Chapter 10: Gases and the Atmosphere Solutions …academic.udayton.edu/.../123_2004/Problemanswers_ch10.pdfChapter 10: Gases and the Atmosphere 315 11. Starting from the earth’s

Chapter 10: Gases and the Atmosphere 313

Chapter 10: Gases and the Atmosphere

Solutions for Chapter 10

Questions for Review and Thought

Review Questions

1. Sections 10.4 and 10.5 give five gas laws. Three of these gas laws relate just P, V, and/or T for a gas sample. They are Boyle’s, Charles’, and the combined gas law. The other two (Avogadro’s law and the ideal gas law) also relate to the quantity of gas present, n, in moles.

Boyle’s law states that the volume (V) of an ideal gas varies inversely with the applied pressure (P) when temperature (T) and the amount of gas (n, moles) are constant.

V∝

1P

(unchanging T and n)

P1V1 = P2V2 (unchanging T and n)

Charles’ law states that the volume (V) of an ideal gas varies directly with the absolute temperature (T) when pressure (P) and the amount of gas (n, moles) are constant.

V∝ T (unchanging P and n)

V1T1

=V2T2

(P and n constant)

The combined gas law states that volume (V) of an ideal gas varies inversely with the applied pressure (P) and directly with the absolute temperature (T) when the amount of gas (n, moles) is constant.

P1V1T1

=P2V2

T2 (n constant)

2. STP is an acronym for “Standard Temperature and Pressure.” When studying gases, standard temperature is exactly 0 °C or 273.15 K, and standard pressure is exactly 1 atmosphere.

3. The volume occupied by one mole of an ideal gas at STP is 22.414 L. (Section 10.5)

4. Pressure is force exerted per unit area. (Section 10.1)

5. Avogadro’s law states that the volume (V) of an ideal gas varies directly with the amount of gas (n, moles) when the temperature (T) and pressure (P) are constant.

V∝ n (unchanging T and P)

Any balanced equation can be interpreted using mole quantities.

O2(g) + 2 H2(g) 2 H2O (g)

One mole of oxygen gas reacts with two moles of hydrogen gas to make two moles of steam. In addition, Avogadro’s Law allows us to also interpret a balanced equation with gas reactants and gas products in terms of gas volumes. For example:


O2(g) + 2 H2(g) 2 H2O (g)

One volume of oxygen gas reacts with two volumes of hydrogen gas to make 2 volumes of steam, if all the volumes are measured at the same temperature and pressure.

6. Dalton’s law and its applications are described in Section 10.8. Dalton’s law of partial pressures states that the total pressure (P) exerted by a mixture of gases is the sum of their partial pressures (p1, p2, p3, etc.), if the volume (V) and temperature (T) are constant.

Ptot = pi∑ (unchanging V and T)

Ptot = p1 + p2 + p3 + ... (V and T constant)

Knowing the percentage of a gas on a mole basis, we can determine its mole fraction.

X i =

n in j∑

=n i

ntot=

mol %100 %

Knowing the mole fraction and the total pressure, we can determine its partial pressure.

pi = XiPtot

A mixture of gases that is 78 % N2 and 22 % O2 on a mole basis, has a total pressure is 720 mmHg.

XO2 =

22 %100 %

= 0.22

pO2 = (0.22) × (720 mmHg) = 1.6 × 102 mmHg

7. The force of the molecules hitting the walls of the container causes the pressure of a gas. If the temperature is constant, then the average speed of the molecules is fixed. If the number of moles of gas is constant, then the number of molecules hitting the walls doesn’t change. So, if the walls are moved farther apart, due to an increase in volume, the molecules will have to travel further to hit the walls, and therefore they will hit the walls less often. If there are fewer collisions with the wall, there is less force, and the pressure goes down. So, an increase in volume causes a decrease in pressure, as stated by Boyle’s law.

8. Gases at low temperature have smaller average kinetic energy, which means they are relatively slow. If the pressure is also high, that means they are relatively close together. When molecules are close together and moving slowly, there is a mu ch better chance that intermolecular forces will be experienced between them. The ideal gas law assumes that the gas particles do not interact. A gas will not behave like an ideal gas if significant interactive forces are experienced.

9. Primary pollutants enter the environment directly from their source. Examples are: particle pollutants, including aerosols and particulates; sulfur dioxide; nitrogen oxides; hydrocarbons. Secondary Pollutants are produced from the chemical reactions of primary pollutants. Examples are: ozone and PAN (peroxyacetylnitrate). See Section 10.12 for details.

The Atmosphere

10. Nitrogen serves to moderate the reactiveness of oxygen by diluting it. Oxygen sustains animal life as a reactant in the conversion of food to energy. Oxygen is produced by plants in the process of photosynthesis.


11. Starting from the earth’s surface, the first layer of the atmosphere is called the “troposphere” and the next layer up is called the “stratosphere.” Many reactions occur in the troposphere, since that is where humans and all other known life on earth live. Tropospheric chemical reactions include combustion, respiration, photosynthesis, creation of primary pollutants, and conversion of primary pollutants to secondary pollutants. Few reactions between molecules occur in the stratosphere, since the air pressure there is very low. Ultraviolet light from the sun does interact with molecules there, breaking bonds and creating free radicals. The ozone layer, and its associated reactions, is in the stratosphere.

12. Define the problem: Convert all the numbers in Table 10.3 to parts per million (ppm) and parts per billion (ppb)

Develop a plan: To accomplish these conversions, we need to describe a relationship between percent and ppm and ppb:

Percent gas in air =

L of gas100 L of air

Parts per million gas in air =

L of gas1,000, 000 L of air

So, 1% = 10,000 ppm. Multiply the number in units of percent by 10,000 to get ppm.

Parts per billion gas in air =

L of gas1,000, 000, 000 L of air

So, 1ppm = 1,000 ppb. Multiply the number in units of ppm by 1,000 to get ppm.

Execute the plan: Molecule ppm ppb

N2 780,840 780,840,000 ↑

O2 209,480 209,480,000

Ar 9,340 9,340,000

CO2 330 330,000

Ne 18.2 18,200 > 1 ppm

H2 10. 10,000

He 5.2 5,200

CH4 2 2,000

↓

Kr 1 1,000 ↑ CO 0.1 100 between Xe 0.08 80 1ppm

O3 0.02 20 and

NH3 0.01 10 1ppb

NO2 0.001 1

↓

SO2 0.0002 0.2 < 1ppb

Check your answer: The numbers are different only by the appropriate factor of 10.

13. Define the problem: Given the mass of the atmosphere and the mass fraction of helium in ppm, determine the mass and moles of helium in the atmosphere

Develop a plan: Use the mass fraction in terms of metric tons as a conversion factor to determine the mass of helium. Use metric and molar mass conversion factors to determine the moles of helium.


Execute the plan:

5.3×1015 metric tons air ×

0.7 metric tons He1, 000 ,000 metric tonsair

= 4 ×109 metric tons He

4× 109 metric tons He ×

1000 kg1 metric ton

×1000 g1 kg

×1 mol He

4.003 g He= 9 ×1014 mol He

Check your answer: Air has a small proportion of He, but the huge quantity of air means there is still a large amount of helium, both in metric tons and in moles. These numbers make sense. (If you use rounded intermediate numbers, the answer is 1 × 1015 mol.

14. Define the problem: Given the mass of a sample of coal, the percentage of sulfur in the coal, the weight

fraction of SO2 in the atmosphere, and (from Question 19) the mass of the atmosphere, determine the mass

of SO2 added to the atmosphere, and the total amount of SO2 in the atmosphere.

Develop a plan: To answer the first question, use metric conversions and the mass fraction as a conversion factor to determine the mass of sulfur. Use mole and molar mass conversion factors to determine the mass

of SO2, assuming that all the sulfur in the coal is converted to SO2 and released into the atmosphere. Convert the mass back to metric tons. To answer the second question, use the mass of the atmosphere

(given in Question 13) and the mass fraction of SO2 in terms of metric tons as a conversion factor to

determine the total mass of SO2.

Execute the plan: First find the mass of S:

3.1×10 9 metric tons coal ×

1000 kg coal1 metric ton coal

×2.5 kg S

100 kg coal×

1000 g S1 kg S

= 7.8 ×1013 g S

Then, find the mass of SO2:

7.8×1013 g S×

1 mol S32.066 g S

×1 mol SO2

1 mol S×

64.0638 g SO21 mol SO2

=1.5 ×1014 g SO2

Then, convert the mass of SO2 back to metric tons:

1.5×1014 g SO2 ×

1 kg SO21000 g SO2

×1 metric ton SO2

1000 kg SO2

= 1.5×10 8 metric tons SO2 was added to the atmosphere in 1980

Get the total mass of SO2 currently in the atmosphere

5.3×1015 metric tons air ×

0.4 metric tons SO21, 000,000,000 metric tons air

= 2 ×10 6 metric tons SO2

Check your answers: It is clear that some of the SO2 presumably released into the atmosphere in 1980 is no

longer there, since the total mass of SO2 is less than what was introduced that year. SO2 is a reactive gas,

getting oxidized to SO3 in the presence of air and then producing sulfuric acid when reacting with rainwater.

This removes the sulfur from the air. See Section 10.12 for more details on acid rain and SO2 as a primary


pollutant.

Properties of Gases

15. Define the problem: Convert a series of pressure quantities into other pressure units.

Develop a plan: Use Table 10.2 to design conversion factors to achieve the conversions.

Execute the plan:

(a) 720 mmHg ×

1 atm760 mmHg

= 0.95 atm

(b) 1.25 atm×

760 mmHg1 atm

= 950. mmHg

(c) 542 mmHg ×

760 torr760 mmHg

= 542 torr

(d) 740 mmHg ×

1 atm760 mmHg

×101.325 kPa

1 atm= 99 kPa

(e) 700 kPa ×

1 atm101.325 kPa

= 7 atm

Check your answers: The unit of atm represents a lot more pressure than the units of kPa, torr, and mmHg, so it makes sense that the numbers of atm are always much smaller than the pressure expressed in these other units. The unit of kPa represents more pressure than the units of torr and mmHg, so it makes sense that the numbers of kPa are always smaller than the pressure expressed in the other units of torr and mmHg. Torr and mmHg are the same size, so their quantities should be identical.

16. Define the problem: Convert a series of pressure quantities into other pressure units.

Develop a plan: Use Table 10.2 to design conversion factors to achieve the conversions.

Execute the plan:

(a) 120 mmHg ×

1 atm760 mmHg

= 0.16 atm

(b) 2 atm×

760 mmHg1 atm

= 2× 103 mmHg (1520 mmHg rounded to 1 significant figure)

(c) 100 kPa ×

1 atm101.325 kPa

×760 mmHg

1 atm= 8× 102 mmHg (750. mmHg with 1 sig. fig.)

(d) 200 kPa ×

1 atm101.325 kPa

= 2 atm (1.97 atm rounded to 1 sig. fig.)

(e) 36 kPa ×

1 atm101.325 kPa

= 0.36 atm

(f) 600 kPa ×

1 atm101.325 kPa

×760 mmHg

1 atm= 5× 103 mmHg (4500 mmHg with 1 sig. fig.)


Check your answers: The unit of atm represents a lot more pressure than the units of kPa, torr, and mmHg, so it makes sense that the numbers of atm are always much smaller than the pressure expressed in these other units. The unit of kPa represents more pressure than the units of torr and mmHg, so it makes sense that the numbers of kPa are always smaller than the pressure expressed in the other units of torr and mmHg. Torr and mmHg are the same size, so their quantities should be identical. These answers make sense.

17. Define the problem: Given the density of mercury, the density of an oil used to construct a barometer, and the atmospheric pressure, determine the height in meters of the oil column in the oil barometer.

Develop a plan: Use Table 10.2 to convert the pressure into mmHg. Pressure on the liquid pushes the

liquid up the barometer until its mass exerts the same force per unit area as the air pressure: Pliquid =

Force/Area = mg2/Area. Since g is a constant, the force per unit area of the liquid counteracting the air

pressure is proportional to just the mass per unit area. Relate the mass per unit area of mercury to the mass per unit area of the oil for 1 atm pressure. Relate the mass of each liquid to its respective density and volume. Relate the volume of each liquid to the dimensions of its respective barometers, including the height. Relate the height of mercury in the mercury barometer to the height of the oil in the oil barometer.

Execute the plan: According to Table 10.2, at 1.0 atm the height of a column of mercury in a mercury barometer is 760 mm. Let m = mass of the liquid, A = cylindrical area of the barometer’s column, d = density of the liquid, V = volume of the liquid, and h = the height of the liquid in the barometer. Now, equate the mass per unit area of each barometer and derives an equation relating their heights:

mHg/AHg = moil/Aoil

dHgVHg/AHg = do ilVoil/Ao il

hV = (Area) × (height) = Ah

dHgAHghHg/AHg = doilAoilhoil/Aoil

dHghHg = doilhoilA

hoil =dHghHg

doil=

13.596 g / cm3

760 mm( )

0.75 g / cm3

×1 m

1000 mm= 14 m

Check your answer: A larger mass of oil will be needed to counterbalance the atmospheric pressure because it is less dense than mercury. A higher column of oil makes sense.

18. If a perfect vacuum is established, atmospheric pressure will push the liquid mercury up the tube 760 millimeters. To whatever extent the vacuum is imperfect, the height will be less than that.

19. With a perfect vacuum at the top of the well, this system would resemble a water barometer. Using the equation derived in the answer to Question 12:


hwater =dHghHgdwater

=13.596 g/ cm3

760 mm( )

1.00 g/ cm3

×1 m

1000 mm×

3.281 feet1 m

= 34 feet

That means atmospheric pressure can only push water up to about 34 feet. So, the well cannot be deeper than that, not even using a high quality vacuum pump.

20. According to Question 19, each 34 feet of water creates 1 atm of pressure.

60 feet×

1 atm34 feet

= 2 atm (1.77 atm rounded to 1 sig. fig.)

At sea level the pressure of the air bubbles will be atmospheric pressure, or 1 atm.

Kinetic-Molecular Theory

21. The five postulates of kinetic-molecular theory are given in Section 10.3:

I. A gas is composed of molecules whose size is much smaller than the distances between them.

II. Gas molecules move randomly at various speeds and in every possible direction.

III. Except when molecules collide, forces of attraction and repulsion between them are negligible.

IV. When collisions occur, they are elastic.

V. The average kinetic energy of gas molecules is proportional to the absolute temperature.

A discussion of nonideal behavior is provided in Section 10.9. Of these five postulates, the ones that becomes false at very high pressures or very low temperatures are postulates III and IV. Slow molecules crowded together are much more likely to interact even when they are not colliding. Collisions may not be elastic under these circumstances, because colliding molecules might stick together due to large enough interactive forces between them.

The postulate that seems most likely to always be correct is postulate II. As long as a substance is a gas, it does tend to disperse to uniformly fill any container it is introduced into, though the rate of dispersal may vary.

22. Define the problem: Given equal volumes of two gases in separate flasks, their molecular identity, their temperatures, and their pressures, compare (a) their average kinetic energy per molecule, (b) their average molecular velocity, and (c) the number of molecules.

Develop a plan: (a) Kinetic energy is proportional to temperature, so compare their temperatures to relate their kinetic energy. (b) Kinetic energy is related to mass and velocity. Since we determine the relative kinetic energy in (a), use that and their molar masses to determine their relative velocity. (c) Use the ideal gas law and the relative P, T, and V to determine which has a larger number (in moles) of molecules.

Execute the plan:

(a) The two samples have the same temperature, so the average kinetic energy per molecule in each sample is the same.

(b) Ekin = 12 mv

2 So,

v2 =

2Em


Because the average kinetic energy of the two samples is the same, only the mass of the molecules

affects the velocity. The molecules with the smaller mass have the faster velocity. Here, H2 (molar

mass = 2.0 g/mol) is lighter than CO2 (molar mass = 44.0 g/mol), so the molecules in the H2 samp le are

moving with a higher velocity than those in the CO2 sample.

(c) Because both the volume and temperature are the same, the only thing affecting the number of moles is the pressure. The ideal gas law (PV = nRT) tells is that the pressure is directly proportional to the

number of moles, if the temperature and volume are fixed. Here, the CO2 sample’s pressure (2 atm) is

twice that of H2 (1 atm), so the CO2 sample has twice as many molecules as the H2 sample.

Check your answer: It makes sense that lightweight things go faster. The postulates of kinetic-molecular theory help us see why these comparisons are sensible.

23. Define the problem: Given the formulas of various atoms and molecules and their common temperature, put their gases in order of increasing average molecular speed.

Develop a plan: Kinetic energy is proportional to temperature. With all of the samples at the same temperature, their average kinetic energies are the same. Kinetic energy is related to mass and velocity. Ekin

= 12 mv2. Velocity is related to kinetic energy and mass:

v2 =

2Em

. Therefore, molecules with smaller mass

have the faster molecular speed. To rank the molecules with increasing speed, rank them from the largest molar mass to the smallest.

Execute the plan: Estimate the molar masses: Kr molar mass 83.8 g/mol, CH4 molar mass = 16.0 g/mol, N2

molar mass is 28.0 g/mol, CH2Cl2 molar mass = 84.9 g/mol.

slowest speed: CH2Cl2 < Kr < N2 < CH4 fastest speed

Check your answer: It makes sense that lightweight things go faster.

24. Define the problem: Given the formulas of molecules, put their gases in order of increasing average molecular speed.

Develop a plan: Kinetic energy is proportional to temperature. Assuming all of the samples are at the same

temperature, their average kinetic energies are the same. Kinetic energy is related to mass and velocity. Ekin

= 12 mv

2. Velocity is related to kinetic energy and mass:

v2 =

2Em

Therefore, molecules with smaller mass

have the faster molecular speed. To rank the molecules with increasing speed, rank them from the largest molar mass to the smallest.

Execute the plan: Estimate the molar masses: SO2 molar mass = 64 g/mol, Cl2 molar mass = 71 g/mol, SOCl2

molar mass = 119 g/mol, Cl2O molar mass = 87 g/mol.

slowest speed: SOCl2 < Cl2O < Cl2 < SO2 :fastest speed


25. Define the problem: Given the formulas of molecules, put their gases in order of decreasing average molecular speed. Predict which one of atmospheric gases might have enough speed to escape into outer space.


Develop a plan: Adapt the method provided in the answers to Questions 23 and 24 : To rank the molecules with decreasing speed, rank them from the smallest molar mass to the largest. The fastest one is probably the one most likely to escape into outer space.

Execute the plan: Rank all the gases in Table 10.3 by increasing molar masses (estimates):

fastest Molar Mass Molar Mass

H2 2 Ar 40.

He 4 CO2 44

CH4 16 NO2 46

NH3 17 O3 48

Ne 20 SO2 64

CO 28.01 Kr 84

N2 28.02 Xe 131

O2 32 slowest

Of the molecules in this table, hydrogen is the fastest, and the most likely to escape.


Gas Behavior and the Ideal Gas Law

26. Define the problem: Given the volume of a sample of air at STP and the volume fraction of CO in ppm, determine the moles of CO.

Develop a plan: Use the volume fraction in terms of liters as a conversion factor to determine the liters of CO. Use the molar volume of a gas at STP as a conversion factor to determine the moles of CO.

Execute the plan:

1.0 L air ×

950 L CO1,000, 000 L air

×1 mol CO

22.414 L CO= 4.2 ×10−5 mol CO

Check your answer: Air has a very small proportion of CO. This small sample of air has a very small amount of CO. This number makes sense.

27. Define the problem: Given the volume and pressure of a sample of gas in one flask and the volume of a flask it is transferred to at the same temperature, determine the new pressure.

Develop a plan: Use Boyle’s law to relate volume to pressure.


Execute the plan:

P2 =

P1V1V2

=(100. mmHg ) × (125 mL)

(200. mL)= 62.5 mmHg

Check your answer: The larger volume should have a smaller pressure. This answer makes sense.

28. Define the problem: Given the volume and pressure of a sample of gas in one flask and the volume of a flask it is transferred to at the same temperature, determine the new pressure.


Develop a plan: Use Boyle’s law to relate volume to pressure.


Execute the plan:

P2 =

P1V1V2

=(75.0 mmHg ) × (256 mL)

(125 mL)=154 mmHg

Check your answer: The smaller volume should have a larger pressure. This answer makes sense.

29. Use Boyle’s law to relate volume to pressure, as described in the answer to Question 27.

V2 =

P1V1P2

=(62 mmHg ) × (100 mL)

(29 mL)=2 ×10 2 mL (214 mL rounded to 1 sig. fig.)


P2 =

P1V1V2

=(735 mmHg )× (3.50 L)

(15.0 L)=172 mmHg


P2 =

P1V1V2

=(2 atm) × (1.0 L)

(4.0 L)= 0.5 atm

32. Define the problem: Given the original volume and temperature of a sample of gas in a syringe (presumably at atmospheric pressure) and the new temperature of the sample, determine the new volume (presumably still at atmospheric pressure).

Develop a plan: Convert the temperatures to Kelvin. Use Charles’ law to relate volume to absolute temperature.

V1T1

=V2T2

(P and n constant)

Execute the plan: (Assume the T1 reading has the same precision as the reading for T2.)

T1 = 20. °C + 273 = 293 K

T2 = 37 °C + 273 = 310. K

V2 = V1 ×

T2T1

= (25.0mL) ×(310. K)(293 K)

= 26.5 mL

Check your answer: Gas at higher temperature should have a larger volume.

33. Define the problem: Given the original volume and temperature of a sample of gas in a syringe (presumably at atmospheric pressure) and the new temperature of the sample, determine the new volume (presumably still at atmospheric pressure).

Develop a plan: Convert the temperatures to Kelvin. Use Charles’ law to relate volume to absolute temperature.


V1T1

=V2T2

(P and n constant)

Execute the plan: (Assume the T2 reading has the same precision as the reading for T1.)

T1 = 23 °C + 273 = 296 K

T2 = –10. °C + 273 = 263 K

V2 = V1 ×

T2T1

= (4.5 L) ×(263 K)(296 K)

= 4.0 L

Check your answer: Gas at lower temperature should have a smaller volume.

34. Use Charles’ law to relate volume to absolute temperature, as described in the answer to Question 32.

(Assume the T1 reading is ± 1 °C.) T1 = 80. °C + 273 = 353 K

T2 = T1 ×

V2V1

= (353 K) ×(1.25 L)(2.50 L)

= 176 K

176 K – 273 = –97 °C

35. Use Charles’ law to relate volume to absolute temperature, as described in the answer to Question 33.

(Assume the T1 reading is ± 1 °C.) T1 = 20. °C + 273 = 393 K

T2 = T1 ×

V2V1

= (393 K)×(8.0 L)(9.0 L)

= 350 K

A gas with a smaller volume should have a lower temperature. Charles’s law tells us that the temperature of this sample will be very small with such a drastic change in the volume.

36. Define the problem: Given the original pressure and temperature of gas in a tire, the assumption that volume is unchanged, and the new temperature, determine the new pressure exerted by the gas in the tire.

Develop a plan: Convert the temperature to Kelvin. Use the combined gas law to relate pressure to absolute temperature.

P1V1T1

=P2V2

T2 (n constant)

At constant volume

P1T1

=P2T2

(V and n constant)

Execute the plan: T1 = 15 °C + 273 = 288 K T2 = 35 °C + 273 = 308 K

P2 = P1 ×

T2T1

= (3.74 atm) ×(308 K)(288 K)

= 4.00 atm

Check your answer: A gas with a higher temperature should exert a higher pressure.

37. Define the problem: Given the original pressure and temperature of gas in a tire, the assumption that volume is unchanged, and the new temperature, determine the new pressure exerted by the gas in the tire.



P1V1T1

=P2V2

T2 (n constant)

At constant volume

P1T1

=P2T2

(V and n constant)

Execute the plan: (Assume the T1 reading is ± 1 °C.)

T1 = 40. °C + 273 = 313 K

T2 = –5 °C + 273 = 268 K

P2 = P1 ×

T2T1

= (3.05 atm)×(268 K)(313 K)

= 2.61 atm

Check your answer: A gas with a lower temperature should exert a lower pressure.

38. Define the problem: Given the original volume, pressure, and temperature of a gas sample, the new temperature, and the new pressure, determine the new volume of the sample.


P1V1T1

=P2V2

T2 (n constant)

Execute the plan: T1 = 22 °C + 273 = 295 K

T2 = 42 °C + 273 = 315 K

V2 = V1 ×

T2T1

×P1P2

= (754 mL)×(315 K)(295 K)

×(165 mmHg )(265 mmHg )

= 501 mL

Check your answer: The temperature fraction:

(315 K)(295 K)

is larger than one, consistent with increasing the

volume due to the increased temperature. The pressure fraction:

(165 mmHg )(265 mmHg )

is smaller than one,

consistent with decreasing the volume due to an increased pressure. Clearly these two effects counteract each other, but this pressure change affects the volume more than the temperature change does.

39. Define the problem: Given the original volume, pressure, and temperature of gas in a balloon, the new temperature, and the new pressure, determine the new volume of the sample.


P1V1T1

=P2V2

T2 (n constant)

Execute the plan: (Assume the T1 reading is ± 1 °C and P2 reading is ± 1 mmHg.)


T1 = 20. °C + 273 = 293 K

T2 = –33 °C + 273 = 240. K

V2 = V1 ×

T2T1

×P1P2

= (1.05 ×103 L)×(240 K)(293 K)

×(745 mmHg )(600. mmHg )

=1.07 ×10 3 L

Check your answer: The temperature fraction:

(240 K)(293 K)

is smaller than one, consistent with decreasing

the volume due to the decreased temperature. The pressure fraction:

(745 mmHg )(600. mmHg )

is larger than one,

consistent with increasing the volume due to a decreased pressure. Clearly these two effects counteract each other, but the pressure change affects the volume slightly more than the temperature change does.

40. Define the problem: Given the mass, identity, temperature, and volume of a gas sample, determine the pressure of the sample.

Develop a plan: Use the ideal gas law:

PV = nRT R = 0.08206

L ⋅atmmol ⋅K

The units of R remind us to determine the moles of gas (using mass and molar mass), to convert the temperature to Kelvin, and to convert the volume to liters.

Execute the plan: (Assume the T reading is ± 1 °C.)

1.55 g Xe ×

1 mol Xe131.29 g Xe

= 0.0118 mol Xe

T = 20. °C + 273 = 293 K

V = 560 mL ×

1 L1000 mL

= 0.56 L

P=

nRTV

=(0.0118 mol) × 0.08206

L ⋅atmmol ⋅K

× (293 K)

(0.56 L)= 0.51 atm

Check your answer: The small fraction of a mole makes sense with the small mass. The units in the pressure calculation cancel properly to give atm. This answer make sense.

41. Define the problem: Given the mass of water turned to gas, the volume of the container, and the temperature, determine the pressure of the water vapor sample.

Develop a plan: Use the ideal gas law:

PV = nRT R = 0.08206

L ⋅atmmol ⋅K

The units of R remind us to determine the moles of gas (using mass and molar mass), and to convert the temperature to Kelvin.

Execute the plan:


1.00 g H2O×

1 mol H2O18.02 g H2O

=0.0555 mol H2O

T = 150. °C + 273 = 423 K

P=

nRTV

=(0.0555 mol) × 0.08206

L ⋅atmmol ⋅K

× (423 K)

(10.0 L)= 0.193 atm

Check your answer: The small fraction of a mole makes sense with the small mass. The units in the pressure calculation cancel properly to give atm. This answer make sense.

42. Define the problem: Given a set of gas samples, determine which has the largest number of molecules and which has the smallest number of molecules.

Develop a plan: Some of the samples are at STP, so use the molar volume of a gas at STP to determine the number of molecules (in units of moles). In the other cases, use the ideal gas law as described in the answer to Question 40. Once all the moles are calculated, identify which sample has the most moles and which sample has the least moles.

Execute the plan: (a) and (b) are at STP. 1 mol of any gas occupies 22.414 L.

1.0 L×

1 mol gas22.414 L

= 0.045 mol gas

(a) 0.045 mol H2

(b) 0.045 mol O2

(c) T = 27 °C + 273 = 300. K

P = 760 mmHg ×

1 atm760 mmHg

= 1.0 atm

n =PVRT

=(1.0 atm) × (1.0 L)

0.08206L ⋅atmmol ⋅K

× (300. K)

= 0.041 mol

(d) (Assume P has as many sig. figs as in (c))

T = 0 °C + 273 = 273 K

P = 800 mmHg ×

1 atm760 mmHg

= 1.1 atm

n =PVRT

=(1.0 atm) × (1.0 L)


× (273 K)

= 0.047 mol

Of these samples, (d) has the most molecules (0.047 moles CO2) and (c) has the least molecules (0.041 moles

H2).

Check your answers: To keep a 1.0-L gas sample at standard temperature and still have larger than standard


pressure suggests that there must be more molecules hitting the walls than a sample at STP. To have a 1.0-L gas sample at higher than standard temperature and still stay at standard pressure suggests that there must be fewer molecules hitting the walls harder, than a sample at STP.

43. Define the problem: Given the volume of an air sample, the concentration of O3 in that sample, the

temperature, and the pressure, determine the number of molecules of O3 in the sample. (All of the molecules in the sample could come in contact with the rubber eventually.)

Develop a plan: Use the volume of the air sample and the concentration of O3 to determine the volume of O3, then use the ideal gas law as described in the answer to Question 41. Then use Avogadro’s number to find the number of molecules from the moles.

Execute the plan:

1.0 cm3 air ×

1 mL

1 cm 3×

1 L1000 mL

×0.02 L O3

1,000,000 Lair= 2.0× 10−11 L O3

T = 25 °C + 273 = 298 K

n =PVRT

=(0.95 atm) × (2.0 ×10−11 L)


× (298 K)

= 7.8× 10−13 mol

7.8×10−13 mol ×

6.022 ×10 23 O3 molecules1 mol O3

= 4.7 ×1011 O3 molecules

Check your answer: 470 billion ozone molecules can attack the rubber. This is a pretty small fraction of the air molecules, but there are still plenty of molecules colliding with the rubber to cause the damage.


Quantities of Gases in Chemical Reactions

44. Define the problem: Given the mass of sucrose, the formula, and the product of a reaction, determine the

maximum volume of CO2 produced at STP. Compare that volume with the typical volume of two loaves of French bread.

Develop a plan: Balance the equation. Determine the moles of sucrose from the molar mass, then use the

stoichiometric relationships given in the balanced equation to determine the moles of CO2 produced. Last,

use the molar volume of a gas at STP to determine the volume of CO2. Estimate the total volume of two loaves of French bread assuming they are cylinders. Compare the two volumes.

Execute the plan: Balance the equation:

C12H22O11(s) + 12 O2(g) 12 CO2 (g) + 11 H2O (l)

2.4 g C12H22O11 ×

1 mol C12H22O11342.2956 g C12H22O11

×12 mol CO2

1 mol C12H22O11×

22.414 L CO2 at STP1 mol CO2

= 1.9 L CO2

Assume one loaf of French bread is a cylinder, 3.0 inches in diameter and 18 inches long.

18" 3"

r = 1.5 in

A = πr2 = π(1.5 in)2 = 7.1 in2

V = Al = (7.1 in2) × (18 in) = 130 in3

130 in3

×

2.54 cm1 in

3×

1 mL

1 cm 3×

1 L1000 mL

=2.1 L

Two loaves would have twice this volume: 2 × (2.1 L) = 4.2 L. The CO2 bubbles produced in the bread are nearly half of its volume.

Check your answers: Slicing open French bread we see that it has a vast “honeycomb” of bubble-shaped

spaces in it. It makes sense that approximately half the loaf’s volume can be associated with the CO2 bubbles formed by the yeast when the bread was rising.

45. Define the problem: Given the description of a chemical reaction and the volume of one reactant at a specified pressure and temperature, determine the volume of the other reactant at a specified pressure and temperature that will cause complete reaction.

Develop a plan: Balance the equation. Avogadro’s Law allows us to interpret a balanced equation with gas reactants in terms of gas volumes, as long as their temperatures and pressures are the same. Use the stoichiometry to relate liters of H2(g) that react with liters of O2(g).

Execute the plan: Balance the equation:

O2(g) + 2 H2(g) 2 H2O (g)


One volume of oxygen gas reacts with two volumes of hydrogen gas to make two volumes of steam, since all the volumes are measured at the same temperature and pressure.

1.5 L H2(g)×

1 L O2 (g)2 L H2 (g)

=0.75 LO2 (g)

Check your answer: Twice as many H2 molecules are needed compared to the number of O2 molecules. So,

it makes sense that the volume of O2 is half the volume of H2.

46. Define the problem: Given the balanced chemical equation for a chemical reaction and the volume of one reactant at a specified pressure and temperature, determine the volume of the other reactant at a specified pressure and temperature that will cause complete reaction, and the volume of one of the products at a specified pressure and temperature that will be produced.

Develop a plan: Avogadro’s Law allows us to interpret a balanced equation with gas reactants and products in terms of gas volumes, as long as their temperatures and pressures are the same. Use the

stoichiometry to relate liters of SiH4(g) that react with liters of O2(g) and liters of H2O(g).

Execute the plan: The balanced equation tells us that one volume of SiH4(g) reacts with two volumes of

O2(g) to make two volumes of H2O(g), since all the volumes are measured at the same temperature and pressure.

5.2 L H2(g)×

2 L O2 (g)1 LSiH4 (g)

=10.4 LO2 (g)

5.2 L H2(g)×

2 L H2O(g)1 LSiH4 (g)

= 10.4 LH2O(g)

Note: Multiplying a number by an exact whole number, n, is like adding that number to itself n times. 2 × (5.2) = 5.2 L + 5.2 L = 10.4 L, that is why we use the addition rule for assessing the significant figures of the results here.

Check your answers: Twice as many O2 molecules are needed compared to the number of SiH4 molecules,

forming twice as many H2O molecules. So, it makes sense that both the volume of O2 and the volume of

H2O are twice the volume of SiH4.

47. Define the problem: Given the balanced chemical equation for a chemical reaction, and the volume of one reactant at a specified pressure and temperature, determine the mass of one of the products produced.

Develop a plan: Use the ideal gas law to determine the moles of the reactant. Use the stoichiometric relationship from the balanced equation to determine moles of product. Then, convert from moles to grams.

Execute the plan: T = 120 °C + 273 = 3.9×102 K

nH 2O =PVRT

=(2.0 atm)× (250 L)

0.08206L ⋅atmmol ⋅ K

× (3.9× 102 K)

= 16 mol H2O

16 mol H2O×

1 mol H21 mol H2O

×2.0158 g H2

1 mol H2= 31 g H2


Check your answer: The units all cancel properly in the calculation of moles of water. The moles of water and hydrogen are equal, and the mass of hydrogen is twice its number of moles.

48. Define the problem: Given the balanced chemical equation for a chemical reaction, the mass of one reactant, and excess other reactant, determine the pressure of the product produced at a specified volume and temperature.

Develop a plan: Convert from grams to moles. Use the stoichiometric relationship from the balanced equation to determine moles of product. Use the ideal gas law to determine the pressure of the product.

Execute the plan:

0.050 g B4H10 ×

1 mol B4H1053.32 g B4H10

×10 mol H2O2 mol B4H10

= 0.0047 mol H2O

T = 30. °C + 273 = 303 K

P=

nH2ORT

V=

(0.0047 mol H2O)× 0.08206L ⋅atmmol ⋅K

× (303 K)

(4.25 L)= 0.027 atm

0.027 atm×

760 mmHg1 atm

= 21 mmHg

Check your answer: The relative quantities of B4H10 and H2O seem sensible. All units cancel properly in the calculation of atmosphere. This is a relatively low pressure for water but the sample is also small.

49. Define the problem: Given the mass of a metallic element and the identity of the gas-phase molecule it is converted to, determine the pressure of the molecular gas at a specified volume and temperature.

Develop a plan: Convert from grams to moles. Use the stoichiometric relationship of the molecular formula to determine moles of molecules. Use the ideal gas law to determine the pressure of the gas.

Execute the plan: 1.0 ×10 3 g U×

1 mol U238.03 g U

×1 mol UF6

1 mol U= 4.2 mol UF6

P=

nUF6 RT

V=

(4.2 mol UF6)× 0.08206L ⋅atmmol ⋅ K

× (335 K)

(3.0× 102 L)= 0.38 atm

Check your answer: The relative quantities of U and UF6 seem sensible. All units cancel properly in the calculation of atmospheres. This is a relatively low pressure, but the container is fairly large.

50. Define the problem: Given the balanced chemical equation for a chemical reaction, the mass of the solid ionic reactant, the formula of the ionic compound and the periodic group that its metal is found in, the pressure of a gas-phase product with a specified volume and temperature, determine the molar mass of the reactant.

Develop a plan: The formula of the ionic compound, Mx(CO3)y , can be used to determine the values of x and y, since M is from Group 2A and the anion carbonate has a known charge. The ideal gas law can be

used to find the moles of CO2 produced. The stoichiometry is then used to find the moles more solid that reacted. Divide the mass by the moles for the molar mass.


Execute the plan: Group 2A metals have 2+ charges and carbonate ion has a 2– charge:

x M2+ + y CO32– Mx(CO3)y

From this equation, we can see that x = y = 1, and the heating of the metal oxide reaction can be simplified to the following:

MCO3(s) MO(s) + CO2(g)

T = 25 °C + 273 = 298 K

69 .8 mmHg ×

1 atm760 mmHg

= 0.0918 atm

285 mL ×

1 L1000 mL

= 0.285 L

nCO2 =PVRT

=(0.0918 atm) × (0.285 L)

0.08206L⋅ atmmol ⋅ K

× (298 K)

= 1.07 ×10−3 mol CO2

1.07 ×10−3 mol CO2 ×

1 mol MCO31 mol CO2

= 1.07 ×10−3 mol MCO3

The given mass (0.158 g) contains 1.07 × 10–3 moles of MCO3

Molar Mass =

0.158 g MCO3

1.07 ×10−3 mol MCO3=148

gmol

Check your answer: If this metal M is really a Group 2A element, it’s molar mass should be similar to one of them. We can subtract the molar mass of one C atom and three O atoms from the molar mass of the solid and get the molar mass of M:

148 g/mol – 12 g/mol – 3 × (16 g/mol) = 88 g/mol M

This is close to that of Sr (molar mass = 87.62 g/mol), a group 2A metal.

51. Define the problem: Given the description of a chemical reaction and the volume of one reactant at a specified pressure and temperature, determine the mass of the product that can be produced.

Develop a plan: Balance the equation. Use the ideal gas law to determine the moles of the reactant. Use the stoichiometric relationship from the balanced equation to determine moles of product. Then, convert from moles to grams.

Execute the plan: Ni(s) + 4 CO(g) Ni(CO)4

T = 25.0 °C + 273.15 = 298.2 K

418 mmHg ×

1 atm760 mmHg

= 0.550 atm

nCO =PVRT

=(0.550 atm) × (1.50 L)


× (298.2 K)

= 0.0337 mol CO


0.0337 mol CO ×

1 mol Ni(CO)44 mol CO

×170.7 g Ni(CO)4

1 mol Ni(CO)4=1.44 g Ni(CO)4

Check your answer: The units all cancel properly in the calculation of moles of Ni(CO)4. A reasonable mass of product is formed considering the molar quantities and the molar mass.

52. (a) This equation is given in Problem-Solving Exercise 10.11:

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

(b) Define the problem: Given the length in miles of a trip, the fuel efficiency of a car, the density of the liquid fuel, and the temperature and pressure, determine the volume of a gas–phase product produced during the trip.

Develop a plan: Use the miles, the fuel efficiency, volume conversions, the density, and the stoichiometry to find the moles of the product. Use the ideal gas law to determine the volume of the product.

Execute the plan:

10 . miles ×

1 gal gasoline32 miles

×3.785 L gasoline

1 gal gasoline×

1000 mL1 L

×1 cm3

1 mL= 1.2 × 103 cm3 gasoline

1.2 ×10 3 cm3 gasoline ×

0.692 g gasoline

1 cm3 gasoline×

1 mol gasoline114.22 g gasoline

×16 mol CO2

2 mol gasoline= 58 mol CO2

T = 25 °C + 273 = 298 K

V =

nCO2 RT

P=

(58 mol CO2) × 0.08206L ⋅atmmol ⋅K

× (298 K)

(1.0 atm)= 1.4 ×10 3 L

Check your answer: This is a large volume of CO2! However, it is not an unreasonable quantity

considering how many gallons of gasoline are used and how much CO2 is generated from each octane

molecule. Comparing the results to the answer in Question 53, less CO2 is generated by methanol as a fuel than octane.

53. (a) The balanced equation: 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)

(b) Define the problem: Given the length in miles of a trip, the fuel efficiency of a car, the density of the liquid fuel, and the temperature and pressure, determine the volume of a gas-phase product produced during the trip.

Develop a plan: Use the miles, the fuel efficiency, volume conversions, the density, and the stoichiometry to find the moles of the product. Use the ideal gas law to determine the volume of the product.

Execute the plan:

10 . miles ×

1 gal fuel20. miles

×3.785 L fuel

1 gal fuel×

1000 mL1 L

×1 cm3

1 mL= 1.9 ×10 3 cm3 fuel


1.9 ×10 3 cm3 fuel ×

0.791 g CH 3OH

1 cm3 fuel×

1 mol CH 3OH32.0417 g CH3OH

×2 mol CO2

2 mol CH3OH= 47 mol CO2

T = 25 °C + 273 = 298 K

V =

nCO2 RT

P=

(47 mol CO2) × 0.08206L ⋅atmmol ⋅K

× (298 K)

(1.0 atm)= 1.1×103 L

Check your answer: This is a large volume of CO2! However, it is not an unreasonable quantity

considering how many gallons of methanol are used and how much CO2 is generated from each

methanol molecule. Comparing the results to the answer in Question 52, less CO2 is generated by methanol as a fuel than octane.

Gas Density and Molar Mass

54. Define the problem: Given the identity of a gaseous compound, the mass of the compound, the volume, and the pressure, determine the temperature of the gas.

Develop a plan: Use the formula and the molar mass to get moles, then use the ideal gas law to get volume.

Execute the plan: 0.425 g SiH4 ×

1 mol SiH432.1171 g SiH 4

= 0.0132 mol SiH 4

580 mL ×

1 L1000 mL

= 0.58 L

T =PVnR

=(0.61 atm)× (0.58 L)

0.0132 mol SiH4( )× 0.08206L⋅atmmol ⋅K

= 330 K (326 K, rounded to 2 sig. figs.)

330 K – 273 = 60 °C

Check your answer: The temp erature is high enough for SiH4 to be a gas. (–112 °C = the boiling point).

55. Define the problem: Given the mass of a gas, the identity of the gas, its temperature, and its pressure, determine the volume occupied by the gas.

Develop a plan: Convert from grams to moles. Use the ideal gas law to determine the volume of the gas.

Execute the plan: 4.4 g CO2 ×

1 mol CO244.0095 g CO2

= 0.10 mol CO2

T = 27 °C + 273 = 300. K 730 mmHg ×

1 atm760 mmHg

= 0.96 atm

V =

nRTP

=(0.10 mol )× 0.08206

L ⋅atmmol ⋅K

× (300. K)

(0.96 atm)= 2.6 L


2.6 L ×

1000 mL1 L

= 2.6 ×103 mL

Check your answer: The molar volume of a gas under these conditions is going to be similar to that at STP (22.4 L/mol), because T is slightly higher than standard T and P is slightly lower than standard P. When using one tenth of a mole of CO2, one would expect approximately one tenth of the molar volume, roughly 2.2 liters. This answer seems right. It is unclear why it needed to be given in milliliters.

56. Define the problem: Given the identity of a gas and its volume, pressure, and temperature, determine the mass of the gas.

Develop a plan: Use the ideal gas law to determine the moles of the gas. Then, convert from moles to grams.

Execute the plan: T = 25 °C + 273 = 298 K

nHe =PVRT

=(1.1 atm)× (5.0 L)


× (298 K)

= 0.22 mol He

0.22 mol He ×

4.0026 g He1 mol He

= 0.90 g He

Note: It is perfectly acceptable to use the equations derived in Section 10.7 to streamline your approach to answering these questions.

mHe =PVMRT

=(1.1 atm) × (5.0 L) ×

4.0026 g He1 mol He


× (298 K)

= 0.90 g He

This can save time, but don't lose sight of what you’re doing. If you find yourself looking through the book for an equation with all the right variables, that is a sign that you are losing sight of what you’re doing.

Check your answer: The units all cancel and the size of the number make sense.

57. Define the problem: Given the partial formula for a hydrocarbon, the percentage of carbon, and the volume, pressure, and temperature of a given mass of the gaseous compound, determine the molecular formula of the gas.

Develop a plan: Use the methods described in Section 3.10 to determine the empirical formula. Use the ideal gas law to determine the moles of gas in the sample. Divide the grams of gas in the sample by the calculated moles of gas in the sample to get the molar mass.

Execute the plan: Select a convenient large sample size: 100.00 g of CxHy, such that the sample contains 92.26 grams of carbon and the rest is hydrogen:

100.00 g CxHy – 92.26 g C = 7.74 g H

Convert each element mass to moles:

92.29 g C ×

1 mol C12.0107 g C

= 7.684 mol C


7.74 g H ×

1 mol H1.0079 g H

= 7.68 mol H

Set up a mole ratio and simplify:

7.684 mol C : 7.68 mol H

1 C : 1 H

The empirical formula is CH and the molecular formula is (CH)n. So, n = x = y

The molar mass of the empirical formula = 13.02 g/mol

Find the molecule’s molar mass:

T = 23 °C + 273 = 296 K

374 mmHg ×

1 atm760 mmHg

= 0.492 atm

185 mL ×

1 L1000 mL

= 0.185 L

nCn Hn =PVRT

=(0.492 atm) × (0.185 L)


× (296 K)

= 3.75 × 10−3 mol CnHn

Molar Mass =

0.293 g Cn Hn

3.75 ×10−3 mol Cn Hn= 78.2 g Cn Hn

Get the value of n by taking a ratio of the molecule’s molar mass and the empirical formula’s molar mass:

n =

78 .2 g / mol13.02 g/ mol

= 6.01

The molecular formula is C6H6.

Check your answer: The mole ratios and molar mass ratios are very close to integer relationships and give the formula for benzene, a common aromatic hydrocarbon compound. All of this gives us confidence that this is the right answer.

58. Define the problem: Given the molar mass of a gas and its pressure and temperature, determine the density of the gas.

Develop a plan: We have insufficient information to take a straightforward approach to this task, so let’s look at what we know: Density is mass per unit volume. If we can calculate the molar volume (the volume per mole) and divide that into the molar mass (M, the grams per mole), we can get the density. Use the ideal gas law to determine the molar volume of the gas.

molar volume =

Vn

=RTP


d =

mV

=MRTP

=MPRT

Execute the plan: T = –23 °C + 273 = 250. K

0.20 mmHg ×

1 atm760 mmHg

= 2.6 ×10−4 atm

d =MPRT

=29.0

gmol

× (2.6× 10−4 atm)


× (250. K)

= 3.7 × 10−4 gL

Check your answer: The units all cancel. The low density makes sense at the very low pressure. These numbers make sense.

59. Define the problem: Given the density of a gas and its pressure and temperature, determine the molar mass of the gas.

Develop a plan: We have insufficient information to take a straightforward approach to this task, so let’s look at what we know: Density is mass per unit volume, so density times volume is mass. If we can calculate the molar volume (the volume per mole) and multiply it by the density, we can get the molar mass (M = grams per mole). Use the ideal gas law to determine the molar volume of the gas.

d ×

Vn

=mn

= M

molar volume =

Vn

=RTP

d ×

RTP

= M

Execute the plan: T = 23.0 °C + 273.15 = 296.2 K

715 mmHg ×

1 atm760 mmHg

= 0.941 atm

M =

dRTP

=2.39

gL

× 0.08206

L ⋅atmmol ⋅K

× (296.2. K)

(0.941 atm)=61.7

gmol

Check your answer: The units all cancel. The molar mass is in the range of typical small gaseous molecules. This answer makes sense.

Partial Pressures of Gases

60. Define the problem: Given the partial pressure of several gases in a sample of the atmosphere, and the total

pressure of the sample, determine the partial pressure of O2, the mole fraction of each gas, and the percent by volume. Compare the percentages to Table 10.3.


Develop a plan: Dalton’s law and its applications are described in Section 10.8. Dalton’s law of partial pressures states that the total pressure (P) exerted by a mixture of gases is the sum of their partial pressures

(p1, p2, p3, etc.), if the volume (V) and temperature (T) are constant.

Ptot = p1 + p2 + p3 + ... (V and T constant)

Because p i = XiPtot, we can use the total pressure and the partial pressure of a component to determine its mole fraction:

X i =

piPtot

According to Avogadro’s law, moles and gas volumes are proportional, so the mole fraction is equal to the volume fraction. To get percent, multiply the volume fraction by 100 %.

Execute the plan:

(a) Ptot = pN2+ pO

2 + pAr + pCO

2 + pH

2O

pO2 = Ptot – pN

2– pAr – pCO

2 – pH

2O

= (740. mmHg) – (575 mmHg) – (6.9 mmHg) – (0.2 mmHg) – (4.0 mmHg) = 154 mmHg

(b) XN 2 =

p N2Ptot

=575 mmHg740. mmHg

= 0.777 XO2 =

pO2Ptot

=154 mmHg740. mmHg

= 0.208

XAr =

pArPtot

=6.9 mmHg740. mmHg

= 0.0093 XCO2 =

pCO2Ptot

=0.2 mmHg740. mmHg

= 0.0003

XH 2O =

pH 2OPtot

=4.0 mmHg

740. mmHg= 0.0054

(c) % N2 = XN2 × 100 % = 0.777 × 100 % = 77.7 %

% O2 = XO2 × 100 % = 0.208 × 100 % = 20.8 %

% Ar = XAr × 100 % = 0.0093 × 100 % = 0.93 %

% CO2 = XCO2 × 100 % = 0.0003 × 100 % = 0.03 %

% H2O = XH2

O × 100 % = 0.0003 × 100 % = 0.54 %

The Table 10.3 figures are slightly different. This sample is wet, whereas the proportions given in Table 10.3 are for dry air.

Check your answers: The percentages are very close to those provided in the table, and the variations are explainable. The sum of the mole fractions is 1, and the sum of the percentages is 100 %.


61. Define the problem: Given the pressure of a gas at a specified volume and temperature, determine the pressure of the molecular gas with a new specified volume and temperature.

Develop a plan: Since temperature remains unchanged, use Boyle’s law to determine the pressure of the gas.

Execute the plan:

P2 =

P1V1V2

=45.6 mmHg( )× (56.0 L)

(2.70 ×104 L)=0.0946 mmHg

Check your answer: The pressure should go down if the volume is increased.

62. Define the problem: Given three containers of gases, with known volume and pressure, determine the total pressure and partial pressures of the gases when the three containers are opened to each other.

Develop a plan: Use Boyle’s law to determine how the pressure of each gas changes with the increase in total volume and then use Dalton’s law to determine the total pressures after mixing. Use the definition of partial pressure to show that the pressures calculated are the partial pressure.

Execute the plan: The gas in one chamber is allowed to diffuse into all three chambers, so its final volume increases to the total of the three volumes: Vtot = 3.00 L + 2.00 + 5.00 L = 10.00 L

Boyle’s law: Pf,gas =

Pi,gas Vi,gasVf ,gas

For O2, Pf,O2 =

PiViVf

=1.46 atm( )× (3.00 L)

(10.00 L)= 0.438 atm

For N2, Pf,N 2 =

0.908 atm( )× (2.00 L)

(10.00 L)= 0.182 atm For Ar,

Pf,Ar =

2.71 atm( )× (5.00 L)

(10.00 L)= 1.36 atm

(a) Pf, tot = p f,O2 + p f,N2

+ p f,Ar = 0.438 atm + 0.182 atm + 1.36 atm = 1.98 atm

(b) Partial pressure is the pressure each gas would cause if it were alone in the container.

Pf ,O2 = pO2 = 0.438 atm,

Pf ,N 2 = pN2 = 0.182 atm, Pf,Ar = pAr = 1.36 atm

Check your answers: All the individual final pressures are less than the initial pressures, which makes semse because the volume is larger. The total pressure is a weighted average of the three pressures, and is influenced most by the gas present in largest quantity (Ar).

63. Define the problem: Given the density of a gas and its pressure and temperature, determine the molar mass of the gas.

Develop a plan: Density (d) is mass (m) per unit volume (V), so dV = m. Therefore, if we calculate the molar volume (the volume per mole) and multiply it by the density, we can get the molar mass (M = grams per mole). Use the ideal gas law to determine the molar volume of the gas.

d ×

Vn

=mn

= M

molar volume =

Vn

=RTP

d ×

RTP

= M

The sum of the molar masses of each component weighted by its mole fraction must be equal to the average


molar mass. The sum of the mole fractions must be equal to 1.

Execute the plan:

(a) T = –63 °C + 273 = 210. K

42 mmHg ×

1 atm760 mmHg

= 0.055 atm

92

g

m3×

1 m100 cm

3

×1 cm3

1 mL×

1000 mL1 L

= 0.092gL

M =

dRTP

=0.092

gL

× 0.08206

L ⋅atmmol ⋅K

× (210. K)

(0.055 atm)= 29

gmol

(b) Use the mo lar mass calculated in (c), The molar mass of N2 is 28.0 g/mol. Molar mass of O2 is 32.0

g/mol. Assuming that air is only N2 and O2, setting the mole fraction of N2 equal to X, means that the

mole fraction of O2 is 1 – X.

X(28.0 g/mol) + (1 – X)(32.0 g/mol) = 29 g/mol

– 4.0 X + 32.0 = 29

– 4.0 X = 3

X = 0.8 = XN2 1 – X = 0.2 = XO2

Check your answers: The small number of significant figures makes it difficulty to compare these numbers in detail. They are approximately the same as described in Table 10.3. These answers make sense.

64. Define the problem: Given two concentrations of a gas as pressure ratios, determine the partial pressure of the gas at a given temperature and pressure.

Develop a plan: Use the ppm pressure ratio values to get the partial pressure:

pi = (pressure ratio) × Ptot

Execute the plan: Ptot = 1 atm at STP

Membrane irritation: Pbenzene =

100 atm1,000, 000 atm

×1 atm= 1×10−4 atm

Fatal narcosis: Pbenzene =

20, 000 atm1,000, 000 atm

×1 atm= 0.02 atm

Check your answers: The larger pressure ratio has a larger partial pressure. These numbers make sense.

65. Define the problem: Given the partial pressure of a gas in a sample of the atmosphere, determine its partial pressure.

Develop a plan: Use pi = XiPtot, to determine its the mole fraction.

Execute the plan: Assume that the Ptot = 760 mmHg, standard pressure.


XH 2O =

pH 2OPtot

=25 mmHg H2O760 mmHg air

= 0.033

The mole fraction of water is not given in Table 10.3, because the air described in that table is dry (i.e., XH2

O

= 0, exactly). This air is humid air. It has significantly more water in it.

Check your answer: The partial pressure is small compared to the total pressure, so the small mole fraction makes sense.

66. Define the problem: Given the mean fraction by weight of water in the atmosphere and the molar mass of “air”, determine the mean mole fraction of water in the atmosphere.

Develop a plan: The mean fraction by weight describes the average number of grams of water per gram of air. Convert grams to moles in both cases, to get the mole fraction. Use p i = XiPtot, to determine its the mole fraction.

Execute the plan:

(a) The molar mass of air is given at 29.2 g/mol. The molar mass of water is 18.0152 g/mol.

XH 2O =

0.0025 g H2O1 g air

×29.2 g air1 mol air

×1 mol H2O

18.0152 g H2O= 0.0041

mol H2Omol air

= 0.0041

(b) Assume that the Ptot = 760 mmHg, standard pressure.

pH2O = XH2OPtot = (0.0041) × (760 mmHg) = 3.1 mmHg

This number represents the mean partial pressure, both humid and dry air, summer and winter, worldwide is included in this average.

Check your answer: The mole fraction is small, so the small partial pressure compared to the total pressure makes sense.

67. Define the problem: Given the balanced equation for a reaction, the mass of a reactant, the volume, temperature and pressure of a product collected over water, and the vapor pressure of water at that same temperature, determine the percent yield of the reaction.

Develop a plan: Calculate the theoretical yield from the mass of reactant. Use Dalton’s law to calculate the pressure of the product gas. Use the idea gas law to determine the moles of product. Convert the moles to grams to get actual yield. Divide the actual yield by the theoretical yield and multiply by 100 % to get percent yield.

Execute the plan: 1 mole CaC2 produces 1 mol C2H2.

2.62 g CaH2 ×

1 mol CaH 264.10 g CaH2

×1 mol C2H21 mol CaH 2

×26.0372 g C2H2

1 mol C2H2= 1.08 g C2H2

Ptot = PC2

H2 + pH

2O

PC2

H2 = Ptot – pH

2O = (735.2 mmHg) – (23.8 mmHg) = 711.4 mmHg


711.4 mmHg ×

1 atm760 mmHg

= 0.9361 atm T = 25.0 °C + 273.15 = 298.2 K

795 mL ×

1 L1000 mL

= 0.795 L

nC2H 2 =PVRT

=(0.9361 atm)× (0.795 L)


× (298.2 K)

= 0.0304 mol C2H2

0.0304 mol C2H2 ×

26.0372 g C2H21 mol C2H2

= 0.792 g C2H2

0.792 g C2H2 actual1.08 g C2H2 theoretical

× 100 % = 73.6 %

Check your answer: The percent yield is a realistic size for collection of a gas.

68. Define the problem: Given the balanced equation for a reaction, and the volume, temperature and pressure of a gas collected over water, determine the mass of product formed.

Develop a plan: Use Table 10.5 and Dalton’s law to calculate the pressure of the product gas. Use the idea gas law to determine the moles of product. Convert the moles to grams to get the mass.

Execute the plan: Table 10.5 gives the vapor pressure of water at 23 °C to be 21.1 mmHg.

Ptot = PO2 + pH

2O

PO2 = Ptot – pH

2O = (750 mmHg) – (21.1 mmHg) = 7.3×102 mmHg

7.3×10 2 mmHg ×

1 atm760 mmHg

= 0.96 atm

T = 23 °C + 273 = 296 K 465 mL ×

1 L1000 mL

= 0.465 L

nO2 =PVRT

=(0.96 atm) × (0.465 L)


× (296 K)

= 0.018 mol O2

0.018 mol O2 ×

31.9988 g O21 mol O2

= 0.59 g O2

Check your answer: The units all cancel and the size of the numerical result makes sense.

The Behavior of Real Gases

69. This is a standard conversion factor problem:


1 mol H2O ×

18.02 g H2O1 mol H2O

×1 mL H2O1.0 gH2O

= 18 mL H2O liquid

1 mol H2O at STP occupies 22.4 L.

Table 10.5 gives the vapor pressure of water at 0 °C to be 4.6 mmHg. At pressures higher than this, water would liquefy. We cannot achieve 1 atm pressure of water vapor at this low temperature, so we cannot achieve the standard state condition for water vapor.

70. The behavior of real gases is discussed in Section 10.9. As external pressure increases, the gas volume decreases, the molecules are squeezed closer together, and the attractions among the molecules get stronger. Figure 10.16 shows that a gas molecule strikes the walls of the container with less force due to the attractive forces between it and its neighbors. This makes the mathematical product PV smaller than the mathematical product nRT.

71. The behavior of real gases is discussed in Section 10.9. At low temperatures, the molecules are moving relatively slowly; however, when the pressure is very low, they are still quite far apart. As ext ernal pressures increases, the gas volume decreases, the slow molecules are squeezed closer together, and the attractions among the molecules get stronger. Figure 10.16 shows that a gas molecule strikes the walls of the container with less force due to the attractive forces between it and its neighbors. This makes the mathematical product PV smaller than the mathematical product nRT.

72. 1 mol He ×

4.003 g He1 mol He

×1 mL He

0.125 g He= 32.0 mL He liquid

1 mol Ne ×

20.18 g Ne1 mol Ne

×1 mL Ne1.20 g Ne

= 16.8 mL Ne liquid

1 mol Ar×

39.95 g Ar1 mol Ar

×1 mL Ar1.40 gAr

= 28.5 mL Ar liquid

1 mol Kr ×

83.8 g Kr1 mol Kr

×1 mL Kr2.42 gKr

= 34.6 mL Kr liquid

1 mol Xe ×

131.3 g Xe1 mol Xe

×1 mL Xe2.95 gXe

= 44.5 mL Xe liquid

The liquids composed of larger atoms occupy more volume in the liquid state. 1 mol of each of these gases at STP occupies 22.4 L.

The noble gas with the largest atom size (Xe) and a boiling point which is the closest to room temperature is most likely to show the largest deviation from ideality at room temperature.

73. Table 10.6 gives the values the van der Waals constants, a and b. The a constant is related to the pressure correction. The smaller the value of a, the closer to ideal the gas is.

aN2 = 1.39 aCO

2 = 3.59

N2 is more like an ideal gas at high pressures.

74. (a) T = 50 °C + 273 = 3.2×102 K


P=

nCO2 RT

V=

(7.0 mol CO2) × 0.08206L ⋅atmmol ⋅K

× (3.2×102 K)

(2.00 L)= 93 atm

(b) Solve the van der Waals equation for P: P=

nRT(V − nb)

−n2a

V2

P=

(7.0 mol) × 0.08206L⋅ atmmol ⋅K

× (3.2× 102 K)

2.00 L − (7.0 mol ) × (0.0427 )[ ] −(7.0 mol )2 × (3.59)

(2.00 L)2= 64 atm

Chemical Reactions in the Atmosphere 75. Introduced in Section 10.10, a free radical is an atom or group of atoms with one or more unpaired electrons;

as a result, it is highly reactive. For example: O2 • O • + • O •

Note: The paired electrons are not shown in these free radical structures, just the unpaired ones. The other four electrons on each of the O atoms are still present in two pairs, even though they are not shown here!

76. Introduced in Section 10.10, a free radical is an atom or group of atoms with one or more unpaired electrons;

as a result, it is highly reactive. Methyl radicals are • CH3. When two of them react with each other, the unpaired electron on each forms a bond pair:

• CH3 + • CH3 CH3–CH3

77. Names related to formulas were described in Chapter 3. Oxidation numbers were described in Section 5.4.

(a) NH3 Ox. #N + 3 × (+1) = 0 Ox. #N = –3

(b) NH4+ Ox. #N + 4 × (+1) = +1 Ox. #N = –3

(c) N2O 2 × Ox. #N + (–2) = 0 Ox. #N = + 1

(d) N2 elemental nitrogen Ox. #N = 0

(e) HNO3 (+1) + Ox. #N + 3 × (–2) = 0 Ox. #N = + 5

(f) HNO2 (+1) + Ox. #N + 2 × (–2) = 0 Ox. #N = + 3

(g) NO2 Ox. #N + 2 × (–2) = 0 Ox. #N = + 4

78. These reactions can be found in Section 10.10.

(a) • O• + O2 O3

(b) O3hν

• O• + O2

(c) •O •+ SO2 SO3

(d) H2O •H + •OH


(e) •NO2 + •NO2 N2O4


(a) •O• + H2O •OH + • OH

(b) •NO2 + ••OH HNO3

(c) RH + •OH •R + H2O

80. (a) An example of a reaction that creates a free radical is the decomposition of ozone into oxygen and a free radical oxygen atom:

O3 O2 + •O•

(b) An example of a reaction that creates a hydroxyl radical is a free radical oxygen atom reacting with a water molecule:

•O• + H2O 2 •OH

Ozone and Ozone Depletion


(a) CF3Clhν

• CF3 + • Cl

(b) • Cl + •O • ClO •

(c) ClO • + • O • •Cl + O2

82. CF4 has no C–Cl bonds, which in CCl4 are readily broken when exposed to UV light. Looking at the bond energies, C–Cl (327 kJ/mol) is much weaker than C–F (486 kJ/mol). In fact, the bond energy of C–F is very close to the bond energy of O=O (498 kJ/mol)!

83. CH3F has no C–Cl bonds, which in CH3Cl are readily broken when exposed to UV light. Looking at the bond energies, C–Cl (327 kJ/mol) is much weaker than C–F (486 kJ/mol). In fact, the bond energy of C–F is very close to the bond energy of O=O (498 kJ/mol)!

84. CFCs must have at least one F and one Cl atom. They must also have four halogens (either F or Cl) attached to the C atom. The total set of possible variations are:

FCCl3, F2CCl2, F3CCl

85. CFCs are not toxic. Refrigerants used before CFCs were very dangerous. One example is NH3, a strong-smelling, reactive chemical. In any web browser, type the keywords “CFCs” or “refrigerants” and you will get a plethora of hits.

86. New CFCs cannot catalyze the destruction of ozone at night, because sunlight is required to initiate the first step and create the • Cl radical. However, additional reactions recreate the • Cl, which can continue to destroy ozone even at night.

87. Ozone is formed from the reaction of oxygen and a free radical oxygen atom.


•O• + O2 O3

Sunlight is not required for this reaction to occur.

Chemistry and Pollution in the Troposphere

88. Primary pollutants are substances that are introduced into the air directly from their source.

• Particle pollutants: pollutants made out of particles:

– aerosols: particles incorporated into water droplets.

– particulates: larger solid particles

• Sulfur dioxide: pollutant produced when sulfur or sulfur compounds are burned in air.

• Nitrogen oxides: pollutant produced when nitrogen and oxygen react at high temperatures.

• Hydrocarbons: pollutants produced from many organic sources; their identity is small

hydrocarbons from CH4 to ones with six or seven carbons.

Secondary pollutants are substances produced from reactions of a primary pollutants.

• Ozone: ozone in the troposphere is produced from a reaction of O2 with O2 in the presence of intense energy (e.g., spark, lightening, etc.)

• Sulfur trioxide: SO3 is produced from the reaction of SO2.

• PAN (peroxyacetylnitrate): produced by the reaction of various free radicals in urban air.

Look at Section 10.12 for the specific ways these pollutants are harmful.

89. Particulates are small enough to stay suspended in the air for a long time, which makes them breathable. Breathing particulate pollution causes lung diseases.

90. Adsorption is the process of firmly attaching to a surface. Absorption is the process of drawing a substance into the bulk of a solid or liquid.

91. Define the problem: Given the percentage of removal of sulfur from coal, the mass of a sample of coal, the percentage of sulfur in the coal, and a chemical reaction showing how sulfur is removed, determine the mass (in metric tons) of one reactant.

Develop a plan: Determine the moles of sulfur in the sample of coal, use stoichiometry to determine the

moles of CaCO3, then convert to grams.

Execute the plan:

4 metric tons coal×

2 metric tonsS total100 metric tons coal

×90 metric tons S removed

100 metric tons S total= 0.07 metric tons S

0.07 metric tons S ×

1000 kg S1 metric ton S

×1000 g S1 kg S

×1 mol S

32.065 g S= 2 ×10 3 mol S

2 ×103 mol S ×

1 mol SO31 mol S

×1 mol CaCO 3

1 mol SO3= 2×103 mol CaCO 3


2 ×103 mol CaCO 3 ×

100.0869 g CaCO31 mol CaCO 3

×1 kg CaCO 3

1000 g CaCO3×

1 metric tons CaCO 31000 kg CaCO3

= 0.2 metric tons CaCO3

Check your answer: A large amount of sulfur needs about three times the amount of CaCO3 by mass.

92. Define the problem: Given the above information and the percentage of sulfur in the coal, determine the mass (in metric tons) of coal burned and determine the number of hours this quantity of coal will burn.

Develop a plan: Determine the moles of sulfur in the product, then determine the mass of coal that contains this amount of sulfur. Use the metric tons of coal per hour to determine the hours one plant would need to burn this quantity.

Execute the plan:

65 ×106 metric tons SO2 ×

1000 kg SO21 metric tons SO2

×1000 g SO21 kg SO2

= 6.5×1013 g SO2

6.5×1013 g SO2 ×

1 mol SO264.07 g SO2

×1 mol S

1 mol SO2×

32.065 g S1 mol S

= 3.3×1013 g S

3.3×1013 g S ×

100 g coal2 g S

×1 kg coal

1000 g coal×

1 metric tons coal1000 kg coal = 1.6× 109 metric tons coal

1.6 ×10 9 metric tons coal ×

1 hr700 metric tons coal

= 2 ×10 6 hr

That’s about 30 decades.

Check your answer: That seems like a long time. It is clear that there are many power plants adding SO2 to the atmosphere for this amount to be produced in one year.

93. Define the problem: Given the dimensions of a room, the concentration of a gaseous product, and a chemical reaction showing how that product is formed, determine the mass of one reactant.

Develop a plan: Determine the volume of the room. Then use the molar volume of a gas at STP to find moles. Use stoichiometry to determine the moles of gasoline, then convert to grams.

Execute the plan: V = (length) × (width) × (height) = (7 m) × (3 m) × (3 m) = 6 × 101 m3

6 ×101 m3 air ×

100 cm1 m

3×

1 mL

1 cm 3×

1 L1000 mL

×1000. L CO

1, 000, 000 L air×

22.414 mol CO1 L = 1×103 mol CO

1×10 3 mol CO ×

1 mol C8H188 mol CO

×114.2278 g C8H18

1 mol C8H18= 2×104 g C8H18

Check your answer: This amount of gasoline represents about 8 gallons, therefore less than a tank of gas would do it. This jives with the common warning against running the engine of a car in an enclosed space because it quickly generates dangerous levels of carbon monoxide gas.

Urban Air Pollution


94. Photochemical reactions require the absorption of a photon of light. An example of a photochemical reaction is given in this equation:

O2hν

•O• + •O•

Not all photons of light have sufficient energy to cause photochemical reactions. Looking at the example above, according to the discussion of Section 10.11, we find that only photons with wavelengths of less than 242 nm are able to break this bond. Therefore, this reaction does not occur if the light is visible light, or even low-energy ultraviolet (with wavelength 242 - 300 nm).

95. The two reactions that account for the production of oxygen free radical in the troposphere are given in Section 10.10:

O2hν

•O • + • O •

•NO2*hν

• NO + • O •

96. The reducing nature of industrial (London) smog is due to a sulfur oxide, SO2. The burning of coal and oil as fuels produce this oxide. Further oxidation happens in air according to this equation:

2 SO2 + O2 2 SO3

97. Two oxidizing gases in photochemical smog are O3 and NO. The energy source for photochemical smog is sunlight.

98. The atmospheric reaction that favors the formation of nitrogen monoxide, NO, is given in this chemical equation:

N2 + O2

heat2 NO

The formation of NO in a comb ustion chamber is similar to the formation of NH3 in a reactor designed to manufacture ammonia, because in both cases a reaction takes elemental nitrogen and makes a compound of nitrogen.

99. Stratospheric ozone is beneficial as a sunscreen. It screens out harmful ultraviolet light that can cause increased incidences of skin cancer in humans. Tropospheric ozone causes diminished respiratory capacity in humans. As a result, it poses a health hazard to people, especially those individuals who are susceptible to lung ailments.

100. Define the problem: Given the concentration of an air pollutant, determine the partial pressure, the mole fraction, and the mass in micrograms contained in a given volume of air at STP. (We are given the time of exposure, too, but that information is not needed to answer the questions asked.)

Determine a plan: Do (b) first by convert the concentration of SO2 from ppm to mole fraction, using

Avogadro’s law to relate volume and moles. Then do (a) by using p i = XiPtot and the standard air

pressure to find the partial pressure of SO2. (c) Use the molar volume of a gas to determine moles of SO2 in the given volume at STP, then convert to micrograms.

Execute the plan: Do (b) first, then (a), then (c).


(b)

0.175 L SO21,000, 000 L air

= 1.75 ×10−7 L SO2L air

A gas-volume ratio is equal to a mole ratio, according to Avogadro’s law, so

XSO2 = 1.75 × 10–7

(a) Assume that the Ptot = 760 mmHg, standard pressure.

pi = XiPtot = (1.75 × 10–7) × (760 mmHg) = = 1.33 × 10–4 mmHg

(c)

1 m3 air ×

100 cm1 m

3×

1 mL

1 cm 3×

1 L1000 mL

×0.175 L SO2

1,000,000 L air=1.75× 10−4 L SO2

1.75× 10−4 L SO2 ×

1 mol SO222.414 L SO 2

×64.0638 g SO2

1 mol SO2×

1 µg SO2

10−6 g SO2=500. µg SO2

Check your answers: The concentration is expressed in small units (ppm), so small partial pressure, the small mole fraction, and the small mass all make sense.

General Questions

101. Define the problem: Given the description of a reaction, the masses of two gaseous reactants, and the volume and temperature of the mixture, determine the partial pressure of the reactants, the total pressure due to the gases in the flask before and after the reaction, the excess reactant, the number of moles of it left over, the partial pressures of the gases in the flask, and the total pressure after the temperature is increased to a given value.

Determine a plan: Balance the equation. Find moles of each reactant and use the ideal gas law to determine the pressures. Use Dalton’s law for total pressures. Find limiting reactant and excess reactant as described in Chapter 4. Use the combined gas law to determine the pressure at a different temperature.

Execute the plan: H2(g) + Cl2(g) 2HCl(g)

(a) 3.0 g H2 ×

1 mol H22.0158 g H2

=1.5 mol H2 140 g Cl2 ×

1 mol Cl 270.906 g Cl 2

=2.0 mol Cl2

T = 28 °C + 273 = 301 K

(We will assume that the volume is 2 sig. figs to prevent serious round-off difficulties.)

pH2

=nRT

V=

(1.5 mol H2) × 0.08206L⋅atmK ⋅mol

× 301 K( )

10. L= 3.7 atm

pCl2

=nRT

V=

(2.0 mol Cl2) × 0.08206L ⋅atmK ⋅mol

× 301 K( )

10. L= 4.9 atm


(b) Ptot,before = pH2 + pCl

2 = (3.7 atm H2) + (4.9 atm Cl2) = 8.6 atm total

(c) The number of moles of gas reactants is equal to the number of moles of gaseous products, so the

total pressure will not change. Ptot, after = 8.6 atm total

(d) The balanced chemical equation shows equal molar quantities of each reactant reacting, so the H2 is

the limiting reactant and Cl2 is the excess reactant, and 1.5 moles of each reactant react. Subtracting

the moles of Cl2 that react from the moles of Cl2 describes how many moles of Cl2 remain in the flask.

2.0 mol Cl2 initial – 1.5 moles Cl2 react = 0.5 moles Cl2 remain

(e) When 1.5 moles of each reactant react, 3.0 moles of HCl are formed.

pHCl =

nRTV

=(3.0 mol HCl )× 0.08206

L ⋅atmK ⋅ mol

× 301 K( )

10. L=7.4 atm

pCl2 = Ptot,after – pHCl = (8.6 atm total) – (7.4 atm HCl) = 1.2 atm Cl2

(f) T1 = 301 K, T2 = 40 °C + 273 = 3.1 × 102 K

P2 = P1 ×

V1V2

×T2T1

= P1 ×T2T1

at constant volume

P2 = P1 ×

T2T1

= (8.6 atm) ×3.1× 102 K

301 K( ) = 8.9 atm

Check your answer: The partial pressure of Cl2 calculated in (e) can also be calculated using the idal gas law:

pCl2

=nRT

V=

(0.5 mol Cl2 ) × 0.08206L⋅atmK ⋅mol

× 301 K( )

10. L=1. atm

We could construct mole fractions of the products and multiply them by the total pressure to determine the partial pressures after the reaction was completed:

XHCl =

3.0 mol HCl3.0 mol HCl +0.5 mol Cl2

= 0.86

pHCl = XHClPtot = (0.86) × (8.6 atm) = 7.4 atm

XCl2 =

0.5 mol Cl23.0 mol HCl +0.5 mol Cl2

=0.14

pCl2 = XCl

2Ptot = (0.14) × (8.6 atm) = 1.2 atm


Ptot,after = pHCl + pCl2 = 7.4 atm + 1.2 atm = 8.6 atm total

These answers are the same as calculated above. The results are self-consistent and reasonable size (i.e., smaller number of moles produce smaller partial pressures and smaller mole fractions).

102. Define the problem: Given the information above, the description of a reaction, and the mass of a gaseous reactant, determine the mass of product produced.

Determine a plan: Use the masses, moles, and the stoichiometry of the balanced equations to determine the masses of the product.

Execute the plan: H2S + O3 SO2 + H2O

2 SO2 + O2 2 SO3

SO3 + H2O H2SO4

100 ×106 metric ton H2S×

1000 kg H2S1 metric ton H2S

×1000 g H2S1 kg H2S

= 1×1014 g H2S

1×1014 g H2S ×

1 mol H2S34.081 g H2S

×1 mol SO21 mol H2S

×2mol SO32 mol SO2

×1 mol H2SO4

1 mol SO3 = 3×1012 mol H2S O4

3× 1012 mol H2SO4 ×

98.078 g H2SO41 mol H2SO4

×1 kg H2SO4

1000 g H2SO4×

1 metric ton H2SO41000 kg H2SO4

=

3× 108 metric tonsH2SO4

Check your answer: 300 million metric tons of H2SO4 makes sense, because the molar mass of H2SO4 is

about three times the molar mass of H2S and the stoichiometry is 1:1, overall.

Applying Concepts

103. We are considering a real gas, N2(g), whose conditions are such that it obeys the ideal gas law exactly.

(a) N2 is a larger molecule than Ne is an atom and therefore N2 is not as much like an ideal gas. If these

conditions allow N2 to behave as an ideal gas, they would certainly be sufficient for Ne to also behave like an ideal gas. Hence, this statement: “A sample of Ne(g) under the same conditions must obey the ideal gas law exactly.” is true.

(b) All collisions are elastic, however, energy can be transferred during an elastic collision from one molecule to another. A faster molecule hitting a slower molecule might make the slow molecule go faster if the first one ends up slower. In addition, each time the molecule hits the wall, there is a split second when its speed is zero as it bounces off the wall and goes flying away in the opposite

direction. Hence, the statement: “The speed at which one particular N2 molecule is moving changes from time to time.” is true.

(c) The average speed of the N2 molecules will be faster than the average speed of the O2 molecules, but ideal gas particles move at varying speeds, so some molecules in each sample will be moving very

slowly and others very quickly. Hence, the statement: “Some of the N2 molecules are moving more

slowly than some of the molecules in a sample of O2(g) under the same conditions.” is true.


(d) The average speed of the N2 molecules will be slower than the average speed of the Ne molecules;

hence, the statement: “Some of the N2 molecules are moving more slowly than some of the molecules in a sample of Ne(g) under the same conditions.” is true.

(e) All collisions are elastic, so collisions must conserve energy. There is no way that both molecules could be going faster, since that implies that energy has increased. Hence, the statement: “When two

N2 molecules collide, it is possible that both may be moving faster after the collision than they were before.” is false.

104. The molar masses of C2H4 and N2 are 26 g/mol and 28 g/mol. The total pressure of the gases is 750

mmHg, and the partial pressure of N2 is 500 mmHg. Dalton’s law tells us that the sum of the partial

pressures is the total pressure, and so the partial pressure of C2H4 is 250 mmHg, or half that of the N2 molecules. The partial pressure is directly proportional to the mole fraction of the molecules in the

container, so there should be half as many C2H4 molecules than N2 molecules. That fact eliminates

choices (b) and (c). So, graph (a) with more N2 molecules than C2H4 molecules (indicated by smaller

vertical rise in the curve) and the lower average speed for the heavier N2 (indicated by the graph curve peaking at a slightly smaller speed value) is the best representation for this mixture.

105. The molar masses of C2H6 and F2 are 30 g/mol and 38 g/mol. That means the average speed of F2 is

somewhat less than that of C2H6. The total pressure of the gases is 720 mmHg, and the partial pressure of

F2 is 540 mmHg. Dalton’s law tells us that the sum of the partial pressures is the total pressure, and so the

partial pressure of C2H6 is 180 mmHg, or one third that of the F2 molecules. The partial pressure is directly proportional to the mole fraction of the molecules, in the container, so there should be one third

as many C2H6 molecules as F2 molecules. So, the graph of number of molecules verses molecular speed

will have the F2 curve peaking at a slightly smaller speed value and the C2H6 curve will have one third of the vertical rise.

F2 Number of Molecules

C2H6

Molecular Speeds (m/s)

106. Boyle’s law: V∝

1P

(unchanging T and n)

V = b

1P

where b is a proportionality constant.


BoyleÕs Law

V=b ( 1

P)

Vslope = b

1

P

Charles’ law: V∝ T (unchanging P and n)

V= cT where c is a proportionality constant.

CharlesÕ LawV=c T

Vslope = c

T Avogadro’s law: V∝ n (unchanging T and P)

V= an where a is a proportionality constant.

AvogadroÕs LawV=a n

Vslope = a

n

107. The speed of the molecules is related to the temperature. In particular, T ∝ mv 2. So, when the

temperature changes by a certain factor, the speed of the molecules (v) changes by the square root of that same factor. We’ll indicate that fact by making the tails of the arrows proportionately shorter. The pressure describes how close together the molecules are. When the pressure is changed by a certain factor, the molecules will be represented as that much closer together. The volume will change the space occupied by the gas, and that will be indicated in the movement of the syringe plunger. For reference, the initial state looks like this:


(a) When the temperature is decreased to one half of its original value, Charles’ law tells us that the volume decreases by half. The molecules are just as far apart as they were (since the pressure is the same), but the plunger level goes down by half, and the tails on the molecules decrease in length by about 0.7 times.

(b) When the pressure decreased to one half of its original value, Boyle’s law tells us that the volume increases by half. The tails on the molecules are the same length, but the molecules are twice as far apart than they were (so we’ll put half as many in this view), and the plunger level goes moves up by half.

(c) When the temperature is tripled and the pressure is doubled, the combined gas law tells us that the volume will have a net increase by 3/2 (three times larger due to the temperature change and half as large due to the pressure change). The molecules are two times closer together than they were (so we’ll add twice as many), the plunger level goes up by three halves, and the tails on the molecules increase in length by about 1.7 times.


108. The initial volume is 40 cm3 and the final volume is 60 cm3. This 2:3 ratio in the gas volumes means for every two molecules of gas reactants there must be three molecules of gas products. The reaction that exhibits this ratio is (c):

2 AB2(g) A2(g) + 2 B2(g)

109. The initial volume is 1.8 L and the final volume is 0.9 L. This 2:1 ratio in the gas volumes means for every two molecules of gas reactants there must be one molecule of gas products. The reactant count is six, so the box that has three product molecules fits these observations. The correct box is (b):

6 AB2(g) 3 A2B4(g)

110. Use methods described in the answer to Questions 26, 3.77, 8.26, and 8.77.

(a) The density times the molar volume gives the molar mass:

1.87 g1 L

×22.414 L

1 mol= 41.9

gmol

(b) 85 .7 g C ×

1 mol C12.0107 g C

= 7.1 mol C 14 .3 g H ×

1 mol H1.0079 g H

= 14.2 mol H

Mole ratio: 7.1 mol C : 14.2 mol H

Simplified: 1C : 2 H

Empirical formula: CH2

Empirical formula molar mass: 14.0 g/mol

(CH2)n n =

41.9 g/ mol14.0 g / mol

= 3

Molecular formula: C3H6

(c) C3H6 must have a ring or a double bond.

111. Use the methods described in the answer to Question 57, and Chapters 3 and 8.

(a) T = 50 °C + 273 = 3.2×102 K

750 mmHg ×

1 atm760 mmHg

= 0.99 atm 125 mL ×

1 L1000 mL

= 0.125 L


nCxH yFz =PVRT

=(0.99 atm)× (0.125 L)


× (3.2× 102 K)

= 4.7 × 10−3 mol CxHy Fz

Molar Mass =

0.298 g CxHyFz

4.7 × 10−3 mol CxHyFz= 63 g / mol CxHyFz

(b) 37.5 g C ×

1 mol C12.0107 g C

= 3.12 mol C 3.15 g H ×

1 mol H1.0079 g H

= 3.13 mol H

59.3 g F ×

1 mol F18.9984 g F

= 3.12 mol F

Set up a mole ratio and simplify:

3.12 mol C : 3.13 mol H : 3.12 mol F

1 C : 1 H : 1 F

The empirical formula is CHF and the molecular formula is (CHF)n.

So, n = x = y = z

The molar mass of the empirical formula = 32.02 g/mol

n =

63 g/ mol32.02 g / mol

= 2.0

The molecular formula is C2H2F2.

(c) C2H2F2 can have cis- and trans-isomers if the F atoms are on different C atoms.

112. No, it is wise not to agree with the assumption that the balloon is defective, at least not immediately, since the gas volume might have dropped due to the lower temperature, not necessarily because the balloon is defective.

Applying Concepts

113. Define the problem: Given the names and masses of three gases and the temperature and volume of the vessel they occupy, determine the total pressure in the vessel.

Develop a plan: Calculate the moles of each substance, using the molar masses. Add the moles of each gas to determine the total moles of gas in the vessel. Use the volume and temperature in the ideal gas law to determine the total pressure.

.. ..

.... ....

C C

F

F

H

H

C C

F

H

F

H

C C

F

H

H

F

..

....

..

....

..

..........


Execute the plan: Carbon dioxide is CO2 and methane is CH4. Vtot = 8.7 L

Find moles of CO2, CH4, and Ar in the sample:

3.44 g CO2 ×

1 mol CO244.009 g CO2

= 0.0782 mol CO2

1.88 g CH4 ×

1 mol CH416.0423 g CH 4

= 0.117 mol CH 4

4.28 g Ar ×

1 mol Ar39.948 g Ar

= 0.107 mol Ar

ntot = 0.0781 mol CO2 + 0.117 mol CH4 + 0.107 mol Ar = 0.302 mol tot

T = 37 °C + 273 = 335 K

P=

n totRTVtot

=(0.302 mol) × 0.08206

L⋅ atmmol ⋅ K

× (335 K)

(8.7 L)= 0.96 atm

Check your answers: The percentages are very close to those provided in the table, and the variations are explainable. The sum of the mole fractions is 1, and the sum of the percentages is 100 %.

114. Define the problem: Ginve the pressure of a gas in a sealed container (with fixed volume and number of moles) immersed in an ice water bath and the pressure of the same gas sample in an oven, determine the temperature of the oven.

Develop a plan: Use gas laws to determine how the pressure of each gas changes with temperature. Using the given temperature and pressures to determine the new temperature.

Execute the plan: The freezing point of water is 0.00°C which is 273.15 K. Pressure and temperature are

related by the idea gas law. PV=nRT. P=

nRV

T. So, P∝T at constant n & V.

So, PiTf = PfTi

Tf =

PiTfPi

=648 mmHg( )× (273.15 K)

(444 mmHg )= 399 K

Tf = 399 K – 273.15 K = 126 K

Check your answer: The temperature of an oven should be higher than room temperature.

115. NOTE: The formula for nitrogylcerine ought to be C3H5(NO3)3.

Define the problem: Given the mass of a reactant and the balanced chemical equation for the explosive decomposition of the reactant, determine the total moles of gases produced, the volume occupied by the product gases at a given temperature and pressure, and the partial pressures of each gas.

Develop a plan: Use molar mass to determine the moles of reactant, and use equation stoichiometry to calculate the moles of each gas phase reactant. (Note: reactants that are not gas phase will not contribute this sum). Add the moles of each gas to determine the total moles of gas in the vessel. Use the pressure


and temperature in the ideal gas law to determine the total volume. Calculate mole fraction with the

following equation: X i =

n intot

then partial pressure is calculated with p i = XiPtot

Develop a plan: Calculate the moles of reactant, using the molar masses.

Execute the plan: 1.00 kg C3H5NO3 ×

1000 g1 kg

×1 mol C3H5(NO3)3

103.0765 g C3H5(NO3)3= 9.70 mol C3H5(NO3)3

(a) Find moles of CO2, N2, and O2 in the sample: (Water is a liquid.)

9.70 mol C3H5(NO3)3 ×

12 mol CO24 mol C3H5(NO3)3

= 29.1 mol CO2

9.70 mol C3H5(NO3)3 ×

6 mol N24 mol C3H5( NO3)3

= 14.6 mol N2

9.70 mol C3H5(NO3)3 ×

1 mol O24 mol C3H5( NO3)3

= 2.43 mol O2

(b) ntot = 29.1 mol CO2 +14.6 mol N2 + 2.43 mol O2 = 46.1 mol tot

T = 25 °C + 273 = 298 K

Vtot =

ntotRTP

=(46.1 mol)× 0.08206

L ⋅atmmol ⋅K

× (298 K)

(1.0 atm)= 1.1×103 L

(c) XCO2 =

nCO2n tot

=29.1 mole CO246.1 mole tot

= 0.631 pCO2 = (0.631)(1.00 atm) = 0.631 atm

XN 2 =

n N2ntot

=14.6 mole N246.1 mole tot

= 0.317 pN2 = (0.317)(1.00 atm) = 0.317 atm

XO2 =

nO2n tot

=2.43 mole O246.1 mole tot

= 0.0527 pCO2 = (0.0527)(1.00 atm) = 0.0527 atm

Check your answers: The decomposition of TNT produces a large volume of gas product, which explains one reason why this decomposition is explosive. The sum of the mole fractions is 1, and the sum of the partial pressures is the total pressure.

116. Define the problem: Given the volume of liquid oxygen and its density, determine the volume of air (a mixture of gases) at STP that is necessary produce this amount of liquid oxygen.

Develop a plan: Use the density and molar mass to determine the moles of the liquid. Then use the molar volume of a gas at STP and the volume percent of oxygen in air to calculate the volume of dry air.

Execute the plan:


200. L O2 ×

1000 mL1 L

×1.14 g O21 mL O2

×1 mol O2

31.9988 g O2= 7.13 ×103 mol O2

The moles of O2 in the liquid must come from the air. Table 10.1 gives the percentage by volume for

oxygen in air.

7.13 ×103 mol O2 ×

22.414 L O21 mol O2 at STP

×100 Lair

20.948 L O2=7.62× 105 L air

Check your answer: Gases have a density much smaller than liquids, so the volume of gas must be much larger than the volume of liquid condensed from it.

117. Define the problem: Given the initial pressure of a gas, the relative amount of it that undergoes reaction, and the stoichiometric relationship between the reactants and products, determine the new pressure in the container.

Develop a plan: Determine the pressure of the gas reactant that reacts. Use the equation stoichiometry interpreted in units of pressure for the gases to determine pressure of the gas products. Then add this number to the pressure of the unreacted reactant gas to calculate the final pressure.

Execute the plan: Half of the reactant reacts, so the pressure of the reactant that undergoes reaction is 0.5×(550. torr) = 275 torr.

275 torr reactant×

2 torr product3 torr reactant

= 183 torr product

The unreacted reactant is has a pressure of 550 torr – 275 torr = 275 torr

The total pressure after this reaction is complete = 275 torr + 183 torr = 458 torr

Check your answer: The reaction reduces the number of gasses, so it makes sense that the pressure after some of the reactant has reacted is less.

118. Define the problem: Given the initial moles, pressure and temperature of a balloon filled with helium gas, determine the balloon volume, determine the mass of air with a give molar mass that would occupy the same volume at the same pressure and temperature, determine the mass of the helium gas, and determine the buoyancy of the balloon.

Develop a plan: Use ideal gas law to determine volume. Use the relationship between density, molar mass, and the ideal gas law to find the density of air, then use the density to determine the mass of air. Use the moles of helium and molar mass to find the mass of helium. Use the ESTIMATION BOX in Section 10.7 for instructions to determine the buoyancy, by calculating the density of the helium gas and determining the difference between the density of air and helium

Execute the plan:

(a) T = 25 °C + 273 = 298 K

VHe =

nHeRTP

=(500. mol) × 0.08206

L⋅atmmol ⋅K

× (298 K)

(1.00 atm)=1.22 ×104 L

(b) The given molar mass for air is 20.9 g/mol.


dair =PMRT

=1.00 atm( )× 29.0

g airmol


× (298 K)

= 1.19gL

air

1.22 ×104 L air ×

1.19 g air1 L air

= 1.45 × 104 g air

(c) 500. mol He ×

4.0026 g He1 mol He

= 2.00× 103 g He

(d) Find the density of the helium.

dHe =PMRT

=1.00 atm( )× 4.0026

g Hemol

0.082057L ⋅atmmol ⋅ K

× (298 K)

= 0.164gL

He

Buoyancy = Dair – dHe= 1.19 g/L – 0.164 g/L = 1.03 g/L

Check your answer: Helium is used for balloons because it makes them float, so it makes sense that the comparable results in this problem show that helium is less dense and creates positive buoyancy in a balloon.

119. Define the problem: Given the volume and pressure of a gas bubble at the bottom of a lake, determine the volume of the same bubble at the surface of the lake with given pressure.

Develop a plan: Use Boyle’s law to determine how the volume of gas changes with the decrease in pressuree.

Execute the plan: Use Boyle’s law:

Vf ,bubble =

Pi,bubble Vi,bubblePf ,bubble

=4.4 atm×

760 torr1 atm

× 1 mm3

(740 torr)= 4.5 mm3

Check your answers: At the lower pressure, it makes sense that the bubble will increase in volume.

120. Define the problem: Given the volume of a container, determine how many moes of gas at a specified pressure and temperature can be stored in it, and determine the amount of liquid with a given density that can be stored in it.

Develop a plan: Use the ideal gas law to determine the moles of the gas. Use the density of the liquid to determine the mass of the liquid and then convert from grams to moles.

Execute the plan: T = 25 °C + 273 = 298 K

ngas =PVRT

=(3.00 atm) × (150 L)

0.082057L⋅atmmol ⋅K

× (298 K)

= 18 mol gas


150 L ×

1000 mL1 L

×0.60 g liquid

1 mL×

1 mol44.0953 g

= 2× 103 mol liquid

Check your answer: The liquid is more dense than the gas so it makes sense that the moles of liquid in the tank is far greater than the moles of gas that would fit in the same tank.

121. (a) When the gas is compressed to a smaller volume, the molecules will be closer together and they will collide with each other more often. The temperature is fixed, so the molecules will hit each other at the same average speed; however, with more collisions the interactions between the molecules will be more noticeable.

(b) When more molecules of the same gas are added to the container, the molecules will be closer together and they will collide with each other more often. The temperature is fixed, so the molecules will hit each other at the same average speed; however, with more collisions the interactions between the molecules will be more noticeable.

(c) When the temperature is increased, the average kinetic energy of the molecules is increased and the molecules will move faster. That means they will collide with each other more often. Because the speed has increased the time these molecules spend in proximity to each other will decrease, so, compared to the situation in (a) and (b), the effect of intermolecular interactions will not be as great.

122. Define the problem: Given parts per billion by volume of a molecular gas in air, determine the partial pressure of the gas in air and determine the number of molecules of this molecular gas in two different volumes.

Develop a plan: Use the relationship between volume and moles and the given ppbv of CH2O to

determine the partial pressure. Use the ppbv of CH2O, the molar volume of a gas and Avogadro’s number

to determine the number of molecules in each of the two volumes.

Execute the plan: (a) There is 1.0 ppbv CH2O, so 1.0 ppbv=

VCH2OVtot

×10 9 , and 1.0× 10−9 =

VCH2OVtot

V is proportional to moles, at constant pressure and temperature, so the volume ratio is the same as the

moles ratio,

VCH2OVtot

=nCH2O

ntot.

1.0× 10−9 =

VCH2OVtot

=nCH2O

n tot = XCH 2O

pCH2O = XH 2O( )Ptot( ) (1.0 × 10–9)(1.00 atm) = 1.0 × 10–9 atm

(b)

1 cm3 ×

1 L

1000 cm3×

1.0 ×10−9 L CH 2O1 L air

×1 mol

22.414 Lat STP×

6.022 ×1023 molecules1 mole

= 2.7 ×1010 molecules

(c)

15.0 ft ×10.0 ft ×8 ft( )×

12 in1 ft

3×

2.54 cm1 in

3×

2.7 ×1010 molecules

1 cm3= 9×1017 molecules

Check your answer: Though 10 trillion molecules seems like a lot, compared to the number of molecules in normal size samples, these are very tiny. It makes sense that the smaller volume has fewer molecules in


it.

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Chapter 10: Gases and the Atmosphere Solutions …academic.udayton.edu/.../123_2004/Problemanswers_ch10.pdfChapter 10: Gases and the Atmosphere 315 11. Starting from the earth’s