# Silberberg7e solution manual ch 05

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
conversion factors to convert pressure in mmHg to units of torr, pascals and lb/in2.
Solution:
Because Pgas < Patm, Pgas = Patm – Δh
Pgas = 753.6 mm Hg – 174.0 mm Hg = 579.6 mm Hg
 1 torr 
Pressure (torr) =  579.6 mm Hg  
 = 579.6 torr
 1 mm Hg 

  1.01325 x 105 Pa 
1 atm
Pressure (Pa) =  579.6 mm Hg  

 
1atm
 760 mm Hg  

= 7.727364 x 104 = 7.727 x 104 Pa
 1 atm
  14.7 lb/in 2 
2
Pressure (lb/in2) =  579.6 mm Hg  
 = 11.21068 = 11.2 lb/in
 
760
mm
Hg
1
atm



5.1B

Plan: Convert the atmospheric pressure to torr. Use the equation for gas pressure in an open-end manometer to
calculate the pressure of the gas. Use conversion factors to convert pressure in torr to units of mmHg, pascals and
lb/in2.
Solution:
Because Pgas > Patm, Pgas = Patm + Δh
Pgas = (0.9475 atm)

760 torr
1 atm

+ 25.8 torr = 745.9 torr
1 mmHg

Pressure (mmHg) = (745.9 torr)
Pressure (Pa) = (745.9 mmHg)

1 torr
1 atm

1.01325 x 105 Pa

760 mmHg
1 atm

Pressure (lb/in2) = (745.9 mmHg)
5.2A

= 745.9 mmHg

760 mmHg

1 atm
14.7 lb/in2
1 atm

= 9.945 x 104 Pa

= 14.4 lb/in2

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
P2 (atm) =

PV
1 1 = (0.871 atm)(105 mL) = 0.260 atm
(352 mL)
V2

P2 (kPa) = (0.260 atm)

101.325 kPa
1 atm

= 26.3 kPa

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5-1

5.2B

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
P
V
1 1
= 2 2
At fixed n and T:
n1T1
n2T2

PV
1 1 = P2V2
1 atm

P2 (atm) = (745 mmHg)
V2 (atm) =

5.3A

PV
1 1 =
P2

760 mmHg
(122 atm)(651 L)
(0.980 atm)

= 0.980 atm
= 8.10 x 104 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 = 23oC
T2 = 100oC
P1 = 0.991 atm
P2 = unknown
n and V remain constant
Converting T1 from oC to K: 23oC + 273.15 = 296 K
Converting T2 from oC to K: 100oC + 273.15 = 373 K
P1 (torr) = (0.991 atm)

760 torr
1 atm

= 753 torr

Arranging the ideal gas law and solving for P2:
P1 V1
n1 T1

=

P2 V2
n2 T2

P2 (torr) = P1

or

P1
T1

=

P2
T2

= (753 torr)

373 K
296 K

= 949 torr

Because the pressure in the tank (949 torr) is less than the pressure at which the safety valve will open (1.00 x 103
torr), the safety valve will not open.
5.3B

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 = 313K
Arranging the ideal gas law and solving for T2:
PV
PV
V1
V
1 1
= 2 2 or
 2
n1T1
n2T2
T1
T2

T2  T1

28.6 L
V2
= (313 K)
= 275 K – 273.15 = 2°C
32.5 L
V1

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5-2

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
P
V
1 1
= 2 2
At fixed V and T:
n1T1
n2T2

P1
P
= 2
n1
n2
 1 mol C2 H 4 
C2 H 4  

28.05 g C2 H 4 
Pn

1 2
=  793 torr 
= 679.714 = 680. torr
P2 =
 1 mol C2 H 4 
n1
 35.0 g C2 H4  

 28.05 g C2 H 4 

 30.0 g

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
Converting m1 (mass) to n1 (moles): (1.26 g N2)

1 mol N2
28.02 g N2

= 0.0450 mol N2 = n1

Converting m2 (mass) to n2 (moles): 0.0450 mol N2 + (1.26 g He)

1 mol He
4.003 g He

= 0.0450 mol N2 + 0.315 mol He = 0.360 mol gas = n2
Arranging the ideal gas law and solving for V2:
P1 V1
n1 T1

=

V2 = V1
5.5A

P2 V2
n2 T2
n2
n1

or

V1
n1

= (1.12 L)

=

V2
n2

0.360 mol
0.0450 mol

= 8.96 L

Plan: Convert the temperatures to kelvin. Examine the ideal gas law, noting the fixed variables and those
PV
PV
variables that change. R is always constant so 1 1 = 2 2 . In this problem, P, V, and T are changing, while n
n1T1
n2T2
remains fixed.
Solution:
T1 = 23oC
T2 = 18oC
P1 = 755 mmHg
P2 = unknown
V1 = 2.55 L
V2 = 4.10 L
n remains constant
Converting T1 from oC to K: 23oC + 273.15 = 296 K
Converting T2 from oC to K: 18oC + 273.15 = 291 K
Arranging the ideal gas law and solving for P2:

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5-3

P1 V1
n1 T1

=

P2 V2
n2 T2

or

P2 (mmHg) = P1
5.5B

T1

V1 T2
V2 T1

=

P2 V2
T2

= (755 mmHg)

n1 T1

=

P2 V2
n2 T2

V2 (L) = V1

or

P1 T2
P2 T1

P1 V1
T1

=

(4.10 L)(296 K)

= 462 mmHg

= (2.2 L)

P2 V2
T2
(0.980 atm )(294 K)

= 1.5 L

(1.40 atm)(301 K)

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
PV
1.37 atm  438 L 
=
= 24.9 mol O2
n=
RT
 0.0821 atm • L 

   273.15  21 K 
mol • K

Mass (g) of O2 = (24.9 mol O2)

5.6B

(2.55 L)(291 K)

Plan: Convert the temperatures to kelvin. Examine the ideal gas law, noting the fixed variables and those
PV
PV
variables that change. R is always constant so 1 1 = 2 2 . In this problem, P, V, and T are changing, while n
n1T1
n2T2
remains fixed.
Solution:
T1 = 28oC
T2 = 21oC
P1 = 0.980 atm
P2 = 1.40 atm
V1 = 2.2 L
V2 = unknown
n remains constant
Converting T1 from oC to K: 28oC + 273.15 = 301 K
Converting T2 from oC to K: 21oC + 273.15 = 294 K
Arranging the ideal gas law and solving for V2:
P1 V1

5.6A

P1 V1

32.00 g O2
1 mol O2

= 796.8 = 797 g O2

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 = 20oC
Converting m (mass) to n (moles): (3950 kg He)

1000 g

1 mol He

1 kg

4.003 g He

= 9.87 x 105 mol = n

Converting T from oC to K: 20oC + 273.15 = 293 K
Converting P from mmHg to atm: (731 mmHg)

1 atm
760 mmHg

= 0.962 atm

PV = nRT
V=

nRT
P

atm • L

(9.87 x 105 mol) 0.0821 mol • K (293 K)
=
= 2.47 x 107 L
(0.962 atm)

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5-4

5.7A

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 V2
=
T1
T2
T2 =

V2T1  2V1  200.15 K 
=
= 400.30 K – 273.15 = 127.15 = 127°C
V1
V1 

5.7B

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 oC to K: 199oC + 273.15 = 472 K
Converting T2 from oC to K: –155oC + 273.15 = 118 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) x 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.
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
. Two calculations are
RT
required, one with T = 0°C = 273 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  = 0.981783 = 0.982 g/L
d=

 0.0821 atm • L 
 273 K   760 torr 

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

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5-5

d=

 44.01 g/mol 1 atm 

= 1.9638566 = 1.96 g/L
 0.0821 atm • L 

  273 K 
mol • K

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 K (24oC + 273.15) and P = 0.950 atm:
d=

(0.950 atm)(46.01 g/mol)
0.0821

atm • L
mol • K

(297 K)

= 1.7926 = 1.79 g/L

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.
PM
dRT
mRT
is rearranged to give M =
or M =
The relationship d 
PV
P
RT
Solution:
Mass (g) of gas = mass of flask + vapor – mass of flask = 68.697 – 68.322 = 0.375 g
T = 95.0°C + 273 = 368 K
 1 atm 
P =  740. torr  
 = 0.973684 atm
 760 torr 
V = 149 mL = 0.149 L
0.0821 atm • L 
 0.375 g  
  368 K 
mRT
mol • K

=
= 78.094 = 78.1 g
M=
PV
 0.973684 atm  0.149 L 

5.9B

Plan: Calculate the mass of the gas by subtracting the mass of the empty glass bulb from the mass of the bulb
PM
containing the gas. The volume, pressure, and temperature of the gas are known. The relationship d 
is
RT
dRT
mRT
or M =
. Use the molar mass of the gas to determine its identity.
rearranged to give M =
PV
P
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 K
P = (733 mmHg)

1 atm
760 mmHg

= 0.965 atm

V = 350. mL = 0.350 L
atm • L

mRT (0.225 g) 0.0821 mol • K (295 K)
=
= 16.1 g/mol
(0.965 atm )(0.350 L)
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.

M=

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 
Moles of He =  5.50 g He  
 = 1.373970 mol He
 4.003 g He 

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5-6

 1 mol Ne 
Moles of Ne = 15.0 g Ne  
 = 0.743310 mol Ne
 20.18 g Ne 

 1 mol Kr 
Moles of Kr =  35.0 g Kr  
 = 0.417661 mol Ke
 83.80 g Kr 
Total number of moles of gas = 1.373970 + 0.743310 + 0.417661 = 2.534941 mol
PA = XA x 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 x Ptotal to calculate the mole fraction of He. Multiply the mole fraction by 100% to
calculate the mole percent of He.
Solution:
PHe = XHe x Ptotal
Mole percent He = XHe (100%) =

PHe
Ptotal

(100%) =

143 atm
204 atm

(100%) = 70.1%

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
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   103 L 
 738.4 torr 1495 mL 
PV
Moles of hydrogen = n =
=


 
RT
 0.0821 atm • L 
 760 torr   1 mL 

273.15
16
K

mol • K

= 0.061186 mol H2
 2.016 g H 2 
Mass (g) of hydrogen =  0.061186 mol H 2  
 = 0.123351 = 0.123 g H2
 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
(730. torr)(307 mL)
1 atm
1L
PV
=
Moles of oxygen = n =
atm • L
1000 mL
0.0821
(293 K) 760 torr
RT
mol • K
= 0.012258 mol O2
Mass (g) of oxygen = (0.012258 mol O2)

32.00 g O2
1 mol O2

= 0.3923 = 0.392 g O2

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5-7

5.12A

Plan: Write a balanced equation for the reaction. Calculate the moles of HCl(g) from the starting amount of
sodium chloride using the stoichiometric ratio from the balanced equation. Find the volume of the HCl(g) from
the molar volume at STP.
Solution:
The balanced equation is H2SO4(aq) + 2NaCl(aq)  Na2SO4(aq) + 2HCl(g).
 103 g   1 mol NaCl   2 mol HCl 
= 2.00205 mol HCl
Moles of HCl =  0.117 kg NaCl  
 1 kg   58.44 g NaCl   2 mol NaCl 




 22.4 L  1 mL 
Volume (mL) of HCl =  2.00205 mol HCl  


 1 mol HCl  103 L 
= 4.4846 x 104 = 4.48 x 104 mL HCl
5.12B

Plan: Write a balanced equation for the reaction. Use the ideal gas law to calculate the moles of CO2(g) scrubbed.
Use the molar ratios from the balanced equation to calculate the moles of lithium hydroxide needed to scrub that
amount of CO2. Finally, use the molar mass of lithium hydroxide to calculate the mass of lithium hydroxide
required.
Solution:
The balanced equation is 2LiOH(s) + CO2(g)  Li2CO3(s) + H2O(l).
Amount (mol) of CO2 scrubbed = n =
Mass (g) of LiOH = 8.33 mol CO2

5.13A

(0.942 atm)(215 L)
PV
=
= 8.3340 = 8.33 mol CO2
atm • L
0.0821
(296 K)
RT
mol • K

2 mol LiOH

23.95 g LiOH

1 mol CO2

1 mol LiOH

= 399.0070 = 399 g LiOH

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.
PV
 0.452 atm 10.0 L 
=
= 0.18653 mol NH3
Moles of ammonia =
RT
 0.0821 atm • L 

   273.15  22  K 
mol • K

 7.50 atm 155 mL   103 L 
PV
=

 = 0.052249 mol HCl
RT
 0.0821 atm • L 
 1 mL 
271
K

mol • K

The HCl is limiting so the moles of ammonia gas left after the reaction would be
0.18653 – 0.052249 = 0.134281 mol NH3.
0.0821 atm • L 
 0.134281 mol  
   273.15  22  K 
nRT
mol • K

Pressure (atm) of ammonia =
=
V
10.0 L 
= 0.325387 = 0.325 atm NH3
Moles of hydrogen chloride =

5.13B

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).

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5-8

Amount (mol) of F2 that reacts = n =

(0.974 atm)(2.48 L)
PV
=
= 0.1011 = 0.101 mol F2
atm • L
0.0821
(291 K)
RT
mol • K

Amount (mol) of IF5 produced from F2 = 0.101 mol F2
Amount (mol) of IF5 produced from I2 = 4.16 g I2

2 mol IF5
5 mol F2

= 0.0404 mol IF5

1 mol I2

2 mol IF5

253.8 g I2

1 mol I2

= 0.0328 mol IF5

Because a smaller number of moles is produced from the I2, I2 is limiting and 0.0328 mol of IF5 are produced.
Volume (L) of IF5 =
5.14A

nRT
P

atm • L

=

(0.0328 mol) 0.0821 mol • K (378 K)
= 1.08867 = 1.09 L
(0.935 atm)

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:
MC2 H6
Rate He

Rate C 2 H 6
MHe

 0.010 mol He 

 1.25 min 

 0.010 mol C H 
2 6

t
C
H
2 6

0.800 t = 2.74078
t = 3.42597 = 3.43 min
5.14B

 30.07 g/mol 
 4.003 g/mol 

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
Rate ofAr
Rate ofunknown gas

Munknown gas

=

Rateof Ar
Rate ofunknown gas

Munknown gas = (MAr)

MAr
2

Munknown gas

=

MAr
Rate ofAr

2

Rate ofunknown gas

Munknown gas = (39.95 g/mol)

13.8 mL/time 2
7.23 mL/time

= 146 g/mol

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5-9

B5.1

Plan: Examine the change in density of the atmosphere as altitude changes.
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 x Ptotal. The mole fraction of argon is given in Table B5.1.
Solution:
Volume percent = mole fraction x 100 = 0.00934 x 100 = 0.934 %
The total pressure at sea level is 1.00 atm = 760 torr.
PAr = XAr x Ptotal = 0.00934 x 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 = 5.14x1015 t 



 1 t   1 kg   28.8 g 
= 1.78472x1020 = 1.78x1020 mol
b) PV = nRT
L•atm 

1.78472 x1020 mol  0.0821
  273  25 K 
nRT
mol•K 

V=
=
= 4.36646x1021 = 4 x 1021 L
P
1 atm 

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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5-10

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:
hH 2 O
d Hg

hHg
d H2O
hH 2 O 

5.7

d H 2O

 10 3 m   1 cm 
 13.5 g/mL 
x hHg = 
  2  = 985.5 = 990 cm H2O
  730 mmHg  
 1.00 g/mL 
 1 mm   10 m 

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:
hH 2 O
d Hg

hHg
d H2O
hH 2 O 

5.8

d Hg

d Hg
d H 2O

 13.5 g/mL 
4
x hHg = 
  755 mmHg  = 10,192.5 = 1.02x10 mm H2O
1.00
g/mL

Plan: Use the conversion factors between pressure units:
1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar
Solution:
 760 mmHg 
a) Converting from atm to mmHg: P(mmHg) =  0.745 atm  
 = 566.2 = 566 mmHg
 1 atm

 1.01325 bar 
b) Converting from torr to bar: P(bar) =  992 torr  
 = 1.32256 = 1.32 bar
 760 torr 

1 atm
c) Converting from kPa to atm: P(atm) =  365 kPa  
 = 3.60227 = 3.60 atm
 101.325 kPa 

 101.325 kPa 
d) Converting from mmHg to kPa: P(kPa) =  804 mmHg  
 = 107.191 = 107 kPa
 760 mmHg 

5.9

Plan: Use the conversion factors between pressure units:
1 atm = 760 mmHg = 760 torr = 101.325 kPa = 1.01325 bar
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 

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5-11

 101.325 kPa 
3
3
b) Converting from atm to kPa: P(kPa) =  27.5 atm  
 = 2.786x10 = 2.79x10 kPa
1
atm

 1.01325 bar 
c) Converting from atm to bar: P(bar) =  6.50 atm  
 = 6.5861 = 6.59 bar
1 atm

 760 torr 
d) Converting from kPa to torr: P(torr) =  0.937 kPa  
 = 7.02808 = 7.03 torr
 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
pressure corresponding to the height difference (Δh) 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 
P(atm) =  715.0 torr  
 = 0.940789 = 0.9408 atm
 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 corresponding to the height difference (Δh) 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 
P(kPa) =  752.2 torr  
 = 100.285 = 100.3 kPa
 760 torr 

5.12

Plan: This is a closed-end manometer. The difference in the height of the Hg (Δh) 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 



5.13

Plan: This is a closed-end manometer. The difference in the height of the Hg (Δh) 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   1.01325x105 Pa 
P(Pa) =  3.56 cm  
= 4746.276 = 4.75x103 Pa
 1 cmHg   10 3 m   760 mmHg 




5.14

Plan: Use the conversion factors between pressure units:
1 atm = 760 mmHg = 760 torr = 1.01325x105 Pa = 14.7 psi

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5-12

Solution:
 1 atm 
a) Converting from mmHg to atm: P(atm) = 2.75x102 mmHg 
 = 0.361842 = 0.362 atm
 760 mmHg 

 1 atm 
b) Converting from psi to atm: P(atm) =  86 psi  
 = 5.85034 = 5.9 atm
 14.7 psi 
1 atm

c) Converting from Pa to atm: P(atm) = 9.15x106 Pa 
 = 90.303 = 90.3 atm
5
 1.01325x10 Pa 

 1 atm 
d) Converting from torr to atm: P(atm) = 2.54x104 torr 
 = 33.42105 = 33.4 atm
 760 torr 

5.15

Plan: 1 atm = 1.01325x105 Pa = 1.01325x105 N/m2. So the force on 1 m2 of ocean is 1.01325x105 N where
kg•m
1 N = 1 2 . Use F = mg to find the mass of the atmosphere in kg/m2 for part a). For part b), convert this mass
s
to g/cm2 and use the density of osmium to find the height of this mass of osmium.
Solution:
a) F = mg
1.01325x105 N = mg
kg•m
1.01325 x 105 2 = (mass) (9.81 m/s2)
s
mass = 1.03287x104 = 1.03x104 kg
2

kg   103 g   102 m 

3
2
b)  1.03287x104 2  
 
 = 1.03287x10 g/cm (unrounded)
1
kg
1
cm
m




g

Height =  1.03287x103
cm 2

3
  1 mL   1 cm 

  22.6 g   1 mL  = 45.702 = 45.7 cm Os



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

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 x 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

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5-13

Solution:

PV
PV
V1
V
1 1
= 2 2 or
= 2
n1T1
n2T2
n1T1
n2T2
T can double as V doubles only if n is fixed.
PV
PV
or P1V1 = P2V2
b) T and n are both fixed and V doubles: 1 1 = 2 2
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 V2
1 1
= 2 2 or
=
n1T1
n2T2
T1
T2
V and T are directly proportional so as V is doubled, T is doubled.
a) P is fixed; both V and T double:

5.20

Plan: Use the relationship

PV
PV
PV n T
1 1
= 2 2 or V2 = 1 1 2 2 .
P2 n1T1
n1T1
n2T2

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 = ⅓V1
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
PV n T
(1)(V1 )(4n1 )(1)
V2 = 1 1 2 2 =
V2 = 4V1
P2 n1T1
(1)(n1 )(1)
5.21

Plan: Use the relationship

PV
PV
PV T
1 1
= 2 2 or V2 = 1 1 2 . R and n are fixed.
P2T1
T1
T2

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 ½. As the temperature of a fixed amount of gas (n is fixed) decreases by a
factor of ½ (from 310 K to 155 K) at constant pressure, the volume decreases by a factor of ½. 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)(V1 )(155 K)
V2 = 1 1 2 =
V2 = 1 4 V1
P2T1
(202 kPa)(310 K)
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5-14

b) As the pressure on a fixed amount of gas (n is fixed) decreases at constant temperature (T is fixed), the
molecules move farther together, 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 T
(2 atm)(V1 )(305 K)
V2 = 1 1 2 =
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 together, 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/ 4P1 )(1)
5.22

Plan: Use the relationship

PV
PV
PV T
1 1
= 2 2 or V2 = 1 1 2 . R and n are fixed.
P2T1
T1
T2

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 = ½ V1
P2T1
(1)(800 K)
T2 = 500°C + 273 = 773 K
b) T1 = 250°C + 273 = 523 K
The temperature increases by a factor of 773/523 = 1.48, so the volume is increased by a factor of 1.48
PV T
(1)(V1 )(773 K)
V2 = 1 1 2 =
V2 = 1.48V1
(Charles’s law).
P2T1
(1)(523 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 = ⅓V1
P2T1
(6 atm)(1)
5.23

Plan: Use the relationship

PV
PV
PV n T
1 1
= 2 2 or V2 = 1 1 2 2 .
P2 n1T1
n1T1
n2T2

Solution:
 1 atm 
a) P1 =  722 torr  
 = 0.950 atm
 760 torr 

T1 =

5
[T
9

(in °F) – 32] =

5
9

[32°F – 32] = 0°C

T1 = 0°C + 273 = 273 K

Both P and T are fixed: P1 = P2 = 0.950 atm; T1 = T2 = 273 K, so the volume remains constant.
PV n T
(1)(V1 )(1)(1)
V2 = 1 1 2 2 =
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
1

(1)(V1 )( n1 )(1)
PV
2
1 1n2T2
V2 = ½V1
=
(1)( n1 )(1)
P2 n1T1
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.
(P )(V )(1)( 1 4 T1 )
PV n T
V2 = 1 1 2 2 = 1 1
V2 =V1
P2 n1T1
( 1 4 P1 )(1)(T1 )

V2 =

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5-15

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)

1 atm
760 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

0.844 atm

= 1.84 L

= (725 mmHg)

10.0 L
7.50 L

= 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
T2 = unknown
T1 = 198°C (convert to K)
n and P remain constant
Converting T from °C to K: T1 = 198°C + 273 = 471K
Arranging the ideal gas law and solving for T2:
PV
PV
V1
V
1 1
= 2 2 or
 2
n1T1
n2T2
T1
T2

T2  T1
5.27

0.966 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 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

= (1.61 L)

V2
= 471 K
V1

 2.50 L 
 9.10 L  = 129.396 K – 273 = –143.604 = –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 = 418 K
T2 = –22°C + 273 = 251 K

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5-16

Arranging the ideal gas law and solving for V2:
PV
PV
V1
V
1 1
= 2 2 or
 2
n1T1
n2T2
T1
T2

V2  V1
5.28

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 oC
T2 = 195 oC
P2 = unknown
P1 = 177 atm
n and V remain constant
Converting T1 from oC to K: 25 oC + 273.15 = 298 K
Converting T2 from oC to K: 195 oC + 273.15 = 468 K
Arranging the ideal gas law and solving for P2:
P1 V1
n1 T1

=

P2 = P1
5.29

T2
 251 K 
= 93 L 
 = 55.844 = 56 L
T1
 418 K 

P2 V2
n2 T2

or

P1
T1

= (177 atm)

=

P2

T2
468 K
298 K

= 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 oC
T2 = unknown
P1 = 110. psi
P2 = 105 psi
n and V remain constant
Converting T1 from oC to K: 30.0 oC + 273.15 = 303.2 K
Arranging the ideal gas law and solving for T2:
P1 V1
n1 T1

=

T2 = T1

P2 V2
n2 T2

or

P1
T1

= (303.2 K)

=

P2
T2

105 psi
110. psi

= 289 K

Converting T2 from K to oC: 289 K - 273.15 = 16 oC
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
V2 = unknown
V1 = 12.5 L
P and T remain constant
Converting m1 (mass) to n1 (moles): (1.92 g He)
Converting m2 (mass) to n2 (moles): (1.07 g He)

1 mol He
4.003 g He
1 mol He
4.003 g He

= 0.480 mol He = n1
= 0.267 mol He = n2

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5-17

Arranging the ideal gas law and solving for V2:
P1 V1
n1 T1

=

V2 = V1
5.31

n2 T2

or

n1 T1

=

n2 = n1

n1

= (12.5 L)

P2 V2
n2 T2

or

=

V2
n2

0.267 mol He
0.480 mol He

= 6.95 L

V1
n1

=

V2
n2

= (1 x 1022 molecules of air)

350 mL
500 mL

= 7 x 1021 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
V2 = unknown
V1 = 25.5 L
T1 = 298 K
T2 = 273 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
T1
T2
n 2T2
 T  P 
V2 = V1  2   1 
 T1   P2 

5.33

V1

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:
n1 = 1 x 1022 molecules of air*
n2 = unknown
V2 = 350 mL
V1 = 500 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

5.32

P2 V2

 273 K   153.3 kPa 
=  25.5 L  

 = 35.3437 = 35.3 L
 298 K   101.325 kPa 

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
T2 = –14°C + 273 = 259 K
T1 = 298 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
T1
T2
n 2T2
 T  P 
 259 K   745 torr 
V2 = V1  2   1  = 3.65 L  

 = 6.4397 = 6.44 L
T
P
 298 K   367 torr 
 1  2 

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5-18

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 = 310 K
PV = nRT
Solving for n:
PV
(0.43158 atm)(5.0 L)
= 0.08479 = 0.085 mol chlorine
n=

L•atm 
RT 
0.0821
(310
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
n = 1.47 x 10–3 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:
T = 26°C + 273 = 299 K
PV = nRT
Solving for P:
L•atm 

1.47x103 mol  0.0821
 299 K 
nRT
mol•K 

=
= 0.48114 atm
P=
V
0.0750 L
 760 torr 
Convert P to units of torr:  0.48114 atm  
 = 365.6664 = 366 torr
 1 atm 

5.36

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 
Converting V from mL to L:
V =  357 mL  
= 0.357 L
 1 mL 

Converting T from °C to K:
T = 45°C + 273 = 318 K
 1 atm 
Converting P from mmHg to atm: P =  699 mmHg  
 = 0.91974 atm
 760 mmHg 
PV = nRT
Solving for n:
PV  0.91974 atm  0.357 L 
n=

= 0.01258 mol ClF3
L•atm 
RT 
0.0821
318
K

mol•K 

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5-19

 92.45 g ClF3 
Mass ClF3 =  0.01258 mol ClF3  
 = 1.163021 = 1.16 g ClF3
 1 mol ClF3 

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 = 388 K
 1 mol N 2 O 
n =  75.0 g N 2 O  
Converting from mass of N2O to moles:
 = 1.70377 mol N2O
 44.02 g N 2 O 
PV = nRT
Solving for P:
nRT
P=

V

5.38

1.70377 mol   0.0821

L•atm 
 388 K 
mol•K 

 3.1 L 

= 17.5075 = 18 atm N2O

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 = 296 K
 1 atm 
Converting P from psi to atm:
P =  99.7 psi  
 = 6.7823 atm
 14.7 psi 
PV = nRT
Solving for n:
 6.7823 atm 1.5 L 
PV
n=

= 0.41863 = 0.42 mol SO2
L•atm 
RT 
0.0821
296
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
T2 = –5°C + 273 = 268 K
T1 = 25°C + 273 = 298 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
T1
T2
n 2T2

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5-20

 T  P 
 268 K  0.98026 atm 
V2 = V1  2   1  =  65 L  

 = 868.219 = 870 L
 298 K  0.066 atm 
 T1   P2 
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.

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 D2.
Solution:
n
nB
4 A particles
3 B particles
a) X A = A =
= 0.25
XB =
=
= 0.1875
ntotal
ntotal
16 total particles
16 total particles

nD2
nC
4 D2 particles
5 C particles
=
= 0.3125
=
= = 0.25
X D2 =
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.
PD2 = 0.25 x 0.75 atm = 0.1875 = 0.19 atm
c) PD2 = X D2 x Ptotal
XC =

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:
1 atm 131.3 g/mol 
PM
d=

= 5.8581 = 5.86 g/L
L•atm 
RT

0.0821
273
K

mol•K 

5.46

Plan: Rearrange the ideal gas law to calculate the density of CFCl3 from its molar mass. Temperature must
be converted to kelvins.

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5-21

Solution:
P = 1.5 atm
T = 120°C + 273 = 393 K
M of CFCl3 = 137.4 g/mol
d = unknown
PV = nRT
Rearranging to solve for density:
1.5 atm 137.4 g/mol 
PM
d=

= 6.385807663 = 6.4 g/L
L•atm 
RT

0.0821
393
K

mol•K 

5.47

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 = 0°C + 273 = 273 K
P = 1 atm
n = unknown
M of AsH3 = 77.94 g/mol
PV = nRT
Solving for n:
PV
1 atm  0.0400 L 

n=
= 1.78465x10–3 = 1.78x10–3 mol AsH3
L•atm 
RT

 0.0821 mol•K   273 K 

Converting moles of AsH3 to mass of AsH3:
 77.94 g AsH 3 
Mass (g) of AsH3 = 1.78465x10 3 mol AsH 3 
 = 0.1391 g AsH3
 1 mol AsH 3 

d=
5.48

 0.1391 g  = 3.4775 = 3.48 g/L
mass
=
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 = 0°C + 273 = 273 K
M = unknown
d = 2.71 g/L
PM
d=
RT
Rearranging to solve for molar mass:
L•atm 
 2.71 g/L   0.0821
 273 K 
dRT
mol•K 

= 20.24668 = 20.2 g/mol
M =
=
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 = 0.206 μL
T = 45°C + 273 = 318 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 

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5-22

 106 L 
–7
 = 2.06x10 L
 1 L 

Converting V from μL to L:

V =  0.206 L  

Converting m from ng to g:

 109 g 
m =  206 ng  
= 2.06x10–7 g
 1 ng 

 m 
PV = 
 RT
M 
Solving for molar mass, M:

L•atm 
 2.06x10 g   0.0821 mol•K
  318 K 

 0.510526 atm   2.06x10 L 
7

mRT

M =
PV
5.50

7

= 51.1390 = 51.1 g/mol

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 = 295 K
P = 747 mm Hg
m = 0.103 g
M = unknown
 1 atm 
Converting P from mmHg to atm: P =  747 mmHg  
 = 0.982895 atm
 760 mmHg 
Converting V from mL to L:

 103 L 
V =  63.8 mL  
= 0.0638 L
 1 mL 

 m 
PV = 
 RT
M 
Solving for molar mass, M:
L•atm 

0.103 g   0.0821

 295 K 
mRT
mol•K 

= 39.7809 = 39.8 g/mol

M =
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 O2 ) 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 = 500. K
P = 1.20 atm
n = unknown
PV = nRT
Solving for n:
1.20 atm  0.600 L 
PV
n=

= 0.017539586 mol Ar
L•atm 
RT 
 0.0821 mol•K   500. K 

For O2:
V = 0.200 L
T = 127°C + 273 = 400. K
P = 501 torr
n = unknown

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5-23

Converting P from torr to atm:

 1 atm 
P =  501 torr  
 = 0.6592105 atm
 760 torr 

PV = nRT
Solving for n:
PV  0.6592105 atm  0.200 L 
n=

= 0.004014680 mol O2
L•atm 
RT

0.0821
400.
K

mol•K 

ntotal = nAr + n O2 = 0.017539586 mol + 0.004014680 mol = 0.021554266 mol
For the mixture of Ar and O2:
V = 400 mL
T = 27°C + 273 = 300. K
P = unknownn
n = 0.021554265 mol
 10 3 L 
Converting V from mL to L:
V =  400 mL  
= 0.400 L
 1 mL 

PV = nRT
Solving for P:
L•atm 

0.021554266 mol   0.0821

 300 K 
nRT
mol•K 

Pmixture =
= 1.32720 = 1.33 atm

V
 0.400 L 
5.52

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 = 308 K
P = 626 mmHg
ntotal = unknown
 1 atm 
Converting P from mmHg to atm: P =  626 mmHg  
 = 0.823684 atm
 760 mmHg 
Converting V from mL to L:

 103 L 
V =  355 mL  
= 0.355 L
 1 mL 

PV = nRT
Solving for ntotal:
PV  0.823684 atm  0.355 L 
ntotal =

= 0.011563655 mol Ne + mol Ar
L•atm 
RT 
0.0821
308
K

mol•K 

 1 mol Ne 
Moles Ne =  0.146 g Ne  
 = 0.007234886 mol Ne
 20.18 g Ne 
Moles Ar = ntotal – nNe = (0.011563655 – 0.007234886) mol = 0.004328769 = 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 = 0°C + 273 = 273 K
P = 1 atm
n = unknown
PV = nRT

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution
in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

5-24

Solving for n:
1 atm  35.5 L 
PV
n=

= 1.583881 mol O2
L•atm 
RT 
0.0821
273
K

mol•K 

P4(s) + 5O2(g)  P4O10(s)
 1 mol P4   123.88 g P4 
Mass P4 = 1.583881 mol O 2  

 = 39.24224 = 39.2 g P4
 5 mol O 2   1 mol P4 
5.54

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.
Solution:
V = 638 mL
T = 128°C + 273 = 401 K
P = 752 torr
n = unknown
 1 atm 
Converting P from torr to atm:
P =  752 torr  
 = 0.9894737 atm
 760 torr 
Converting V from mL to L:

 103 L 
V =  638 mL  
= 0.638 L
 1 mL 

PV = nRT
Solving for n:
PV  0.9894737 atm  638 L 
n=

= 0.0191751 mol O2
L•atm 
RT 
 0.0821 mol•K   401 K 

2KClO3(s)  2KCl(s) + 3O2(g)
 2 mol KClO3   122.55 g KClO3 
Mass (g) of KClO3 =  0.0191751 mol O 2  
 = 1.5666 = 1.57 g KClO3

 3 mol O 2   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 = 0°C + 273 = 273 K
P = 1 atm
n = unknown
PV = nRT
Solving for n:
1 atm 83.0 L 
PV
n=

= 3.7031584 mol H2
L•atm 
RT 
0.0821
273
K

mol•K 

P4(s) + 6H2(g)  4PH3(g)
 1 mol P4   4 mol PH 3 
PH3 from P4 =  37.5 g P4  

 = 1.21085 mol PH3
 123.88 g P4   1 mol P4 
 4 mol PH 3 
PH3 from H2 =  3.7031584 mol H 2  
 = 2.4687723 mol PH3
 6 mol H 2 
P4 is the limiting reactant because it forms less PH3.

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution
in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

5-25

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