1T1
(12.28 g/2.016 g) ! 400.0 K
8. The amount of gas that occupies 60.82 L at 31.0 ºC and 367.0 mmHg is __________
mol.
(a). 1.18
(b). 0.850
(c). 894
(d). 11.6
Explanation: This problem involves using the ideal gas law and rearranging it to find
the number of moles (n) of the gas. To use the ideal gas law convert the temperature
to Kelvin and the pressure to atm before using these numbers.
PV = nRT or n =
PV (367.0 mm Hg/760 mm Hg) ! 60.82 L
=
= 1.18 moles
RT
0.0821 Latm/molK ! 304.15 K
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CHE141
Chapter 10
9. The pressure of a sample of 6.022 g of CH4 gas in a 30.0 L vessel at 402.5 K is
__________ atm.
(a). 2.42
(b). 6.62
(c). 0.414
(d). 12.4
Explanation: This problem involves using the ideal gas law and rearranging it to find
the pressure (P) of the gas. To use the ideal gas law convert the grams of methane to
moles (n), before using this number.
PV = nRT and P =
nRT (6.022 g/16.043 g) ! 0.0821 Latm/molK ! 402.5 K
=
= 0.414 atm
V
30.0 L
10. A 0.325 L flask filled with gas at 0.914 atm and 19.0 ºC contains __________ mol of
gas.
(a) 1.24 x 10-2
(b). 1.48 x 10-2
(c). 9.42
(d). 12.4
Explanation: This problem involves using the ideal gas law and rearranging it to find
the number of moles (n) of the gas. To use the ideal gas law convert the temperature
to Kelvin before using the number.
PV = nRTand n =
PV
0.914 atm " 0.325 L
=
= 1.24 " 10! 2 moles
RT 0.0821 Latm/molK " 292.15 K
11. A gas in a 350.0 mL container has a pressure of 695.0 torr at 29.0 ºC. Calculate the
number of moles of the gas in the flask.
(a). 1.48 x 10-2
(b).1.29 x 10-2
(c). 9.42
(d). 12.4
Explanation: This problem involves using the ideal gas law and rearranging it to find
the number of moles (n) of the gas. To use the ideal gas law convert the temperature
to Kelvin, the pressure to atm and volume to liters before using these numbers.
1 atm
" 0.350 L
PV
760 torr
PV = nRT and n =
=
= 1.29 " 10! 2 moles
RT 0.0821 Latm/molK " 302.15 K
695 torr "
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CHE141
Chapter 10
12. Calculate the volume occupied by a sample of gas (1.30 mol) at 22.0 ºC and 2.50 atm.
(a). 0.079
(b). 0.94
(c). 12.8
(d). 31
Explanation: This problem involves using the ideal gas law and rearranging it to find
the volume (V) of the gas. To use the ideal gas law convert the temperature to
Kelvin, the before using this number.
PV = nRT and V =
nRT 1.30 mol ! 0.0821 Latm/molK ! 299.15 K
=
= 12.8 L
P
2.50 atm
13. What is the volume of 0.65 mol of an ideal gas at 365.0 torr and 97.0 ºC?
(a). 0.054
(b). 9.5
(c). 11
(d).41.1
Explanation: This problem involves using the ideal gas law and rearranging it to find
the volume (V) of the gas. To use the ideal gas law convert the temperature to Kelvin
and the pressure to atm before using these numbers.
PV = nRT and V =
nRT 0.65 mol ! 0.0821 Latm/molK ! 370.15 K
=
= 41.1 L
1 atm
P
365.0 torr !
760 torr
14. The density of ammonia gas in a 4.32 L container at 837.5 torr and 45.0 ºC is
__________g/L.
(a). 3.86
(b).0.718
(c). 0.432
(d). 0.194
Explanation: The density of a gas can be calculated by using a formula that relates
the density (g/L), pressure (P in atm), molar mass (M in g/mol) and temperature (T)
of the gas. The molar mass of NH3 is 17.031 g/mol. The pressure must be converted
to atm and the temperature to Kelvin before using these numbers.
PM
d=
=
RT
1 atm
! 17.031 g/mol
760 torr
= 0.718 g/L
0.0821 Latm/molK ! 318.15 K
837.5 torr !
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CHE141
Chapter 10
15. The density of N2O at 1.53 atm and 45.2 ºC is __________g/L.
(a). 18.2
(b). 1.76
(c). 0.288
(d).2.58
Explanation: The density of a gas can be calculated by using a formula that relates
the density (g/L), pressure (P in atm), molar mass (M in g/mol) and temperature (T)
of the gas. The molar mass of N2O is 44.013 g/mol. The temperature must be
converted to Kelvin before using this number.
d=
PM
1.53 atm ! 44.013 g/mol
=
= 2.58 g/L
RT 0.0821 Latm/molK ! 318.35 K
16. The molar mass of a gas that has a density of 6.70 g/L at STP is __________ g/mol.
(a). 496
(b).150
(c). 73.0
(d). 3.35
Explanation: The molar mass of a gas is related to its density (in g/L), pressure (in
atm) and its temperature (in K).
M=
dRT 6.70 g/L ! 0.0821L atm/molK ! 273.15 K
=
= 150. g/mol
P
1.00 atm
17. What is the volume of hydrogen gas that can be produced by the reaction of 4.33 g of
zinc with excess sulfuric acid at 38.0 ºC and 760.0 torr?
(a). 1.69
(b). 2.71 x 10-4
(c). 3.69 x 104
(d). 2.84
Explanation: This problem uses a combination of stoichiometry and the ideal gas
law. Need to write a balanced chemical equation for the reaction and then calculate
the number of moles of hydrogen that can be produced from the moles of the limiting
reagent (Zinc in this case).
Zn + H2SO4
ZnSO4 + H2
4.33 g Zn "
1 mol Zn 1 mol H 2
"
= 6.62 " 10! 2 moles H 2
65.39 g 1 mol Zn
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CHE141
Chapter 10
The volume occupied by these moles of H2 can now be calculated using the ideal gas
law:
V=
nRT 6.62 ! 10"2 ! 0.0821 Latm/molK ! 311.15 K
=
= 1.69 L
1 atm
P
760.0 torr !
760 torr
18. What is the volume of HCl gas required to react with excess magnesium metal to
produce 6.82 L of hydrogen gas at 2.19 atm and 25.0 ºC?
(a). 6.82
(b). 2.19
(c). 13.6
(d). 4.38
Explanation: This problem uses a combination of stoichiometry and the ideal gas
law. The number of moles of the hydrogen can be calculated from the ideal gas law
and then used with the stoichiometry of the reaction.
n=
PV
2.19 atm ! 6.82 L
=
= 0.610 moles H 2
RT 0.0821 Latm/molK ! 298.15 K
The balanced equation for this reaction is:
Mg + 2HCl
MgCl2 + H2
The equation shows that 2 moles of HCl are needed for every mole of H2 produced.
Thus (0.610 x 2 = 1.22 moles) of HCl would be required for this reaction. The
volume occupied by these moles of HCl can then be calculated by using the ideal gas
law:
nRT 1.22 ! 0.082 1Latm/molK ! 298.15 K
V=
=
= 13.6 L
P
2.19 atm
19. The Mond process produces pure nickel metal via the thermal decomposition of
nickel tetracarbonyl:
Ni(CO)4 (l)
Ni (s) + 4CO(g)
What volume (L) of CO is formed from the complete decomposition of 444.0 g of
Ni(CO)4 at 752.0 torr and 22.0 ºC?
(a). 0.356
(b). 63.7
(c). 255.
(d). 20.2
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 10
Explanation: This problem uses a combination of stoichiometry and the ideal gas
law. Using the reaction stoichiometry from the above balanced equation, need to
calculate the number of moles of CO produced, and then calculate the volume
occupied using the ideal gas law.
444.0 g Ni(CO) 4 !
V=
1 mole
4 moles CO
!
= 10.40 moles of CO
170.733 g 1 mole Ni(CO) 4
nRT 10.40 ! 0.0821 Latm/molK ! 295.15 K
=
= 255. L
1 atm
P
752.0 torr !
760 torr
20. Ammonium nitrite undergoes thermal decomposition to produce only gases:
NH4NO2 (s)
N2 (g) + 2H2O (g)
What volume (L) of gas is produced by the decomposition of 35.0 g of NH4NO2 (s) at
525.2 ºC and 1.50 atm?
(a). 47
(b). 160
(c). 15
(d).72.
Explanation: This problem uses a combination of stoichiometry and the ideal gas
law. Using the reaction stoichiometry from the above balanced equation, need to
calculate the number of moles of gases formed and then calculate the volume
occupied by these gases. It is not necessary to calculate the moles of the 2 products
individually.
1 mole 3 moles (1N 2 + 2H 2O)
!
= 1.64 moles
64.044 g
1 mole
nRT 1.64 ! 0.0821 Latm/molK ! 798.65 K
V=
=
= 72. L
P
1.50 atm
35.0 g NH 4 NO 2 !
21. The pressure in a 12.2 L vessel that contains 2.34 g of carbon dioxide, 1.73 g of sulfur
dioxide and 3.33 g of argon, all at 42.0 ºC is __________ mmHg.
(a). 263.
(b). 134
(c). 395
(d). 116
Explanation: Since this is a mixture of gases need to find the total number of moles
of gases (nt) and then calculate the pressure of this gas mixture.
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CHE141
Chapter 10
2.34 g CO 2
1.73 g SO 2
3.33 g Ar
+
+
= 0.163 moles
44.009 g/mol 64.063 g/mol 39.948 g/mol
nRT 0.163 ! 0.0821 Latm/molK ! 315.15 K 760 mm
P=
=
!
= 263. mm
V
12.2 L
1 atm
22. A sample of He gas (3.0 L) at 5.6 atm and 25.0 ºC was combined with 4.5 L of Ne
gas at 3.6 atm and 25.0 ºC at constant temperature in a 9.0 L flask. The total pressure
in the flask was __________ atm. Assume the initial pressure in the flask was 0.00
atm.
(a). 2.6
(b). 9.2
(c). 1.0
(d).3.7
Explanation: Calculate the number of moles of each of the gases and find the total
number of moles of the gas mixture. The pressure of this gas mixture can now be
calculated using the ideal gas law.
n=
PV
RT
The number of moles of He using this formula = 0.686 and of Ne = 0.662. The moles
of gas mixture = 1.348. Now using the appropriate rearrangement of the ideal gas
law the pressure of the gas mixture = 3.7 atm.
23. A flask contains a mixture of He and Ne at a total pressure of 2.6 atm. There are 2.0
mol of He and 5.0 mol of Ne in the flask. The partial pressure of He is __________
atm.
(a). 9.1
(b). 6.5
(c). 1.04
(d).0.74
Explanation: Calculate the mole fraction of He and then the partial pressure of He.
The partial pressure of a gas in a mixture is directly proportional to its mole fraction.
n He
2.0
=
= 0.286.
Total moles 2.0 + 5.0
Partial pressure p He = 0.286 ! 2.6atm = 0.74atm
÷He =
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CHE141
Chapter 10
24. SO2 (5.00 g) and CO2 (6.00 g) were placed in a 750.0 mL container at 50.0 ºC. The
total pressure in the container was ___________ atm.
(a). 0.192
(b). 4.02
(c). 2.76
(d).6.78
Explanation: Since each gas behaves as if the other was absent, need to calculate the
partial pressure of each gas and then add them together to find the total pressure.
5.00 g SO 2
= 0.078 moles
64.063 g/mol
nRT 0.0780 ! 0.0821 Latm/molK ! 323.15 K
PSO 2 =
=
= 2.76 atm
V
0.750 L
6.00 g CO 2
= 0.136 moles
44.009 g/mol
nRT 0.136 ! 0.0821 Latm/molK ! 323.15 K
PCO 2 =
=
= 4.82 atm
V
0.750 L
Total pressure = PSO 2 + PCO 2 = 6.78 atm
25. CO (5.00 g) and CO2 (5.00g) were placed in a 750.0 mL container at 50.0 ºC. The
total pressure in the container was __________ atm.
(a). 10.3
(b). 4.02
(c). 6.31
(d). 0.292
Explanation: Since each gas behaves as if the other was absent, need to calculate the
partial pressure of each gas and then add them together to find the total pressure.
5.00 g CO
= 0.178 moles
28.01 g/mol
nRT 0.178 ! 0.0821 Latm/molK ! 323.15 K
PSO 2 =
=
= 6.31 atm
V
0.750 L
5.00 g CO 2
= 0.114 moles
44.009 g/mol
nRT 0.114 ! 0.0821 Latm/molK ! 323.15 K
PCO 2 =
=
= 4.02 atm
V
0.750 L
Total pressure = PSO 2 + PCO 2 = 10.3 atm
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 10
26. In ideal gas equation calculations, expressing pressure in Pascals (Pa), necessitates
the use of the gas constant, R, equal to __________.
(a). 0.08206 atm L mol-1K-1
(b).8.314 J mol-1K-1
(c). 62.36 L torr mol-1K-1
(d). 1.987 cal mol-1K-1
Explanation: When the unit Pa is used for pressure the volume is measured in m3.
The product of these 2 units (PV) then has units of energy which is Joule.
27. Of the following, __________ is a correct statement of Boyle’s law.
(a). PV = constant
P
(b).
= constant
V
V
(c).
= constant
P
V
(d).
= constant
T
Explanation: According to Boyle’s law at constant temperature, the pressure and
volume of an ideal gas are inversely related (P a 1/V) to each other. To convert this
proportionality to an equality the product PV must be a constant.
28. Of the following, __________ is a valid statement of Charles’ law.
P
= constant
T
V
(b).
= constant
T
(c). PV = constant
(d). V = constant x n
(a).
Explanation: According to Charles’s law, at constant pressure, the temperature and
volume of an ideal gas are directly related (V a T) to each other. To convert this
proportionality to an equality the V/T must be a constant.
Copyright © 2006 Dr. Harshavardhan D. Bapat
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CHE141
Chapter 10
29. Which one of the following is a valid statement of Avogadro’s law?
P
= constant
T
V
(b). = constant
T
(c). PV = constant
(d).V = constant x n
(a).
Explanation: According to Avogadro’s law, at constant pressure and temperature the
volume of an ideal gas is directly related (V a n) to the number of moles. This
proportionality can also be expressed as V x constant = n.
30. The molar volume of a gas at STP is __________ L.
(a). 0.08206
(b). 273.15
(c). 1.00
(d).22.4
Explanation: This is a fact.
31. Standard temperature and pressure (STP), in the context of gases, refers to
__________.
(a). 298 K and 1 atm
(b).273.15 K and 1 atm
(c). 298 K and 1 torr
(d). 273 K and 1 pascal
Explanation: This is a fact.
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