CHAPTER 5 GASES AND THE KINETICMOLECULAR THEORY
FOLLOW–UP PROBLEMS
5.1A
Plan: Use the equation for gas pressure in an open-end manometer to calculate the pressure of the gas. Use
2
conversion factors to convert pressure in mmHg to units of torr, pascals and lb/in .
Solution:
Because Pgas < Patm, Pgas = Patm – h
Pgas = 753.6 mmHg – 174.0 mmHg = 579.6 mmHg
 1 torr 

 = 579.6 torr
Pressure (torr) = 579.6 mmHg 
1 mmHg 


 1 atm 1.01325105 Pa 

 = 7.727364 × 104 = 7.727 × 104 Pa

Pressure (Pa) = 579.6 mmHg 

1atm

 760 mmHg 
 1 atm 14.7 lb/in 2 
2
 = 11.21068 = 11.2 lb/in2

Pressure (lb/in ) = 579.6 mmHg 
 760 mmHg 
 1 atm 



5.1B

Plan: Convert the atmospheric pressure to torr. Use the equation for gas pressure in an open-end manometer to calculate
2
the pressure of the gas. Use conversion factors to convert pressure in torr to units of mmHg, pascals and lb/in .
Solution:
Because Pgas > Patm, Pgas = Patm + h


Pgas = (0.9475 atm) 
+ 25.8 torr = 745.9 torr



Pressure (mmHg) = (745.9 torr) 


Pressure (Pa) = (745.9 mmHg) 


2
Pressure (lb/in ) = (745.9 mmHg) 

5.2B
Copyright

= 745.9 mmHg


 5






2

 = 9.94452 × 104 = 9.945 × 104 Pa


2
 = 14.427 = 14.4 lb/in
Plan: Given in the problem is an initial volume, initial pressure, and final volume for the argon gas. The final
pressure is to be calculated. The temperature and amount of gas are fixed. Rearrange the ideal gas law to the
appropriate form and solve for P2. Once solved for, P2 must be converted from atm units to kPa units.
Solution:
P1 = 0.871 atm; V1 = 105 mL
P2 = unknown V2 = 352 mL
PV
PV
1 1
= 2 2
At fixed n and T:
n1T1
n2T2
PV
1 1 = P2V2
PV
P2 (atm) = 1 1 =
= 0.259815 = 0.260 atm
V2
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5-1

P2 (kPa) = (0.260 atm) 

5.2A

 = 26.3 kPa

Plan: Given in the problem is an initial volume, initial pressure, and final pressure for the oxygen gas. The final
volume is to be calculated. The temperature and amount of gas are fixed. Convert the final pressure to atm units.
Rearrange the ideal gas law to the appropriate form and solve for V2.
Solution:
P1 = 122 atm;
V1 = 651 L
P2 = 745 mmHg
V2 = unknown
PV
P2V2
1 1
=
At fixed n and T:
n1T1
n2T2
PV
1 1 = P2V2

P2 (atm) = (745 mmHg) 

PV
V2 (atm) = 1 1 =
P2
5.3A

 = 0.980263 atm

4
4
= 8.1021 × 10 = 8.10 × 10 L
Plan: Convert the temperatures to kelvin units and the initial pressure to units of torr. Examine the ideal gas law,
PV
PV
noting the fixed variables and those variables that change. R is always constant so 1 1 = 2 2 . In this problem,
n1T1
n2T2
P and T are changing, while n and V remain fixed.
Solution:
T1 = 23°C
T2 = 100°C
P1 = 0.991 atm
P2 = unknown
n and V remain constant
Converting T1 from °C to K: 23°C + 273.15 = 296.15 K
Converting T2 from °C to K: 100°C + 273.15 = 373.15 K


P1 (torr) = (0.991 atm) 
= 753.16 torr


Arranging the ideal gas law and solving for P2:
P1 V1
P2 V2
P1
P2
n1 T1
n 2 T2
T1
T2
P2 (torr) = P1
T2
T1




 = 948.98 = 949 torr

Because the pressure in the tank (949 torr) is less than the pressure at which the safety valve will open
3
(1.00 × 10 torr), the safety valve will not open.
5.3B
Copyright
Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly
proportional to the absolute temperature of the gas. The temperature must be lowered to reduce the volume of a
gas. Arrange the ideal gas law, solving for T2 at fixed n and P. Temperature must be converted to kelvin units.
Solution:
V1 = 32.5 L
V2 = 28.6 L
T1 = 40°C (convert to K)
T2 = unknown
n and P remain constant
Converting T from °C to K: T1 = 40 °C + 273.15 = 313.15K
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5-2
Arranging the ideal gas law and solving for T2:
PV
PV
V1
V
1 1
= 2 2 or
 2
n1T1
n2 T2
T1
T2
V2


T2  T1 = (313.15 K) 
= 275.572 K – 273.15 = 2.422 = 2°C


V1
5.4A
Plan: In this problem, the amount of gas is decreasing. Since the container is rigid, the volume of the gas will not
change with the decrease in moles of gas. The temperature is also constant. So, the only change will be that the
pressure of the gas will decrease since fewer moles of gas will be present after removal of the 5.0 g of ethylene.
Rearrange the ideal gas law to the appropriate form and solve for P2. Since the ratio of moles of ethylene is equal
to the ratio of grams of ethylene, there is no need to convert the grams to moles. (This is illustrated in the solution
by listing the molar mass conversion twice.)
Solution:
P1 = 793 torr; P2 = ?
mass1 = 35.0 g;
mass2 = 35.0 – 5.0 = 30.0 g
PV
P2V2
1 1
=
At fixed V and T:
n1T1
n2T2
P1
P
= 2
n1
n2
 1 mol C2 H 4 

2 4

2 4
P1 n2


793
torr

P2 =
=
= 679.714 = 680. torr
 1 mol C2 H 4 
n1


35.0 g C2 H 4 

 28.05 g C 2 H 4 
30.0 g C H  28.05 g C H

5.4B

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
PV
PV
so 1 1 = 2 2 . In this problem, n and V are changing, while P and T remain fixed.
n1T1
n2T2
Solution:
m1 = 1.26 g N2
m2 = 1.26 g N2 + 1.26 g He
V1 = 1.12 L
V2 = unknown
P and T remain constant


2 
 = 0.044968 mol N2 = n1
Converting m1 (mass) to n1 (moles): (1.26 g N2) 


2

Converting m2 (mass) to n2 (moles): 0.044968 mol N2 + (1.26 g He) 




= 0.044968 mol N2 + 0.31476 mol He = 0.35973 mol gas = n2
Arranging the ideal gas law and solving for V2:
P1 V1
P2 V2
V1
V2
n1 T1
n2 T2
n1
n2
V2 = V1
Copyright
P2
P1




 = 8.9597 = 8.96 L

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5-3
5.5A
Plan: Convert the temperatures to kelvin. Examine the ideal gas law, noting the fixed variables and those variables
PV
PV
that change. R is always constant so 1 1 = 2 2 . In this problem, P, V, and T are changing, while n remains
n1T1
n2T2
fixed.
Solution:
T1 = 23°C
T2 = 18°C
P1 = 755 mmHg
P2 = unknown
V1 = 2.55 L
V2 = 4.10 L
n remains constant
Converting T1 from °C to K: 23°C + 273.15 = 296.15 K
Converting T2 from °C to K: 18°C + 273.15 = 291.15 K
Arranging the ideal gas law and solving for P2:
PV
P2V2
P1V1 P2V2
1 1
n 1 T1 n 2 T2
T1
T2
P2 (mmHg) = P1
5.5B



V1T2
V2T1

 = 461.645 = 462 mmHg

Plan: Convert the temperatures to kelvin. Examine the ideal gas law, noting the fixed variables and those variables
PV
PV
that change. R is always constant so 1 1 = 2 2 . In this problem, P, V, and T are changing, while n remains
n1T1
n2T2
fixed.
Solution:
T1 = 28°C
T2 = 21°C
P1 = 0.980 atm
P2 = 1.40 atm
V1 = 2.2 L
V2 = unknown
n remains constant
Converting T1 from °C to K: 28°C + 273.15 = 301.15 K
Converting T2 from °C to K: 21°C + 273.15 = 294.15 K
Arranging the ideal gas law and solving for V2:
P1V1
P2V2
P1V1 P2V2
n 1 T1
n 2 T2
T1
T2
V2 (L) = V1
5.6A



P1T2
P2T1

 = 1.5042 = 1.5 L

Plan: From Sample Problem 5.6 the temperature of 21°C and volume of 438 L are given. The pressure is 1.37 atm
and the unknown is the moles of oxygen gas. Use the ideal gas equation PV = nRT to calculate the number of
moles of gas. Multiply moles by molar mass to obtain mass.
Solution:
PV = nRT
n=



Mass (g) of O2 = (24.8475 mol O2) 

Copyright

1.37 atm 438 L
PV
=
= 24.8475 mol O2
 0.0821 atm  L 
RT

273.15  21 K 
 mol  K

2
2

 = 795.12 = 795 g O2

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5-4
5.6B
Plan: Convert the mass of helium to moles, the temperature to kelvin units, and the pressure to atm units. Use the
ideal gas equation PV = nRT to calculate the volume of the gas.
Solution:
P = 731 mmHg
V = unknown
m = 3950 kg He
T = 20°C



 = 9.8676 × 105 mol = n
Converting m (mass) to n (moles): (3950 kg He) 





Converting T from °C to K: 20°C + 273.15 = 293.15 K

Converting P from mmHg to atm: (731 mmHg) 


 = 0.962 atm

PV = nRT
nRT
V=
P
5.7A
Copyright
5



 
mol  K 
7
7
= 2.4687 × 10 = 2.47 × 10 L
Plan: Balance the chemical equation. The pressure is constant and, according to the picture, the volume approximately
doubles. The volume change may be due to the temperature and/or a change in moles. Examine the balanced reaction
for a possible change in number of moles. Rearrange the ideal gas law to the appropriate form and solve for the variable
that changes.
Solution:
The balanced chemical equation must be 2CD  C2 + D2
Thus, the number of mole of gas does not change (2 moles both before and after the reaction). Only the
temperature remains as a variable to cause the volume change. Let V1 = the initial volume and 2V1 = the final
volume V2.
T1 = (–73 + 273.15) K = 200.15 K
PV
PV
1 1
= 2 2
At fixed n and P:
n1T1
n2T2
V1
V
= 2
T1
T2
T2 =
5.7B

2V1 200.15 K 
V2 T1
=
= 400.30 K – 273.15 = 127.15 = 127°C
V1
V1 
Plan: The pressure is constant and, according to the picture, the volume approximately decreases by a factor of 2
(the final volume is approximately one half the original volume). The volume change may be due to the temperature
change and/or a change in moles. Consider the change in temperature. Examine the balanced reactions for a possible
change in number of moles. Think about the relationships between the variables in the ideal gas law in order to
determine the effect of temperature and moles on gas volume.
Solution:
Converting T1 from °C to K: 199°C + 273.15 = 472.15 K
Converting T2 from °C to K: –155°C + 273.15 = 118.15 K
According to the ideal gas law, temperature and volume are directly proportional. The temperature decreases by a
factor of 4, which should cause the volume to also decrease by a factor of 4. Because the volume only decreases
by a factor of 2, the number of moles of gas must have increased by a factor of 2 (moles of gas and volume are
also directly proportional).
1/4 (decrease in V from the decrease in T) × 2 (increase in V from the increase in n)
= 1/2 (a decrease in V by a factor of 2)
Thus, we need to find a reaction in which the number of moles of gas increases by a factor of 2.
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5-5
In equation (1), 3 moles of gas yield 2 moles of gas.
In equation (2), 2 moles of gas yield 4 moles of gas.
In equation (3), 1 mole of gas yields 3 moles of gas.
In equation (4), 2 moles of gas yield 2 moles of gas.
Because the number of moles of gas doubles in equation (2), that equation best describes the reaction in the
figure in this problem.
5.8A
Plan: Density of a gas can be calculated using a version of the ideal gas equation, d = (PM )/(RT). Two
calculations are required, one with T = 0°C = 273.15 K and P = 380 torr and the other at STP which is defined as
T = 273 K and P = 1 atm.
Solution:
Density at T = 273 K and P = 380 torr:
380 torr 44.01 g/mol  1 atm 

d=
 = 0.98124 = 0.981 g/L
 760 torr 
 0.0821 atm  L 

 273.15 K
 mol  K






Density at T = 273 K and P = 1 atm. (Note: The 1 atm is an exact number and does not affect the significant
figures in the answer.)
44.01 g/mol1 atm
d=
 0.0821 atm  L 

 273.15 K
 mol  K



= 1.96249 = 1.96 g/L
The density of a gas increases proportionally to the increase in pressure.
5.8B
Plan: Density of a gas can be calculated using a version of the ideal gas equation, d = PM /RT
Solution:
Density of NO2 at T = 297.15 K (24°C + 273.15) and P = 0.950 atm:
d=
= 1.7917 = 1.79 g/L
atm  L 

mol  K
Nitrogen dioxide is more dense than dry air at the same conditions (density of dry air = 1.13 g/L).
5.9A



Plan: Calculate the mass of the gas by subtracting the mass of the empty flask from the mass of the flask
containing the condensed gas. The volume, pressure, and temperature of the gas are known.
dRT
mRT
The relationship d = PM /RT is rearranged to give M =
or M =
P
PV
Solution:
Mass (g) of gas = mass of flask + vapor – mass of flask = 68.697 – 68.322 = 0.375 g
T = 95.0°C + 273.15 = 368.15 K
 1 atm 
P = 740 torr 
 = 0.973684 atm
 760 torr 
V = 149 mL = 0.149 L
M=
Copyright
mRT
=
PV
 0.0821 atm  L 
368.15 K 
mol  K

= 78.125 = 78.1 g
0.973684 atm0.149 L 
0.375 g 
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5-6
5.9B
Plan: Calculate the mass of the gas by subtracting the mass of the empty glass bulb from the mass of the bulb
containing the gas. The volume, pressure, and temperature of the gas are known. The relationship d = PM /RT is
dRT
mRT
rearranged to give M =
or M =
. Use the molar mass of the gas to determine its identity.
P
PV
Solution:
Mass (g) of gas = mass of bulb + gas – mass of bulb = 82.786 – 82.561 = 0.225 g
T = 22°C + 273.15 = 295.15 K


 = 0.965 atm
P = (733 mmHg) 


V = 350 mL = 0.350 L

atm  L 



mRT
mol
K
M=
=
= 16.1426 = 16.1 g/mol
PV
Methane has a molar mass of 16.04 g/mol. Nitrogen monoxide has a molar mass of 30.01 g/mol. The gas that has
a molar mass that matches the calculated value is methane.
5.10A
Plan: Calculate the number of moles of each gas present and then the mole fraction of each gas. The partial
pressure of each gas equals the mole fraction times the total pressure. Total pressure equals 1 atm since the
problem specifies STP. This pressure is an exact number, and will not affect the significant figures in the answer
Solution:
 1 mol He 
 = 1.373970 mol He
Moles of He = 5.50 g He
 4.003 g He 
 1 mol Ne 
 = 0.743310 mol Ne
Moles of Ne = 15.0 g Ne
 20.18 g Ne 
 1 mol Kr 
 = 0.417661 mol Kr
Moles of Kr = 35.0 g Kr
 83.80 g Kr 
Total number of moles of gas = 1.373970 + 0.743310 + 0.417661 = 2.534941 mol
PA = XA × Ptotal
1.37397 mol He 
PHe = 
 1 atm  = 0.54201 = 0.542 atm He
 2.534941 mol 
 0.74331 mol Ne 
PNe = 
1 atm  = 0.29323 = 0.293 atm Ne
 2.534941 mol 
 0.41766 mol Kr 
PKr = 
1 atm  = 0.16476 = 0.165 atm Kr
 2.534941 mol 
5.10B
Plan: Use the formula PA = XA × Ptotal to calculate the mole fraction of He. Multiply the mole fraction by 100% to
calculate the mole percent of He.
Solution:
PHe = XHe × Ptotal
P 



Mole percent He = XHe (100%) =  He 
(100%) = 70.1%


 total 
5.11A
Plan: The gas collected over the water will consist of H2 and H2O gas molecules. The partial pressure of the water
can be found from the vapor pressure of water at the given temperature given in the text. Subtracting this partial
pressure of water from total pressure gives the partial pressure of hydrogen gas collected over the water. Calculate
Copyright
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5-7
the moles of hydrogen gas using the ideal gas equation. The mass of hydrogen can then be calculated by
converting the moles of hydrogen from the ideal gas equation to grams.
Solution:
From the table in the text, the partial pressure of water is 13.6 torr at 16°C.
P = 752 torr – 13.6 torr = 738.4 = 738 torr H2
The unrounded partial pressure (738.4 torr) will be used to avoid rounding error.
 1 atm 
738.4 torr 1495 mL
103 L 
PV

Moles of hydrogen = n =
=



 760 torr 
 0.0821 atm  L 
RT
 1 mL 

 273.15  16 K
 mol  K





 
= 0.061186 mol H2
 2.016 g H 2 
 = 0.123351 = 0.123 g H2
Mass (g) of hydrogen = 0.061186 mol H 2 
 1 mol H 2 
5.11B
Plan: The gas collected over the water will consist of O2 and H2O gas molecules. The partial pressure of the water
can be found from the vapor pressure of water at the given temperature given in the text. Subtracting this partial
pressure of water from total pressure gives the partial pressure of oxygen gas collected over the water. Calculate
the moles of oxygen gas using the ideal gas equation. The mass of oxygen can then be calculated by converting
the moles of oxygen from the ideal gas equation to grams.
Solution:
From the table in the text, the partial pressure of water is 17.5 torr at 20°C.
P = 748 torr – 17.5 torr = 730.5 = 730 torr O2
PV





Moles of oxygen = n =
=





atm

L

RT



mol  K 
= 0.012252 mol O2


2
 = 0.39207 = 0.392 g O2
Mass (g) of oxygen = (0.012252 mol O2) 


2 
5.12A
Plan: Write a balanced equation for the reaction. Calculate the moles of CO(g) from the ideal gas equation. Then,
use the stoichiometric ratio from the balanced equation and the molar mass of SiO2 to determine the mass of SiO2.
Solution:
The balanced equation is SiO2(s) + 2C(s)  Si(s) + 2CO(g).
(0.975 atm)(44.8 L)
PV
Moles of CO 

 1.785  1.79 mol CO
RT (0.0821 atm  L/mol  K)(298 K)
1 mol SiO2 
 60.09 g SiO2 

Mass(g) of SiO2  (1.79 mol CO) 

  53.78  53.8 g SiO 2
 2 mol CO 
 1 mol SiO 2 
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5-8
5.12B
Plan: Write a balanced equation for the reaction. Calculate the moles of H2(g) from the starting amount of CuO(s)
using the stoichiometric ratio from the balanced equation. Find the volume of the H2(g) from the ideal gas equation.
Solution:
The balanced equation is CuO(s) + H2(g)  Cu(s) + H2O(g).
 1 mol CuO 
 1 mol H 2 

Amount (mol) of H 2 ( g)  (35.5 g CuO) 

  0.46626  0.446 mol H 2 ( g)
 79.55 g CuO 
1 mol CuO 
Volume of H 2 ( g) 
5.13A
(498 K)  1 atm 
nRT
  18.1 L H ( g)

 (0.446 mol)(0.0821 atm  L/mol  K)
2
P
(765 torr)  760 torr 
Plan: Balance the equation for the reaction. Determine the limiting reactant by finding the moles of each reactant
from the ideal gas equation, and comparing the values. Calculate the moles of remaining excess reactant. This is
the only gas left in the flask, so it is used to calculate the pressure inside the flask.
Solution:
The balanced equation is NH3(g) + HCl(g)  NH4Cl(s).
The stoichiometric ratio of NH3 to HCl is 1:1, so the reactant present in the lower quantity of moles is the limiting
reactant.
Moles of ammonia =



0.452 atm 10.0 L
PV
=
= 0.18653 mol NH3
 0.0821 atm  L 
RT


 mol  K  273.15  22 K

Moles of hydrogen chloride =

 


7.50 atm 155 mL
PV
=
 0.0821 atm  L 
RT


 mol  K  271.15 K


103 L 

 = 0.052220 mol HCl

 1 mL 
The HCl is limiting so the moles of ammonia gas left after the reaction would be
0.18653 – 0.052220 = 0.134310 mol NH3.
 0.0821 atm  L 
 273.15  22 K
0.134310 mol 
 mol  K

nRT
Pressure (atm) of ammonia =
=
V
10.0 L




 

= 0.325458 = 0.325 atm NH3
5.13B
Copyright
Plan: Balance the equation for the reaction. Use the ideal gas law to calculate the moles of fluorine that react.
Determine the limiting reactant by determining the moles of product that can be produced from each of the
reactants and comparing the values. Use the moles of IF5 produced and the ideal gas law to calculate the volume
of gas produced.
Solution:
The balanced equation is I2(s) + 5F2 (g)  2IF5(g).
PV
Amount (mol) of F2 that reacts = n =
=
= 0.10105 mol F2

atm  L 
RT



mol  K 
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5-9

Amount (mol) of IF5 produced from F2 = 0.10105 mol F2 


Amount (mol) of IF5 produced from I2 = 4.16 g I2 

5
2
2
2





 = 0.040421 mol IF5

5
2

 = 0.032782 mol IF5

Because a smaller number of moles is produced from the I2, I2 is limiting and 0.032782 mol of IF5 are produced.



nRT
Volume (L) of IF5 =
P
5.14A
atm  L 

mol  K 
= 1.08850 = 1.09 L
Plan: Graham’s law can be used to solve for the effusion rate of the ethane since the rate and molar mass of
helium are known, along with the molar mass of ethane. In the same way that running slower increases the time to
go from one point to another, so the rate of effusion decreases as the time increases. The rate can be expressed as
1/time.
Solution:
Rate He
Rate C 2 H 6

Molar mass C2 H 6
Molar m ass He
 0.010 mol He 



 1.25 min 

 0.010 mol C 2 H6 



tC H


2
30.07 g/mol
4.003 g/mol
6
0.800 t = 2.74078
t = 3.42597 = 3.43 min
5.14B
Plan: Graham’s law can be used to solve for the molar mass of the unknown gas since the rates of both gases and
the molar mass of argon are known. Rate can be expressed as the volume of gas that effuses per unit time.
Solution:
Rate of Ar = 13.8 mL/time
Rate of unknown gas = 7.23 mL/time
Mass of Ar = 39.95 g/mol
 Molar mass




Molar mass


Ar



 Molar mass

 Molar mass


as 
2

Munknown gas = (M Ar) 

Ar






2


Munknown gas = (39.95 g/mol) 

 = 146 g/mol

2
CHEMICAL CONNECTIONS BOXED READING PROBLEMS
B5.1
Copyright
Plan: Examine the change in density of the atmosphere as altitude changes.
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5-10
Solution:
The density of the atmosphere decreases with increasing altitude. High density causes more drag on the aircraft.
At high altitudes, low density means that there are relatively few gas particles present to collide with the aircraft.
B5.2
Plan: The conditions that result in deviations from ideal behavior are high pressure and low temperature. At high
pressure, the volume of the gas decreases, the distance between particles decreases, and attractive forces between gas
particles have a greater effect. A low temperature slows the gas particles, also increasing the affect of attractive
forces between particles.
Solution:
Since the pressure on Saturn is significantly higher and the temperature significantly lower than that on Venus,
atmospheric gases would deviate more from ideal gas behavior on Saturn.
B5.3
Plan: To find the volume percent of argon, multiply its mole fraction by 100. The partial pressure of argon gas can
be found by using the relationship PAr = XAr × Ptotal. The mole fraction of argon is given in Table B5.1.
Solution:
Volume percent = mole fraction × 100 = 0.00934 × 100 = 0.934%
The total pressure at sea level is 1.00 atm = 760 torr.
PAr = XAr × Ptotal = 0.00934 × 760 torr = 7.0984 = 7.10 torr
B5.4
Plan: To find the moles of gas, convert the mass of the atmosphere from t to g and divide by the molar mass of air.
Knowing the moles of air, the volume can be calculated at the specified pressure and temperature by using the
ideal gas law.
Solution:
1000 kg 1000 g 
 1 mol 
a) Moles of gas = 5.14 1015 t 

 28.8 g 
 1 t  1 kg 

20
20
= 1.78472 × 10 = 1.78 × 10 mol
b) PV = nRT
nRT
V=
=
P
1.7847210
20
L  atm 

mol 0.0821
273.15  25 K

mol  K 


1 atm
 
21
21
= 4.36866 × 10 = 4 × 10 L
END-OF-CHAPTER PROBLEMS
5.1
Plan: Review the behavior of the gas phase vs. the liquid phase.
Solution:
a) The volume of the liquid remains constant, but the volume of the gas increases to the volume of the larger
container.
b) The volume of the container holding the gas sample increases when heated, but the volume of the container
holding the liquid sample remains essentially constant when heated.
c) The volume of the liquid remains essentially constant, but the volume of the gas is reduced.
5.2
The particles in a gas are further apart than those are in a liquid.
a) The greater empty space between gas molecules allows gases to be more compressible than liquids.
b) The greater empty space between gas molecules allows gases to flow with less resistance (hindrance) than
liquids.
c) The large empty space between gas molecules limits their interaction, allowing all mixtures of gases to be
solutions.
d) The large empty space between gas molecules increases the volume of the gas, therefore decreasing the density.
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5-11
5.3
The mercury column in the mercury barometer stays up due to the force exerted by the atmosphere on the mercury
in the outer reservoir just balancing the gravitational force on the mercury in the tube. Its height adjusts according
to the air pressure on the reservoir. The column of mercury is shorter on a mountaintop as there is less atmosphere
to exert a force on the mercury reservoir. On a mountaintop, the air pressure is less, so the height of mercury it
balances in the barometer is shorter than at sea level where there is more air pressure.
5.4
The pressure of mercury is its weight (force) per unit area. The weight, and thus the pressure, of the mercury
column is directly proportional to its height.
5.5
When the mercury level in the arm attached to the flask is higher than the level in the other arm, the pressure in
the flask is less than the pressure exerted in the other arm. This is an impossible situation for a closed-end
manometer as the flask pressure cannot be less than the vacuum in the other arm.
5.6
Plan: The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the
densities of the two liquids. Convert the height in mm to height in cm.
Solution:
hH2 O
dHg

hHg
dH2 O
hH O 
2
dHg
dH O
2
5.7


Plan: The ratio of the heights of columns of mercury and water are inversely proportional to the ratio of the
densities of the two liquids.
Solution:
hH2 O
dHg

hHg
dH2 O
hH O 
2
5.8
 hHg
13.5 g/mL 
103 m 
 1 cm 



=
 730 mmHg 
 2  = 985.5 = 990 cm H2O
1.00 g/mL 
10 m 
 1 mm 
13.5 g/mL 

 755 mmHg = 10,192.5 = 1.02 × 104 mm H O
 hHg = 
2
1.00 g/mL 
dH O

2
dHg


Plan: Use the conversion factors between pressure units:
1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar
Solution:
 760 mmHg 

 = 566.2 = 566 mmHg
a) Converting from atm to mmHg: P(mmHg) = 0.745 atm 
 1 atm 


1.01325 bar 

 = 1.32256 = 1.32 bar
b) Converting from torr to bar: P(bar) = 992 torr 
 760 torr 


 1 atm 

 = 3.60227 = 3.60 atm
c) Converting from kPa to atm: P(atm) = 365 kPa 
101.325 kPa 


101.325 kPa 

 = 107.191 = 107 kPa
d) Converting from mmHg to kPa: P(kPa) = 804 mmHg 
 760 mmHg 

5.9
Copyright

Plan: Use the conversion factors between pressure units:
1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar
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5-12
Solution:
a) Converting from cmHg to atm:
102 m 
 1 mm 
 1 atm 

P(atm) = 76.8 cmHg 
 3 
 = 1.01053 = 1.01 atm
 1 cm 
10 m 
 760 mmHg 


101.325 kPa 

 = 2.786 × 103 = 2.79 × 103 kPa
b) Converting from atm to kPa: P(kPa) = 27.5 atm 

 1 atm



1.01325 bar 

 = 6.5861 = 6.59 bar
c) Converting from atm to bar: P(bar) = 6.50 atm 
 1 atm 


 760 torr 

 = 7.02808 = 7.03 torr
d) Converting from kPa to torr: P(torr) = 0.937 kPa 
101.325 kPa 

5.10

Plan: This is an open-end manometer. Since the height of the mercury column in contact with the gas is higher
than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. Therefore the
h) between the two arms is subtracted from the atmospheric
pressure. Since the height difference is in units of cm and the barometric pressure is given in units of torr, cm
must be converted to mm and then torr before the subtraction is performed. The overall pressure is then given in
units of atm.
Solution:
102 m 
 1 mm 
 1 torr 

2.35 cm 
 3 
 = 23.5 torr
 1 cm 
10 m 
1 mmHg 


738.5 torr – 23.5 torr = 715.0 torr
 1 atm 

 = 0.940789 = 0.9408 atm
P(atm) = 715.0 torr 
 760 torr 

5.11

Plan: This is an open-end manometer. Since the height of the mercury column in contact with the gas is higher
than the column in contact with the air, the gas is exerting less pressure on the mercury than the air. Therefore the
pressure
h) between the two arms is subtracted from the atmospheric
pressure. Since the height difference is in units of cm and the barometric pressure is given in units of mmHg, cm
must be converted to mm before the subtraction is performed. The overall pressure is then given in units of kPa.
Solution:
102 m 
 1 mm 

1.30 cm 
 3  = 13.0 mmHg
 1 cm 
10 m 


765.2 mmHg – 13.0 mmHg = 752.2 mmHg
101.325 kPa 

 = 100.285 = 100.3 kPa
P(kPa) = 752.2 torr 
 760 torr 

5.12
Plan: This is a closed-end manometer.
h) equals the gas pressure. The
height difference is given in units of m and must be converted to mmHg and then to atm.
Solution:
 1 mmHg 
 1 atm 

P(atm) = 0.734 mHg  3

 = 0.965789 = 0.966 atm
10 mHg 
 760 mmHg 

Copyright


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5-13
5.13
Plan: This is a closed-end manometer.
h) equals the gas pressure.
The height difference is given in units of cm and must be converted to mmHg and then to Pa.
Solution:
102 mHg 
1 mmHg 
1.01325105 Pa 

3
P(Pa) = 3.56 cm 
 3 
 = 4746.276 = 4.75 × 10 Pa
 1 cmHg 
 10 m 
 760 mmHg 

5.14

Plan: Use the conversion factors between pressure units:
5
1 atm = 760 mmHg = 760 torr = 1.01325 × 10 Pa = 14.7 psi
Solution:
 1 atm 

 = 0.361842 = 0.362 atm
a) Converting from mmHg to atm: P(atm) = 2.75102 mmHg 
 760 mmHg 


 1 atm 

 = 5.85034 = 5.9 atm
b) Converting from psi to atm: P(atm) = 86 psi 
14.7 psi 



1 atm

 = 90.303 = 90.3 atm
c) Converting from Pa to atm: P(atm) = 9.1510 6 Pa 
1.0132510 5 Pa 


 1 atm 

 = 33.42105 = 33.4 atm
d) Converting from torr to atm: P(atm) = 2.54 10 4 torr 
 760 torr 


5
5.15
5
2
2
5
Plan: 1 atm = 1.01325 × 10 Pa = 1.01325 × 10 N/m . So the force on 1 m of ocean is 1.01325 × 10 N, where
1N=1
kg  m
2
. Use F = mg to find the mass of the atmosphere in kg/m for part (a). For part (b), convert this
s2
2
mass to g/cm and use the density of osmium to find the height of this mass of osmium.
Solution:
a) F = mg
5
1.01325 × 10 N = mg
1.0132510 5
kg  m
2
= (mass) (9.81 m/s )
s2
4
4
mass = 1.03287 × 10 = 1.03 × 10 kg
2
103 g 102 m 



4 kg 
3
2




b) 1.0328710 2 

 = 1.03287 × 10 g/cm (unrounded)

m  1 kg 
 1 cm 
3
1 cm 
g  1 mL 


Height = 1.03287103

 22.6 g  1 mL  = 45.702 = 45.7 cm Os

cm 2 



5.16
The statement is incomplete with respect to temperature and mass of sample. The correct statement is: At constant
temperature and moles of gas, the volume of gas is inversely proportional to the pressure.
5.17
a) Charles’s law: At constant pressure, the volume of a fixed amount of gas is directly proportional to its kelvin
temperature. Variable: volume and temperature; Fixed: pressure and moles
b) Avogadro’s law: At fixed temperature and pressure, the volume occupied by a gas is directly proportional to
the moles of gas. Variable: volume and moles; Fixed: temperature and pressure
c) Amontons’s law: At constant volume, the pressure exerted by a fixed amount of gas is directly proportional to
the kelvin temperature. Variable: pressure and temperature; Fixed: volume and moles
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5-14
5.18
Plan: Examine the ideal gas law; volume and temperature are constant and pressure and moles are variable.
Solution:
RT
PV = nRT
R, T, and V are constant
P=n
V
P = n × constant
At constant temperature and volume, the pressure of the gas is directly proportional to the amount of gas in moles.
5.19
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
PV
PV
so 1 1 = 2 2 .
n1T1
n2T2
Solution:
PV
PV
V1
V
= 2
a) P is fixed; both V and T double: 1 1 = 2 2 or
n1T1
n2T2
n1T1
n2T2
T can double as V doubles only if n is fixed.
PV
PV
b) T and n are both fixed and V doubles: 1 1 = 2 2
or P1V1 = P2V2
n1 T1
n2 T2
P and V are inversely proportional; as V doubles, P is halved.
c) T is fixed and V doubles. n doubles since one mole of reactant gas produces a total of 2 moles of product gas.
PV
PV
PV
PV
1 1
1 1
= 2 2
= 2 2
or
n1 T1
n2 T2
n1
n2
V and n can both double only if P is fixed.
d) P is fixed and V doubles. n is fixed since 2 moles of reactant gas produce 2 moles of product gas.
PV
PV
V1
V
1 1
= 2 2 or
= 2
n1T1
n2 T2
T1
T2
V and T are directly proportional so as V is doubled, T is doubled.
5.20
Plan: Use the relationship
PV
PV
PV n T
1 1
= 2 2 or V2 = 1 1 2 2 .
n1T1
n2T2
P2 n1T1
Solution:
a) As the pressure on a fixed amount of gas (n is fixed) increases at constant temperature (T is fixed), the
molecules move closer together, decreasing the volume. When the pressure is tripled, the volume decreases to
one-third of the original volume at constant temperature (Boyle’s law).
PV n T
(P )(V )(1)(1)
V2 = 1 1 2 2 = 1 1
V2 = V1
P2 n1T1
(3P1 )(1)(1)
b) As the temperature of a fixed amount of gas (n is fixed) increases at constant pressure (P is fixed), the gas
molecules gain kinetic energy. With higher energy, the gas molecules collide with the walls of the container with
greater force, which increases the size (volume) of the container. If the temperature is increased by a factor of 3.0
(at constant pressure) then the volume will increase by a factor of 3.0 (Charles’s law).
PV n T
(1)(V1 )(1)(3T1 )
V2 = 1 1 2 2 =
V2 = 3V1
P2 n1T1
(1)(1)(T1 )
c) As the number of molecules of gas increases at constant pressure and temperature (P and T are fixed), the force
they exert on the container increases. This results in an increase in the volume of the container. Adding 3 moles of
gas to 1 mole increases the number of moles by a factor of 4, thus the volume increases by a factor of 4
(Avogadro’s law).
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5-15
V2 =
5.21
5.22
PV
nT
(1)(V1 )(4n1 )(1)
1 1 2 2
=
P2 n1T1
(1)(n1 )(1)
Plan: Use the relationship
PV
PV
PV T
1 1
= 2 2 or V2 = 1 1 2 . R and n are fixed.
T1
T2
P2 T1
Solution:
a) As the pressure on a fixed amount of gas (n is fixed) doubles from 101 kPa to 202 kPa at constant temperature,
the volume decreases by a factor of . As the temperature of a fixed amount of gas (n is fixed) decreases by a
factor of (from 310 K to 155 K) at constant pressure, the volume decreases by a factor of . The changes in
pressure and temperature combine to decrease the volume by a factor of 4.
P1 = 760 torr = 101 kPa
T1 = 37°C + 273 = 310 K
PV
T
(101
kPa)(
V
)(155
K)
1
V2 = 1 1 2 =
V2 = 1 4 V1
P2 T1
(202 kPa)(310 K)
b) As the pressure on a fixed amount of gas (n is fixed) decreases at constant temperature (T is fixed), the
molecules move farther apart, increasing the volume. When the pressure is reduced by a factor of 2, the volume
increases by a factor of 2 at constant temperature (Boyle’s law).
T2 = 32°C + 273 = 305 K
P2 = 101 kPa = 1 atm
PV
(2 atm)(V1 )(305 K)
1 1T2
V2 =
=
V2 = 2V1
P2T1
(1 atm)(305 K)
c) As the pressure on a fixed amount of gas (n is fixed) decreases at constant temperature (T is fixed), the
molecules move farther apart, increasing the volume. When the pressure is reduced by a factor of 4, the volume
increases by a factor of 4 at constant temperature (Boyle’s law).
PV T
(P )(V )(1)
V2 = 1 1 2 = 1 1
V2 =4V1
P2T1
(1/ 4 P1 )(1)
PV
PV
PV T
Plan: Use the relationship 1 1 = 2 2 or V2 = 1 1 2 . R and n are fixed.
T1
T2
P2 T1
Solution:
a) The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2 (Charles’s law).
PV T
(1)(V1 )(400 K)
V2 = 1 1 2 =
V2 = V1
P2 T1
(1)(800 K)
b) T1 = 250°C + 273.15 = 523.15 K
T2 = 500°C + 273.15 = 773.15 K
The temperature increases by a factor of 773.15/523.15 = 1.48, so the volume is increased by a factor of 1.48
PV
(1)(V1 )(773.15 K)
1 1T2
=
V2 = 1.48V1
P2T1
(1)(523.15 K)
c) The pressure is increased by a factor of 3, so the volume decreases by a factor of 3 (Boyle’s law).
PV T
(2 atm)(V1 )(1)
V2 = 1 1 2 =
V2 = V1
P2T1
(6 atm)(1)
PV
PV
PV n T
Plan: Use the relationship 1 1 = 2 2 or V2 = 1 1 2 2 .
n1T1
n2T2
P2 n1T1
Solution:
 1 atm 

 = 0.950 atm
a) P1 = 722 torr 
 760 torr 
V2 =
(Charles’s law).
5.23
V2 = 4V1

T1 =
5
9

[T (in °F) – 32] =
5
9
[32°F – 32] = 0°C
T1 = 0°C + 273.15 = 273.15 K
Both P and T are fixed: P1 = P2 = 0.950 atm; T1 = T2 = 273.15 K, so the volume remains constant.
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5-16
PV
(1)(V1 )(1)(1)
1 1 n2 T2
=
V2 = V1
P2 n1T1
(1)(1)(1)
b) Since the number of moles of gas is decreased by a factor of 2, the volume would be decreased by a factor
of 2 (Avogadro’s law).
1
(1)(V1 )( n1 )(1)
PV n T
2
V2 = V1
V2 = 1 1 2 2 =
P2 n1T1
(1)( n1 )(1)
c) If the pressure is decreased by a factor of 4, the volume will increase by a factor of 4 (Boyle’s law). If the
temperature is decreased by a factor of 4, the volume will decrease by a factor of 4 (Charles’s law). These two
effects offset one another and the volume remains constant.
PV n T
(P )(V )(1)(14 T1 )
V2 = 1 1 2 2 = 1 1
V2 = V1
P2 n1T1
( 14 P1 )(1)(T1 )
V2 =
5.24
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
PV
PV
so 1 1 = 2 2 . In this problem, P and V are changing, while n and T remain fixed.
n1T1
n2T2
Solution:
V1 = 1.61 L
V2 = unknown
P1 = 734 torr
P2 = 0.844 atm
n and T remain constant


Converting P1 from torr to atm: (734 torr) 

= 0.966 atm

Arranging the ideal gas law and solving for V2:
PV
PV
1 1
= 2 2
or P1V1 = P2V2
n1 T1
n2 T2
V2 = V1
5.25
Copyright



 = 1.84 L

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
PV
PV
so 1 1 = 2 2 . In this problem, P and V are changing, while n and T remain fixed.
n1T1
n2T2
Solution:
V1 = 10.0 L
V2 = 7.50 L
P1 = 725 mmHg
P2 = unknown
n and T remain constant
Arranging the ideal gas law and solving for P2:
PV
PV
1 1
= 2 2
or P1V1 = P2V2
n1 T1
n2 T2
P2 = P1
5.26
P1
P2
V1
V2



 = 967 mmHg

Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly
proportional to the absolute temperature of the gas. The temperature must be lowered to reduce the volume of a
gas. Arrange the ideal gas law, solving for T2 at fixed n and P. Temperature must be converted to kelvin.
Solution:
V1 = 9.10 L
V2 = 2.50 L
T1 = 198°C (convert to K)
T2 = unknown
n and P remain constant
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5-17
Converting T from °C to K: T1 = 198°C + 273.15 = 471.15K
Arranging the ideal gas law and solving for T2:
PV
PV
V1
V
1 1
= 2 2 or
 2
n1T1
n2 T2
T1
T2
T2  T1
5.27
T2
 251.15 K 
= 93 L 
 = 55.858 = 56 L
 418.15 K 
T1
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
PV
PV
so 1 1 = 2 2 . In this problem, P and T are changing, while n and V remain fixed.
n1T1
n2T2
Solution:
T1 = 25 °C
T2 = 195 °C
P1 = 177 atm
P2 = unknown
n and V remain constant
Converting T1 from °C to K: 25 °C + 273.15 = 298.15 K
Converting T2 from °C to K: 195 °C + 273.15 = 468.15 K
Arranging the ideal gas law and solving for P2:
P1 V1
P2 V2
P1
P2
n 1 T1
n 2 T2
T1
T2
P2 = P1
5.29
 2.50 L 

 9.10 L  = 129.437 K – 273.15 = –143.713 = –144°C
Plan: This is Charles’s law: at constant pressure and with a fixed amount of gas, the volume of a gas is directly
proportional to the absolute temperature of the gas. If temperature is reduced, the volume of gas will also be
reduced. Arrange the ideal gas law, solving for V2 at fixed n and P. Temperature must be converted to kelvins.
Solution:
V1 = 93 L
V2 = unknown
T1 = 145°C (convert to K)
T2 = –22°C
n and P remain constant
Converting T from °C to K: T1 = 145°C + 273.15 = 418.15 K
T2 = –22°C + 273.15 = 251.15 K
Arranging the ideal gas law and solving for V2:
PV
PV
V1
V
1 1
= 2 2 or
 2
n1T1
n2 T2
T1
T2
V2  V1
5.28
V2
= 471.15 K
V1
T2

T1




 = 277.92 = 278 atm

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
PV
PV
so 1 1 = 2 2 . In this problem, P and T are changing, while n and V remain fixed.
n1T1
n2T2
Solution:
T1 = 30.0 °C
T2 = unknown
P1 = 110. psi
P2 = 105 psi
n and V remain constant
Converting T1 from °C to K: 30.0 °C + 273.15 = 303.15 K
Arranging the ideal gas law and solving for T2:
P1 V1
P2 V2
P1
P2
n 1 T1
n 2 T2
T1
T2
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5-18
T2 = T1



P2

P1

 = 289.37 = 289 K

Converting T2 from K to °C: 289.37 K  273.15 = 16.22 = 16 °C
5.30
Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
PV
PV
so 1 1 = 2 2 . In this problem, n and V are changing, while P and T remain fixed.
n1T1
n2T2
Solution:
m1 = 1.92 g He
m2 = 1.92 g – 0.850 g = 1.07 g He
V1 = 12.5 L
V2 = unknown
P and T remain constant


 = 0.47964 mol He = n1
Converting m1 (mass) to n1 (moles): (1.92 g He) 



Converting m2 (mass) to n2 (moles): (1.07 g He) 


 = 0.26730 mol He = n2

Arranging the ideal gas law and solving for V2:
P1 V1
P2 V2
V1
V2
n1 T1
n2 T2
n1
n2
V2 = V1 n 2
n1
5.31



 = 6.9661 = 6.97 L

Plan: Examine the ideal gas law, noting the fixed variables and those variables that change. R is always constant
PV
PV
so 1 1 = 2 2 . In this problem, n and V are changing, while P and T remain fixed.
n1T1
n2T2
Solution:
22
n1 = 1 × 10 molecules of air*
n2 = unknown
V1 = 500 mL
V2 = 350 mL
P and T remain constant
*The number of molecules of any substance is directly proportional to the moles of that substance, so we can use
number of molecules in place of n in this problem.
Arranging the ideal gas law and solving for n2:
P1 V1
P2 V2
V1
V2
n1 T1
n2 T2
n1
n2
n2 = n1
5.32
Copyright

V2
22
= (1 × 10 molecules of air) 

V1

21
 = 7 × 10 molecules of air

Plan: Since the volume, temperature, and pressure of the gas are changing, use the combined gas law.
Arrange the ideal gas law, solving for V2 at fixed n. STP is 0°C (273 K) and 1 atm (101.325 kPa)
Solution:
P1 = 153.3 kPa
P2 = 101.325 kPa
V1 = 25.5 L
V2 = unknown
T1 = 298 K
T2 = 273 K
n remains constant
Arranging the ideal gas law and solving for V2:
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5-19
PV
PV
1 1
= 2 2
n1T1
n 2 T2
PV
PV
1 1
= 2 2
T1
T2
or
 273 K 
153.3 kPa 
 T  P 



V2 = V1  2  1  = 25.5 L 
101.325 kPa  = 35.3437 = 35.3 L
 P2 
 298 K 
 T1 


5.33

Plan: Since the volume, temperature, and pressure of the gas are changing, use the combined gas law. Arrange the
ideal gas law, solving for V2 at fixed n. Temperature must be converted to kelvins.
Solution:
P1 = 745 torr
P2 = 367 torr
V1 = 3.65 L
V2 = unknown
T1 = 298 K
T2 = –14°C + 273.15 = 259.15 K
n remains constant
Arranging the ideal gas law and solving for V2:
PV
PV
PV
PV
1 1
1 1
= 2 2 or
= 2 2
n1T1
n 2 T2
T1
T2
 T  P 
V2 = V1  2  1  = 3.65 L
 P2 
 T1 
 259.15 K 

 298 K 
 745 torr 

 = 6.4434 = 6.44 L
 367 torr 
5.34
Plan: Given the volume, pressure, and temperature of a gas, the number of moles of the gas can be calculated using the
ideal gas law, solving for n. The gas constant, R = 0.0821 L atm/mol K, gives pressure in atmospheres and
temperature in kelvin. The given pressure in torr must be converted to atmospheres and the temperature converted to
kelvins.
Solution:
P = 328 torr (convert to atm)
V = 5.0 L
T = 37°C
n = unknown
 1 atm 
Converting P from torr to atm:
P = 328 torr 
= 0.43158 atm
 760 torr 
Converting T from °C to K:
T = 37°C + 273.15 = 310.15 K
PV = nRT
Solving for n:
PV
(0.43158 atm)(5.0 L)
n=

= 0.084745 = 0.085 mol chlorine
RT 0.0821 L  atm  (310.15 K)


mol  K 
5.35
Plan: Given the volume, moles, and temperature of a gas, the pressure of the gas can be calculated using the ideal
gas law, solving for P. The gas constant, R = 0.0821 L atm/mol K, gives volume in liters and temperature in
Kelvin. The given volume in mL must be converted to L and the temperature converted to kelvins.
Solution:
V = 75.0 mL
T = 26°C
–3
n = 1.47 × 10 mol
P = unknown
103 L 
 = 0.0750 L
Converting V from mL to L:
V = 75.0 mL
 1 mL 
Converting T from °C to K:
PV = nRT
Solving for P:
Copyright
T = 26°C + 273.15 = 299.15 K
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5-20

L  atm 
mol 0.0821
299.15 K 

mol  K 
= 0.48138 atm
0.0750 L
 760 torr 
Convert P to units of torr: 0.48138 atm 
= 365.848 = 366 torr
 1 atm 
nRT
P=
=
V
5.36
1.4710
3

Plan: Solve the ideal gas law for moles and convert to mass using the molar mass of ClF3.
The gas constant, R = 0.0821 L atm/mol K, gives volume in liters, pressure in atmospheres, and temperature in
kelvin so volume must be converted to L, pressure to atm, and temperature to K.
Solution:
V = 357 mL
T = 45°C
P = 699 mmHg
n = unknown
103 L 
 = 0.357 L
Converting V from mL to L:
V = 357 mL 
 1 mL 
Converting T from °C to K:
T = 45°C + 273.15 = 318.15 K
 1 atm 
 = 0.91974 atm
Converting P from mmHg to atm: P = 699 mmHg
 760 mmHg 
PV = nRT
Solving for n:
0.91974 atm 0.357 L 
PV
n=

= 0.012571 mol ClF3

RT 0.0821 L  atm 318.15 K 


mol  K 
 92.45 g ClF3 
 = 1.16216 = 1.16 g ClF3
Mass ClF3 = 0.012571 mol ClF3  
 1 mol ClF 
3
5.37
Plan: Solve the ideal gas law for pressure; convert mass to moles using the molar mass of N2O.
The gas constant, R = 0.0821 L atm/mol K, gives temperature in kelvin so the temperature must be converted to
units of kelvins.
Solution:
V = 3.1 L
T = 115°C
n = 75.0 g (convert to moles)
P = unknown
Converting T from °C to K:
T = 115°C + 273.15 = 388.15 K
 1 mol N 2 O 
 = 1.70377 mol N2O
Converting from mass of N2O to moles:
n = 75.0 g N 2 O
 44.02 g N 2 O 
PV = nRT
Solving for P:
nRT
P=

V
5.38
Copyright

L  atm 
1.70377 mol0.0821
388.15 K 

mol  K 
= 17.5143 = 18 atm N2O
3.1 L 
Plan: Solve the ideal gas law for moles. The gas constant, R = 0.0821 L atm/mol K, gives pressure in
atmospheres, and temperature in kelvin so pressure must be converted to atm and temperature to K.
Solution:
V = 1.5 L
T = 23°C
P = 85 + 14.7 = 99.7 psi
n = unknown
Converting T from °C to K:
T = 23°C + 273.15 = 296.15 K
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5-21
Converting P from psi to atm:
 1 atm 
 = 6.7823 atm
P = 99.7 psi
14.7 psi 
PV = nRT
Solving for n:
6.7823 atm 1.5 L 
PV
n=

= 0.418421 = 0.42 mol SO2
RT 0.0821 L  atm 296.15 K 


mol  K 
5.39
Plan: Assuming that while rising in the atmosphere the balloon will neither gain nor lose gas molecules, the number
of moles of gas calculated at sea level will be the same as the number of moles of gas at the higher altitude (n is
fixed). Volume, temperature, and pressure of the gas are changing. Arrange the ideal gas law, solving for V2 at fixed
n. Given the sea-level conditions of volume, pressure, and temperature, and the temperature and pressure at the
higher altitude for the gas in the balloon, we can set up an equation to solve for the volume at the higher altitude.
Comparing the calculated volume to the given maximum volume of 835 L will tell us if the balloon has reached its
maximum volume at this altitude. Temperature must be converted to kelvins and pressure in torr must be converted
to atm for unit agreement.
Solution:
P1 = 745 torr
P2 = 0.066 atm
V1 = 65 L
V2 = unknown
T1 = 25°C + 273.15 = 298.15 K
T2 = –5°C + 273.15 = 268.15 K
n remains constant
 1 atm 
Converting P from torr to atm:
P = 745 torr 
= 0.98026 atm
 760 torr 
Arranging the ideal gas law and solving for V2:
PV
PV
PV
PV
1 1
1 1
= 2 2 or
= 2 2
n1T1
n 2 T2
T1
T2
 T  P 
 268.15 K 
 0.98026 atm  = 868.268 = 870 L
V2 = V1  2  1  = 65 L
 P2 
 T1 
 298.15 K  0.066 atm 
The calculated volume of the gas at the higher altitude is more than the maximum volume of the balloon. Yes, the
balloon will reach its maximum volume.
Check: Should we expect that the volume of the gas in the balloon should increase? At the higher altitude, the pressure
decreases; this increases the volume of the gas. At the higher altitude, the temperature decreases, this decreases the
volume of the gas. Which of these will dominate? The pressure decreases by a factor of 0.98/0.066 = 15. If we label the
initial volume V1, then the resulting volume is 15V1. The temperature decreases by a factor of 298/268 = 1.1, so the
resulting volume is V1/1.1 or 0.91V1. The increase in volume due to the change in pressure is greater than the decrease in
volume due to change in temperature, so the volume of gas at the higher altitude should be greater than the volume at
sea level.
5.40
Air is mostly N2 (28.02 g/mol), O2 (32.00 g/mol), and argon (39.95 g/mol). These “heavy” gases dominate the
density of dry air. Moist air contains H2O (18.02 g/mol). The relatively light water molecules lower the density of
the moist air.
5.41
The molar mass of H2 is less than the average molar mass of air (mostly N2, O2, and Ar), so air is denser. To
collect a beaker of H2(g), invert the beaker so that the air will be replaced by the lighter H2. The molar mass of
CO2 is greater than the average molar mass of air, so CO2(g) is more dense. Collect the CO2 holding the beaker
upright, so the lighter air will be displaced out the top of the beaker.
5.42
Gases mix to form a solution and each gas in the solution behaves as if it were the only gas present.
Copyright
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5-22
5.43
PA = XA PT The partial pressure of a gas (PA) in a mixture is directly proportional to its mole fraction (XA).
5.44
Plan: Calculate the mole fraction of each gas; the partial pressure of each gas is directly proportional to its mole
fraction so the gas with the highest mole fraction has the highest partial pressure. Use the relationship between
partial pressure and mole fraction to calculate the partial pressure of gas D 2.
Solution:
4 A particles
3 B particles
n
nB
a) X A = A =
= 0.25
XB =
=
= 0.1875
16 total particles
16 total particles
ntotal
ntotal
nD2
4 D2 particles
5 C particles
nC
XC =
=
= 0.3125
X D2 =
=
= = 0.25
16 total particles
ntotal
16 total particles
ntotal
Gas C has the highest mole fraction and thus the highest partial pressure.
b) Gas B has the lowest mole fraction and thus the lowest partial pressure.
c) PD2 = X D2 x Ptotal
PD2 = 0.25 × 0.75 atm = 0.1875 = 0.19 atm
5.45
Plan: Rearrange the ideal gas law to calculate the density of xenon from its molar mass at STP. Standard
temperature is 0°C (273 K) and standard pressure is 1 atm. Do not forget that the pressure at STP is exact and will
not affect the significant figures.
Solution:
P = 1 atm
T = 273 K
M of Xe = 131.3 g/mol
d = unknown
PV = nRT
Rearranging to solve for density:
d = PM /RT 
5.46
Plan: Rearrange the ideal gas law to calculate the density of CFCl3 from its molar mass. Temperature must
be converted to kelvins.
Solution:
P = 1.5 atm
T = 120°C + 273.15 = 393.15 K
M of CFCl3 = 137.4 g/mol
d = unknown
PV = nRT
Rearranging to solve for density:
d = PM /RT 
5.47
Copyright
1 atm131.3 g/mol
= 5.8581 = 5.86 g/L

L  atm 
0.0821

273
K


mol  K 
1.5 atm 137.4 g/mol
= 6.3852 = 6.4 g/L

L  atm 
0.0821

393.15
K


mol  K 
Plan: Solve the ideal gas law for moles. Convert moles to mass using the molar mass of AsH3 and divide this mass
by the volume to obtain density in g/L. Standard temperature is 0°C (273 K) and standard pressure is 1 atm. Do
not forget that the pressure at STP is exact and will not affect the significant figures.
Solution:
V = 0.0400 L
T = 273 K
P = 1 atm
n = unknown
M of AsH3 = 77.94 g/mol
PV = nRT
Solving for n:
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5-23



1 atm 0.0400 L
PV
–3
–3
= 1.78465 × 10 = 1.78 × 10 mol AsH3


L  atm 
RT
0.0821
273 K

mol  K 
Converting moles of AsH3 to mass of AsH3:
 77.94 g AsH 
3

3
Mass (g) of AsH3 = 1.7846510 mol AsH3 
 = 0.1391 g AsH3
 1 mol AsH3 
n=
d=
5.48






0.1391 g
mass
=
= 3.4775 = 3.48 g/L
volume
0.0400 L
Plan: Solve the density form of the ideal gas law for molar mass. Temperature must be converted to kelvins.
Compare the calculated molar mass to the molar mass values of the noble gases to determine the identity of the gas.
Solution:
P = 3.00 atm
T = 273 K
d = 2.71 g/L
M = unknown
d = PM /RT
Rearranging to solve for molar mass:
L  atm 

2.71 g/L 0.0821
273 K 

dRT
mol  K 
=
M =
= 20.24668 = 20.2 g/mol
P
3.00 atm




Therefore, the gas is Ne.
5.49
Plan: Rearrange the formula PV = (m/M )RT to solve for molar mass. Convert the mass
in ng to g and volume in L to L. Temperature must be in kelvin and pressure in atm.
Solution:
V
T = 45°C + 273.15 = 318.15 K
P = 380 torr
m = 206 ng
M = unknown
 1 atm 
Converting P from torr to atm:
P = 380 torr 
= 0.510526 atm
 760 torr 
106 L 
 = 2.06 × 10–7 L
Converting V
V = 0.206 L
 1 L 
109 g 
 = 2.06 × 10–7 g
m = 206 ng
 1 ng 
Converting m from ng to g:
PV = (m/M )RT
Solving for molar mass, M :

mRT
M =

PV
Copyright
L  atm 
318.15 K 
mol  K 
= 51.163 = 51.2 g/mol
0.510526 atm 2.06 107 L 
2.06 107 g0.0821
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5-24
5.50
Plan: Rearrange the formula PV = (m/M )RT to solve for molar mass. Compare the calculated molar mass
to that of N2, Ne, and Ar to determine the identity of the gas. Convert volume to liters, pressure to atm, and
temperature to kelvin.
Solution:
V = 63.8 mL
T = 22°C + 273.15 = 295.15 K
P = 747 mm Hg
m = 0.103 g
M = unknown
 1 atm 
 = 0.982895 atm
Converting P from mmHg to atm: P = 747 mmHg
 760 mmHg 
Converting V from mL to L:
103 L 
 = 0.0638 L
V = 63.8 mL
 1 mL 
PV = (m/M )RT
Solving for molar mass, M :

L  atm 
295.15 K 
0.103 g0.0821

mRT
mol
 K 
M =
= 39.8011 = 39.8 g/mol

PV
0.982895 atm 0.0638 L
The molar masses are N2 = 28 g/mol, Ne = 20 g/mol, and Ar = 40 g/mol.
Therefore, the gas is Ar.
5.51
Plan: Use the ideal gas law to determine the number of moles of Ar and of O2. The gases are combined
(ntotal = nAr + n O 2 ) into a 400 mL flask (V) at 27°C (T). Use the ideal gas law again to determine the total pressure
from ntotal, V, and T. Pressure must be in units of atm, volume in units of L and temperature in K.
Solution:
For Ar:
V = 0.600 L
T = 227°C + 273.15 = 500.15 K
P = 1.20 atm
n = unknown
PV = nRT
Solving for n:
1.20 atm0.600 L
PV
n=

= 0.01753433 mol Ar
RT 0.0821 L  atm 500.15 K 

mol  K 
For O2:
V = 0.200 L
T = 127°C + 273.15 = 400.15 K
P = 501 torr
n = unknown
 1 atm 
Converting P from torr to atm:
P = 501 torr 
= 0.6592105 atm
 760 torr 
PV = nRT
Solving for n:
0.6592105 atm 0.200 L
PV
n=

= 0.00401318 mol O2

RT 0.0821 L  atm  400.15 K 


mol  K 
ntotal = nAr + n O 2 = 0.01753433 mol + 0.00401318 mol = 0.02154751 mol
For the mixture of Ar and O2:
V = 400 mL
T = 27°C + 273.15 = 300.15 K
P = unknownn
n = 0.02154751 mol
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5-25
103 L 
 = 0.400 L
V = 400 mL
 1 mL 
Converting V from mL to L:
PV = nRT
Solving for P:
Pmixture
5.52
nRT
=

V

L  atm 
0.02154751 mol0.0821
300.15 K 

mol  K 
= 1.32745 = 1.33 atm
0.400 L
Plan: Use the ideal gas law, solving for n to find the total moles of gas. Convert the mass of
Ne to moles and subtract moles of Ne from the total number of moles to find moles of Ar. Volume
must be in units of liters, pressure in units of atm, and temperature in kelvins.
Solution:
V = 355 mL
T = 35°C + 273.15 = 308.15 K
P = 626 mmHg
ntotal = unknown
 1 atm 
 = 0.823684 atm
Converting P from mmHg to atm: P = 626 mmHg
 760 mmHg 
103 L 
 = 0.355 L
V = 355 mL 
 1 mL 
Converting V from mL to L:
PV = nRT
Solving for ntotal:
0.823684 atm0.355 L 
PV
ntotal =

= 0.011558029 mol Ne + mol Ar
RT 0.0821 L  atm 308.15 K 


mol  K 
 1 mol Ne 
 = 0.007234886 mol Ne
Moles Ne = 0.146 g Ne
 20.18 g Ne 
Moles Ar = ntotal – nNe = (0.011558029 – 0.007234886) mol = 0.004323143 = 0.0043 mol Ar
5.53
Plan: Use the ideal gas law, solving for n to find the moles of O2. Use the molar ratio from the
balanced equation to determine the moles (and then mass) of phosphorus that will react with the oxygen.
Standard temperature is 0°C (273 K) and standard pressure is 1 atm.
Solution:
V = 35.5 L
T = 273 K
P = 1 atm
n = unknown
PV = nRT
Solving for n:
1 atm 35.5 L 
PV
n=

= 1.583881 mol O2
RT 0.0821 L  atm 273 K 


mol  K 
P4(s) + 5O2(g)  P4O10(s)
 1 mol P4 
123.88 g P4 

Mass P4 = 1.583881 mol O2 

 = 39.24224 = 39.2 g P4
 5 mol O2 
 1 mol P4 

5.54
Copyright

Plan: Use the ideal gas law, solving for n to find the moles of O2 produced. Volume must be in units
of liters, pressure in atm, and temperature in kelvins. Use the molar ratio from the balanced equation to determine
the moles (and then mass) of potassium chlorate that reacts.
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5-26
Solution:
V = 638 mL
P = 752 torr
Converting P from torr to atm:
Converting V from mL to L:
T = 128°C + 273.15 = 401.15 K
n = unknown
 1 atm 
P = 752 torr 
= 0.9894737 atm
 760 torr 
103 L 
 = 0.638 L
V = 638 mL
 1 mL 
PV = nRT
Solving for n:
0.9894737 atm0.638 L
PV
n=

= 0.0191679 mol O2
RT 0.0821 L  atm  401.15 K 

mol  K 
2KClO3(s)  2KCl(s) + 3O2(g)
 2 mol KClO3 
122.55 g KClO3 

Mass (g) of KClO3 = 0.0191679 mol O2 

 = 1.5660 = 1.57 g KClO3
 3 mol O2 
 1 mol KClO3 

5.55

Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. To find the mass of PH3,
write the balanced equation and use molar ratios to find the number of moles of PH3 produced by each reactant.
The smaller number of moles of product indicates the limiting reagent. Solve for moles of H2 using the ideal gas
law.
Solution:
Moles of hydrogen:
V = 83.0 L
T = 273 K
P = 1 atm
n = unknown
PV = nRT
Solving for n:
1 atm83.0 L
PV
n=

= 3.7031584 mol H2
RT 0.0821 L  atm 273 K 


mol  K 
P4(s) + 6H2(g)  4PH3(g)
 4 mol PH3 
 = 2.4687723 mol PH3
PH3 from H2 = 3.7031584 mol H2 
 6 mol H2 
 1 mol P4 
 4 mol PH3  = 1.21085 mol PH
PH3 from P4 = 37.5 g P4 
3
 1 mol P4 
123.88 g P4 
P4 is the limiting reactant because it forms less PH3.
 1 mol P4 
 4 mol PH3  33.99 g PH3 

Mass PH3 = 37.5 g P4 


 = 41.15676 = 41.2 g PH3
123.88 g P4 
 1 mol P4  1 mol PH3 

5.56
Copyright

Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. To find the mass of NO,
write the balanced equation and use molar ratios to find the number of moles of NO produced by each reactant.
Since the moles of gas are directly proportional to the volumes of the gases at the same temperature and pressure,
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5-27
the limiting reactant may be found by comparing the volumes of the gases. The smaller volume of product
indicates the limiting reagent. Then use the ideal gas law to convert the volume of NO produced to moles and then
to mass.
Solution:
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)
 4 L NO 
 = 35.6 L NO
Mol NO from NH3 = 35.6 L NH3 
 4 L NH 
3
 4 L NO 
 = 32.4 L NO
Mol NO from O2 = 40.5 L O2 
 5 L O2 
O2 is the limiting reactant since it forms less NO.
Converting volume of NO to moles and then mass:
V = 32.4 L
T = 273 K
P = 1 atm
n = unknown
PV = nRT
Solving for n:
1 atm32.4 L
PV
n=

= 1.44557 mol NO

RT 0.0821 L  atm 273 K 


mol  K
 30.01 g NO 

 = 43.38156 = 43.4 g NO
Mass (g) of NO = 1.44557 mol NO
 1 mol NO 
5.57
Plan: First, write the balanced equation. The moles of hydrogen produced can be calculated from the ideal gas
law. The problem specifies that the hydrogen gas is collected over water, so the partial pressure of water vapor
must be subtracted from the overall pressure given. Table 5.2 reports pressure at 26°C (25.2 torr) and 28°C
(28.3 torr), so take the average of the two values to obtain the partial pressure of water at 27°C. Volume must be
in units of liters, pressure in atm, and temperature in kelvins. Once the moles of hydrogen produced are known,
the molar ratio from the balanced equation is used to determine the moles of aluminum that reacted.
Solution:
V = 35.8 mL
T = 27°C + 273.15 = 300.15 K
Ptotal = 751 mmHg
n = unknown
Pwater vapor = (28.3 + 25.2) torr/2 = 26.75 torr = 26.75 mmHg
Phydrogen = Ptotal – Pwater vapor = 751 mmHg – 26.75 mmHg = 724.25 mmHg
 1 atm 
 = 0.952960526 atm
Converting P from mmHg to atm: P = 724.25 mmHg
 760 mmHg 
Converting V from mL to L:
103 L 
 = 0.0358 L
V = 35.8 mL 
 1 mL 
PV = nRT
Solving for n:
PV 0.952960526 atm 0.0358 L 
n=

= 0.001384447 mol H2

L  atm 
RT
0.0821
300.15
K



mol  K 
2Al(s) + 6HCl(aq)  2AlCl3(aq) + 3H2(g)
 2 mol Al 
 26.98 g Al 

Mass (g) of Al = 0.001384447 mol H2 

 = 0.024902 = 0.0249 g Al
 3 mol H 2 
 1 mol Al 
Copyright
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5-28
5.58
Plan: First, write the balanced equation. Convert mass of lithium to moles and use the molar ratio from the
balanced equation to find the moles of hydrogen gas produced. Use the ideal gas law to find the volume of that
amount of hydrogen. The problem specifies that the hydrogen gas is collected over water, so the partial pressure
of water vapor must be subtracted from the overall pressure given. Table 5.2 reports the vapor pressure of water at
18°C (15.5 torr). Pressure must be in units of atm and temperature in kelvins.
Solution:
2Li(s) + 2H2O(l)  2LiOH(aq) + H2(g)
 1 mol Li 1 mol H 2 
 = 0.0605100 mol H2

Moles H2 = 0.84 g Li
 6.941 g Li  2 mol Li 
Finding the volume of H2:
V = unknown
T = 18°C + 273.15 = 291.15 K
Ptotal = 725 mmHg
n = 0.0605100 mol
Pwater vapor = 15.5 torr = 15.5 mmHg
Phydrogen = Ptotal – Pwater vapor = 725 mmHg – 15.5 mmHg = 709.5 mmHg
 1 atm 
 = 0.933552631 atm
Converting P from mmHg to atm: P = 709.5 mmHg
 760 mmHg 
PV = nRT
Solving for V:
nRT
V=
=
P
5.59

L  atm 
0.0605100 mol0.0821
291.15 K 

mol  K 
= 1.5493 = 1.5 L H2
0.933552631 atm 
Plan: Rearrange the ideal gas law to calculate the density of the air from its molar mass. Temperature must
be converted to kelvins and pressure to atmospheres.
Solution:
P = 744 torr
T = 17°C + 273.15 = 290.15 K
or
T = 60°C + 273.15 = 333.15 K
M of air = 28.8 g/mol
d = unknown
 1 atm 
Converting P from torr to atm:
P = 744 torr 
= 0.978947368 atm
 760 torr 
PV = nRT
Rearranging to solve for density:
At 17°C
0.978947368 atm 28.8 g/mol
d = PM /RT 
= 1.18355 = 1.18 g/L

L  atm 
0.0821
290.15 K 

mol  K 
At 60.0°C
0.978947368 atm 28.8 g/mol
d = PM /RT 
= 1.03079 = 1.03 g/L

L  atm 
0.0821
(333.15 K)


mol  K 
5.60
Copyright
Plan: Solve the ideal gas law for molar volume, n/V. Pressure must be converted to atm and temperature to K.
Solution:
P = 650. torr
T = –25°C + 273.15 = 248.15 K
n/V = unknown
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5-29
 1 atm 
P = 650 torr 
 = 0.855263157 atm
 760 torr 

Converting P from torr to atm:

PV = nRT
Solving for n/V:
0.855263157 atm 
n
P
=

= 0.041978 = 0.0420 mol/L
V RT 0.0821 L  atm 248.15 K 

mol  K 
5.61
Plan: The problem gives the mass, volume, temperature, and pressure of a gas; rearrange the formula
PV = (m/M )RT to solve for the molar mass of the gas. Temperature must be in kelvin and pressure in atm. The
problem also states that the gas is a hydrocarbon, which by, definition, contains only carbon and hydrogen atoms.
We are also told that each molecule of the gas contains five carbon atoms so we can use this information and the
calculated molar mass to find out how many hydrogen atoms are present and the formula of the compound.
Solution:
V = 0.204 L
T = 101°C + 273.15 = 374.15 K
P = 767 torr
m = 0.482 g
M = unknown
 1 atm 
Converting P from torr to atm:
P = 767 torr 
= 1.009210526 atm
 760 torr 
PV = (m/M )RT
Solving for molar mass, M :

L  atm 
374.15 K 
0.482 g0.0821

mRT
mol
 K 
M =
= 71.9157 g/mol (unrounded)

PV
1.009210526 atm 0.204 L
The mass of the five carbon atoms accounts for [5(12 g/mol)] = 60 g/mol; thus, the hydrogen atoms must make up
the difference (72 – 60) = 12 g/mol. A value of 12 g/mol corresponds to 12 H atoms. (Since fractional atoms are
not possible, rounding is acceptable.) Therefore, the molecular formula is C5H12.
5.62
Plan: Solve the ideal gas law for moles of air. Temperature must be in units of kelvins. Use Avogadro’s
number to convert moles of air to molecules of air. The percent composition can be used to find the number
of molecules (or atoms) of each gas in that total number of molecules.
Solution:
V = 1.00 L
T = 25°C + 273.15 = 298.15 K
P = 1.00 atm
n = unknown
PV = nRT
Solving for n:
Moles of air = n =



1.00 atm 1.00 L
PV
= 0.04085282 mol


L  atm 
RT
0.0821
298.15
K


mol  K 


Converting moles of air to molecules of air:
 6.022 1023 molecules 
 = 2.4601568 × 1022 molecules
Molecules of air = 0.04085282 mol

1 mol

 78.08% N 2 molecules 
Molecules of N2 = 2.46015681022 air molecules



100% air
22
22
= 1.920890 × 10 = 1.92 × 10 molecules N2
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5-30
 20.94% O 2 molecules 
Molecules of O2 = 2.46015681022 air molecules



100% air
21
21
= 5.151568 × 10 = 5.15 × 10 molecules O2
 0.05% CO 2 molecules 
Molecules of CO2 = 2.46015681022 air molecules



100% air
19
19
= 1.230078 × 10 = 1 × 10 molecules CO2
 0.93% Ar molecules 
Molecules of Ar = 2.46015681022 air molecules



100% air
20
20
= 2.287946 × 10 = 2.3 × 10 molecules Ar
5.63
Plan: Since you have the pressure, volume, and temperature, use the ideal gas law to solve for n, the total moles of
gas. Pressure must be in units of atmospheres and temperature in units of kelvins. The partial pressure of SO2 can
be found by multiplying the total pressure by the volume fraction of SO 2.
Solution:
a) V = 21 L
T = 45°C + 273.15 = 318.15 K
P = 850 torr
n = unknown
 1 atm 
Converting P from torr to atm:
P = 850 torr 
= 1.118421053 atm
 760 torr 
PV = nRT
1.118421053 atm 21 L 
PV
Moles of gas = n =

= 0.89919 = 0.90 mol gas
RT 0.0821 L  atm 318.15 K 

mol  K 
b) The equation PSO2  X SO2  Ptotal can be used to find partial pressure. The information given in ppm is a way of
expressing the proportion, or fraction, of SO2 present in the mixture. Since n is directly proportional to V, the
3
volume fraction can be used in place of the mole fraction, XSO2. There are 7.95 × 10 parts SO2 in a million parts of
3
6
–3
mixture, so volume fraction = (7.95 × 10 /1 × 10 ) = 7.95 × 10 .
–3
PD2 = volume fraction x Ptotal = (7.95 × 10 ) (850 torr) = 6.7575 = 6.76 torr
5.64
Plan: First, write the balanced equation. Convert mass of P4S3 to moles and use the molar ratio from the balanced
equation to find the moles of SO2 gas produced. Use the ideal gas law to find the volume of that amount of SO2.
Pressure must be in units of atm and temperature in kelvins.
Solution:
P4S3(s) + 8O2(g)  P4O10(s) + 3SO2(g)
 1 mol P4 S3 
3 mol SO2 

 = 0.010906 mol SO2
Moles SO2 = 0.800 g P4 S3 
1 mol P S 
 220.06 g P S 
4 3
Finding the volume of SO2:
V = unknown
P = 725 torr
Converting P from torr to atm:
4 3
T = 32°C + 273.15 = 305.15 K
n = 0.010906 mol
 1 atm 
P = 725 torr 
= 0.953947368 atm
 760 torr 
PV = nRT
Solving for V:
nRT
V=
=
P
Copyright

L  atm 
0.010906 mol0.0821
305.15 K 

mol  K 
= 0.2864193 L
0.953947368 atm 
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5-31
Converting V from L to mL:
 1 mL 

V = 0.2864193 L 3  = 286.4193 = 286 mL SO2
10 L 
5.65
Plan: The moles of Freon-12 produced can be calculated from the ideal gas law. Volume must be in units of L,
pressure in atm, and temperature in kelvins. Then, write the balanced equation. Once the moles of Freon-12
produced is known, the molar ratio from the balanced equation is used to determine the moles and then grams of
CCl4 that reacted.
Solution:
3
V = 16.0 dm
T = 27°C + 273.15 = 300.15 K
Ptotal = 1.20 atm
n = unknown
 1 L 
3
Converting V from dm to L:
V = 16.0 dm 3 
= 16.0 L
1 dm 3 
PV = nRT
Solving for n:
1.20 atm16.0 L
PV

= 0.7791476 mol Freon-12

RT 0.0821 L  atm 300.15 K 


mol  K 
CCl4(g) + 2HF(g)  CF2Cl2(g) + 2HCl(g)
 1 mol CCl 4 
153.81 g CCl4 

Mass of Freon-12 (CF2Cl2) = 0.7791476 mol CF2 Cl2 


 1 mol CCl 4 
1 mol CF2 Cl2 
2
= 119.8407 = 1.20 × 10 g CCl4
Moles of Freon-12 = n =
5.66
Plan: First, write the balanced equation. Given the amount of xenon hexafluoride that reacts, we can find the
number of moles of silicon tetrafluoride gas formed by using the molar ratio in the balanced equation. Then, using
the ideal gas law with the moles of gas, the temperature, and the volume, we can calculate the pressure of the
silicon tetrafluoride gas. Temperature must be in units of kelvins.
Solution:
2XeF6(s) + SiO2(s)  2XeOF4(l) + SiF4(g)
 1 mol XeF6 
 1 mol SiF4  = 0.0040766 mol SiF
Moles SiF4 = n = 2.00 g XeF6 
4
 2 mol XeF6 
 245.3 g XeF6 
Finding the pressure of SiF4:
V = 1.00 L
P = unknown
PV = nRT
Solving for P:
T = 25°C + 273.15 = 298.15 K
n = 0.0040766 mol

nRT
Pressure SiF4 = P =
=
V
5.67
Copyright
0.0040766 mol SiF4 0.0821
1.00 L
L  atm 
298.15 K 
mol  K 
= 0.099788 = 0.0998 atm SiF4
Plan: Use the ideal gas law with T and P constant; then volume is directly proportional to moles.
Solution:
PV = nRT. At constant T and P, V n. Since the volume of the products has been decreased to the original
volume, the moles (and molecules) must have been decreased by a factor of as well. Cylinder A best represents
the products as there are 2 product molecules (there were 4 reactant molecules).
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5-32
5.68
Plan: Write the balanced equation. Since the amounts of 2 reactants are given, this is a limiting reactant problem.
To find the volume of SO2, use the molar ratios from the balanced equation to find the number of moles of SO 2
produced by each reactant. The smaller number of moles of product indicates the limiting reagent. Solve for moles
of SO2 using the ideal gas law.
Solution:
Moles of oxygen:
V = 228 L
T = 220°C + 273.15 = 493.15 K
P = 2.0 atm
n = unknown
PV = nRT
Solving for n:
2.0 atm228 L
PV

Moles of O2 = n =
= 11.2627 mol O2

RT 0.0821 L  atm  493.15 K 


mol  K 
2PbS(s) + 3O2(g)  2PbO(g) + 2SO2(g)
 2 mol SO2 
 = 7.5085 mol SO2
Moles SO2 from O2 = 11.2627 mol O2 
 3 mol O 
2
10 g 

 1 mol PbS 

 2 mol SO2  = 15.6707 mol SO (unrounded)
Moles SO2 from PbS = 3.75 kg PbS 

2


 1 kg 
 239.3 g PbS  2 mol PbS 


3
O2 is the limiting reagent because it forms less SO2.
Finding the volume of SO2:
V = unknown
T = 0°C + 273.15 = 273.15 K
Ptotal = 1 atm
n = 7.5107 mol
PV = nRT
Solving for V:

L  atm 
 7.5085 mol0.0821
273.15 K 


nRT
2
mol

K
V=
= 168.38 = 1.7 × 10 L SO2
=
P
1 atm 
5.69
Plan: First, write the balanced equation. Given the amount of xenon HgO that reacts (20.0% of the given amount),
we can find the number of moles of oxygen gas formed by using the molar ratio in the balanced equation. Then,
using the ideal gas law with the moles of gas, the temperature, and the volume, we can calculate the pressure of
the silicon tetrafluoride gas. Temperature must be in units of kelvins.
Solution:
2HgO(s)  2Hg(l) + O2(g)
 20.0%  1 mol HgO  1 mol O2 
 = 0.01846722 mol O2
Mole O2 = n = 40.0 g HgO 


 100%  216.6 g HgO  2 mol HgO 
Finding the pressure of O2:
V = 502 mL
T = 25°C + 273.15 = 298.15 K
P = unknown
n = 0.01846722 mol
103 L 
 = 0.502 L
Converting V from mL to L:
V = 502 mL 
 1 mL 


PV = nRT
Solving for P:

nRT
Pressure O2 = P =
=
V
Copyright
0.01846722 mol O2 0.0821
0.502 L
L  atm 
298.15 K 
mol  K 
= 0.900484 = 0.900 atm O2
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5-33
5.70
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.
5.71
At STP (or any identical temperature and pressure), the volume occupied by a mole of any gas will be identical.
One mole of krypton has the same number of particles as one mole of helium and, at the same temperature, all of
the gas particles have the same average kinetic energy, resulting in the same pressure and volume.
5.72
The rate of effusion is much higher for a gas than its rate of diffusion. Effusion occurs into an evacuated space,
whereas diffusion occurs into another gas. It is reasonable to expect that a gas will escape faster into a vacuum than
it will into a space already occupied by another gas. The ratio of the rates of effusion and diffusion for two gases will
be the same since both are inversely proportional to the square root of their molar masses.
5.73
a) PV = nRT Since the pressure, volume, and temperature of the two gases are identical, n must be the same
for the two gases. Since the molar mass of O2 (32.0 g/mol) is greater than the molar mass of H2 (2.02 g/mol), a
given number of moles of O2 has a greater mass than the same number of moles of H2. mass O2 > mass H2
b) d = PM /RT The pressure and temperature are identical and density is directly proportional to molar mass M.
Since the molar mass of O2 (32.0 g/mol) is greater than the molar mass of H2 (2.02 g/mol), the density of O2 is
greater than that of H2. dO > dH
2
2
c) The mean free path is dependent on pressure. Since the two gases have the same pressure, their mean free paths
are identical.
d) Kinetic energy is directly proportional to temperature. Since the two gases have the same temperature, their
average moelcular kinetic energies are identical.
2
e) Kinetic energy = mass × speed . O2 and H2 have the same average kinetic energy at the same temperature
and mass and speed are inversely proportional. The lighter H2 molecules have a higher speed than the heavier O2
molecules. average speed H2 > average speed O2
1
f) Rate of effusion 
H2 molecules with the lower molar mass have a faster effusion time than O2
molar mass
molecules with a larger molar mass. effusion time H2 < effusion time O2
5.74
Copyright
Plan: The molar masses of the three gases are 2.016 for H2 (Flask A), 4.003 for He (Flask B), and 16.04 for CH4
(Flask C). Since hydrogen has the smallest molar mass of the three gases, 4 g of H2 will contain more gas
molecules (about 2 mole’s worth) than 4 g of He or 4 g of CH4. Since helium has a smaller molar mass than
methane, 4 g of He will contain more gas molecules (about 1 mole’s worth) than 4 g of CH 4 (about 0.25 mole’s
worth).
Solution:
a) PA > PB > PC. The pressure of a gas is proportional to the number of gas molecules (PV = nRT). So, the gas
sample with more gas molecules will have a greater pressure.
b) EA = EB = EC. Average kinetic energy depends only on temperature. The temperature of each gas sample is
273 K, so they all have the same average kinetic energy.
c) rateA > rateB > rateC. When comparing the speed of two gas molecules, the one with the lower mass travels
faster.
d) total EA > total EB > total EC. Since the average kinetic energy for each gas is the same (part b) of this
problem), the total kinetic energy would equal the average times the number of molecules. Since the hydrogen
flask contains the most molecules, its total kinetic energy will be the greatest.
e) dA = dB = dC. Under the conditions stated in this problem, each sample has the same volume, 5 L, and the same
mass, 4 g. Thus, the density of each is 4 g/5 L = 0.8 g/L.
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5-34
f) Collision frequency (A) > collision frequency (B) > collision frequency (C). The number of collisions
depends on both the speed and the distance between gas molecules. Since hydrogen is the lightest molecule it has
the greatest speed and the 5 L flask of hydrogen also contains the most molecules, so collisions will occur more
frequently between hydrogen molecules than between helium molecules. By the same reasoning, collisions will
occur more frequently between helium molecules than between methane molecules.
5.75
Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses
(Graham’s law).
Solution:
Rate H2
molar mass UF6
352.0 g/mol
=
=
= 13.2137 = 13.21
Rate UF6
molar mass H2
2.016 g/mol
5.76
Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses
(Graham’s law).
Solution:
Rate O2
molar mass Kr
83.80 g/mol
=
=
= 1.618255 = 1.618
Rate Kr
molar mass O2
32.00 g/mol
5.77
Plan: Recall that the heavier the gas, the slower the molecular speed. The molar mass of Ar is 39.95 g/mol while
the molar mass of He is 4.003 g/mol.
Solution:
a) The gases have the same average kinetic energy because they are at the same temperature. The heavier
Ar atoms are moving more slowly than the lighter He atoms to maintain the same average kinetic energy.
Therefore, Curve 1 with the lower average molecular speed, better represents the behavior of Ar.
b) A gas that has a slower molecular speed would effuse more slowly, so Curve 1 is the better choice.
c) Fluorine gas exists as a diatomic molecule, F2, with M = 38.00 g/mol. Therefore, F2 is much closer in mass to
Ar (39.95 g/mol) than He (4.003 g/mol), so Curve 1 more closely represents the behavior of F2.
5.78
Plan: Recall that the lower the temperature, the lower the average kinetic energy and the slower the molecular
speed.
Solution:
a) At the lower temperature, the average molecular speed is lower so Curve 1 represents the gas at the lower
temperature.
b) When a gas has a higher kinetic energy, the molecules have a higher molecular speed. Curve 2 with the larger
average molecular speed represents the gas when it has a higher kinetic energy.
c) If a gas has a higher diffusion rate, then the gas molecules are moving with a higher molecular speed as in
Curve 2.
5.79
Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses
(Graham’s law). Then use the ratio of effusion rates to find the time for the F2 effusion. Effusion rate and time
required for the effusion are inversely proportional.
Solution:
M of He = 4.003 g/mol
M of F2 = 38.00 g/mol
Rate He
molar mass F2
38.00 g/mol
=
=
= 3.08105 (unrounded)
Rate F2
molar mass He
4.003 g/mol
Rate He
Rate F2
Copyright
=
time F2
time He
time F2
3.08105
=
1.00
4.85 min He
Time F2 = 14.9431 = 14.9 min
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5-35
5.80
Plan: Effusion rate and time required for the effusion are inversely proportional. Therefore, time of effusion for
a gas is directly proportional to the square root of its molar mass. The ratio of effusion times and the molar mass
of H2 are used to find the molar mass of the unknown gas.
Solution:
M of H2 = 2.016 g/mol
Time of effusion of H2 = 2.42 min
Time of effusion of unknown = 11.1 min
rate H2
time unknown
molar mass unknown
=
=
rate unknown
time H2
molar mass H 2
11.1 min
molar mass unknown
=
2.42 min
4.586777 =
2.016 g/mol
molar mass unknown
2.016 g/mol
molar mass unknown
2.016 g/mol
Molar mass unknown = 42.41366 = 42.4 g/mol
21.03852196 =
5.81
Plan: White phosphorus is a molecular form of the element phosphorus consisting of some number, x, of
phosphorus atoms; the number of atoms in a molecule determines the molar mass of the phosphorus molecule.
Use the relative rates of effusion of white phosphorus and neon (Graham’s law) to determine the molar mass of
white phosphorus. From the molar mass of white phosphorus, determine the number of phosphorus atoms, x, in
one molecule of white phosphorus.
Solution:
M of Ne = 20.18 g/mol
Rate Px
molar mass Ne
= 0.404 =
Rate Ne
molar mass Px
20.18 g/mol
molar mass Px
0.404 =
2
(0.404) =
20.18 g/mol
molar mass Px
0.163216 =
20.18 g/mol
molar mass Px
Molar mass Px = 123.6398 g/mol
123.6398 g 
1 mol P 


 = 3.992244 = 4 mol P/mol Px or 4 atoms P/molecule Px
 mol P  30.97 g P 
x
Thus, 4 atoms per molecule, so Px = P4.
5.82
Copyright
Plan: Use the equation for root mean speed (urms) to find this value for He at 0°C and 30°C and for Xe at
30°C. The calculated root mean speed is then used in the kinetic energy equation to find the average kinetic
energy for the two gases at 30°C. Molar mass values must be in units of kg/mol and temperature in kelvins.
Solution:
a) 0°C = 273 K
30°C + 273 = 303 K
R = 8.314 J/mol K
1 J = kg m /s
2
 4.003 g He  1 kg 
 = 0.004003 kg/mol
M of He = 


103 g 
mol
2
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5-36
3RT
Molar mass
urms =
J 

38.314
273 K

mol  K 
0.004003 kg/mol

urmsHe (at 0°C) =
2
J
  kg  m /s


2

 = 1.3042 × 103 = 1.30 × 103 m/s

2
J

 = 1.3740 × 103 = 1.37 × 103 m /s

131.3 g Xe  1 kg 
 = 0.1313 kg/mol
M of Xe = 

 mol 103 g 
2 2
1 J = kg m /s
b) 30°C + 273 = 303 K
R = 8.314 J/mol K
3RT
urms =
Molar mass
J 

38.314
303 K

mol  K 
0.1313 kg/mol

urmsXe (at 30°C) =
2


J 

38.314
303 K

mol  K 
0.004003 kg/mol

urmsHe (at 30°C) =
  kg  m /s
  kg  m /s


2
J
2

 = 239.913 m/s (unrounded)

3
Rate He/Rate Xe = (1.3740 × 10 m/s)/(239.913 m/s) = 5.727076 = 5.73
He molecules travel at almost 6 times the speed of Xe molecules.
c) Ek =
EHe =
EXe =
1
mu2
2
1 

3
 0.004003 kg/mol  1.3740 10


2 
1 
 
2
 

 0.1313 kg/mol   239.913 m/s
2
m/s 1 J/kg  m2 /s2  = 3778.58 = 3.78 × 103 J/mol
2
1 J/kg  m2 /s2  = 3778.70 = 3.78 × 10
3
J/mol
 3778.58 J 
1 mol

–21
–21

d) 
= 6.2746 × 10 = 6.27 × 10 J/He atom
 mol  6.0221023 atoms 
5.83
Plan: Use Graham’s law: the rate of effusion of a gas is inversely proportional to the square root of the molar
mass. When comparing the speed of gas molecules, the one with the lowest mass travels the fastest.
Solution:
a) M of S2F2 = 102.12 g/mol; M of N2F4 = 104.02 g/mol; M of SF4 = 108.06 g/mol
SF4 has the largest molar mass and S2F2 has the smallest molar mass: rate SF4 < rate N2 F4 < rate S2 F2
molar mass N 2 F4
RateS2 F2
104.02 g/mol
=
b)
=
= 1.009260 = 1.0093:1
molar mass S2 F2
102.12 g/mol
Rate N2 F4
c)
Rate X
Rate SF4
0.935 =
= 0.935 =
molar mass SF4
molar mass X
108.06 g/mol
molar mass X
2
(0.935) =
0.874225 =
108.06 g/mol
molar mass X
108.06 g/mol
molar mass X
Molar mass X = 123.60662 = 124 g/mol
Copyright
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5-37
5.84
5.85
5.86
5.87
5.88
5.89
5.90
Plan: To find the ratio of effusion rates, calculate the inverse of the ratio of the square roots of the molar masses
(Graham’s law).
Solution:
Rate
235
Rate
238
UF6
UF6

Molar Mass
Molar Mass
238
UF6
235
UF6

352.04 g/mol
349.03 g/mol
 1.0043
Interparticle attractions cause the real pressure to be less than ideal pressure, so it causes a negative deviation.
The size of the interparticle attraction is related to the constant a. According to Table 5.4, aN2 = 1.39,
aKr = 2.32, and aCO2 = 3.59. Therefore, CO2 experiences a greater negative deviation in pressure than the other
two gases: N2 < Kr < CO2.
Particle volume causes a positive deviation from ideal behavior. Thus, VReal Gases > VIdeal Gases. The particle volume is
related to the constant b. According to Table 5.4, bH2 = 0.0266, bO 2 = 0.0318, and bCl 2 = 0.0562. Therefore, the
order is H2 < O2 < Cl2.
Nitrogen gas behaves more ideally at 1 atm than at 500 atm because at lower pressures the gas molecules are
farther apart. An ideal gas is defined as consisting of gas molecules that act independently of the other gas
molecules. When gas molecules are far apart they act more ideally, because intermolecular attractions are less
important and the volume of the molecules is a smaller fraction of the container volume.
SF6 behaves more ideally at 150°C. At higher temperatures, intermolecular attractions become less important and
the volume occupied by the molecules becomes less important.
Plan: To find the total force, the total surface area of the can is needed. Use the dimensions of the can to find the
surface area of each side of the can. Do not forget to multiply the area of each side by two. The surface area of
2
2
the can in cm must be converted to units of in .
Solution:
Surface area of can = 2(40.0 cm)(15.0 cm) + 2(40.0 cm)(12.5 cm) + 2(15.0 cm)(12.5 cm)
3
2
= 2.575 × 10 cm
2
 1 in  14.7 lb 
3
3
Total force = 2.575103 cm 2 
 
 = 5.8671 × 10 = 5.87 × 10 lbs
 2.54 cm   1 in 2 
Plan: Use the ideal gas law to find the number of moles of O2. Moles of O2 is divided by 4 to find moles of
Hb since O2 combines with Hb in a 4:1 ratio. Divide the given mass of Hb by the number of moles of Hb to obtain
molar mass, g/mol. Temperature must be in units of kelvins, pressure in atm, and volume in L.
Solution:
V = 1.53 mL
T = 37°C + 273.15 = 310.15 K
P = 743 torr
n = unknown
103 L 
 = 1.53 × 10–3 L
Converting V from mL to L:
V = 1.53 mL 
 1 mL 
Converting P from torr to atm:
PV = nRT
Solving for n:
 1 atm 
P = 743 torr 
= 0.977631578 atm
 760 torr 
PV 0.977631578 atm 1.5310 L 
–5
= 5.874240 × 10 mol O2

L  atm 

RT
0.0821
310.15 K 

mol  K 
3
Moles of O2 = n =
 1 mol Hb 
–5
Moles Hb = 5.874240105 mol O2 
 = 1.46856 × 10 mol Hb (unrounded)
 4 mol O2 
1.00 g Hb
4
4
Molar mass hemoglobin =
= 6.80939 × 10 = 6.81 × 10 g/mol
5
1.46856 10 Hb
Copyright
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5-38
5.91
Plan: First, write the balanced equations. Convert mass of NaHCO3 to moles and use the molar ratio from each
balanced equation to find the moles of CO2 gas produced. Use the ideal gas law to find the volume of that amount
of CO2. Temperature must be in kelvins.
Solution:
Reaction 1: 2NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)
 1 mol NaHCO3 
 1 mol CO2  = 5.95167 × 10–3 mol CO
Moles CO2 = 1.00 g NaHCO3 
2


 84.01 g NaHCO 
2 mol NaHCO 
3
Finding the volume of CO2:
V = unknown
P = 0.975 atm
PV = nRT
Solving for V:
3
T = 200°C + 273.15 = 473.15 K
–3
n = 5.95167 × 10 mol

L  atm 
5.95167 103 mol 0.0821
473.15 K 

nRT
mol
 K 
Volume of CO2 = V =
= 0.2371244 L
=
P
0.975 atm 
Converting V from L to mL:
 1 mL 

V = 0.2371244 L 3  = 237.1244 = 237 mL CO2 in Reaction 1
10 L 


Reaction 2: NaHCO3(s) + H (aq)  H2O(l) + CO2(g) + Na (aq)
 1 mol NaHCO3 
1 mol CO2 

 = 1.1903 × 10–2 mol CO2
Moles CO2 = 1.00 g NaHCO3 
1 mol NaHCO 
 84.01 g NaHCO 
+
+
3
Finding the volume of CO2:
V = unknown
P = 0.975 atm
PV = nRT
Solving for V:
3
T = 200°C + 273.15 = 473.15 K
–2
n = 1.1903 × 10 mol

L  atm 
1.1903102 mol 0.0821
473.15 K 

nRT
mol
 K 
Volume of CO2 = V =
= 0.4742352 L
=
P
0.975 atm 
Converting V from L to mL:
 1 mL 

V = 0.4742352 L 3  = 474.2352 = 474 mL CO2 in Reaction 2
10 L 

5.92
Plan: Use the relationship

PV
PV
PV n T
1 1
= 2 2 or V2 = 1 1 2 2 ; P is fixed while V and T change. n2 is 0.75n1 since
n1T1
n2 T2
P2 n1T1
one-fourth of the gas leaks out. Only the initial and final conditions need to be considered.
Solution:
P1 = 1.01 atm
P2 = 1.01 atm
(Thus, P has no effect, and does not need to be included.)
T1 = 305 K
T2 = 250 K
n1 = n1
n2 = 0.75n1
V1 = 600. L
V2 = ?
600 L0.75n1 250 K
Vn T
V2 = 1 2 2 =
= 368.852 = 369 L
n1T1
n1  305 K

5.93
Copyright

Plan: Convert the mass of Cl2 to moles and use the ideal gas law and van der Waals equation to
find the pressure of the gas.
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5-39
Solution:
103 g  1 mol Cl2 
 = 8.3921016 mol

a) Moles Cl2: 0.5950 kg Cl2 
 70.90 g Cl2 
 1 kg 
V = 15.50 L
n = 8.3921016 mol
Ideal gas law: PV = nRT
T = 225°C + 273.15 = 498.15 K
P = unknown
Solving for P:
nRT
PIGL =
=
V
L  atm 
498.15 K 
8.3921016 mol0.0821 mol
 K 
15.50 L

n2 a 
b) van der Waals equation:  P  2 V  nb  nRT

V 
= 22.1433 = 22.1 atm
Solving for P:
nRT
n2 a
atm  L2
PVDW =
From Table 5.4: a = 6.49
;
 2
V  nb V
mol 2
n = 8.3921016 mol from part (a)
b = 0.0562
L
mol
2
2
2
PVDW


atm  L 
L  atm 

8.3921016 mol Cl 0.0821 mol
498.15 K  8.3921016 mol Cl  6.49 mol 

K
=

L 

15.50 L 8.3921016 mol Cl 0.0562
15.50 L


mol 
2
2
2
2
= 20.93573 = 20.9 atm
5.94
Plan: Rearrange the formula PV = (m/M )RT to solve for molar mass. Convert the volume in mL to L.
Temperature must be in kelvin. To find the molecular formulas of I, II, III, and IV, assume 100 g of each sample
so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles
by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to
whole numbers to determine the empirical formula. The empirical formula mass and the calculated molar mass
will then relate the empirical formula to the molecular formula. For gas IV, use Graham’s law to find the molar
mass
Solution:
a) V = 750.0 mL
T = 70.00°C + 273.15 = 343.15 K
m = 0.1000 g
P = 0.05951 atm (I); 0.07045 atm (II); 0.05767 atm (III)
M = unknown
103 L 
 = 0.7500 L
Converting V from mL to L:
V = 750.0 mL 
 1 mL 
PV = (m/M )RT
Solving for molar mass, M :
mRT
Molar mass I = M =

PV
Copyright

L  atm 
343.15 K 

mol  K 
= 63.0905 = 63.09 g I/mol
0.05951 atm 0.7500 L 
0.1000 g0.08206
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5-40

L  atm 
343.15 K 

mol  K 
= 53.293 = 53.29 g II/mol
0.07045 atm 0.7500 L 
0.1000 g0.08206
mRT
Molar mass II = M =

PV

L  atm 
0.1000 g0.08206
343.15 K 

mRT
mol  K 
Molar mass III = M =
= 65.10349 = 65.10 g III/mol

PV
0.05767 atm 0.7500 L 
b) % H in I = 100% – 85.63% = 14.37% H
% H in II = 100% – 81.10% = 18.90% H
% H in III = 100% – 82.98% = 17.02% H
Assume 100 g of each so the mass percentages are also the grams of the element.
I
 1 mol B 

 = 7.921369 mol B (unrounded)
Moles B = 85.63 g B 
10.81 g B 


 1 mol H 

 = 14.25595 mol H (unrounded)
Moles H = 14.37 g H 
1.008 g H 


 7.921369 mol B
14.25595 mol H 


 = 1.00

 = 1.7997




7.921369
mol
B

 7.921369 mol B

The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be
multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying each value by 5
gives (1.00 × 5) = 5 for B and (1.7997 × 5) = 9 for H. The empirical formula is B5H9, which has a
formula mass of 63.12 g/mol. The empirical formula mass is near the molecular mass from part (a)
(63.09 g/mol). Therefore, the empirical and molecular formulas are both B5H9.
II
 1 mol B 

 = 7.50231 mol B (unrounded)
Moles B = 81.10 g B 
10.81 g B 


 1 mol H 

 = 18.750 mol H (unrounded)
Moles H = 18.90 g H 
1.008 g H 

III

 7.50231 mol B
 18.750 mol H 


 = 1.00
 = 2.4992



7.50231
mol
B

 7.50231 mol B

The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be
multiplied by the smallest value to get near whole numbers. This value is 2. Multiplying each value by 2
gives (1.00 × 2) = 2 for B and (2.4992 × 2) = 5 for H. The empirical formula is B2H5, which has a
formula mass of 26.66 g/mol. Dividing the molecular formula mass from part (a) by the empirical
formula mass gives the relationship between the formulas: (53.29 g/mol)/(26.66 g/mol) = 2. The
molecular formula is two times the empirical formula, or B4H10.
 1 mol B 

 = 7.6762 mol B (unrounded)
Moles B = 82.98 g B 
10.81 g B 


 1 mol H 

 = 16.8849 mol H (unrounded)
Moles H = 17.02 g H 
1.008 g H 

Copyright

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5-41
c)
 7.6762 mol B
16.8849 mol H 


 = 1.00
 = 2.2



 7.6762 mol B
 7.6762 mol B 
The hydrogen value is not close enough to a whole number to round. Thus, both amounts need to be
multiplied by the smallest value to get near whole numbers. This value is 5. Multiplying each value by 5
gives (1.00 × 5) = 5 for B and (2.2 × 5) = 11 for H. The empirical formula is B5H11, which has a formula
mass of 65.14 g/mol. The empirical formula mass is near the molecular mass from part (a). Therefore,
the empirical and molecular formulas are both B5H11.
Rate SO2
molar mass IV
=
Rate IV
molar mass SO2
 250.0 mL 



molar mass IV
13.04 min 
= 0.657318 =
 350.0 mL 
64.06 g/mol




12.00 min 
molar mass IV
2
0.657318 =
64.06 g/mol
Molar mass IV = 27.6782 = 27.68 g/mol
% H in IV = 100% – 78.14% = 21.86% H
 1 mol B 

Moles B = 78.14 g B 
 = 7.22849 mol B (unrounded)
10.81 g B 


 1 mol H 

 = 21.6865 mol H (unrounded)
Moles H = 21.86 g H 
1.008 g H 


 7.22849 mol B
 21.6865 mol H 


 = 1.00
 = 3.00


 7.22849 mol B
 7.22849 mol B 
The empirical formula is BH3, which has a formula mass of 13.83 g/mol. Dividing the molecular
formula mass by the empirical formula mass gives the relationship between the formulas:
(27.68 g/mol)/(13.83 g/mol) = 2. The molecular formula is two times the empirical formula, or B2H6.
5.95
Copyright
Plan: Calculate the mole fraction of each gas; the partial pressure of each gas is directly proportional to its mole
fraction so the gas with the highest mole fraction has the highest partial pressure. Remember that kinetic energy is
directly proportional to kelvin temperature.
Solution:
n
a) X A = A
ntotal
3 A particles
4 A particles
5 A particles
I. XA =
= 0.33; II. XA =
= 0.33; III. XA =
= 0.33
9 total particles
12 total particles
15 total particles
The partial pressure of A is the same in all 3 samples since the mole fraction of A is the same in all samples.
3 B particles
3 B particles
3 B particles
b) I. XB =
= 0.33; II. XB =
= 0.25; III. XB =
= 0.20
9 total particles
12 total particles
15 total particles
The partial pressure of B is lowest in Sample III since the mole fraction of B is the smallest in that sample.
c) All samples are at the same temperature, T, so all have the same average kinetic energy.
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5-42
5.96
Plan: Use the relationship
PV
PV
PV T
1 1
= 2 2 or V2 = 1 1 2 . R and n are fixed. Temperatures must be converted to
T1
T2
P2 T1
kelvin.
Solution:
a) T1 = 200°C + 273.15 = 473.15 K;
PV T
(2 atm)(V1 )(373.15 K)
V2 = 1 1 2 =
= 1.577 V1
P2T1
(1 atm)(473.15 K)
b) T1 = 100°C + 273.15 = 373.15 K;
PV T
(1 atm)(V1 )(573.15 K)
V2 = 1 1 2 =
= 0.5120 V1
P2T1
(3 atm)(373.15 K)
c) T1 = –73°C + 273.15 = 200.15 K;
PV T
(3 atm)(V1 )(400.15 K)
V2 = 1 1 2 =
= 0.9996V1
P2 T1
(6 atm)(200.15 K)
d) T1 = 300°C + 273.15 = 573.15 K;
PV T
(0.2 atm)(V1 )(423.15 K)
V2 = 1 1 2 =
= 0.3691V1
P2T1
(0.4 atm)(573.15 K)
5.97
Copyright
T2 = 100°C + 273.15 = 373.15 K
Increase
T2 = 300°C + 273.15 = 573.15 K
Decrease
T2 = 127°C + 273.15 = 400.15 K
Unchanged
T2 = 150°C + 273.15 = 423.15 K
Decrease
Plan: Partial pressures and mole fractions are calculated from Dalton’s law of partial pressures: PA = XA × Ptotal.
Remember that 1 atm = 760 torr. Solve the ideal gas law for moles and then convert to molecules using
Avogadro’s number to calculate the number of O2 molecules in the volume of an average breath.
Solution:
a) Convert each mole percent to a mole fraction by dividing by 100%. Ptotal = 1 atm = 760 torr
PNitrogen = XNitrogen × Ptotal = 0.786 × 760 torr = 597.36 = 597 torr N2
POxygen = XOxygen × Ptotal = 0.209 × 760 torr = 158.84 = 159 torr O2
PCarbon Dioxide = XCarbon Dioxide × Ptotal = 0.0004 × 760 torr = 0.304 = 0.3 torr CO2
PWater = XWater × Ptotal = 0.0046 × 760 torr = 3.496 = 3.5 torr H2O
b) Mole fractions can be calculated by rearranging Dalton’s law of partial pressures:
P
XA = A and multiply by 100 to express mole fraction as percent
Ptotal
Ptotal = 1 atm = 760 torr
569 torr
XNitrogen =
× 100% = 74.8684 = 74.9 mol% N2
760 torr
104 torr
XOxygen =
× 100% = 13.6842 = 13.7 mol% O2
760 torr
40 torr
X Carbon Dioxide =
× 100% = 5.263 = 5.3 mol% CO2
760 torr
47 torr
XWater =
× 100% = 6.1842 = 6.2 mol% H2O
760 torr
c) V = 0.50 L
T = 37°C + 273.15 = 310.15 K
P = 104 torr
n = unknown
 1 atm 
Converting P from torr to atm:
P = 104 torr 
= 0.136842105 atm
 760 torr 
PV = nRT
Solving for n:
PV 0.136842105 atm 0.50 L 
n=

= 0.0026870 mol O2
RT 0.0821 L  atm 310.15 K 


mol  K 
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5-43
 6.022 1023 molecules O 
2
Molecules of O2 = 0.0026870 mol O 2 


1 mol O 2

21
21
= 1.6181 × 10 = 1.6 × 10 molecules O2
5.98
Plan: Convert the mass of Ra to moles and then atoms using Avogadro’s number. Convert from number of Ra
atoms to Rn atoms produced per second and then to Rn atoms produced per day. The number of Rn atoms is
converted to moles and then the ideal gas law is used to find the volume of this amount of Rn.
Solution:
23
 1 mol Ra 
 6.022 10 Ra atoms 

21
Atoms Ra = 1.0 g Ra 

 = 2.664602 × 10 Ra atoms
 226 g Ra 
1 mol Ra




1.37310 4 Rn atoms 
 = 3.65849855 × 1010 Rn atoms/s
Atoms Rn produced/s = 2.664602 10 21 Ra atoms 
 1.0 1015 Ra atoms 


 3.658498551010 Rn atoms  3600 s  24 h 
1 mol Rn




Moles Rn produced/day = 

23








s
 h  day  6.02210 Rn atoms 
–9
= 5.248992 × 10 mole Rn/day
PV = nRT
Solving for V (at STP):

L  atm 
5.248992 109 mol 0.0821
273 K 

nRT
mol
 K 
Volume of Rn = V =
=
P
1 atm 
–7
–7
= 1.17647 × 10 = 1.2 × 10 L Rn

5.99

Plan: For part (a), since the volume, temperature, and pressure of the gas are changing, use the combined gas law.
For part (b), use the ideal gas law to solve for moles of air and then moles of N2.
Solution:
a) P1 = 1450. mmHg
P2 = 1 atm
V1 = 208 mL
V2 = unknown
T1 = 286 K
T2 = 298 K
 1 atm 
 = 1.9079 atm
Converting P1 from mmHg to atm: P1 = 1450. mmHg
 760 mmHg 
Arranging the ideal gas law and solving for V2:
PV
PV
1 1
= 2 2
T1
T2
 T  P 
V2 = V1  2  1  = 208 L 
 P2 
 T1 
 298 K 

 286 K 
b) V = 208 mL
P = 1450 mmHg = 1.9079 atm
Converting V from mL to L:
1.9079 atm 
2

= 413.494 mL = 4 × 10 mL
 1 atm 
T = 286 K
n = unknown
103 L 
 = 0.208 L
V = 208 mL
 1 mL 
PV = nRT
Solving for n:
Moles of air = n =
Copyright
1.9079 atm 0.208 L
PV

= 0.016901 mol air
RT 0.0821 L  atm 286 K 

mol  K 
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5-44
 77% N 2 
Mole of N2 = 0.016901 mol
 = 0.01301 = 0.013 mol N2
 100% 
5.100
Plan: The amounts of both reactants are given, so the first step is to identify the limiting reactant. Write the balanced
equation and use molar ratios to find the number of moles of NO2 produced by each reactant. The smaller number of
moles of product indicates the limiting reagent. Solve for volume of NO2 using the ideal gas law.
Solution:
Cu(s) + 4HNO3(aq)  Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
 8.95 g Cu  1 mol Cu 
2 mol NO2 




Moles NO2 from Cu = 4.95 cm3 

3
 1 mol Cu  = 1.394256 mol NO2
 cm
 63.55 g Cu 



 68.0% HNO3 1 cm3 1.42 g 1 mol HNO3 
 2 mol NO2 




Moles NO2 from HNO3 = 230.0 mL 

3
 4 mol HNO 

100%
 1 mL  cm  63.02 g 
3
= 1.7620 mol NO2
Since less product can be made from the copper, it is the limiting reactant and excess nitric acid will be left after
the reaction goes to completion. Use the calculated number of moles of NO 2 and the given temperature and
pressure in the ideal gas law to find the volume of nitrogen dioxide produced. Note that nitrogen dioxide is the
only gas involved in the reaction.
V = unknown
T = 28.2°C + 273.15 = 301.35 K
P = 735 torr
n = 1.394256 mol NO2
 1 atm 
Converting P from torr to atm:
P = 735 torr 
= 0.967105 atm
 760 torr 
PV = nRT
Solving for V:

L  atm 
1.394256 mol0.0821
301.35 K 

nRT
mol
 K 
= 35.668.6 = 35.7 L NO2
V=

P
0.967105 atm 

5.101

Plan: Solve the ideal gas law for moles of air and then convert to molecules using Avogadro’s number. Volume
must be converted to liters and temperature to kelvins.
Solution:
a) V = 1200 mL
T = 37°C + 273.15 = 310.15 K
P = 1.0 atm
n = unknown
103 L 
 = 1.2 L
Converting V from mL to L:
V = 1200 mL
 1 mL 
PV = nRT
Solving for n:
1.0 atm1.2 L
PV

= 0.047127 = 0.047 mol air

RT 0.0821 L  atm 310.15 K 

mol  K 
 6.022 1023 molecules 

 = 2.837971 × 1022 = 2.8 × 1022 molecules air
b) Molecules of air = 0.047127 mol air 


1 mol air

Moles of air = n =
Copyright
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5-45
5.102
Plan: The amounts of two reactants are given, so the first step is to identify the limiting reactant.
Write the balanced equation and use molar ratios to find the number of moles of Br2 produced by each reactant.
The smaller number of moles of product indicates the limiting reagent. Solve for volume of Br2 using the ideal gas
law.
Solution:
5NaBr(aq) + NaBrO3(aq) + 3H2SO4(aq)  3Br2(g) + 3Na2SO4(aq) + 3H2O(g)
 1 mol NaBr 
 3 mol Br2 

Moles Br2 from NaBr = 275 g NaBr 

 = 1.60365 mol Br2 (unrounded)
102.89 g NaBr 
 5 mol NaBr 


 1 mol NaBrO3 
3 mol Br2 



Moles Br2 from NaBrO3 = 175.6 g NaBrO3 
1 mol NaBrO  = 3.491285 mol Br2
150.89 g NaBrO3 
3
The NaBr is limiting since it produces a smaller amount of Br2.
Use the ideal gas law to find the volume of Br2:
V = unknown
T = 300°C + 273.15 = 573.15 K
P = 0.855 atm
n = 1.60365 mol Br2
PV = nRT
Solving for V:

L  atm 
1.60365 mol0.0821
573.15 K 

nRT
mol
 K 
Volume (L) of Br2 = V =
= 88.258 = 88. L Br2

P
0.855 atm 
Plan: First, write the balanced equation. Convert mass of NaN3 to moles and use the molar ratio from the balanced
equation to find the moles of nitrogen gas produced. Use the ideal gas law to find the volume of that amount of
nitrogen. The problem specifies that the nitrogen gas is collected over water, so the partial pressure of water vapor
must be subtracted from the overall pressure given. Table 5.2 reports the vapor pressure of water at 26°C (25.2
torr). Pressure must be in units of atm and temperature in kelvins.
Solution:
2NaN3(s)  2Na(s) + 3N2(g)
 1 mol NaN3 
 3 mol N2  = 1.15349 mol N
Moles N2 = 50.0 g NaN3 
2


 65.02 g NaN 
2 mol NaN 

5.103

3
3
Finding the volume of N2:
V = unknown
T = 26°C + 273.15 = 299.15 K
Ptotal = 745.5 mmHg
n = 1.15319 mol
Pwater vapor = 25.2 torr = 25.2 mmHg
Pnitrogen = Ptotal – Pwater vapor = 745.5 mmHg – 25.2 mmHg = 720.3 mmHg
 1 atm 
 = 0.9477632 atm
Converting P from mmHg to atm: P = 720.3 mmHg
 760 mmHg 
PV = nRT
Solving for V:
nRT
V=
=
P
5.104
Copyright

L  atm 
1.15349 mol0.0821
299.15 K 

mol  K 
= 29.8914 = 29.9 L N2
0.9477632 atm 
Plan: Use the percent composition information to find the empirical formula of the compound. Assume 100 g of
sample so the percentages are numerically equivalent to the masses of each element. Convert each of the masses
to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and
convert to whole numbers to determine the empirical formula. Rearrange the formula PV = (m/M )RT to solve for
molar mass.The empirical formula mass and the calculated molar mass will then relate the empirical formula to
the molecular formula.
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5-46
Solution:
Empirical formula:
Assume 100 g of each so the mass percentages are also the grams of the element.
 1 mol C 

Moles C = 64.81 g C 
 = 5.39634 mol C
12.01 g C 


 1 mol H 

 = 13.49206 mol H
Moles H = 13.60 g H 
1.008 g H 


 1 mol O 

 = 1.349375 mol O
Moles O = 21.59 g O 
16.00 g O 


 5.39634 mol C 
13.749206 mol H 


 = 4
 = 10




1.349375
mol
O

 1.349375 mol O 

Empirical formula = C4H10O (empirical formula mass = 74.12 g/mol)
Molecular formula:
V = 2.00 mL
T = 25°C + 273.15 = 298.15 K
m = 2.57 g
P = 0.420 atm
M = unknown
1.349375 mol O 

 = 1.00

1.349375 mol O 
PV = (m/M )RT
Solving for molar mass, M :
Molar mass = M =
mRT

PV

L  atm 
298.15 K 

mol  K 
= 74.89 g/mol
0.420 atm 2.00 L 
2.57 g0.0821
Since the molar mass (74.89 g/mol ) and the empirical formula mass (74.12 g/mol) are similar, the empirical and
molecular formulas must both be: C4H10O
5.105
3+
–
Plan: The empirical formula for aluminum chloride is AlCl3 (Al and Cl ). The empirical formula mass is
(133.33 g/mol). Calculate the molar mass of the gaseous species from the ratio of effusion rates (Graham’s law).
This molar mass, divided by the empirical weight, should give a whole-number multiple that will yield the
molecular formula.
Solution:
Rate unknown
Rate He
0.122 =
= 0.122 =
molar mass He
molar mass unknown
4.003 g/mol
molar mass unknown
0.014884 =
4.003 g/mol
molar mass unknown
Molar mass unknown = 268.9465 g/mol
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5-47
The whole-number multiple is 268.9465/133.33, which is about 2. Therefore, the molecular formula of the
gaseous species is 2 × (AlCl3) = Al2Cl6.
5.106
Plan: First, write the balanced equation. Convert mass of C8H18 to moles and use the molar ratio from the balanced
equation to find the total moles of gas produced. Use the ideal gas law to find the volume of that amount of gas. For
part (b), use the molar ratio from the balanced equation to find the moles of oxygen that react with the C8H18. Use the
composition of air to calculate the amount of N2 and Ar that remains after the O2 is consumed and use the ideal gas
law to find the volume of that amount of gas. The volume of the gaseous exhaust is the sum of the volume of gaseous
products and the residual air (N2 and Ar) that does not react.
Solution:
2C8H18(l) + 25O2(g)  16CO2(g) + 18H2O (g)
 1 mol C8 H18 
 34 mol gas 

a) Moles gaseous products = 100. g C8 H18 

 = 14.883558 mol gas
114.22 g C8 H18 
 2 mol C8 H18 
Finding the volume of gaseous product:
V = unknown
T = 350°C + 273.15 = 623.15 K
Ptotal = 735 torr
n = 14.883558 mol
 1 atm 
Converting P from torr to atm:
P = 735 torr 
= 0.9671053 atm
 760 torr 
PV = nRT


Solving for V:

L  atm 
14.883558 mol0.0821
623.15 K 

mol  K 
= 787.352 = 787 L gas
0.9671053 atm 
 1 mol C H 
25 mol O2 
8 18 



b) Moles O2 = 100 g C8 H18 

 2 mol C H  = 10.94379 mol O2
114.22 g C8 H18 
8 18 
nRT
Volume gas = V =
=
P


 78% + 1% N 2 and Ar 

 = 41.1695 mol Ar + N
Moles other gases = 10.94379 mol O2 
2

21% O2


Finding the volume of Ar + N2:
V = unknown
T = 350°C + 273.15 = 623.15 K
Ptotal = 735 torr = 0.9671053 atm
n = 41.1695 mol
PV = nRT
Solving for V:

L  atm 
41.1695 mol 0.0821
623.15 K 

nRT
mol
 K 
Volume gas = V =
= 2177.90 L residual air
=
P
0.9671053 atm 
3
Total volume of gaseous exhaust = 787.352 L + 2177.90 L = 2965.25 = 2.97 × 10 L

5.107
Copyright

Plan: First, write the balanced equation for the reaction: 2SO2 + O2  2SO3. The total number of moles of gas will
change as the reaction occurs since 3 moles of reactant gas forms 2 moles of product gas. From the volume,
temperature, and pressures given, we can calculate the number of moles of gas before and after the reaction using
the ideal gas law. For each mole of SO3 formed, the total number of moles of gas decreases by 1/2 mole. Thus,
twice the decrease in moles of gas equals the moles of SO3 formed.
Solution:
Moles of gas before and after reaction:
V = 2.00 L
T = 800 K
Ptotal = 1.90 atm
n = unknown
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5-48
PV = nRT
Initial moles = n =
Final moles = n =



1.90 atm 2.00 L
PV
=
= 0.05785627 mol

L  atm 
RT
0.0821
800.
K

mol  K 





1.65 atm 2.00 L
PV
=
= 0.050243605 mol

L  atm 
RT
0.0821
800.
K


mol  K 


Moles of SO3 produced = 2 × decrease in the total number of moles
= 2 × (0.05785627 mol – 0.050243605 mol)
–2
= 0.01522533 = 1.52 × 10 mol
Check: If the starting amount is 0.0578 total moles of SO2 and O2, then x + y = 0.0578 mol,
where x = mol of SO2 and y = mol of O2. After the reaction:
(x – z) + (y – 0.5z) + z = 0.0502 mol
Where z = mol of SO3 formed = mol of SO2 reacted = 2(mol of O2 reacted).
Subtracting the two equations gives:
x – (x – z) + y – (y – 0.5z) – z = 0.0578 – 0.0502
z = 0.0152 mol SO3
The approach of setting up two equations and solving them gives the same result as above.
5.108
Plan: Use the density of C2H4 to find the volume of one mole of gas. Then use the van der Waals equation with
1.00 mol of gas to find the pressure of the gas (the mole ratio is 1:1, so the number of moles of gas remains the
same).
Solution:
3
 28.05 g C2 H 4 
 1 mL 
10 L  = 0.130465 L = 0.130 L
a) 1 mole C2 H 4 




 1 mole C2 H 4 
 0.215 g  1 mL 
V = 0.130465 L
T = 10°C + 273.15 = 283.15 K + 950 K = 1233.15 K
Ptotal = unknown
n = 1.00 mol
atm  L2
From Table 5.4 for CH4: a = 2.25
;
mol 2
2 

 P  n a V  nb  nRT

V 2 
Pressure of CH4 = PVDW =
b = 0.0428
L
mol
nRT
n2 a
 2
V  nb V
atm  L2 
2
L  atm 

1.00 mol 2.25

1.00 mol0.0821
1233.15
K


mol 2 


mol  K 

PVDW =
= 1022.681 atm = 1023 atm
2
0.130465 L 1.00 mol 0.0428 L/mol
0.130465 L
b) n =
1022.681 atm 0.130465 L 
PV
=
= 1.31788 = 1.32 mol

L  atm 
RT
0.0821
1233.15 K 


mol  K 
This value is smaller than that shown in Figure 5.23 for CH4. The temperature in this situation is very high
(1233 K). At high temperatures, the gas particles have high kinetic energy. Thus the gas particles have the energy
to overcome the effects of intermolecular attraction and the gas behaves more ideally.
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5-49
5.109
Plan: First, write the balanced equation. Convert mass of ammonium nitrate to moles and use the molar ratio from the
balanced equation to find the moles of gas produced. Use the ideal gas law to find the volume of that amount of gas.
Solution:
103 g 
 1 mol NH 4 NO3 
 7 mol gas 

Moles gas = 15.0 kg NH 4 NO3 



 1 kg 
 80.05 g NH 4 NO3 
 2 mol NH 4 NO3 
= 655.840 mol gas
Finding the volume of gas:
V = unknown
T = 307°C + 273.15 = 580.15 K
P = 1.00 atm
n = 655.840 mol
PV = nRT
Solving for V:

L  atm 
655.840 mol0.0821
580.15 K 

nRT
4
4
mol  K 
V=
= 3.12379 × 10 = 3.12 × 10 L
=
P
1.00 atm

5.110

Plan: Multiply the molarity and volume of the I2 solution to find initial moles of I2 added. Multiply molarity and
2–
volume of the S2O3 solution to obtain moles of that solution and use the molar ratio in the balanced equation to
find moles of excess I2. Subtract the excess I2 from the initial moles of added I2 to find the moles of I2 reacted with
the SO2; the molar ratio betweeen SO2 and I2 gives the moles of SO2 present. Use the ideal gas law to find the
moles of air which is compared to the moles of SO2 present.
Solution:
2
2
The balanced equation is: I2(aq) + 2S2O3 (aq)  2I(aq) + S4O6 (aq).
103 L  0.01017 mol I2 

 = 2.034 × 10–4 mol I initial

Initial moles of I2 = 20.00 mL 

2
 1 mL 
L



103 L  0.0105 mol S2 O32  1 mol I 2 

2–



Moles I2 reacting with S2O3 = 11.37 mL 


 2 mol S O 2 
L
 1 mL 
2 3


–5
2–
= 5.96925 × 10 mol I2 reacting with S2O3 (not reacting with SO2)
–4
–5
Moles I2 reacting with SO2 = 2.034 × 10 mol – 5.96925 × 10 mol
–4
= 1.437075 × 10 mol I2 reacted with SO2
+
The balanced equation is: SO2(aq) + I2(aq) + 2H2O(l)  HSO4(aq) + 2I(aq) + 3H (aq).
1 mol SO 
4
–4
2
Moles SO2 = 1.43707510 mol I 2 
 = 1.437075 × 10 mol SO2
 1 mol I2 


Moles of air:
V = 500 mL = 0.500 L
T = 38°C + 273.15 = 311.15 K
Ptotal = 700 torr
Converting P from torr to atm:
PV = nRT
Moles air = n =
n = unknown
 1 atm 
P = 700 torr 
 = 0.921052631 atm
 760 torr 




0.921052632 atm 0.500 L
PV
=
= 0.018028 mol air (unrounded)

L  atm 
RT
0.0821
311.15
K


mol  K 

1.43707510 mol SO 2
4
Volume % SO2 = mol % SO2 =
Copyright
0.018028 mol air
100 = 0.797146 = 0.797%
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5-50
5.111
Plan: First, write the balanced equation. The moles of CO that react in part (a) can be calculated from the ideal gas
law. Volume must be in units of L, pressure in atm, and temperature in kelvins. Once the moles of CO that react are
known, the molar ratio from the balanced equation is used to determine the mass of nickel that will react with the
3
CO. For part (b), assume the volume is 1 m . Use the ideal gas law to solve for moles of Ni(CO)4, which equals the
3
moles of Ni, and convert moles to grams using the molar mass. For part (c), the mass of Ni obtained from 1 m
(part b) can be used to calculate the amount of CO released. Use the ideal gas law to calculate the volume of CO.
The vapor pressure of water at 35°C (42.2 torr) must be subtracted from the overall pressure (see Table 5.2).
Solution:
a) Ni(s) + 4CO(g)  Ni(CO)4(g)
3
V = 3.55 m
T = 50°C + 273.15 = 323.15 K
P = 100.7 kPa
n = unknown
 1L 
3
Converting V from m to L:
V = 3.55 m 3  3 3  = 3550 L
10 m 
 1 atm

Converting P from kPa to atm:
P = 100.7 kPa 
= 0.993831729 atm
101.325 kPa 
PV = nRT
Solving for n:
PV 0.993831729 atm 3550 L 

= 132.98232 mol CO
RT 0.0821 L  atm 323.15 K 

mol  K 
 1 mol Ni 
 58.69g Ni 

3
Mass Ni = 132.98232 mol CO

 = 1951.183 = 1.95 × 10 g Ni
 4 mol CO 
 1mol Ni 
Moles of CO = n =
b) Ni(s) + 4 CO(g)  Ni(CO)4(g)
3
V=1m
T = 155°C + 273.15 = 428.15 K
P = 21 atm
n = unknown
 1L 
3
Converting V from m to L:
V = 1 m 3  3 3  = 1000 L
10 m 
PV = nRT
Solving for n:
21 atm1000 L
PV

Moles of Ni(CO)4 = n =
= 597.42059 mol Ni(CO)4
RT 0.0821 L  atm  428.15 K 

mol  K 
 1 mol Ni 
 58.69 g Ni 

Mass Ni = 597.42059 mol Ni(CO)4 


1 mol Ni(CO)4 
 1 mol Ni 
4
4
= 3.50626 × 10 = 3.5 × 10 g Ni
The pressure limits the significant figures.
 1 mol Ni  4 mol CO 
 = 2389.68238 mol CO

c) Moles CO = 3.50626 10 4 g Ni
 58.69 g Ni  1 mol Ni 
Finding the volume of CO:
V = unknown
T = 35°C + 273.15 = 308.15 K
Ptotal = 769 torr
n = 2389.68238 mol
Pwater vapor = 42.2 torr
PCO = Ptotal – Pwater vapor = 769 torr – 42.2 torr = 726.8 torr
 1 atm 
Converting P from torr to atm:
P = 726.8 torr 
= 0.956315789 atm
 760 torr 
PV = nRT
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5-51
Solving for V:
nRT
V=
=
P

L  atm 
2389.68238 mol0.0821
308.15 K 

mol  K 
= 63218.4995 L CO
0.956315789 atm 
103 m 3 
 = 63.2184995 = 63 m3 CO
V = 63218.4995 L 
 1 L 
The answer is limited to two significant figures because the mass of Ni comes from part (b).
3
Converting V from L to m :
5.112
Plan: Use the percent composition information to find the empirical formula of the compound. Assume 100 g of
sample so the percentages are numerically equivalent to the masses of each element. Convert each of the masses
to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and
convert to whole numbers to determine the empirical formula. Rearrange the formula PV = (m/M )RT to solve for
molar mass.The empirical formula mass and the calculated molar mass will then relate the empirical formula to
the molecular formula.
Solution:
Empirical formula:
Assume 100 g of each so the mass percentages are also the grams of the element.
 1 mol Si 

 = 1.17515 mol Si
Moles Si = 33.01 g Si 
 28.09 g Si 


 1 mol F 

Moles F = 66.99 g F 
 = 3.525789 mol F
19.00 g F 


1.17515 mol Si 

 = 1

1.17515 mol Si 
 3.525789 mol F 

 = 3

 1.17515 mol Si 
Empirical formula = SiF3 (empirical formula mass = 85.1 g/mol)
Molecular formula:
V = 0.250 L
T = 27°C + 273.15 = 300.15 K
m = 2.60 g
P = 1.50 atm
M = unknown
PV = (m/M )RT
Solving for molar mass, M :

L  atm 
300.15 K 
2.60 g0.0821

mRT
mol
 K 
Molar mass = M =
= 170.853 g/mol

PV
1.50 atm 0.250 L 
The molar mass (170.853 g/mol ) is twice the empirical formula mass (85.1 g/mol), so the molecular formula must
be twice the empirical formula, or 2 × SiF3 = Si2F6.
5.113
Copyright
a) A preliminary equation for this reaction is 4CxHyNz + nO2  4CO2 + 2N2 + 10H2O.
Since the organic compound does not contain oxygen, the only source of oxygen as a reactant is oxygen gas. To
form 4 volumes of CO2 would require 4 volumes of O2 and to form 10 volumes of H2O would require 5 volumes
of O2. Thus, 9 volumes of O2 was required.
b) Since the volume of a gas is proportional to the number of moles of the gas we can equate volume and moles.
From a volume ratio of 4CO2:2N2:10H2O we deduce a mole ratio of 4C:4N:20H or 1C:1N:5H for an empirical
formula of CH5N.
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5-52
5.114
6
6
a) There is a total of 6 × 10 blue particles and 6 × 10 black particles. When equilibrium is reached after
opening the stopcocks, the particles will be evenly distributed among the three containers. Therefore, container B
6
6
will have 2 × 10 blue particles and 2 × 10 black particles.
6
6
b) The particles are evenly distributed so container A has 2 × 10 blue particles and 2 × 10 black particles.
6
6
6
c) There are 2 × 10 blue particles and 2 × 10 black particles in C for a total of 4 × 10 particles.


750 torr
 = 500 torr
Final pressure in C = 4 106 particles
6
 6 10 particles 
6
6
6
d) There are 2 × 10 blue particles and 2 × 10 black particles in B for a total of 4 × 10 particles.


750 torr
 = 500 torr
Final pressure in B = 4 106 particles
6
 6 10 particles 
5.115
Plan: Write the balanced equation for the combustion of n-hexane. For part (a), assuming a 1.00 L sample of air at
STP, use the molar ratio in the balanced equation to find the volume of n-hexane required to react with the oxygen in
1.00 L of air. Convert the volume n-hexane to volume % and divide by 2 to obtain the LFL. For part (b), use the LFL
calculated in part (a) to find the volume of n-hexane required to produce a flammable mixture and then use the ideal
gas law to find moles of n-hexane. Convert moles of n-hexane to mass and then to volume using the density.
Solution:
a) 2C6H14(l) + 19O2(g)
(g) + 14H2O(g)
2
For a 1.00 L sample of air at STP:
 20.9 L O2  2 L C6 H14 
 = 0.0220 L C6H14
Volume of C6H14 vapor needed = 1.00 L air 

 100 L air 
 19 L O2 
C H volume
0.0220 L C 6 H14
Volume % of C6H14 = 6 14
100 =
100 = 2.2% C6H14
air volume
1.00 L air
LFL = 0.5(2.2%) = 1.1% C6H14
 1 L 1.1% C 6 H14 
b) Volume of C6H14 vapor = 1.000 m 3 air  3 3 
 = 11.0 L C6H14
10 m  100% air 
V = 11.0 L
P = 1 atm
PV = nRT
Solving for n:
T = 0°C + 273 = 273 K
n = unknown
1 atm11.0 L
PV

= 0.490780 mol C6H14
RT 0.0821 L  atm 273 K 


mol  K 
 86.17 g C6 H14 
1 mL

Volume of C6H14 liquid = 0.490780 mol C6 H14 

 1 mol C H 
0.660 g C H
Moles of C6H14 = n =
6
5.116
Copyright
14
6

 = 64.0765 = 64 mL C6H14

14 
Plan: To find the factor by which a diver’s lungs would expand, find the factor by which P changes from 125 ft to
the surface, and apply Boyle’s law. To find that factor, calculate Pseawater at 125 ft by converting the given depth
from ft-seawater to mmHg to atm and adding the surface pressure (1.00 atm).
Solution:
 2 
12 in 
 2.54 cm 10 m  1 mm  = 3.81 × 104 mmH2O
P(H2O) = 125 ft 
 1 in  1 cm 103 m 
 1 ft 
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5-53
P(Hg):
hH2 O
hHg
=
dHg
3.8110 4 mmH 2 O
dH 2 O
hHg
=
13.5 g/mL
1.04 g/mL
hHg = 2935.1111 mmHg
 1 atm 
 = 3.861988 atm (unrounded)
P(Hg) = 2935.11111 mmHg
 760 mm Hg 
Ptotal = (1.00 atm) + (3.861988 atm) = 4.861988 atm (unrounded)
Use Boyle’s law to find the volume change of the diver’s lungs:
P 1V 1 = P 2V 2
V2
P
V2
4.861988 atm
= 1
=
= 4.86
V1
P2
V1
1 atm
To find the depth to which the diver could ascend safely, use the given safe expansion factor (1.5) and
the pressure at 125 ft, P125, to find the safest ascended pressure, Psafe.
P125/Psafe = 1.5
Psafe = P125/1.5 = (4.861988 atm)/1.5 = 3.241325 atm (unrounded)
Convert the pressure in atm to pressure in ft of seawater using the conversion factors above. Subtract this distance
from the initial depth to find how far the diver could ascend.
 760 mmHg 
h(Hg): 4.861988  3.241325 atm 
= 1231.7039 mmHg
 1 atm 
hH2 O
hH2 O
dHg
13.5 g/mL
=
=
hH2 O = 15988.464 mm
1231.7039 mmHg
1.04 g/mL
hHg
dH 2 O
103 m 1.094 yd  3 ft 

 = 52.4741 ft

 1 mm  1 m 1 yd 
Therefore, the diver can safely ascend 52.5 ft to a depth of (125 – 52.4741) = 72.5259 = 73 ft.
15988.464 mmH 2 O
5.117
Plan: Write a balanced equation. Convert mass of CaF2 to moles and use the molar ratio from the balanced
equation to find the moles of gas produced. Use the ideal gas law to find the temperature required to store that
amount of HF gas at the given conditions of temperature and pressure.
Solution:
CaF2(s) + H2SO4(aq)  2HF(g) + CaSO4(s)
 1 mol CaF2 
 2 mol HF 

Moles HF gas = 15.0 g CaF2 
1 mol CaF  = 0.3842213 mol HF
 78.08 g CaF2 
2


Finding the temperature:
V = 8.63 L
P = 875 torr
Converting P from torr to atm:
PV = nRT
Solving for T:

T = unknown
n = 0.3842213 mol
 1 atm 
P = 875 torr 
= 1.151312789 atm
 760 torr 


1.151315789 atm 8.63 L
PV
=
= 314.9783 K

L  atm 
nR
0.3842213 mol HF 0.0821

mol  K 
The gas must be heated to 315 K.
T=
5.118
Copyright
Plan: First, write the balanced equation. According to the description in the problem, a given volume of peroxide
solution (0.100 L) will release a certain number of “volumes of oxygen gas” (20). Assume that 20 is exact.
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5-54
A 0.100 L solution will produce (20 × 0.100 L) = 2.00 L O2 gas. Use the ideal gas law to convert this volume of
O2 gas to moles of O2 gas and convert to moles and then mass of H2O2 using the molar ratio in the balanced
equation.
Solution:
2H2O2(aq)  2H2O(l) + O2(g)
V = 2.00 L
T = 0°C + 273 = 273 K
P = 1 atm
n = unknown
PV = nRT
Solving for n:
1 atm2.00 L
PV
–2

Moles of O2 = n =
= 8.92327 × 10 mol O2

L

atm

RT 0.0821
273 K 

mol  K 
 2 mol H O 
34.02 g H2 O2 
2 2

 = 6.071395 = 6.07 g H2O2
Mass H2O2 = 8.92327102 mol O2 
 1 mol H O 
 1 mol O2 
2 2
5.119
Plan: The moles of gas may be found using the ideal gas law. Multiply moles of gas by Avogadro’s number
to obtain the number of molecules.
Solution:
V = 1 mL = 0.001 L
T = 500 K
–8
P = 10 mmHg
n = unknown
 1 atm 
 = 1.315789 × 10–11 atm
Converting P from mmHg to atm: P = 108 mmHg
 760 mmHg 
PV = nRT
Solving for n:
11
PV 1.315789 10 atm 0.001 L
–16
= 3.2053337 × 10 mol gas

L  atm 

RT
0.0821
500 K 

mol  K 
 6.022 10 23 molecules 

8
8
Molecules = 3.20533371016 mol
 = 1.93025 × 10 = 10 molecules
1 mol


–8
(The 10 mmHg limits the significant figures.)
Moles of gas = n =
5.120
Plan: Use the equation for root mean speed (urms) to find this value for O2 at 0°C. Molar mass values must be
in units of kg/mol and temperature in kelvins. Divide the root mean speed by the mean free path to obtain the
collision frequency.
Solution:
 32.00 g O2  1 kg 
 = 0.03200 kg/mol
a) 0°C = 273 K
M of O2 = 

 mol
103 g 
2
R = 8.314 J/mol K
3RT
urms =
Molar mass
J 

38.314
273 K

mol  K 
0.03200 kg/mol

urms =
b) Collision frequency =
Copyright
2
1 J = kg m /s
  kg  m /s


2
J
2

 = 461.2878 = 461 m/s

urms
461.2878 m/s
9
9 –1
=
= 7.2873 × 10 = 7.29 × 10 s
8
mean free path
6.3310 m
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5-55
5.121
3
Plan: Convert the volume of the tubes from ft to L. Use the ideal gas law to find the moles of gas that will
occupy that volume at the given conditions of pressure and temperature. From the mole fraction of
propylene and the total moles of gas, the moles of propylene can be obtained. Convert moles to mass in
grams and then pounds using the molar mass and scale the amount added in 1.8 s to the amount in 1 h.
Solution:
3
3 
3 

102 m   1 L 

3  12 in   2.54 cm  


Volume of tubes = 100 ft 
 3 3  = 2831.685 L

3 
3 


 1 in 3 

1
cm

 1 ft  

 10 m
V = 2831.685 L
T = 330°C + 273.15 = 603.15 K
P = 2.5 atm
n = unknown
PV = nRT
2.5 atm 2831.685 L 
PV
Moles of gas = n =
=
= 142.9606 mol gas

L  atm 
RT
0.0821
603.15
K



mol  K 
moles of propylene
Xpropylene =
moles of mixture
x moles of propylene
0.07 =
142.996 moles of mixture
Moles of propylene = 10.00724 moles
 42.08 g C3 H6 
 1 lb C3 H6  1 
 3600 s 
Pounds of propylene = 10.00724 moles C3 H6 


 1 h 
 1 mole C H 
453.6 g C H 1.8 s 
3
6
3
6
= 1856.723 = 1900 lb/h C3H6
5.122
Plan: Use the ideal gas law to calculate the molar volume, the volume of exactly one mole of gas, at the
temperature and pressure given in the problem.
Solution:
V = unknown
P = 90 atm
PV = nRT
T = 730. K
n = 1.00 mol
Solving for V:
nRT
V=
=
P
5.123

L  atm 
1.00 mol0.0821
730. K 

mol  K 
= 0.66592 = 0.67 L/mol
90 atm 
Plan: The diagram below describes the two Hg height levels within the barometer. First, find the pressure of the
N2. The PN2 is directly related to the change in column height of Hg. Then find the volume occupied by the N 2.
The volume of the space occupied by the N2(g) is calculated from the length and cross-sectional area of the
barometer. To find the mass of N2, use these values of P and V (T is given) in the ideal gas law to find moles
which is converted to mass using the molar mass of nitrogen.
vacuum
(74.0 cm)
with N2
(64.0 cm)
Solution:
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5-56
102 mHg 
 1 mmHg 
 1 atm  = 0.1315789 atm
Pressure of the nitrogen = 74.0 cmHg  64.0 cmHg

 760 mmHg 

3

 10 mHg 
 1 cmHg 

 1 mL 103 L 
 = 0.0432 L
Volume of the nitrogen = 1.00 10 2 cm  64.0 cm 1.20 cm 2 

1 cm 3  1 mL 
V = 0.0432 L
T = 24°C + 273.15 = 297.15 K
P = 0.1315789 atm
n = unknown
PV = nRT
Solving for n:
0.1315789 atm 0.0432 L 
PV
–4
Moles of N2 = n =
=
= 2.32998 × 10 mol N2

L  atm 
RT
0.0821
297.15 K 

mol  K 
 28.02 g N 
–3
–3
2
Mass of N2 = 2.32998104 mol N2 
 = 6.52859 × 10 = 6.53 × 10 g N2
 1 mol N2 
5.124
5.125
Plan: Use the ideal gas law to find the moles of ammonia gas in 10.0 L at this pressure and temperature.
Molarity is moles per liter. Use the moles of ammonia and the final volume of solution (0.750 L) to get the
molarity.
Solution:
V = 10.0 L
T = 33°C + 273.15 = 306.15 K
P = 735 torr
n = unknown
 1 atm 
Converting P from torr to atm:
P = 735 torr 
= 0.9671053 atm
 760 torr 
PV = nRT
Solving for n:
0.9671053 atm 10.0 L
PV
Moles of ammonia = n =

= 0.384766 mol
RT 0.0821 L  atm 306.15 K 

mol  K 
mol ammonia
0.384766 mol
Molarity = M =
= 0.513021 = 0.513 M
=
liters of solution
0.750 L
Plan: Use the ideal gas law to determine the total moles of gas produced. The total moles multiplied by the
fraction of each gas gives the moles of that gas which may be converted to metric tons.
Solution:
3
3
V = 1.5 × 10 m
T = 298 K
P = 1 atm
n = unknown
 1L 
3
6
Converting V from m to L:
V = 1.5103 m 3  3 3  = 1.5 × 10 L
10 m 
PV = nRT
Solving for n:
Moles of gas/day = n =
1 atm 1.5106 L 
PV
5
= 6.13101 × 10 mol/day

RT 0.0821 L  atm 298 K 

mol  K 
 6.1310110 5 mol 
 365.25 day  = 2.23935 × 107 mol/yr
Moles of gas/yr = 
 1 yr

day



 2.23935107 mol  44.01 g CO2  1 kg 

 3  13 t  = 482.519 = 4.83 × 102 t CO2/yr

Mass CO2 = 0.4896


 10 kg 
yr

 1 mol CO2 10 g 
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5-57
 2.2393510 7 mol  28.01 g CO  1 kg 
 1 t  = 9.15773 = 9.16 t CO/yr


Mass CO = 0.0146

10 3 kg 
3


yr

 1 mol CO 10 g 

 2.23935107 mol 18.02 g H2 O  1 kg 

 3  13 t  = 149.70995 = 1.50 × 102 t H2O/yr

Mass H2O = 0.3710
10 kg 
yr

 1 mol H2 O 10 g 
 2.23935107 mol  64.06 g SO2  1 kg 

 3  13 t  = 169.992 = 1.70 × 102 t SO2/yr

Mass SO2 = 0.1185
10 kg 
yr

 1 mol SO2 10 g 
 2.23935107 mol  64.12 g S2  1 kg 

 3  13 t  = 0.4307614 = 4 × 10–1 t S2/yr

Mass S2 = 0.0003


 10 kg 
yr

 1 mol S2 10 g 
 2.23935107 mol  2.016 g H2  1 kg 

 3  13 t  = 0.21218 = 2.1 × 10–1 t H2/yr

Mass H2 = 0.0047


 10 kg 
yr

 1 mol H2 10 g 
 2.2393510 7 mol  36.46 g HCl  1 kg 
1t 


 3  = 0.6531736 = 6 × 10–1 t HCl/yr
Mass HCl = 0.0008
3

10 kg 

yr
 1 mol HCl 10 g 
 2.23935107 mol  34.08 g H2 S 1 kg 

 3  13 t  = 0.228951 = 2 × 10–1 t H2S/yr

Mass H2S = 0.0003






yr
 1 mol H2 S 10 g 10 kg 
5.126
Plan: Use the molar ratio from the balanced equation to find the moles of H 2 and O2 required to form 28.0 moles
of water. Then use the ideal gas law in part (a) and van der Waals equation in part (b) to find the pressure needed
to provide that number of moles of each gas.
Solution:
a) The balanced chemical equation is: 2H2(g) + O2(g)  2H2O(l)
 2 mol H2 
 = 28.0 mol H2
Moles H2 = 28.0 mol H2 O
 2 mol H O 
2
 1 mol O2 
 = 14.0 mol O2
Moles O2 = 28.0 mol H2 O
 2 mol H2 O 
V = 20.0 L
P = unknown
PV = nRT
Solving for P:
nRT
PIGL of H2 =
=
V
T = 23.8°C + 273.15 =296.95 K
n = 28.0 mol H2; 14.0 mol O2
L  atm 
28.0 mol0.0821 mol
296.95 K 
K
= 34.131 = 34.1 atm H2
20.0 L

L  atm 
14.0 mol 0.0821
296.95 K 

nRT
mol  K 
PIGL of O2 =
=
= 17.0657 = 17.1 atm O2
V
20.0 L
b) V = 20.0 L
T =23.8°C + 273.15 =296.95 K
P = unknown
n = 28.0 mol H2; 14.0 mol O2
Van der Waals constants from Table 5.4:
atm  L2
L
H2: a = 0.244
; b = 0.0266
mol 2
mol
atm  L2
L
O2: a = 1.36
;
b = 0.0318
mol 2
mol

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
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in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
5-58
2 

 P  n a V  nb  nRT

V 2 
PVDW =
nRT
n2 a
 2
V  nb V
atm  L2 
2
L  atm 

28.0 mol 0.244

28.0 mol0.0821
296.95
K


mol 2 


mol  K 

PVDW of H2 =
= 34.9734 = 35.0 atm H2
20.0 L  28.0 mol 0.0266 L/mol
20.0 L2
atm  L2 
2
L  atm 



14.0
mol

1.36



14.0 mol0.0821
296.95 K 
mol 2 


mol  K 

PVDW of O2 =
= 16.7878 = 16.8 atm O2
20.0 L 14.0 mol 0.0318 L/mol
20.0 L2
c) The van der Waals value for hydrogen is slightly higher than the value from the ideal gas law. The van der
Waals value for oxygen is slightly lower than the value from the ideal gas law.
5.127
Plan: Deviations from ideal gas behavior are due to attractive forces between particles which reduce the pressure
of the real gas and due to the size of the particle which affects the volume. Compare the size and/or attractive
forces between the substances. The greater the size and/or the stronger the attractive forces, the greater the
deviation from ideal gas behavior.
Solution:
a) Xenon would show greater deviation from ideal behavior than argon since xenon is a larger atom than argon.
The electron cloud of Xe is more easily distorted so intermolecular attractions are greater. Xe’s larger size also
means that the volume the gas occupies becomes a greater proportion of the container’s volume at high pressures.
b) Water vapor would show greater deviation from ideal behavior than neon gas since the attractive forces
between water molecules are greater than the attractive forces between neon atoms. We know the attractive forces
are greater for water molecules because it remains a liquid at a higher temperature than neon (water is a liquid at
room temperature while neon is a gas at room temperature).
c) Mercury vapor would show greater deviation from ideal behavior than radon gas since the attractive forces
between mercury atoms is greater than that between radon atoms. We know that the attractive forces for mercury
are greater because it is a liquid at room temperature while radon is a gas.
d) Water is a liquid at room temperature; methane is a gas at room temperature. Therefore, water molecules have
stronger attractive forces than methane molecules and should deviate from ideal behavior to a greater extent than
methane molecules.
5.128
Plan: Use the molarity and volume of the solution to find the moles of HBr needed to make the solution.
Then use the ideal gas law to find the volume of that number of moles of HBr gas at the given conditions.
Solution:
1.20 mol HBr 
Moles of HBr in the hydrobromic acid: 
3.50 L  = 4.20 mol HBr


L
V = unknown
P = 0.965 atm
PV = nRT
Solving for V:
5.129
Copyright
T = 29°C + 273.15 =302.15 K
n = 4.20 mol

L  atm 
4.20 mol0.0821
302.15 K 

nRT
mol
 K 
V=
= 107.9662 = 108 L HBr
=
P
0.965 atm 
Plan: Convert the mass of each gas to moles using the molar mass. Calculate the mole fraction of CO. Use the
relationship between partial pressure and mole fraction (PA = XA × Ptotal) to calculate the partial pressure of CO.
Solution:
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5-59
 1 mol CO 
 = 0.24991 mol CO
Moles CO = 7.0 g CO
 28.01 g CO 
 1 mol SO2 
 = 0.156104 mol SO2
Moles SO2 = 10.0 g SO2 
 64.06 g SO2 
mol CO
0.24991 mol CO
=
= 0.615521
mol CO + mol SO 2
0.24991 mol CO + 0.156104 mol SO 2
PCO = XCO × Ptotal = 0.615521 × 0.33 atm = 0.20312 = 0.20 atm CO
XCO =
5.130
Plan: First, balance the equation. The pressure of N2 is found by subtracting the pressure of O2 from the total
pressure. The pressure of the remaining 15% of O2 is found by taking 15% of the original O2 pressure. The
molar ratio between O2 and SO2 in the balanced equation is used to find the pressure of the SO2 that is produced.
Since pressure is directly proportional to moles of gas, the molar ratio may be expressed as a pressure ratio.
Solution:
4FeS2(s) + 11O2(g)  8SO2(g) + 2Fe2O3(s)
Pressure of N2 = 1.05 atm – 0.64 atm O2 = 0.41 atm N2
Pressure of unreacted O2 = (0.64 atm O2) [(100 – 85)%/100%] = 0.096 atm O2
 8 atm SO2 
 = 0.46545 = 0.47 atm SO2
Pressure of SO2 produced = 0.64 atm O2 
 11 atm O2 
Total Pressure = (0.41 atm) + (0.096 atm) + (0.46545 atm) = 0.97145 = 0.97 atm total
5.131
Plan: V and T are not given, so the ideal gas law cannot be used. The total pressure of the mixture is given.
Use PA = XA × Ptotal to find the mole fraction of each gas and then the mass fraction. The total mass of the two
gases is 35.0 g.
Solution:
Ptotal = Pkrypton + Pcarbon dioxide = 0.708 atm
The NaOH absorbed the CO2 leaving the Kr, thus Pkrypton = 0.250 atm.
Pcarbon dioxide = Ptotal – Pkrypton = 0.708 atm – 0.250 atm = 0.458 atm
Determining mole fractions: PA = XA × Ptotal
PCO2
0.458 atm
Carbon dioxide: X =
=
= 0.64689
Ptotal
0.708 atm
P
0.250 atm
Krypton: X = Kr =
= 0.353107
Ptotal
0.708 atm

 83.80 g Kr  
 0.353107

 mol  

Relative mass fraction = 
 = 1.039366
 0.64689 44.01 g CO2 



mol

35.0 g = x g CO2 + (1.039366 x) g Kr
35.0 g = 2.039366 x
Grams CO2 = x = (35.0 g)/(2.039366) = 17.16219581 = 17.2 g CO2
Grams Kr = 35.0 g – 17.162 g CO2 = 17.83780419 = 17.8 g Kr
5.132
As the car accelerates, the air in the car is pressed toward the back. The helium balloon floats on “top” of this
denser air, which pushes it toward the front of the car.
5.133
Plan: Convert molecules of OH to moles and use the ideal gas law to find the pressure of this
3
number of moles of OH in 1 m of air. The mole percent is the same as the pressure percent as long as the other
conditions are the same.
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5-60
Solution:
3
3

1 mol
 2.51012 molecules 
10 m 
–15


Moles of OH = 


 = 4.151445 × 10 mol/L
3
23




m
 6.022 10 molecules 
 1 L 
V = 1.00 L
P = unknown
PV = nRT
Solving for P:
T = 22°C + 273.15 =295.15 K
–15
n = 4.151445 × 10 mol

L  atm 
4.1514451015 mol 0.0821
295.15 K 

nRT
mol  K 
Pressure of OH = P =
=
V
1.00 L
–13
–13
=1.005970 × 10 = 1.0 × 10 atm OH
1.005970 1013 atm
–11
–11
Mole percent =
100 = 1.005459 × 10 = 1.0 × 10 %
1.00 atm

5.134

Plan: Write the balanced equations. Use the ideal gas law to find the moles of SO2 gas and then use the molar
ratio between SO2 and NaOH to find moles and then molarity of the NaOH solution.
Solution:
The balanced chemical equations are:
SO2(g) + H2O(l)  H2SO3(aq)
H2SO3(aq) + 2NaOH(aq)  Na2SO3(aq) + 2H2O(l)
Combining these equations gives:
SO2(g) + 2NaOH(aq)  Na2SO3(aq) + H2O(l)
V = 0.200 L
P = 745 mmHg
T = 19°C + 273.15 =292.15 K
n = unknown
Converting P from mmHg to atm:
 1 atm 
 = 0.980263 atm
P = 745 mmHg
 760 mmHg 
PV = nRT
Solving for n:
0.980263 atm0.200 L
PV
–3

= 8.17379 × 10 mol SO2
RT 0.0821 L  atm 292.15 K 

mol  K 
 2 mol NaOH 
 = 0.0163476 mol NaOH
Moles of NaOH = 8.17379103 mol SO2 
 1 mol SO2 
mol NaOH
0.0163476 mol NaOH  1 mL 
 3  = 1.63476 = 1.63 M NaOH
M NaOH =
=
10 L 
volume of NaOH
10.0 mL
Moles of SO2 = n =
5.135
Plan: Rearrange the formula PV = (m/M )RT to solve for molar mass. The mass of the flask and water and the density of
water are used to find the volume of the flask and thus the gas. The mass of the condensed liquid equals the mass of the
gas in the flask. Pressure must be in units of atmospheres, volume in liters, and temperature in kelvins.
Solution:
Mass of water = mass of flask filled with water – mass of flask = 327.4 g – 65.347 g = 262.053 g
3


1 mL
10 L  = 0.2628415 L
Volume of flask = volume of water = 262.053 g water 

 0.997 g water  1 mL 
Mass of condensed liquid = mass of flask and condensed liquid – mass of flask = 65.739 g – 65.347 g = 0.392 g
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5-61
V = 0.2628415 L
P = 101.2 kPa
M = unknown
Converting P from kPa to atm:
T = 99.8°C + 273.15 = 372.95 K
m = 0.392 g
 1 atm


 = 0.998766 atm
P = 101.2 kPa 
101.325 kPa 


PV = (m/M )RT
Solving for molar mass, M :

L  atm 
0.392 g0.0821
372.95 K 

mRT
mol  K 
M =
= 45.72166 = 45.7 g/mol

PV
0.998766 atm 0.2628415 L 
5.136
Plan: Write the balanced chemical equations. Convert mass of hydride to moles and use the molar ratio from the
balanced equation to find the moles of hydrogen gas produced. Use the ideal gas law to find the volume of that
amount of hydrogen. Pressure must be in units of atm and temperature in kelvins.
Solution:
LiH(s) + H2O(l)  LiOH(aq) + H2(g)
MgH2(s) + 2H2O(l)  Mg(OH)2(s) + 2H2(g)
LiOH is water soluble; however, Mg(OH)2 is not water soluble.
Lithium hydride LiH:
 1 kg 103 g 
 1 mol LiH  1 mol H 2  = 57.0746 mol H
Moles H2 = 1.00 lb LiH 

2
 7.946 g LiH 1 mol LiH 

 2.205 lb  1 kg 

Finding the volume of H2:
V = unknown
P = 750 torr
Converting P from torr to atm:
T = 27°C + 273.15 = 300.15 K
n = 57.0746 mol
 1 atm 
P = 750. torr 
= 0.98684 atm
 760 torr 
PV = nRT
Solving for V:

L  atm 
57.0746 mol0.0821
300.15 K 

nRT
mol  K 
V=
= 1425.2029 = 1430 L H2 from LiH
=
P
0.98684 atm 
Magnesium hydride, MgH2
 1 kg 103 g  1 mol MgH2 
 2 mol H2  = 34.44852 mol H

Moles H2 = 1.00 lb MgH2 

2





 2.205 lb  1 kg  26.33 g MgH 
1 mol MgH 
2
Finding the volume of H2:
V = unknown
P = 750 Torr = 0.98684 atm
PV = nRT
Solving for V:
nRT
V=
=
P
Copyright
2
T = 27°C + 273.15 = 300.15 K
n = 34.44852 mol

L  atm 
34.44852 mol0.0821
300.15 K 

2
mol  K 
= 860.20983= 8.60 × 10 L H2 from MgH2
0.98684 atm 
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5-62
5.137
5.138
Plan: Use the equation for root mean speed (urms). Molar mass values must be in units of kg/mol and temperature
in kelvins.
Solution:







urms Ne =
3 8.314 J /mol  K 370 K 10 3 g  kg  m 2 /s2 
 = 676.24788 = 676 m/s Ne


 kg 
J

20.18 g/mol
urms Ar =
3 8.314 J /mol  K 370 K 10 3 g  kg  m 2 /s2 
 = 480.6269 = 481 m/s Ar


 kg 
J

39.95 g/mol
urms He =
3 8.314 J /mol  K 370 K 103 g  kg  m 2 /s2 
3


 = 1518.356 = 1.52 × 10 m/s He




kg
J



4.003 g/mol










Plan: Use the ideal gas law to find the number of moles of CO2 and H2O in part (a). The molar mass is then
used to convert moles to mass. Temperature must be in units of kelvins, pressure in atm, and volume in L. For part
(b), use the molar ratio in the balanced equation to find the moles and then mass of C6H12O6 that produces the
number of moles of CO2 exhaled during 8 h.
Solution:
a) V = 300 L
T = 37.0°C + 273.15 = 310.15 K
P = 30.0 torr
n = unknown
 1 atm 
Converting P from torr to atm:
P = 30.0 torr 
= 0.0394737 atm
 760 torr 
PV = nRT
Solving for n:
0.0394737 atm300 L
PV

= 0.4650653 mol
RT 0.0821 L  atm 310.15 K 

mol  K 
 44.01 g CO2 
 = 20.4675 = 20.5 g CO2
Mass (g) of CO2 = 0.4650653 mol CO2 
 1 mol CO2 
Moles of CO2 = moles of H2O = n =
18.02 g H2 O 
 = 8.3805 = 8.38 g H2O
Mass (g) of H2O = 0.4650653 mol H2 O
 1 mol H2 O 
b) C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(g)
 0.4650653 mol CO 2 
Moles of CO2 exhaled in 8 h = 
8 h  = 3.720523 mol CO2


h
1 mol C6 H12 O6 180.16 g C6 H12 O6 


Mass (g) of C6H12O6 = 3.720523 mol CO2 
 1 mol C6 H12 O6 
 6 mol CO2 
2
= 111.7149 = 1 × 10 g C6H12O6 (= body mass lost)
(This assumes the significant figures are limited by the 8 h.)
5.139
Copyright
Plan: For part (a), the number of moles of water vapor can be found using the ideal gas law. Convert moles of
water to mass using the molar mass and adjust to the 1.6% water content of the kernel. For part (b), use the ideal
gas law to find the volume of water vapor at the stated set of condition.
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5-63
Solution:
 75% H 2 O 
 = 0.1875 mL = 1.875 × 10–4 L
a) Volume of water in kernel = 0.25 mL kernel
100% kernel 
–4
V = 1.875 × 10 L
P = 9.0 atm
PV = nRT
Solving for n:
T = 170°C + 273.15 = 443.15 K
n = unknown
9.0 atm 1.875 104 L
PV
–5
= 4.638204 × 10 mol

L

atm


RT 0.0821
 443.15 K 

mol  K 
18.02 g H O 100% 

2 
 = 0.052238 = 0.052 g
Mass (g) of = 4.638204 105 mol H 2 O 

 1 mol H 2 O  1.6% 
b) V = unknown
T = 25°C + 273.15 = 298.15 K
–5
P = 1.00 atm
n = 4.639775 × 10 mol
PV = nRT
Solving for V:

L  atm 
298.15 K 
4.638204 104 mol0.0821
nRT
mol
 K 
V=
= 0.00113534 L = 1.13534 mL = 1.1 mL
=
P
1.00 atm
Moles of H2O = n =

5.140

Plan: For part (a), find the SO2 volume in 4 GL of flue gas and take 95% of that as the volume of SO 2 removed.
The ideal gas law is used to find the number of moles of SO2 in that volume. The molar ratio in the balanced
equation is used to find the moles and then mass of CaSO4 produced. For part (b), use the molar ratio in the
balanced equation to calculate the moles of O2 needed to produce the amount of CaSO4 found in part (a). Use the
ideal gas law to obtain the volume of that amount of O2 and adjust for the mole fraction of O2 in air.
Solution:
–10
–7
a) Mole fraction SO2 = 1000(2 × 10 ) = 2 × 10
9
10 L 

2 107  95%  = 760 L
Volume of SO2 removed = 4 GL

 1 GL 
100% 
V = 760 L
P = 1.00 atm
PV = nRT
Solving for n:
T = 25°C + 273.15 = 298.15 K
n = unknown
1.00 atm760 L
PV

= 31.04814mol

RT 0.0821 L  atm 298.15 K 


mol  K 
The balanced chemical equations are:
CaCO3(s) + SO2(g)  CaSO3(s) + CO2(g)
2CaSO3(s) + O2(g)  2CaSO4(s)
1 mol CaSO4 
136.14 g CaSO 4 
 1 kg 

Mass (kg) of CaSO4 = 31.04814 mol SO2 

 3  = 4.2269 = 4 kg CaSO4
 1 mol SO2 
 1 mol CaSO4 
10 g 
b) Moles of O2 =
103 g  1 mol CaSO4 
 1 mol O2 

4.2269 kg CaSO4 

 = 15.52407 mol O2
 1 kg 136.14 g CaSO4 
 2 mol CaSO4 
V = unknown
T = 25°C + 273.15 = 298.15 K
Moles of SO2 = n =

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5-64
P = 1.00 atm
PV = nRT
n = 15.52407 mol
Solving for V:

L  atm 
15.52407 mol0.0821
298.15 K 

mol  K 
= 380 L O2
1.00 atm 
 1 
Volume of air = 380 L O 2 
 = 1818.1818 = 2000 L air
 0.209 
nRT
Volume of O2 = V =
=
P
5.141
Plan: Use the ideal gas law to find the moles of gas occupying the tank at 85% of the 85.0 atm ranking. Then use
van der Waals equation to find the pressure of this number of moles of gas.
Solution:
a) V = 850. L
T = 298 K
 80% 
P = 85.0 atm 
n = unknown
 = 68.0 atm
100% 
PV = nRT
Solving for n:
Moles of Cl2 = n =
68.0 atm 850. L 
PV
3
3

= 2.36248 × 10 = 2.36 × 10 mol Cl2
RT 0.0821 L  atm 298 K 


mol  K 
b) V = 850 L
P = unknown
Van der Waals constants from Table 5.4:
a = 6.49
atm  L2
;
mol 2
b = 0.0562
T =298 K
3
n = 2.36248 × 10 mol Cl2
L
mol
2 

 P  n a V  nb  nRT
V 2 

PVDW =
nRT
n2 a
 2
V  nb V

L  atm 
mol Cl 2 0.08206
 298 K

mol  K 

L 
850 L  2.3624810 3 mol Cl 2 0.0562


mol 
2.3624810
PVDW =
3





2.3624810

3
2
atm  L2 

mol Cl 2 6.49
mol 2 


850 L
2
= 30.4134 = 30.4 atm
c) The engineer did not completely fill the tank. She should have filled it to (80.0%/100%)(85.0 atm) = 68 atm,
but only filled it to 30.4 atm.
5.142
Plan: Solve the density form of the ideal gas law for molar mass. Temperature must be converted to kelvins and
pressure to atmospheres.
Solution:
P = 102.5 kPa
T = 10.0°C + 273.15 = 283.15 K
d = 1.26 g/L
M = unknown
 1 atm


 = 1.011596 atm
Converting P from kPa to atm:
P = 102.5 kPa 
101.325 kPa 

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
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5-65
d = (PM )/(RT)
Rearranging to solve for molar mass:
L  atm 

1.26 g/L 0.0821
283.15 K 

dRT
mol  K 
=
M =
= 28.9550 = 29.0 g/mol
P
1.011596 atm



5.143
Plan: Use the relationship

PV
PV
PV n T
1 1
= 2 2 or V2 = 1 1 2 2 . R is fixed.
n1T1
n2 T2
P2 n1T1
Solution:
a) As the pressure on a fixed amount of gas (n is fixed) increases at constant temperature (T is fixed), the
molecules move closer together, decreasing the volume. When the pressure is increased by a factor of 2, the
volume decreases by a factor of 2 at constant temperature (Boyle’s law).
PV T
(P )(V )(1)
V2 = 1 1 2 = 1 1
V2 = V1
P2T1
(2 P1 )(1)
Cylinder B has half the volume of the original cylinder.
b) The temperature is decreased by a factor of 2, so the volume is decreased by a factor of 2 (Charles’s law).
PV T
(1)(V1 )(200 K)
V2 = 1 1 2 =
V2 = V1
P2T1
(1)(400 K)
Cylinder B has half the volume of the original cylinder.
c) T1 = 100°C + 273.15 = 373.15 K
T2 = 200°C + 273.15 = 473.15 K
The temperature increases by a factor of 473/373 = 1.27, so the volume is increased by a factor of 1.27
(Charles’s law).
PV T
(1)(V1 )(473.15 K)
V2 = 1 1 2 =
V2 = 1.27V1
P2T1
(1)(373.15 K)
None of the cylinders show a volume increase of 1.27.
d) As the number of molecules of gas increases at constant pressure and temperature (P and T are fixed), the force
they exert on the container increases. This results in an increase in the volume of the container. Adding 0.1 mole
of gas to 0.1 mole increases the number of moles by a factor of 2, thus the volume increases by a factor of 2
(Avogadro’s law).
PV n T
(1)(V1 )(0.2)(1)
V2 = 1 1 2 2 =
V2 = 2V1
P2 n1T1
(1)(0.1)(1)
Cylinder C has a volume that is twice as great as the original cylinder.
e) Adding 0.1 mole of gas to 0.1 mole increases the number of moles by a factor of 2, thus increasing the volume
by a factor of 2. Increasing the pressure by a factor of 2 results in the volume decreasing by a factor of . The two
volume changes cancel out so that the volume does not change.
PV n T
(P )(V )(0.2)(1)
V2 = 1 1 2 2 = 1 1
V2 = V1
P2 n1T1
(2 P1 )(0.1)(1)
Cylinder D has the same volume as the original cylinder.
5.144
Copyright
Plan: Use both the ideal gas law and van der Waals equation to solve for pressure. Convert mass of ammonia to
moles and temperature to kelvin.
Solution:
Ideal gas law:
V = 3.000 L
T = 0°C + 273.15 =273.15 K or 400°C + 273.15 =673.15 K
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5-66
P = unknown
 1 mol NH3 
 = 3.0005872 mol
n = 51.1 g NH3 
17.03 g NH3 
PV = nRT
Solving for P:
nRT
PIGL of NH3 at 0°C =
=
V
L  atm 
3.0005872 mol0.0821 mol
273.15 K 
K
3.000 L
= 22.43000 = 22.4 atm

L  atm 
3.0005872 mol 0.0821
673.15 K 

nRT
mol
 K 
PIGL of NH3 at 400°C =
=
= 55.27643 = 55.3 atm
V
3.000 L
van der Waals equation:
V = 3.000 L
T = 0°C + 273.15 =273.15 K or 400°C + 273.15 =673.15 K
 1 mol NH3 
 = 3.0005872 mol
P = unknown
n = 51.1 g NH3 
17.03 g NH 


3
van der Waals constants from Table 5.4:
atm  L2
L
; b = 0.0371
mol 2
mol
2 

 P  n a V  nb  nRT
V 2 

a = 4.17
PVDW =
nRT
n2 a
 2
V  nb V
atm  L2 
2
L  atm 

3.0005872 mol 4.17

273.15
K
3.0005872 mol0.0821


mol 2 


mol  K 

PVDW of NH3 at 0°C =
3.000 L  3.0005872 mol 0.0371 L/mol
3.000 L2
= 19.10964 = 19.1 atm NH3
atm  L2 
2
L  atm 


3.0005872 mol0.0821
673.15 K  3.0005872 mol 4.17

mol 2 


mol  K 

PVDW of NH3 at 400°C =
3.000 L  3.0005872 mol 0.0371 L/mol
3.000 L2
= 53.2350 = 53.2 atm NH3
5.145
Copyright
Plan: Since the mole fractions of the three gases must add to 1, the mole fraction of methane is found by
subtracting the sum of the mole fractions of helium and argon from 1. Pmethane = Xmethane Ptotal is used to calculate
the pressure of methane and then the ideal gas law is used to find moles of gas. Avogadro’s number is needed to
convert moles of methane to molecules of methane.
Solution:
Xmethane = 1.00 – (Xargon + Xhelium) = 1.00 – (0.35 + 0.25) = 0.40
Pmethane = Xmethane Ptotal = (0.40)(1.75 atm) = 0.70 atm CH4
V = 6.0 L
T = 45°C + 273.15 =318.15 K
P = 0.70 atm
n = unknown
PV = nRT
Solving for n:
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5-67
0.70 atm6.0 L
PV

= 0.1607956 mol

RT 0.0821 L  atm 318.15 K 


mol  K 
 6.0221023 CH molecules 
4

Molecules of CH4 = 0.1607956 mol CH 4 
1 mol CH4


Moles of CH4 = n =
22
22
= 9.68311 × 10 = 9.7 × 10 molecules CH4
5.146
Plan: For part (a), convert mass of glucose to moles and use the molar ratio from the balanced equation to find the
moles of CO2 gas produced. Use the ideal gas law to find the volume of that amount of CO2. Pressure must be in
units of atm and temperature in kelvins. For part (b), use the molar ratios in the balanced equation to calculate the
moles of each gas and then use Dalton’s law of partial pressures to determine the pressure of each gas.
Solution:
a) C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g)
 1 mol C6 H12 O6 
6 mol CO2 

 = 0.666075 mol CO2
Moles CO2: 20.0 g C6 H12 O6 
1 mol C H O 
180.16 g C H O 
6
Finding the volume of CO2:
V = unknown
P = 780. torr
Converting P from torr to atm:
12
6
6
12
6
T = 37°C + 273.15 = 310.15 K
n = 0.666075 mol
 1 atm 
P = 780 torr 
 = 1.0263158 atm
 760 torr 


PV = nRT
Solving for V:
nRT
V=
=
P

L  atm 
0.666075 mol 0.0821
310.15 K 

mol  K 
= 16.5256 = 16.5 L CO2
1.0263158 atm 
This solution assumes that partial pressure of O2 does not interfere with the reaction conditions.
 1 mol C6 H12 O6 

6 mol


b) Moles CO2 = moles O2 = 10.0 g C6 H12 O6 

1 mol C H O 
180.16 g C H O 
6
12
6
6
12
6
= 0.333037 mol CO2 = mol O2
At 37°C, the vapor pressure of water is 48.8 torr. No matter how much water is produced, the partial pressure of
H2O will still be 48.8 torr. The remaining pressure, 780 torr – 48.8 torr = 731.2 torr is the sum of partial pressures
for O2 and CO2. Since the mole fractions of O2 and CO2 are equal, their pressures must be equal, and must be onehalf of 731.2 torr.
Pwater = 48.8 torr
2
(731.2 torr)/2 = 365.6 = 3.7 × 10 torr Poxygen = Pcarbon dioxide
5.147
Plan: Use the equations for average kinetic energy and root mean speed (urms) to find these values for N2 and H2.
Molar mass values must be in units of kg/mol and temperature in kelvins.
Solution:
The average kinetic energies are the same for any gases at the same temperature:
3
3
Average kinetic energy = 3/2RT = (3/2)(8.314 J/mol K) (273 K) = 3.40458 × 10 = 3.40 × 10 J for both gases
rms speed:
T = 0°C + 273 = 273 K
Copyright
 28.02 g N 2  1 kg 
 = 0.02802 kg/mol
M of N2 = 


103 g 
mol
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5-68
 2.016 g H 2  1 kg 
 = 0.002016 kg/mol
M of H2 = 


103 g 
mol
R = 8.314 J/mol K
urmsN2 =
urms H2 =
5.148

2
2
1 J = kg m /s


3 8.314 J/mol  K 273 K  kg  m 2 /s2 
2
2

 = 4.92961 × 10 = 4.93 × 10 m/s N2


J


0.02802 kg/mol





3 8.314 J /mol  K 273K  kg  m 2 /s2 
 = 1.83781 × 103 = 1.84 × 103 m/s H

2


J


0.002016 kg/mol


Plan: Use the relationship between mole fraction and partial pressure, PA = XA Ptotal, to find the mole fraction of
each gas in parts (a) and (b). For parts (c) and (d), use the ideal gas law to find the moles of air in 1000 L of air at
these conditions and compare the moles of each gas to the moles of air. Mass and molecules must be converted to
moles.
Solution:
a) Assuming the total pressure is 1 atm = 760 torr.
PA = XA Ptotal
PBr2
0.2 torr
–4
6
XBr2 =
=
= 2.6315789 × 10 × (10 ) = 263.15789 = 300 ppmv Unsafe
Ptotal
760 torr
PCO2
0.2 torr
–4
6
=
b) XCO2 =
= 2.6315789 × 10 × (10 ) = 263.15789 = 300 ppmv Safe
Ptotal
760 torr
6
(0.2 torr CO2/760 torr)(10 ) = 263.15789 = 300 ppmv CO2 Safe
 1 mol Br2 
 = 2.5031 × 10–6 mol Br2 (unrounded)
c) Moles Br2 = 0.0004 g Br2 
159.80 g Br2 
Finding the moles of air:
V = 1000 L
T = 0°C + 273 =273 K
P = 1.00 atm
n = unknown
PV = nRT
1.00 atm1000 L
PV

Moles of air = n =
= 44.616 mol air (unrounded)
RT 0.0821 L  atm 273 K 

mol  K 
6
–6
6
Concentration of Br2 = mol Br2/mol air(10 ) = [(2.5031 × 10 mol)/(44.616 mol)] (10 )
= 0.056103 = 0.06 ppmv Br2 Safe


1 mol CO2
 = 0.046496 mol CO2
d) Moles CO2 = 2.81022 molecules CO2 
23
 6.02210 molecules CO2 
6
6
Concentration of CO2 = mol CO2/mol air(10 ) = [(0.046496 mol)/(44.616 mol)] (10 ) = 1042.1
3
= 1.0 × 10 ppmv CO2 Safe
5.149
Copyright
Plan: For part (a), use the ideal gas law to find the moles of NO in the flue gas. The moles of NO are converted to
moles of NH3 using the molar ratio in the balanced equation and the moles of NH3 are converted to volume using
the ideal gas law. For part (b), the moles of NO in 1 kL of flue gas is found using the ideal gas law; the molar ratio
in the balanced equation is used to convert moles of NO to moles and then mass of NH 3.
Solution:
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5-69
a) 4NH3(g) + 4NO(g) + O2(g)  4N2(g) + 6H2O(g)
Finding the moles of NO in 1.00 L of flue gas:
V = 1.00 L
T = 365°C + 273.15 =638.15 K
–5
P = 4.5 × 10 atm
n = unknown
PV = nRT
Solving for n:
4.5105 atm1.00 L
PV
–7
Moles of NO = n =
= 8.5891 × 10 mol NO

RT 0.0821 L  atm 638.15 K 


mol  K 
 4 mol NH 

3
–7
Moles of NH3 = 8.5891107 mol NO
 = 8.5891 × 10 mol NH3
 4 mol NO 
Volume of NH3:
V = unknown
T =365°C + 273.15 =638.15 K
–7
P = 1.00 atm
n = 8.5891 × 10 mol
PV = nRT
Solving for V:

L  atm 
8.5891107 mol0.0821
638.15 K 
nRT
–5
–5
mol  K 
V=
= 4.50001 × 10 = 4.5 × 10 L NH3
=
P
1.00 atm 
b) Finding the moles of NO in 1.00 kL of flue gas:
V = 1.00 kL = 1000 L
T = 365°C + 273.15 =638.15 K
–5
P = 4.5 × 10 atm
n = unknown
PV = nRT
Solving for n:
5
PV 4.5000110 atm 1000 L 
–4
Moles of NO = n =
= 8.5891 × 10 mol NO

L  atm 

RT
0.0821
638.15 K 

mol  K 
 4 mol NH 

3
–4
Moles of NH3 = 8.5891104 mol NO
 = 8.5891 × 10 mol NH3
 4 mol NO 
17.03 g NH 
3
Mass of NH3 = 8.5891104 mol NH3 
 = 0.014627 = 0.015 g NH3
 1 mol NH3 
5.150
Plan: Use Graham’s law to compare effusion rates.
Solution:
molar mass Xe
131.3 g/mol
2.55077
Rate Ne
=
=
=
enrichment factor (unrounded)
molar mass Ne
20.18 g/mol
1
Rate Xe
Thus XNe =
5.151
moles of Ne
2.55077 mol
= 0.71837 = 0.7184
=
moles of Ne + moles of Xe
2.55077 mol + 1 mol
Plan: To find the number of steps through the membrane, calculate the molar masses to find the ratio of effusion
rates. This ratio is the enrichment factor for each step.
Solution:
Rate 235 UF
352.04 g/mol
molar mass 238 UF6
6
=
=
235
Rate 238 UF
molar mass UF6
349.03 g/mol
6
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5-70
= 1.004302694 enrichment factor
235
Therefore, the abundance of UF6 after one membrane is 0.72% × 1.004302694
235
N
Abundance of UF6 after “N” membranes = 0.72% × (1.004302694)
235
N
Desired abundance of UF6 = 3.0% = 0.72% × (1.004302694)
Solving for N:
N
3.0% = 0.72% × (1.004302694)
N
4.16667 = (1.004302694)
N
ln 4.16667 = ln (1.004302694)
ln 4.16667 = N × ln (1.004302694)
N = (ln 4.16667)/(ln 1.004302694)
N = 1.4271164/0.004293464 = 332.39277 = 332 steps
5.152
Plan: Use van der Waals equation to calculate the pressure of the gas at the given conditions.
Solution:
V = 22.414 L
T = 273.15 K
P = unknown
n = 1.000 mol
Van der Waals constants from Table 5.4:
atm  L2
L
a = 1.45
; b = 0.0395
mol 2
mol
2 

 P  n a V  nb  nRT

V 2 
PVDW =
PVDW =
nRT
n2 a
 2
V  nb V
L  atm 
1.000 mol CO0.08206 mol
273.15 K 
K
L 

22.414 L  1.000 mol CO 0.0395

mol 




L 

1.000 mol CO 1.45 atm
mol 
2
2
2

22.414 L
2
= 0.998909977 = 0.9989 atm
5.153
Plan: The amount of each gas that leaks from the balloon is proportional to its effusion rate. Using 35% as the rate
for H2, the rate for O2 can be determined from Graham’s law.
Solution:
molar mass H 2
2.016 g/mol
rate O2
Rate O 2
=
=
=
molar mass O2
32.00 g/mol
Rate H 2
35
0.250998008 =
rate O2
35
Rate O2 = 8.78493
Amount of H2 that leaks = 35%; 100–35 = 65% H2 remains
Amount of O2 that leaks = 8.78493%; 100–8.78493 = 91.21507% O2 remains
O2
91.21507
=
= 1.40331 = 1.4
H2
65
5.154
Copyright
Plan: For part (a), put together the various combinations of the two isotopes of Cl with P and add the masses.
Multiply the abundances of the isotopes in each combination to find the most abundant for part (b). For part (c),
use Graham’s law to find the effusion rates.
Solution:
a) Options for PCl3:
All values are g/mol
P
First Cl
Second Cl
Third Cl
Total
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5-71
31
35
35
35
136
31
37
35
35
138
31
37
37
35
140
31
37
37
37
142
35
37
b) The fraction abundances are Cl = 75%/100% = 0.75, and Cl = 25%/100% = 0.25.
The relative amount of each mass comes from the product of the relative abundances of each Cl isotope.
Mass 136 = (0.75) (0.75) (0.75) = 0.421875 = 0.42 (most abundant)
Mass 138 = (0.25) (0.75) (0.75) = 0.140625 = 0.14
Mass 140 = (0.25) (0.25) (0.75) = 0.046875 = 0.047
Mass 142 = (0.25) (0.25) (0.25) = 0.015625 = 0.016
c)
Rate P 37 Cl3
=
Rate P 35 Cl3
molar mass P 35 Cl 3
37
molar mass P Cl3
=
136 g/mol
142 g/mol
= 0.978645 = 0.979
5.155
Plan: Use the combined gas law for parts (a) and (b). For part (c), use the ideal gas law to find the moless of air
represented by the pressure decrease and convert moles to mass using molar mass.
Solution:
PV
PV
a) 1 1 = 2 2
T1
T2
35.0 psi318 K 
P1 V1T2
PT
= 1 2 =
= 37.7288 = 37.7 psi
V2 T1
T1
295 K
b) New tire volume = V2 = V1(102%/100%) = 1.02V1
PV
PV
1 1
= 2 2
T1
T2
P2 =
P2 =
35.0 psiV1 318 K 
PV
PT
1 1T2
= 36.9890 = 37.0 psi
= 1 2 =
V2T1
T1
1.02V1 295 K 
 1 atm 
 = 0.135306 atm
c) Pressure decrease = 36.9890 – 35.0 psi = 1.989 psi; 1.989 psi
14.7 psi 
V = 218 L
T = 295 K
P = 0.135306 atm
n = unknown
PV = nRT
Solving for n:
0.135306 atm 218 L
PV
Moles of air lost = n =

= 1.217891 mol

RT 0.0821 L  atm 295 K 


mol  K 
 28.8 g 1 min 

Time = 1.217891 mol
= 23.384 = 23 min
 1 mol  1.5 
5.156
Copyright
Plan: Rearrange the ideal gas law to calculate the density of O2 and O3 from their molar masses. Temperature must
be converted to kelvins and pressure to atm.
Solution:
P = 760 torr = 1.00 atm
T = 0°C + 273 = 273 K
M of O2 = 32.00 g/mol
d = unknown
M of O3 = 48.00 g/mol
PV = nRT
Rearranging to solve for density:
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5-72
1.00 atm 32.00 g/mol
PM

= 1.42772 = 1.43 g O2/Ls

L  atm 
RT
0.0821

273
K


mol  K 
1.00 atm 48.00 g/mol
PM
d of O3 =

= 2.141585576 = 2.14 g O3/L

L  atm 
RT
0.0821
273 K 


mol  K 
b) dozone/doxygen = (2.141585576)/(1.42772) = 1.5
The ratio of the density is the same as the ratio of the molar masses.
d of O2 =
5.157
Plan: Since the amounts of two reactants are given, this is a limiting reactant problem. Write the balanced
equation and use molar ratios to find the number of moles of IF7 produced by each reactant. The mass of I2 is
converted to moles using its molar mass and the moles of F2 is found using the ideal gas law. The smaller number
of moles of product indicates the limiting reagent. Determine the moles of excess reactant gas and the moles of
product gas and use the ideal gas law to solve for the total pressure.
Solution:
Moles of F2:
V = 2.50 L
T = 250 K
P = 350 torr
n = unknown
 1 atm 
Converting P from torr to atm:
P = 350 torr 
 = 0.460526315 atm
 760 torr 
PV = nRT
Solving for n:
PV 0.460526315 atm 2.50 L 
n=

= 0.056093339 mol F2

L  atm 
RT
0.0821
250. K 


mol  K 
7F2(g) + I2(s)  2IF7(g)
 2 mol IF7 

 = 0.016026668 mol IF (unrounded)
Moles IF7 from F2 = 0.056093339 mol F2 
7
 7 mol F2 
 1 mol I2 
 2 mol IF7  = 0.019700551 mol IF (unrounded)
Moles IF7 from I2 = 2.50 g I2 
7


1 mol I2 
 253.8 g I 2 
F2 is limiting. All of the F2 is consumed.
Mole I2 remaining = original amount of moles of I2 – number of I2 moles reacting with F2
 1 mol I 2 
 1 mol I2 
 = 1.83694 × 10–3 mol I
 – 0.056093339 mol F2 
Mole I2 remaining = 2.50 g I2 
2
 253.8 g I2 
 7 mol F2 
–3
Total moles of gas = (0 mol F2) + (0.016026668 mol IF7) + (1.83694 × 10 mol I2)
= 0.0178636 mol gas
V = 2.50 L
T = 550. K
P = unknown
n = 0.0178636 mol
PV = nRT
Solving for P:

L  atm 
0.0178636 mol0.0821
550. K 

nRT
mol  K 
P (atm) =
=
= 0.322652 atm
V
2.50 L
 760 torr 
P (torr) = 0.322652 atm 
= 245.21552 = 245 torr
 1 atm 
–3
Piodine (torr) = Xiodine Ptotal = [(1.83694 × 10 mol I2)/(0.0178636 mol)] (245.21552 torr) = 25.215869 = 25.2 torr
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5-73