CHAPTER 5
The Gaseous State
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
5.1. First, convert to atm (57 kPa = 57 103 Pa).
1 atm
57 103 Pa = 0.562 = 0.56 atm
1.01325 105 Pa
Next, convert to mmHg.
760 mmHg
57 103 Pa 5
= 427.5 = 4.3 102 mmHg
1.01325 10 Pa
5.2. Application of Boyle’s law gives
Pi 1.00 atm
Vf = Vi = 20.0 L = 24.096 = 24.1 L
Pf 0.830 atm
5.3. First, convert the temperatures to the Kelvin scale.
Ti = (19 + 273) = 292 K
Tf = (25 + 273) = 298 K
Following is the data table.
Vi = 4.38 dm3 Pi = 101 kPa Ti = 292 K
Vf = ? Pf = 101 kPa Tf = 298 K
Apply Charles’s law to obtain
Tf 298 K
Vf = Vi = 4.38 dm3 = 4.470 = 4.47 dm3
Ti 292 K
5.4. First, convert the temperatures to kelvins.
Ti = (24 + 273) = 297 K
Tf = (35 + 273) = 308 K
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 5: The Gaseous State 159
Following is the data table.
Vi = 5.41 dm3 Pi = 101.5 kPa Ti = 297 K
Vf = ? Pf = 102.8 kPa Tf = 308 K
Apply both Boyle’s law and Charles’s law combined to get
Pi T 101.5 kPa 308 K
Vf = Vi f = 5.41 dm3 = 5.539 = 5.54 dm3
Pf Ti 102.8 kPa 297 K
5.5. Use the ideal gas law, PV = nRT, and solve for n:
PV V
n= = P
RT RT
Note that everything in parentheses is constant. Therefore, you can write
n = constant P
Or, expressing this as a proportion,
nP
5.6. First, convert the mass of O2 to moles of O2 (molar mass 32.00 g/mol) and convert temperature to
kelvins.
T = 23 + 273 = 296 K
1000 g 1 mol O2
3.03 kg O2 = 94.688 mol O2
1 kg 32.00 g O2
Summarize the data in a table.
Variable Value
P ?
V 50.0 L
T 296 K
n 94.688 mol
Solve the ideal gas equation for P, and substitute the data to get
nRT (94.688)(0.08206 L atm/K mol)(296 K)
P= = = 46.00 = 46.0 atm
V 50.0 L
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 160 Chapter 5: The Gaseous State
5.7. The data given are
Variable Value
1 atm
P 752 mmHg = 0.98947 atm
760 mmHg
V 1 L (exact number)
T (21 + 273) = 294 K
n ?
Using the ideal gas law, solve for n, the moles of helium.
PV (0.98947 atm)(1 L)
n= = = 0.04101 mol
RT (0.08206 L atm/K mol)(294 K)
Now convert mol He to grams.
4.00 g He
0.04101 mol He = 0.16404 g He
1 mol He
Therefore, the density of He at 21°C and 752 mmHg is 0.164 g/L.
The difference in mass between one liter of air and one liter of helium is
Mass air − mass He = 1.188 g − 0.16404 g = 1.02396 = 1.024 g difference
5.8. Tabulate the values of the variables.
Variable Value
P 0.862 atm
V 1 L (exact number)
T (25 + 273) = 298 K
n ?
From the ideal gas law, PV = nRT, you obtain
PV (0.862 atm)(1 L)
n= = = 0.03525 mol
RT (0.08206 L atm/K mol)(298 K)
Dividing the mass of the vapor by moles gives you the mass per mole (the molar mass).
grams vapor 2.26 g
Molar mass = = = 64.114 g/mol
moles vapor 0.03525 mol
Therefore, the molecular weight is 64.1 amu.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 5: The Gaseous State 161
5.9. First, determine the number of moles of Cl2 from the mass of HCl (molar mass 36.46 g/mol) and
from the stoichiometry of the chemical equation:
1 mol HCl 5 mol Cl2
9.41 g HCl = 0.080653 mol Cl2
36.46 g HCl 16 mol HCl
Tabulate the values of the variables:
Variable Value
1 atm
P 787 mmHg = 1.0355 atm
760 mmHg
T (40 + 273) K = 313 K
n 0.080653 mol
V ?
Rearrange the ideal gas law to obtain V:
nRT (0.080653 mol)(0.08206 L atm/K mol)(313 K)
V= = = 2.001 = 2.00 L
P 1.0355 atm
5.10. Each gas obeys the ideal gas law. In each case, convert grams to moles and substitute into the
ideal gas law to determine the partial pressure of each.
1 mol O2
1.031 g O2 = 0.0322188 mol O2
32.00 g O2
nRT (0.0322188)(0.08206 L atm/K mol)(291 K)
P= = = 0.076936 atm
V 10.0 L
1 mol CO2
0.572 g CO2 = 0.012997 mol CO2
44.01 g CO 2
nRT (0.012997)(0.08206 L atm/K mol)(291 K)
P= = = 0.031036 atm
V 10.0 L
The total pressure is equal to the sum of the partial pressures:
P = PO2 + PCO2 = 0.076936 + 0.031036 = 0.10797 = 0.1080 atm
The mole fraction of oxygen in the mixture is
PO2 0.07694 atm
Mole fraction O2 = = = 0.7122 = 0.712
P 0.10802 atm
5.11. Determine the number of moles of O2 from the mass of KClO3 and from the stoichiometry of the
chemical reaction.
1 mol KClO3 3 mol O2
1.300 g KClO3 = 0.0159184 mol O 2
122.5 g KClO3 2 mol KClO3
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
The Gaseous State
■ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
5.1. First, convert to atm (57 kPa = 57 103 Pa).
1 atm
57 103 Pa = 0.562 = 0.56 atm
1.01325 105 Pa
Next, convert to mmHg.
760 mmHg
57 103 Pa 5
= 427.5 = 4.3 102 mmHg
1.01325 10 Pa
5.2. Application of Boyle’s law gives
Pi 1.00 atm
Vf = Vi = 20.0 L = 24.096 = 24.1 L
Pf 0.830 atm
5.3. First, convert the temperatures to the Kelvin scale.
Ti = (19 + 273) = 292 K
Tf = (25 + 273) = 298 K
Following is the data table.
Vi = 4.38 dm3 Pi = 101 kPa Ti = 292 K
Vf = ? Pf = 101 kPa Tf = 298 K
Apply Charles’s law to obtain
Tf 298 K
Vf = Vi = 4.38 dm3 = 4.470 = 4.47 dm3
Ti 292 K
5.4. First, convert the temperatures to kelvins.
Ti = (24 + 273) = 297 K
Tf = (35 + 273) = 308 K
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 5: The Gaseous State 159
Following is the data table.
Vi = 5.41 dm3 Pi = 101.5 kPa Ti = 297 K
Vf = ? Pf = 102.8 kPa Tf = 308 K
Apply both Boyle’s law and Charles’s law combined to get
Pi T 101.5 kPa 308 K
Vf = Vi f = 5.41 dm3 = 5.539 = 5.54 dm3
Pf Ti 102.8 kPa 297 K
5.5. Use the ideal gas law, PV = nRT, and solve for n:
PV V
n= = P
RT RT
Note that everything in parentheses is constant. Therefore, you can write
n = constant P
Or, expressing this as a proportion,
nP
5.6. First, convert the mass of O2 to moles of O2 (molar mass 32.00 g/mol) and convert temperature to
kelvins.
T = 23 + 273 = 296 K
1000 g 1 mol O2
3.03 kg O2 = 94.688 mol O2
1 kg 32.00 g O2
Summarize the data in a table.
Variable Value
P ?
V 50.0 L
T 296 K
n 94.688 mol
Solve the ideal gas equation for P, and substitute the data to get
nRT (94.688)(0.08206 L atm/K mol)(296 K)
P= = = 46.00 = 46.0 atm
V 50.0 L
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, 160 Chapter 5: The Gaseous State
5.7. The data given are
Variable Value
1 atm
P 752 mmHg = 0.98947 atm
760 mmHg
V 1 L (exact number)
T (21 + 273) = 294 K
n ?
Using the ideal gas law, solve for n, the moles of helium.
PV (0.98947 atm)(1 L)
n= = = 0.04101 mol
RT (0.08206 L atm/K mol)(294 K)
Now convert mol He to grams.
4.00 g He
0.04101 mol He = 0.16404 g He
1 mol He
Therefore, the density of He at 21°C and 752 mmHg is 0.164 g/L.
The difference in mass between one liter of air and one liter of helium is
Mass air − mass He = 1.188 g − 0.16404 g = 1.02396 = 1.024 g difference
5.8. Tabulate the values of the variables.
Variable Value
P 0.862 atm
V 1 L (exact number)
T (25 + 273) = 298 K
n ?
From the ideal gas law, PV = nRT, you obtain
PV (0.862 atm)(1 L)
n= = = 0.03525 mol
RT (0.08206 L atm/K mol)(298 K)
Dividing the mass of the vapor by moles gives you the mass per mole (the molar mass).
grams vapor 2.26 g
Molar mass = = = 64.114 g/mol
moles vapor 0.03525 mol
Therefore, the molecular weight is 64.1 amu.
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Chapter 5: The Gaseous State 161
5.9. First, determine the number of moles of Cl2 from the mass of HCl (molar mass 36.46 g/mol) and
from the stoichiometry of the chemical equation:
1 mol HCl 5 mol Cl2
9.41 g HCl = 0.080653 mol Cl2
36.46 g HCl 16 mol HCl
Tabulate the values of the variables:
Variable Value
1 atm
P 787 mmHg = 1.0355 atm
760 mmHg
T (40 + 273) K = 313 K
n 0.080653 mol
V ?
Rearrange the ideal gas law to obtain V:
nRT (0.080653 mol)(0.08206 L atm/K mol)(313 K)
V= = = 2.001 = 2.00 L
P 1.0355 atm
5.10. Each gas obeys the ideal gas law. In each case, convert grams to moles and substitute into the
ideal gas law to determine the partial pressure of each.
1 mol O2
1.031 g O2 = 0.0322188 mol O2
32.00 g O2
nRT (0.0322188)(0.08206 L atm/K mol)(291 K)
P= = = 0.076936 atm
V 10.0 L
1 mol CO2
0.572 g CO2 = 0.012997 mol CO2
44.01 g CO 2
nRT (0.012997)(0.08206 L atm/K mol)(291 K)
P= = = 0.031036 atm
V 10.0 L
The total pressure is equal to the sum of the partial pressures:
P = PO2 + PCO2 = 0.076936 + 0.031036 = 0.10797 = 0.1080 atm
The mole fraction of oxygen in the mixture is
PO2 0.07694 atm
Mole fraction O2 = = = 0.7122 = 0.712
P 0.10802 atm
5.11. Determine the number of moles of O2 from the mass of KClO3 and from the stoichiometry of the
chemical reaction.
1 mol KClO3 3 mol O2
1.300 g KClO3 = 0.0159184 mol O 2
122.5 g KClO3 2 mol KClO3
© 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
