chapter 9 – the gaseous statechemweb/chem1000/docs/online course docs...chapter 9 – the gaseous...

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Chapter 9 – The Gaseous State 9.1 (a) effusion; (b) Boyle’s law; (c) combined gas law; (d) ideal gas; (e) pressure; (f) Dalton’s law of partial

pressures; (g) molar volume; (h) ideal gas constant, R 9.2 (a) diffusion; (b) Charles’s law; (c) Gay-Lussac’s law of combining volumes; (d) ideal gas law; (e)

barometer; (f) kinetic-molecular theory; (g) standard temperature and pressure (STP); (h) Avogadro’s hypothesis

9.3 The y-axis units and label are missing (in both the graph and the data tables), x-axis units are missing, the

points should not be connected by lines (a straight “best fit” line that comes closest to all points should be used), gridlines should be uniformly labeled (for example, 10, 20, 30, 40, 50, etc.), the x-axis maximum should be 60 instead of 120, a graph title is missing.

9.4 Each graph axis should have labels and units. In addition rather than “connecting the dots”, scientific

graphs use “trend lines” which show the trend of the data. Notice the trend line is a “best-fit” line and does not necessarily touch any of the data points. A title should also be included.

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9.5 There is a trend, but it is not linear because the data clearly are curved in an upward direction.

9.6

After graphing the data draw a best-fit line through the data points. Determining the equation that fits the line requires determining the variables m and b in the equation for a straight line:

y = mx + b You can do this by using two points on the best fit line. By taking the points from the best-fit line you are getting the equation that best fits all the data. 100ºC 67 g/100 mL 40ºC 34 g/100 mL Next solve the set of equations:

67 = m(100) + b 34 = m(40) + b m = 0.55 g/(ºC·100 mL) b = 12 g/100 mL Equation: y = 0.55x + 12

9.7 The values for ∆h represent the pressure applied to the gas in excess of 1 atm. The length of the gas

column is proportional to the volume of the gas. When the pressure is high, the volume is small. As the pressure decreases, the volume increases. Gas pressure and volume are inversely related.

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9.8 Pressure and volume are inversely related. Because the slope is constant we can state that

Pressure (∆h) = 1kV

× or PV = k

where k is a constant.

9.9 You can estimate the vapor pressure by drawing a line vertically from the temperature and seeing where the

intersection point lies on the vapor pressure axis (horizontal line). 22ºC 21 torr 38ºC 50 torr

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9.10 There appears to be a linear trend in the data. At 0.92 g/mL the velocity is approximately 1360 m/s.

9.11 (a) 5 1 3x + = ; subtract 1 from both sides

5 2x = ; divide both sides by five

x = 25

= 0.4

(b) 0.412 / 2.00x = ; multiply both sides by x

0.412 2.00 x= ⋅ ; divide both sides by 2

x = 0.4122.00

= 0.206

(c) 2 232x x= − ; add x2 to both sides

22 32x⋅ = ; divide both sides by 2 2 16x = ; take the square root of both sides

x = 2 16x = = ±4

(d) 2 6x x= − ; add x to both sides 3 6x = ; divide both sides by 3

x = 63

= 2

9.12 (a) 14 16 44x + = ; subtract 16 from both sides

14 28x = ; divide both sides by 14

x = 2814

= 2.0

(b) 1 3 53

x + = ; subtract 3 from both sides

1 23

x = ; multiply both sides by 3

x = 2 3× = 6 (c) ( )12 4 12x x= − ; divide both sides by four

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3 12x x= − ; add x to both sides 4 12x = ; divide both sides by 4

x = 124

= 3

(d) 3 15 4 12x x+ = + ; subtract 3x from both sides

15 12x= + ; subtract 12 from both sides x = 15 12− = 3

9.13 For temperature conversions we use the equation:

1.8 32F CT T° °= + To convert from Fahrenheit to Celsius we rearrange the equation:

( )321.8

FC

TT °°

−=

It is worth noting that 1.8 and 32 are exact in this equation and that the significant figures are determined by the precision of the temperature you are converting:

(a) ( )212 32 180. 100. C

1.8 1.8CT°−

= = = °

(b) ( )80.0 32 48.0 26.7 C

1.8 1.8CT°−

= = = °

(c) ( )32.0 32 0.0 0.0 C

1.8 1.8CT°−

= = = °

(d) ( )40.0 32 72.0 40.0 C

1.8 1.8CT°− − −

= = = − °

9.14 For Celsius to Fahrenheit temperature conversions we use the equation:

1.8 32F CT T° °= + It is worth noting that 1.8 and 32 are exact in this equation and that the significant figures are determined by the precision of the temperature you are converting:

(a) ( )1.8 37.0 32 66.6 32 98.6 CFT° = × + = + = °

(b) ( )1.8 212 32 381.6 32 414 CFT° = × + = + = °

(c) ( )1.8 100.0 32 180.0 32 212.0 CFT° = × + = + = °

(d) ( )1.8 40.0 32 72.0 32 40.0 CFT° = × − + = − + = − ° 9.15 Solve the equation algebraically and then calculate for the unknown value.

(a) ( )( )

( )( )1.00 1.00

24.4.500 0.08206

PVTnR

= = =

(b) ( )( )( )( )

0.750 3.000.116

0.08206 237PVnRT

= = =

(c) ( )( )( )

( )1.50 0.08206 455

17.23.25

nRTVP

= = =

(d) ( )( )( )

( )2.67 0.08206 322

4.7015.0

nRTPV

= = =

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9.16 Solve the equation algebraically and then calculate for the unknown value.

(a) ( )( )

( )( )3.55 1.75

369.205 0.08206

PVTnR

= = =

(b) ( )( )

( )( )1.00 22.5

0.9200.08206 298

PVnRT

= = =

(c) ( )( )( )

( )3.00 0.08206 535

10500.125

nRTVP

= = =

(d) ( )( )( )

( )1.57 0.08206 343

7.076.25

nRTPV

= = =

9.17 When compared to other states of matter, gases have low densities and are very compressible. They also

take the shapes of their containers. We describe a gas in terms of its pressure, temperature, volume, and the number of moles of the gas in the sample.

9.18 Unlike liquids and solids, gases take the shapes of and fill their containers. Atoms and molecules in the gas

phase are very much more mobile than they are in the solid or liquid phases. Compared to solids and liquids, gases have low densities and are very compressible.

9.19 The density of warm air is lower than the density of cool air. 9.20 Warm air rises because it is less dense than cool air. 9.21 As temperature decreases, gas molecules move more slowly. As a result, they collide with less force, and a

larger number of molecules occupy a unit volume. As shown in the figure, at lower temperatures the gas density is higher, but the molecular velocity is lower.

9.22 (a) When a gas is heated, either the pressure or volume (or both) can change. If the volume changes, there

should be fewer molecules in the image, but the piston should have moved upward. The gas molecules should also be moving faster as a consequence of the temperature increase. (b) The piston should have moved up to show the expansion of the gas. The image incorrectly shows that the molecules increase in size. When the volume of gas increases, the molecules do not increase in size. (c) The sample density should decrease (because the volume increased), but the size of molecules does not change. (d) Although the piston moved up, the density of molecules did not go down as it should have.

9.23 Gas pressure is the amount of force exerted by the gas particles divided by the area over which the force is

exerted. forcearea

P =

9.24 When a gas molecule strikes a wall, the wall experiences a force. When we add together the forces of all

the molecules striking the wall and divide by the area of the wall, we determine the pressure of the gas.

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9.25 We measure absolute pressure with a barometer (Figure 9.11). Essentially, a barometer allows us to compare the pressure exerted by a column of a liquid (usually mercury) to the pressure exerted by the atmosphere. A tire gauge compares the pressure of the air inside a tire to atmospheric pressure.

9.26 (a) We commonly use many different units for measuring pressure. These would include atmospheres,

torr, pascals, kilopascals, pounds per square inch, and inches, centimeters and millimeters of mercury. (b) Here are several different pressure units and their relationships to 1 atmosphere (atm).

1 atm = 29.9 in Hg = 76 cm Hg = 760 mm Hg = 760 torr = 101,325 Pascal = 101.325 kPa = 14.7 lb/in2 (also known as pounds per square inch, or psi)

Because each of these values is equal to 1 atm, they are also equal to each other. When we convert from inches of mercury to torr, for example, we can use the relationship 760 torr = 29.9 in Hg.

9.27 If the temperature doesn’t change, the velocity of the gas molecules remains constant. Because the volume

is larger, the density of molecules is lower.

9.28 (a) If the volume of a gas sample increases, the density of gas molecules should decrease because we are

not adding any gas molecules so those that are present can spread out over a larger volume. In this image, the density of particles is the same (which is incorrect) and the molecules increase in size (which is also incorrect). When a gas expands, the space between the molecules increases but the molecules do not change size. (b) The density of the particles did not change; it should have decreased. (c) The density appears to have changed, but the piston did not move. In order for the volume to increase, the piston must move up (as is shown in figures (a) and (b)). (d) The piston should have moved up and the gas molecules should have remained the same size.

9.29 We use the following relationships to convert between the various pressure units:

1 atm = 29.9 in Hg = 76 cm Hg = 760 mm Hg = 760 torr = 101,325 Pa = 101.325 kPa = 14.7 lb/in2

For example, to convert between pascals and mm Hg, use the relationship: 760 mm Hg = 101,325 Pa.

(a) Pressure in atm = 745 torr 1 atm760 torr

× = 0.980 atm

(b) Pressure in torr = 1.23 atm 760 torr1 atm

× = 935 torr

(c) Pressure in atm = 90.1 mm Hg 1 atm760 mm Hg

× = 0.119 atm

(d) Pressure in Pa = 0.643 kPa 1000 Pa1 kPa

× = 643 Pa

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(e) Pressure in mm Hg = 51.35 10 Pa×760 mm Hg101,325 Pa

× = 1.01 × 103 mm Hg

(f) Pressure in torr = 47.51 10 Pa×760 torr

101,325 Pa× = 563 torr;

(g) Pressure in Pa = 798 torr 101,325 Pa760 torr

× = 1.06 × 105 Pa;

(h) Pressure in mm Hg = 29.3 cm 10 mmHg1 cm

× = 293 mm Hg

9.30 We use the following relationships to convert between the various pressure units:

1 atm = 29.9 in Hg = 76 cm Hg = 760 mm Hg = 760 torr = 101,325 Pa = 101.325 kPa = 14.7 lb/in2

For example, to convert between pascals and millimeters of mercury, use the relationship: 760 mm Hg = 101,325 Pa.

(a) Pressure in torr = 1.15 atm 760 torr1 atm

× = 874 torr

(b) Pressure in atm = 968 torr 1 atm760 torr

× = 1.27 atm

(c) Pressure in atm = 52.50 10 Pa×1 atm

101,325 Pa× = 2.47 atm

(d) Pressure in torr = 695 mm Hg 760 torr760 mm Hg

× = 695 torr

Note that the units torr and mm Hg are interchangeable: 1 torr = 1 mm Hg.

(e) Pressure in Pa = 0.953 atm 101,325 Pa1 atm

× = 9.66 × 104 Pa

(f) Pressure in mm Hg = 653 torr 760 mm Hg760 torr

× = 653 mm Hg

(g) Pressure in Pa = 1545 mm Hg 101,325 Pa760 mm Hg

× = 2.060 × 105 Pa

(h) Pressure in kPa = 3.73 kPa 1 atm101.325 kPa

× = 3.68 × 10−2 atm

9.31 The conversion from inches of mercury to pascals is quite lengthy. The first half of the conversion

involves converting from inches to millimeters (English to metric conversion). The second half involves using the relationships among pressure units to convert from millimeters of mercury to pascals. The problem solving map looks like:

2.54 cm 1 in 10 mm 1 cmPressure in Hg cm Hg mm Hg= =→ →

760 mm Hg 1 atm 1 atm 101,325 Pamm Hg atm Pa= =→ →

Pressure in Pa = 30.24 in 2.54 cm×

1 in10 mm

×1 cm

1 atm×

760 mm101,325 Pa

1 atm× = 1.024 × 105 Pa

An alternate, but important, conversion method involves using the definitions of pressure given in the text. We know that 1 atm = 29.9 in Hg and also that 1 atm = 101,325 Pa. This means that 29.9 in Hg = 101,325 Pa. Using this factor, we can convert from inches to pascals in one step (although with less precision):

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Pressure in Pa = 30.24 in Hg 101,325 Pa29.9 in Hg

× = 1.02 × 105 Pa

9.32 The conversion from kilopascals to inches of mercury is quite lengthy. The first half of the conversion is

converting from kilopascals (kPa) to millimeters of mercury (mm Hg) (pressure conversions). The second half involves converting from millimeters of mercury to inches (English to metric conversion). The problem solving map looks like:

1000 Pa 1kPa 101,325 Pa 1 atm 760 mm Hg 1 atm 10 mm 1 cmkPa Pa atm mm Hg2.54 cm 1 incm Hg in Hg

= = = =→ → → →=→

Pin Hg = 101.19 kPa 1000 Pa×

1 kPa1 atm

×101,325 Pa

760 mm Hg×

1 atm1 cm Hg

×10 mm Hg

1 in Hg2.54 cm Hg

× = 29.881 in Hg

The conversion factor 2.54 cm/in is exact.

An alternate, but important, conversion method involves using the definitions of pressure given in the text. We know that 1 atm = 29.9 in Hg and also that 1 atm = 101,325 Pa. This means that 29.9 in Hg = 101,325 Pa. To convert to kilopascals, we use the definition: 1 kPa = 1000 Pa:

Pressure in Hg = 101.19 kPa 1000 Pa×

1 kPa29.9 in Hg

101,325 Pa× = 29.9 in Hg

9.33 Boyle’s law tells us that, at constant temperature, as the pressure of a gas increases the volume decreases.

This is an inverse relationship, so that if the pressure increases by a factor of three, the volume decreases to one third (1/3) of its original volume, assuming the temperature remains constant.

9.34 Boyle’s law is an inverse relationship: As the volume of a gas sample increases, the pressure of the gas

decreases, assuming that the temperature is kept constant. When the volume of a gas increases by four times its original volume, its pressure decreases to one fourth its original pressure (at constant temperature). Mathematically, we can write Boyle’s law as P1V1 = P2V2.

9.35 If the container volume decreases, the particles will collide with the container walls more frequently,

causing an increase in pressure. If the container size is decreased to half its original volume, and the sample size remains constant, we expect to find twice as many molecules in the same space. The velocity of the molecules will not change as long as the temperature remains constant, so the pressure on the container walls will double.

9.36 If the volume of the container is increased, the particles will collide with the container wall less frequently

causing a decrease in pressure. If the container volume is tripled, the pressure of the gas molecules against the container walls decreases to one third of its original pressure. This is because, at constant temperature, the velocity of the molecules does not change, but the frequency of their collisions with the container walls decreases. Because there are 8 gas molecules in the original picture there should be about 8 ´ 1/3 = 2 2/3 molecules in the new picture.

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9.37 (a) 3.60 L; (b) 26.7 mL; (c) 0.392 mL. Boyle’s law describes the relationship of pressure and volume

changes (P & V) on a gas sample, assuming that the sample size (number of moles) and temperature are kept constant. Boyle’s law is an inverse relationship: As the volume of a gas sample increases, the pressure of the gas decreases, assuming that the temperature is kept constant. For example, in part (a) because the pressure increases by a factor of 5.00 atm/2.00 atm (units must be the same), we conclude that the volume will decrease by a factor of 2.00 atm/5.00 atm. We can compute the final volume as follows:

Final volume = 3.00atm 6.00L5.00atm

× = 3.60 L

Boyle’s law also states that P1V1 = P2V2. For part (a) we have:

V1 P1 P2 V2 6.00 L 3.00 atm 5.00 atm ?

We can rearrange Boyle’s law and solve for V2 as follows:

1 12

2=

PVVP

V2 = (3.00 atm )(6.00L)5.00 atm

= 3.60 L

By either applying the proportionality, or using Boyle’s law, we obtain the same result. After you complete your calculation, you should make sure to evaluate your answer to ensure that it is reasonable. The effect of the pressure increase should be a volume decrease. (b) The pressure increases, so we should see a volume decrease, and we do.

V2 = (60.0 torr )(40 mL)90.0 torr

= 26.7 mL

(c) The pressure increases, so we should see a volume decrease, and we do.

V2 = (40.0 torr )(2.50mL)255 torr

= 0.392 mL

9.38 Boyle’s law describes the relationship of pressure and volume changes (P and V) on a gas sample,

assuming that the sample size (number of moles) and temperature are kept constant. Boyle’s law is an inverse relationship: As the volume of a gas sample increases, the pressure of the gas decreases, assuming that the temperature is kept constant. Before we apply Boyle’s law, we must make certain that the pressure units are the same. Mathematically, we can write Boyle’s law as P1V1 = P2V2. For these problems, we are solving for V2:

1 12

2=

PVVP

(a) Using Boyle's law, we get the new volume:

9−11

V2 = (2.50 atm )(2.50L)725 atm

= 0.00862 L

Because the pressure increases, the volume should decrease, and it does. (b) With a decrease in pressure we expect an increase in the volume, V2.

V2 = (825 torr )(6.25L)456 torr

= 11.3 L

(c) V2 = (50.0 torr )(450mL)30.0 torr

= 7.5 × 102 mL

9.39 According to Boyle's law, if the volume of a gas sample increases by a factor of 1512 mL/405 mL, its

pressure decreases by a factor of 205 mL/1512 mL. For part (a) we can write:

P2 = 405 mL1512 mL

602 torr× = 161 torr

We can obtain the same results using the mathematical expression of Boyle's law, P1V1 = P2V2, solving for P2:

1 12

2=

PVPV

Note: When we solve these problems, the volume units must agree so that they cancel properly.

(a) P2 = (602 torr)(405 mL )1512 mL

= 161 torr

Because the sample volume increases, we expect that the pressure will decrease, and it does.

(b) Before we can use Boyle's law, we must convert 1.50 L into mL.

Volume in mL = 1.50 L 1000 mL1 L

× = 1.50 × 103 mL

P2 = 3(0.00100 torr)(1.50 10 mL× )

15.0 mL = 0.100 torr

Because the volume decreases, we expect the pressure to increase, and it does.

(c) P2 = (0.832 atm)(805 L )37.5 L

= 17.9 atm

Because the volume decreases, we expect pressure to increase, and it does. 9.40 According to Boyle's law, a decrease in the volume of a gas results in an increase in pressure.

Mathematically, we express Boyle's law as P1V1 = P2V2. We are looking for the final pressure, so we solve the equation for P2:

1 12

2=

PVPV

If the units of V1 and V2 are the same, we need not worry about unit conversions. (a) Prior to using Boyle's law, we must convert 155 mL to L so that the volume units agree.

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Volume in L = 155 mL 1 L1000 mL

= 0.155 L

P2 = (845 torr)(0.155 L )1.55 L

= 84.5 torr

We see that the calculated pressure is lower than the initial pressure as a result of an increase in volume. This is in agreement with the inverse relationship between pressure and volume shown by Boyle’s law.

(b) P2 = (5.30 atm)(2.85 L )4.50 L

= 3.36 atm

Because the volume increases, the pressure decreases as predicted by Boyle’s law.

(c) Prior to using Boyle's law, we must convert 5500 mL to L so that the volume units agree.


= 5.5 L

P2 = (755 torr)(2.00 L )5.5 L

= 275 torr

Because the volume increases, the pressure decreases as predicted by Boyle’s law. 9.41 It is helpful to organize the data in table form.

P1 V1 P2 V2 1.25 atm 925 L ? 6.35 L

Based on Boyle's law, we predict that if the volume of the gas sample decreases by a factor of 6.35 L/925 L, the pressure should increase by a factor of 925 L/6.35 L, assuming the sample size and temperature are kept constant. Solving Boyle's law for P2 we have:

1 12

2=

PVPV

P2 = 1.25atm 925 L×6.35 L

=182 atm

As we predicted, the pressure is higher by a factor of 925 L/6.35 L. 9.42 It is helpful to organize the data in table form.

P1 V1 P2 V2 0.945 atm 186 mL 1.76 atm ?

Based on Boyle's law, we predict that if the pressure of a gas sample increases by a factor of 1.76 atm/0.945 atm, its volume will decrease by a factor of 0.945 atm/1.76 atm, assuming that the sample size and temperature are kept constant. Solving Boyle's law for V2 we have:

1 12

2=

PVVP

V2 = (0.945 atm )(186 mL)1.76 atm

= 99.9 mL

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As we predicted, the volume decreased by a factor of 0.945 atm/1.76 atm. 9.43 Organize the data in table form.

P1 V1 P2 V2 1.00 atm 0.550 L 725 torr ?

Notice that the units of pressure for P1 and P2 differ. Because 1 atm = 760 torr, we can substitute 1.00 atm with 760 torr in the Boyle's law expression:

1 12

2=

PVVP

= (760 torr )(0.550L)725 torr

= 0.577 L

The volume of H2 required is larger than 0.550 L because it is being collected at a lower pressure. 9.44 Organize the data in table form.

P1 V1 P2 V2 2.00 atm 2.50 L 735 torr ?

Notice that the units of pressure for P1 and P2 differ. We must convert 735 torr to atm before using Boyle's law.

Pressure in atm = 735 torr 1 atm760 torr

× = 0.967 atm

1 12

2=

PVVP

= (2.00 atm )(2.50L)0.967 atm

= 5.17 L

The larger volume the gas occupies at the lower pressure is consistent with Boyle's law. 9.45 Charles’s law states that if the pressure and sample size of a gas are kept constant, the volume and

temperature (in kelvins) of the gas are directly proportional to each other (volume increases when temperature increases; volume decreases when temperature decreases). If the temperature increases, the gas will occupy a larger volume.

9.46 Charles’s law indicates the direct proportionality between the absolute temperature of a gas and its volume,

provided the pressure and sample size are kept constant. If the temperature increases, the volume of the gas will increase. If the temperature decreases, the volume of the gas will decrease. Mathematically, we can write Charles’s law as:

1 2

1 2=

V VT T

9.47 The velocity of the gas particles increases as temperature increases. This means they strike the walls of the container with greater force. If the container volume does not increase, the gas pressure increases. To maintain a constant pressure, as stated in the problem, the volume of the container will increase.

9.48 The velocity of the gas particles decreases as the temperature decreases. This means that they strike the

container walls with less force. If the container does not decrease in volume, the pressure decreases. To maintain a constant pressure, the volume of the container will decrease.

9.49 In Charles's law problems, we must always express temperature in kelvins. We can state Charles’s law

mathematically as:

1 2

1 2=

V VT T

9−14

Solving for V2: 12 2

1= ×

VV TT

We see that if the temperature, T2, increases, the volume, V2 must also increase. (a) Both temperatures are given in Celsius. The conversion is: TK = T°C + 273.15 K.

T1 = 30.0°C + 273.15 = 303.2 K

T2 = 0.0°C + 273.15 = 273.2 K

26.00 L

303.2 KV = 273.2 K× = 5.41 L

Because the temperature decreases, the volume also decreases.

(b) Both temperatures are given in Celsius. The conversion is: TK = T°C + 273.15 K.

T1 = −60.0°C + 273.15 = 213.15 K (4 sig figs)

T2 = 401.0°C + 273.15 = 674.15 K (4 sig figs)

2212 mL

213.15 KV = 674.15 K× = 671 mL

Because temperature increases, the volume also increases.

(c) 247.5 L212 K

V = 337 K× = 75.5 L

Because the temperature increases, the volume also increases. 9.50 When we solve Charles's law problems, we must always express temperature in kelvins. We can state

Charles’s law mathematically as:

1 2

1 2=

V VT T

Solving for V2: 12 2

1= ×

VV TT

We see that if the temperature, T2, increases, the volume, V2 must also increase. (a) Both temperatures are given in Celsius. The conversion is: TK = T°C + 273.15 K.

T1 = 0.0°C + 273.15 = 273.2 K

T2 = 100.0°C + 273.15 = 373.2 K

2224 L

273.2 K=V 373.2 K× = 306 L

Because the temperature increases, the volume also increases.

(b) 2152 mL45 K

=V 450 K× = 1.5 × 103 mL

Because the temperature increases, the volume also increases.

(c) Both temperatures are given in Celsius. The conversion is: TK = T°C + 273.15 K.

9−15

T1 = 45°C + 273.15 = 318 K

T2 = 450°C + 273.15 = 7.2 × 102 K

2156 L318 K

=V 27.2 10 K× × = 3.5 × 102 mL

Because the temperature increases, the volume also increases. 9.51 When we solve Charles's law problems, we must always express temperature in kelvins. We can state

Charles’s as:

1 2

1 2=

V VT T

Solving for T2: 12 2

1= ×

TT VV

(a) Although we are asked to give the final temperature in Celsius, we must solve for T2 in kelvins and then express that temperature in degrees Celsius.

T1 = 0.0°C + 273.15 = 273.2 K

2 140.0 mLT =273.2 K70.0 mL

× = 546 K

Because the volume doubles, the temperature also doubles. Convert T2 to Celsius:

T°C = TK − 273.15 K = 546 K − 273.15 = 273°C

(b) The units for volume must be the same so they will cancel. First, we convert 85 mL to liters and −37°C to kelvins before we apply Charles’s law:

Volume in liters = 85 mL l L1000 mL

× = 0.085 L

T1 = −37°C + 273.15 = 236 K

2 0.085 LT =236 K2.55 L

× = 7.9 K

Because the volume decreases, the temperature also decreases.

T°C = TK − 273.15 K = 7.9 K − 273.15 = −265.2°C

(c) 2 135 LT =165 K87.5 L

× = 255 K

T°C = TK − 273.15 K = 255 K − 273.15 = −19°C

9.52 When we solve Charles's law problems, we must express the temperature in kelvins. We can state

Charles’s law mathematically as:

1 2

1 2=

V VT T


1= ×

TT VV

(a) The units for volume must be the same so they will cancel. First, we convert 100.0°C to kelvins:

9−16

T1 = 100.0°C + 273.15 = 373.2 K

2 100.0 mLT =373.2 K

250.0 mL× = 149.2 K


T°C = TK − 273.15 K = 149.2 K − 273.15 = −124.0°C

(b) First, we convert 27.5°C to kelvins before we use Charles’s law:

T1 = 27.5°C + 273.15 = 300.7 K

2 148 mL=T 300.7 K125 mL

× = 356 K

Because the volume increases, the temperature also increases.

T°C = TK − 273.15 K = 356 K − 273.15 = 83°C

(c) 2 57.2 L=T 300 K13.7 L

× = 1 × 103 K (to 1 significant figure).

T°C = TK − 273.15 K = 1 × 103 − 273.15 = 1 × 103 °C (to one significant figure)

9.53 It is often helpful to create a table like the one shown below and input the data from the problem.

V1 T1 V2 T2 0.150 mL 24.2°C ? 62.5°C

Because only temperature and volume are changing, we use Charles’s law to solve for the new volume:

12 2

1= ×

VV TT

Before doing the calculation, we must convert the temperatures to kelvins:

T1 = 24.2°C + 273.15 = 297.4 K

T2 = 62.5°C + 273.15 = 335.7 K

20.150 mL297.4 K

V = × 335.7 K = 0.169 mL

Because the temperature increases, the volume of the bubble also increases. 9.54 It is often helpful to create a table like the one shown below and input the data from the problem.

V1 T1 V2 T2 79.0 L 27.0°C 32.0 L ?

Because only temperature and volume are changing, we use Charles’s law and solve for temperature:

12 2

1= ×

TT VV

Convert 27.0°C to kelvins:

T1 = 27.0°C + 273.15 = 300.2 K

9−17

2 32.0 L=T 300.2 K79.0 L

× = 121 K


T°C = TK − 273.15 K = 121 K − 273.15 = −152°C

9.55 Nothing happens to the particles if the temperature, pressure, and volume are constant. If the tank is sealed

so that no gas molecules can escape, then the pressure does not change. If the tank is opened, then some of the gas molecules will leave the tank to maintain an equilibrium pressure with the air outside the tank.

9.56 The pressure at sea level is higher than at 9000 ft. As the external pressure on the container increases, the

volume of the container will decrease as long as the temperature stays constant. 9.57 We know that the temperature and pressure of a fixed volume of gas are directly related, so we can write an

equation relating the change in pressure that occurs when the temperature of a gas sample changes. The equation is similar to that for Charles’s law, except that we substitute pressure for volume (recall that volume and temperature are also directly proportional).

1 2

1 2=

P PT T

Solving for P2: 12 2

1= ×

PP TT

(a) Convert temperatures to kelvins:

T1 = 0.0°C + 273.15 = 273.2 K

T2 = 105.0°C + 273.15 = 378.2 K

2 378.2 K=P 302 torr273.2 K

× = 418 torr

The temperature increases, so the pressure also increases.

(b) Convert temperatures to kelvins:

T1 = 25.0°C + 273.15 = 298.2 K

T2 = 0.0°C + 273.15 = 273.2 K

2 273.2 K=P 735 torr298.2 K

× = 673 torr

The temperature decreases, so the pressure also decreases.

(c) 2 373 K=P 3.25 atm273 K

× = 4.44 atm

Because the temperature increases, the pressure also increases. 9.58 We know that temperature and pressure of a fixed volume of a gas are directly related, so we can write an

equation relating the change in pressure of a gas sample when the temperature of the gas sample changes. The equation is similar to that for Charles’s law, except that we substitute pressure for volume (recall that volume and temperature are also directly proportional).

1 2

1 2=

P PT T

9−18


1= ×

PP TT

(a) 2 315 KP =255 torr225 K

× = 357 torr

Because the temperature decreases, the pressure also decreases.

(b) Convert temperatures to kelvins:

T1 = 25.0°C + 273.15 = 298.2 K

2 206 KP =895 torr298.2 K

× = 618 torr

Because the temperature decreases, the pressure also decreases.

(c) Convert temperatures to kelvins:

T1 = 150°C + 273.15 = 420 K

T2 = 23°C + 273.15 = 296 K

2 296 KP =2.74 atm420 K

× = 1.9 atm

Because the temperature decreases, the pressure also decreases. 9.59 We know that temperature and pressure of a fixed volume of a gas are directly related, so we can write an

equation that relates the change in pressure of a gas sample with a change in temperature. The equation is similar to Charles’s law, except that we substitute pressure for volume (recall that volume and temperature are also directly proportional).

1 2

1 2=

P PT T


1= ×

TT PP

(a) Convert temperature to kelvins:

T1 = 30.0°C + 273.15 = 303.2 K

2 915 torrT =303.2 K

1525 torr× = 182 K

Convert T2 to degrees Celsius:

T°C = TK − 273.15 K = 182 K − 273.15 = −91°C

(b) Convert temperature to kelvins:

T1 = 250.0°C + 273.15 = 523.2 K

The pressures need to be expressed in the same units. Convert 1042 torr to atm:


× = 1.371 atm

2 1.371 atmT =523.2 K

0.70 atm× = 1.0 × 103 K

9−19


T°C = TK − 273.15 K = 1.0 × 103 K − 273.15 = 750°C

(c) 2 1000.0 torrT =355 K

500.0 torr× = 7.10 × 102 K


T°C = TK − 273.15 K = 7.10 × 102 K − 273.15 = 437°C 9.60 We know that the temperature and pressure of a fixed volume of a gas are directly related, so we can write

an equation relating the change in pressure associated with a change in the temperature of a gas sample. The equation is similar to Charles’s law, except that we substitute pressure for volume (recall that volume and temperature are also directly proportional).

1 2

1 2=

P PT T


1= ×

TT PP

(a) Convert temperature to kelvins:

T1 = 25.0°C + 273.15 = 298.2 K

2 735 torr=T 298.2 K243 torr

× = 902 K


T°C = TK − 273.15 K = 902 K − 273.15 = 629°C

(b) 2 1.20 atm=T 205 K2.35 atm

× = 105 K


T°C = TK − 273.15 K = 105 K − 273.15 = −168°C

(c) The pressures must be in the same units. Convert 875 torr to atm:


× = 1.15 atm

2 0.85 atm=T 375 K1.15 atm

× = 2.8 × 102 K


T°C = TK − 273.15 K = 2.8 × 102 K − 273.15 = 0°C (There are no significant figures in the answer.)

9.61 We know that the temperature and pressure of a fixed volume of a gas are directly related, so we can write

an equation relating the pressure change associated with a change in the temperature of a gas sample. The equation is similar to Charles’s law, except that pressure replaces volume (recall that volume and temperature are also directly proportional).

1 2

1 2=

P PT T

9−20


1= ×

PP TT

P1 T1 P2 T2 7.25 atm 18.5°C ? 37.2°C

Convert temperatures to Celsius and calculate P2:

T1 = 18.5°C + 273.15 = 291.7 K

T2 = 37.2°C + 273.15 = 310.4 K

2 310.4 K=P 7.25 atm291.7 K

× = 7.71 atm

9.62 We know that the temperature and pressure of a fixed volume of a gas are directly related, so we can write

an equation relating the pressure change associated with a change in the temperature of a gas sample. The equation is similar to Charles’s law, except that pressure replaces volume (recall that volume and temperature are also directly proportional).

1 2

1 2=

P PT T

P1 T1 P2 T2 6.75 atm 25.0°C 1.25 atm ?

Convert temperature to kelvins:

T1 = 28.0°C + 273.15 = 298.2 K

2 1.25 atmT =298.2 K

6.75 atm× = 55.2 K


T°C = TK − 273.15 K = 55.2 K − 273.15 = −217.9°C

9.63 When we use the combined gas law, it helps if we solve the equation for the desired variable before we substitute numbers into the expression. When we rearrange the equation, we try to keep the state 1 and state 2 variables separate (it helps us to keep from mixing up the variables). We must also make sure that the units are consistent (i.e. that the pressures are in the same units and the temperatures are in kelvins).

Combined gas law: 1 1 2 2

1 2=

PV P VT T

(a) The missing value on the data table is V2:

1 1 22

1 2= ×

PV TVT P

P1 V1 T1 P2 V2 T2 0.50 atm 2.50 L 20.0°C 760.0 torr ? 0.0°C We need to express the pressures in common units and convert the temperatures to kelvins. Since 760.0 torr is 1.000 atm, we can simply substitute 1.000 atm for P2. First, we convert temperatures to kelvins and then solve for V2:

T1 = 20.0°C + 273.15 = 293.2 K

9−21

T2 = 0.0°C + 273.15 = 273.2 K

2(0.50 atm

=V )(2.50 L)293.2 K

273.2 K×

1.000 atm = 1.2 L

(b) The value missing from the data table is T2:

12 2 2

1 1= ×

TT P VPV

P1 V1 T1 P2 V2 T2 0.250 atm 125 L 25.0°C 100.0 torr 62.0 L ? Temperature and pressure both need to be converted to appropriate units:

T1 = 25.0°C + 273.15 = 298.2 K

P2 (atm) = 100.0 torr 1 atm760 torr

× = 0.1316 atm

2298.2 K

(0.250 atm=T

)(125 L(0.1316 atm

)× )(62.0 L ) = 77.8 K (or −195.3°C)

(c) The value missing from the data table is P2:

1 1 22

1 2= ×

PV TPT V

P1 V1 T1 P2 V2 T2 200.0 torr 455 mL 300.0 K ? 200.0 mL 327°C First, we convert T2 to kelvins:

T2 = 327°C + 273.15 = 6.00 × 102 K

2(200.0 torr)(455 mL

=P )(300.0 K

26.00 10 K)

××

200.0 mL = 9.10 × 102 torr

9.64 When we use the combined gas law, it helps if we solve the equation for the desired variable before we substitute numbers into the expression. When we rearrange the equation, we try to keep the state 1 and state 2 variables separate (it helps to keep us from mixing up the variables). We must also make certain that the units are consistent (i.e. that pressures are in the same units and temperatures are in kelvins.

Combined gas law: 1 1 2 2

1 2=

PV P VT T

(a) The value missing from the data table is P2:

1 1 22

1 2= ×

PV TPT V

P1 V1 T1 P2 V2 T2 900.0 torr 601 mL −10.0°C ? 1200.0 mL 0.0°C Convert temperatures to kelvins:

T1 = −10.0°C + 273.15 = 263.2 K

T2 = 0.0°C + 273.15 = 273.2 K

9−22

2(900.0 torr)(601 mL

=P )(263.2 K

273.2 K)

×1200.0 mL

= 468 torr

(b) The value missing from the data table is T2:

12 2 2

1 1= ×

TT P VPV

P1 V1 T1 P2 V2 T2 75.0 atm 237 mL 147 K 150.0 atm 474 mL ?

2147 K

(75.0 atm=T

)(237 mL(150.0 atm

)× )(474 mL ) = 588 K

(c) The value missing from the data table is V2:

1 1 22

1 2= ×

PV TVT P

P1 V1 T1 P2 V2 T2 760.0 torr 1.12 L 0.0°C 700.0 torr ? 25.0°C The pressures and temperature units need to be changed. Because 760.0 torr = 1.000 atm, we can make this substitution. Convert temperatures to kelvins:

T1 = 0.0°C + 273.15 = 273.2 K

T2 = 25.0°C + 273.15 = 298.2 K

2(760.0 torrV =

)(1.12 L)273.2 K

298.2 K×

700.0 torr = 1.33 L

9.65 We want to calculate the volume of oxygen at STP (V1). Standard temperature and pressure are 273.15 K

(0°C) and 1 atm (by definition). The tank will hold 0.500 L at 3.50 atm and 24.5°C. This can be considered the final state of the gas.

P1 V1 T1 P2 V2 T2 1 atm ? 273.15 K 3.50 atm 0.500 L 24.5°C

Convert 24.5°C to kelvins:

T2 = 24.5°C + 273.15 = 297.7 K

Solve the combined gas law for V1.

2 2 11

2 1= ×

P V TVT P

= (3.50 atm )(0.500 L)297.7 K

273.15 K×

1 atm = 1.61 L

9.66 The initial state of the gas is 22.0 L at 25.0°C and 725 torr, and we are trying to determine the pressure (P2)

when the gas is heated to 134°C and compressed to 4.5 L.

P1 V1 T1 P2 V2 T2 725 torr 22.0 L 25.0°C ? 4.50 L 134°C

Convert temperatures to kelvins:

9−23

T1 = 25.0°C + 273.15 = 298.2 K

T2 = 134°C + 273.15 = 407 K

Solve the combined gas law for P2.

1 1 22

1 2= ×

PV TPT V

= (725 torr)(22.0 L )298.2 K

407 K×

4.50 L = 4.84 × 103 torr

9.67 Gay-Lussac’s Law states that volumes of gases react in simple, whole number ratios when the volumes of

the reactants and products are measured at the same temperature and pressure. These ratios correspond to the coefficients in the balanced chemical equation for the reaction.

9.68 The ratio of the volumes of the gaseous reactant and product corresponds to the ratio of the stoichiometric

coefficients for these substances in the balanced chemical reaction. 9.69 The molar volume of all gases is approximately 22.414 L/mol at STP. 9.70 The volume of 1.00 mol of H2 gas at STP is 22.4 L. This is approximately true for all gases. 9.71 At STP, 1.00 mol of any gas occupies 22.414 L. We determine the mass of each gas sample from the

number of moles and the molar mass of the gas.

(a) Moles CH4 = 8.62 L 1 mol22.414 L

× = 0.385 mol CH4

Given that the molar mass of CH4 (16.04 g/mol) we can calculate the mass of CH4:

Mass CH4 = 0.385 mol 16.04 g1 mol

× = 6.17 g CH4

(b) Convert mL to L:

Volume in L = 350.0 mL 1 L1000 mL

× = 0.3500 L

Moles Xe = 0.3500 L 1 mol22.414 L

× = 1.562 × 10−2 mol Xe

We use the molar mass of Xe (131.3 g/mol) to calculate the mass of Xe gas in the sample:

Mass Xe = 21.562 10 mol−×131.3 g1 mol

× = 2.050 g Xe

(c) Moles CO = 48.1 L 1 mol22.414 L

× = 2.15 mol CO

We use the molar mass of CO (28.01 g/mol) to calculate the mass of CO gas in the sample:

Mass CO = 2.15 mol 28.01 g1 mol

× = 60.1 g CO

9.72 At STP, 1.00 mol of any gas occupies 22.414 L. We determine the mass of each gas sample from the number of moles and the molar mass of the gas. (a) Convert mL to L:

9−24


× = 0.135 L

Moles H2 = 0.135 L 1 mol22.414 L

× = 6.02 × 10−3 mol H2

We use the molar mass of H2 (2.016 g/mol) to calculate the mass of H2 gas in the sample:

Mass H2 = 36.02 10 mol−×2.016 g1 mol

× = 0.0121 g H2

(b) Moles N2 = 8.96 L 1 mol22.414 L

× = 0.400 mol N2

We use the molar mass of N2 (28.02 g/mol) to calculate the mass of N2 gas in the sample:

Mass N2 = 0.400 mol 28.02 g1 mol

× = 11.2 g N2

(c) Moles He = 0.75 L 1 mol22.414 L

× = 0.033 mol He

We use the molar mass of He (4.003 g/mol) to calculate the mass of He gas in the sample:

Mass He = 0.033 mol 4.003 g1 mol

× = 0.13 g He

9.73 Avogadro’s law tells us that because both balloons have the same volume they contain the same number of

gas particles (and, therefore, the same number of moles). Because argon atoms are more massive than helium atoms, the balloon containing argon has the greater mass, and, therefore, the greater density (the volumes of the balloons are the same).

9.74 Avogadro’s law tells us that both balloons contain the same number of gas molecules. The CO2-filled

balloon has the greater mass, because CO2 molecules are more massive than O2 molecules. Because the volumes of the balloons are the same and the mass of the CO2 balloon is greater, the density of the CO2 balloon is also greater.

9.75 We use the molar mass of each gas to calculate the number of moles of gas in each sample. Because the

gases are at STP, one mole of each gas occupies 22.414 L. (a) 5.8 g NH3 (17.03 g/mol)

Moles NH3 = 5.8 g 1 mol17.03 g

× = 0.34 mol NH3

Volume NH3 = 0.34 mol 22.414 L1 mol

× = 7.6 L NH3

(b) 48 g O2 (32.00 g/mol)

Moles O2 = 48 g 1 mol32.00 g

× = 1.5 mol O2

Volume O2 = 1.5 mol 22.414 L1 mol

× = 34 L O2

9−25

(c) 10.8 g He (4.003 g/mol)

Moles He = 10.8 g 1 mol4.003 g

× = 2.70 mol He

Volume He = 2.7 mol 22.414 L1 mol

× = 60.5 L He

9.76 We use the molar mass of each gas to calculate the number of moles of gas in each sample. Because the gases are at STP, one mole of each gas occupies 22.414 L.

(a) 2.2 g CO2 (44.01 g/mol)

Moles CO2 = 2.2 g 1 mol44.01 g

× = 0.050 mol CO2

Volume CO2 = 0.050 mol 22.414 L1 mol

× = 1.1 L CO2

(b) 5.6 g N2 (28.02 g/mol)

Moles N2 = 5.6 g 1 mol28.02 g

× = 0.20 mol N2

Volume N2 = 0.20 mol 22.414 L1 mol

× = 4.5 L N2

(b) 145 g Ar (39.95 g/mol)

Moles Ar = 145 g 1 mol39.95 g

× = 3.6 mol Ar

Volume Ar = 3.6 mol 22.414 L1 mol

× = 81 L Ar

9.77 We convert the mass of helium (425 g, given in the figure) to the equivalent number of moles, and then to volume in liters, using the molar mass of helium (4.003 g/mol) and the molar volume of a gas at STP (22.414 L/mol). Once we have determined the volume, we can determine approximately how many balloons we can fill from the tank.

Moles He = 425 g 1 mol4.003 g

× = 106 mol He

Volume He = 106 mol 22.414 L1 mol

× = 2.38 × 103 L He

Because each balloon holds 1.0 L of gas, we can estimate that we can fill approximately 2380 balloons. 9.78 We convert the mass of CO2 (370 g, given in the figure) to the equivalent number of moles, and then to

volume in liters, using the molar mass of CO2 (44.01 g/mol) and the molar volume of a gas at STP (22.414 L/mol). Once we determine the volume of CO2, we can estimate the number of balloons we can fill using the volume of each balloon.

Moles CO2 = 370 g 1 mol44.01 g

× = 8.4 mol CO2

Volume CO2 = 8.4 mol 22.414 L1 mol

× = 1.9 × 102 L CO2

9−26

Because each balloon holds 0.75 L, we can estimate the number of balloons we can fill from the CO2 in the tank:

Balloons = 21.9 10 L×1 balloon0.75 L

× = 2.5 × 102 balloons

9.79 Convert 25.0°C to kelvins: T1 = 25.0°C + 273.15 = 298.2 K. For STP, T2 = 273.15 K and P2 = 1 atm.

Solve the combined gas law for V2.

1 1 22

1 2

PV TVT P

= × = (1.00 atm )(12.0 L)298.2 K

273.15 K×

1 atm = 11.0 L

9.80 To solve this problem using the combined gas law, we can first calculate the volume of the H2 at STP

(1 atm, 273.15 K). Then we can use the combined gas law to calculate the volume the He occupies at 30.0°C and 0.975 atm:

1 1 22

1 2= ×

PV TVT P

Volume of gas at STP = 1.50 mol 22.414 L1 mol

× = 33.6 L

P1 V1 T1 P2 V2 T2 1 atm 33.6 L 273.15 K 0.975 atm ? 30.0°C

Convert T2 to kelvins:

T2 = 30.0°C + 273.15 = 303.2 K

2(1 atm

=V )(33.6 L)273.15 K

303.2 K×

0.975 atm = 38.3 L

9.81 An ideal gas is any gas whose behavior is described by the five postulates of kinetic-molecular theory. Ideal gases have elastic collisions (they bounce and/or collide without losing energy), travel in straight lines (because they are not attracted to other molecules), and occupy zero volume. In addition, the average kinetic energy of an ideal gas is directly proportional to the temperature of the gas.

9.82 (a) The ideal gas law is: PV = nRT. (b) The ideal gas law is used to relate the variables that describe the

state of a gas. These variables are pressure, volume, number of moles, and temperature. Given any three of these variables, we can calculate the fourth.

9.83 To calculate the volume occupied by each gas, we first determine the number of moles of each gas and

convert the temperature of the gases to kelvins. Whenever we use the ideal gas law, we must make certain that the units of P, V, n, and T match the units of R that we are using (R = 0.08206 L∙atm/(mol∙K)).

For each gas, the temperature is T = 100.0°C + 273.15 K = 373.15 K (four significant figures)

Rearranging the ideal gas law for volume: nRTV =P

(a) 5.8 g NH3 (17.03 g/mol)

Moles NH3 = 5.8 g 1 mol17.03 g

× = 0.34 mol NH3

9−27

0.34 molV =

( ) L atm0.08206 ⋅mol K⋅

373.15 K

( )15.0 atm

= 0.69 L NH3

(b) 48 g O2 (32.00 g/mol)

Moles O2 = 48 g 1 mol32.00 g

× = 1.5 mol O2

1.5 molV =

( ) L atm0.08206 ⋅mol K⋅

373.2 K

( )15.0 atm

= 3.1 L O2

(c) 10.8 g He (4.003 g/mol)

Moles He = 10.8 g 1 mol4.003 g

× = 2.70 mol He

2.70 molV =

( ) L atm0.08206 ⋅mol K⋅

373.2 K

( )15.0 atm

= 5.51 L He

9.84 To calculate the volume occupied by each gas, we first determine the number of moles of each gas and convert the temperature of the gases to kelvins. Anytime we use the ideal gas law, we must make certain that the units P, V, n, and T match the units of R that we are using (R = 0.08206 L∙atm/(mol∙K)).

For each gas, the temperature is T = 75.0°C + 273.15 K = 348.2 K

Rearranging the ideal gas law for volume: nRTV =P

(a) 2.2 g CO2 (44.01 g/mol)

Moles CO2 = 2.2 g 1 mol44.01 g

× = 0.050 mol CO2

0.050 molV =

( ) L atm0.08206 ⋅mol K⋅

348.2 K

( )3.5 atm

= 0.41 L CO2

(b) 5.6 g N2 (32.00 g/mol)

Moles N2 = 5.6 g 1 mol28.02 g

× = 0.20 mol N2

0.20 molV =

( ) L atm0.08206 ⋅mol K⋅

348.2 K

( )3.5 atm

= 1.6 L N2

(c) 7.5 g Ar (39.95 g/mol)

Moles Ar = 7.5 g 1 mol39.95 g

× = 0.19 mol Ar

9−28

0.19 molV =

( ) L atm0.08206 ⋅mol K⋅

348.2 K

( )3.5 atm

= 1.5 L Ar

9.85 Rearranging the ideal gas law for the number of moles in the sample: PVn =RT

Because R = 0.08206 L∙atm/(mol∙K), we must express pressure in atmospheres and, of course, temperature in kelvins.


× = 0.950 atm

T = 87.5°C + 273.15 = 360.7 K

(a) 7.62 L CH4 (16.04 g/mol)

(0.950 atm=

PVn =RT

)(7.62 L )L0.08206 atm⋅mol K⋅

360.7 K

( ) = 0.245 mol

Mass CH4 = 0.245 mol 16.04 g1 mol

× = 3.92 g CH4

(b) 135 mL H2 (2.016 g/mol) First convert mL to L:


× = 0.135 L

(0.950 atm=

PVn =RT

)(0.135 L )L0.08206 atm⋅mol K⋅

360.7 K

( ) = 4.33 × 10−3 mol

Mass H2 = 34.33 10 mol−×2.016 g1 mol

× = 8.74 × 10−3 g H2

(c) 8.96 L N2 (28.02 g/mol)

(0.950 atm=

PVn =RT

)(8.96 L )L0.08206 atm⋅mol K⋅

360.7 K

( ) = 0.288 mol

Mass N2 = 0.288 mol 28.02 g1 mol

× = 8.06 g N2

9.86 Rearranging the ideal gas law for the number of moles of gas in the sample: PVn =RT

Because R = 0.08206 L∙atm/(mol∙K), we must express pressure in atmospheres, and, of course, temperature in kelvins.


× = 1.04 atm (2 significant figures)

9−29

T = 54.0°C + 273.15 = 327.2 K (a) 150.0 mL Xe (131.3 g/mol)

Convert 150.0 mL Xe to liters:

Volume in L = 150.0 mL 1 L1000 mL

× = 0.1500 L

(1.04 atm=

PVn =RT

)(0.1500 L )L0.08206 atm⋅mol K⋅

327.2 K

( ) = 5.8 × 10−3 mol

Mass Xe = 35.8 10 mol−×131.3 g1 mol

× = 0.76 g Xe

(b) 38.1 L CO (28.01 g/mol)

(1.04 atm=

PVn =RT

)(38.1 L )L0.08206 atm⋅mol K⋅

327.2 K

( ) = 1.5 mol

Mass CO = 1.5 mol 28.01 g1 mol

× = 41 g CO

(c) 2.5 L O2 (32.00 g/mol)

(1.04 atm=

PVn =RT

)(2.5 L )L0.08206 atm⋅mol K⋅

327.2 K

( ) = 0.097 mol

Mass O2 = 0.097 mol 32.00 g1 mol

× = 3.1 g O2

9.87 The balloons sink or float depending on how their densities compare to the density of air. Recall that if the

balloons (or any containers) have equal volume, pressure, and temperature, they contain the same number of particles (Avogadro’s Law or ideal gas law). Because each CO2 molecule is heavier than each He atoms, and there are equal numbers of each, the CO2-filled balloon will have more mass and, therefore, higher density. Why doesn’t CO2 float in air? Air is less dense than CO2. Air is primarily a mixture of N2 and O2, both of which have lower masses than CO2. The CO2-filled balloon sinks because CO2 gas is more dense than air.

9.88 The larger balloon is at the higher temperature. The velocity of the N2 molecules is greater in the warmer

balloon, so they collide with the walls of the balloon more often and with greater force, causing the volume to increase. The density of particles is lower in the larger balloon.

9−30

9.89 At STP, one mole of any gas occupies 22.414 L. If we know the molar mass of a gas, we can calculate its density (d = m/V, the mass of one mole divided by the volume of one mole). (a) NH3 (17.03 g/mol).

d = 17.03 g22.414 L

= 0.7598 g/L

(b) N2 (28.02 g/mol)

d = 28.02 g22.414 L

= 1.250 g/L

(c) N2O (44.02 g/mol)

d = 44.02 g22.414 L

= 1.964 g/L

9.90 At STP, one mole of gas occupies 22.414 L. If we know the molar mass of the gas, we can calculate its

density (d = m/V, the mass of one mole divided by the volume of one mole). (a) NO (30.01 g/mol)

d = 30.01 g22.414 L

= 1.339 g/L

(b) NO2 (46.01 g/mol)

d = 46.01 g22.414 L

= 2.053 g/L

(c) O2 (32.00 g/mol)

d = 32.00 g22.414 L

= 1.428 g/L

9.91 Density is the mass of a substance divided by its volume. For gas samples, it is convenient for us to

determine density by dividing the molar mass of the gas by its volume, which we calculate using the ideal gas law.

T = 25°C + 273.15 = 298.2 K


× = 0.967 atm

n = 1.00 mol (we choose this for convenience)

1.00 mol=

nRTV =P

( ) L atm0.08206 ⋅mol K⋅

( )298.2 K

0.967 atm

= 25.3 L

To calculate the density, we divide the molar mass by the volume it occupies:

(a) NH3 (17.03 g/mol) – One mole of NH3 has a mass of 17.03 g and occupies a volume of 25.3 L. The density is:

17.03 g25.3 L

=md =V

= 0.673 g/L

9−31

(b) N2 (28.02 g/mol) – One mole of N2 has a mass of 28.02 g and occupies a volume of 25.3 L. The density is:

28.02 g25.3 L

=md =V

= 1.11 g/L

(c) N2O (44.02 g/mol) – One mole of N2O has a mass of 44.02 g and occupies a volume of 25.3 L. The density is:

44.02 g25.3 L

=md =V

= 1.74 g/L

9.92 Density is the mass of a substance divided by volume. For gas samples, it is convenient for us to determine

density by dividing the molar mass of the gas by its volume, which we calculate using the ideal gas law. T = 25°C + 273.15 = 298.2 K


× = 0.967 atm

n = 1.00 mol

1.00 mol=

nRTV =P

( ) L atm0.08206 ⋅mol K⋅

( )298.2 K

0.967 atm

= 25.3 L

To calculate the density, divide the molar mass by the volume it occupies: (a) NO (30.01 g/mol) – One mole of NO has a mass of 30.01 g and occupies a volume of 25.3 L. The

density is:

30.01 g25.3 L

=md =V

= 1.19 g/L

(b) NO2 (46.01 g/mol) – One mole of NO2 has a mass of 46.01 g and occupies a volume of 25.3 L. The density is:

46.01 g25.3 L

=md =V

= 1.82 g/L

(c) O2 (32.00 g/mol) – One mole of O2 has a mass of 32.00 g and occupies a volume of 25.3 L. The density is:

32.00 g25.3 L

=md =V

= 1.26 g/L

9.93 According to the ideal gas law, equal-molar amounts of every gas occupy the same volume at the same

pressure and temperature. Using this law, we can calculate the number of moles of gas in the container. First, we convert pressure and temperature to units consistent with the units in the gas constant:


× = 1.1 atm

Temperature in K = 50.0°C + 273.15 = 323.2 K

9−32

1.1 atm= =

PVnRT

( ) 5.00 L( )L0.08206 atm⋅mol K⋅

323.2 K

( ) = 0.21 mol

Now we can determine the mass of each gas using its molar mass: (a) H2 (2.016 g/mol)

Mass H2 = 0.21 mol 2.016 gmol

× = 0.42 g H2

(b) CH4 (16.04 g/mol)

Mass CH4 = 0.21 mol 16.04 gmol

× = 3.3 g CH4

(c) SO2 (64.06 g/mol)

Mass SO2 = 0.21 mol 64.06 gmol

× = 13 g SO2

9.94 According to the ideal gas law, equal-molar amounts of every gas occupy the same volume at the same

pressure and temperature. Using this law, we can calculate the number of moles of gas in the container. First, we convert pressure and temperature to units that are consistent with the units in the gas constant:


× = 0.86 atm


0.86 atm= =

PVnRT

( ) 2.50 L( )L0.08206 atm⋅mol K⋅

298.2 K

( ) = 0.087 mol

Now, we can determine the mass of each gas using its molar mass: (a) O2 (32.00 g/mol)

Mass O2 = 0.087 mol 32.00 gmol

× = 2.8 g O2

(b) CO2 (44.01 g/mol)

Mass CO2 = 0.087 mol 44.01 gmol

× = 3.8 g CO2

(c) He (4.003 g/mol)

Mass He = 0.087 mol 4.003 gmol

× = 0.35 g He

9.95 Dalton’s law of partial pressures states that the total pressure in a container is the sum of the pressures

exerted by each of the individual gases in the container. Each gas exerts a pressure on the walls of the container as if it were the only gas in the container. Gas molecules behave in this way because they are not strongly attracted to each other.

9−33

9.96 Suppose we have a closed container that contains CO2(g) and some liquid water. The total pressure inside the container is actually the sum of the pressures of the CO2(g) and H2O(g) (water vapor). To determine the pressure of just the CO2(g) we can subtract the pressure of the water from the total pressure in the container. The vapor pressure of water depends only on the temperature so we can determine that pressure from a table in the textbook or from another reference book. Mathematically, we express Dalton’s law of partial pressures for this system as:

2total CO waterP P P= + . The pressure of CO2 is: 2CO total waterP = P P−

9.97 According to Dalton’s law of partial pressures, the total pressure (728 torr) is the sum of the pressure

exerted by the oxygen gas and water vapor (20.0 torr). To calculate the partial pressure of oxygen gas we subtract the water vapor pressure from the total pressure. Mathematically, Dalton’s law of partial pressures looks like:

2total O waterP P P= +

2O total waterP = P P− = 728 torr – 20.0 torr = 708 torr 9.98 According to Dalton’s law of partial pressures, the total pressure (742 torr) is a sum of the pressure exerted

by the hydrogen gas and water vapor (15.5 torr). To calculate the partial pressure of hydrogen gas we subtract the water vapor pressure from the total pressure. Mathematically, Dalton’s law of partial pressures looks like:

2total H waterP P P= +

2H total waterP = P P− = 742 torr – 15.5 torr = 727 torr 9.99 (a) We use molar mass to calculate the number of moles of N2 (28.02 g/mol).

Moles N2 = 78.0 g 1 mol28.02 g

× = 2.78 mol N2

(b) We use molar mass to calculate the number of moles of Ne (20.18 g/mol).

Moles Ne = 42.0 g 1 mol20.18 g

× = 2.08 mol Ne

(c) To calculate the partial pressure of N2 we use the ideal gas law (PV = nRT). We know the number of moles of N2 (part (a)) and temperature (50.0°C), but the volume is missing. To calculate the volume, we take advantage of a concept from Dalton’s law of partial pressures; that each gas occupies the total volume of the container. Because we know the total pressure, number of moles of N2, and temperature, we can calculate the total volume.

Ptotal = 3.75 atm

ntotal = 2.78 mol + 2.08 mol = 4.86 mol

T = 50.0°C + 273.15 = 323.2 K

Vtotal = ?

totaltotal

4.86 molnRTV =P

=( ) L atm0.08206 ⋅

mol K⋅323.2 K

( )3.75 atm

= 34.4 L

Using the total volume, calculate the partial pressure of N2 using the moles of N2.

9−34

22

NN

total

2.78 moln RT

P =V

=( ) L0.08206 atm

mol⋅

K⋅323.2 K

( )34.4 L

= 2.15 atm

(d) To calculate the partial pressure of Ne, we can do exactly as we did in (c) or we can use Dalton’s law of partial pressures. Either method gives the same result. According to Dalton's law of partial pressures, the total pressure is the sum of the pressures of Ne and N2. Mathematically, we can express Dalton's law of partial pressure as:

2total N NeP = P + P

2Ne total NP = P P− = 3.75 atm − 2.15 atm = 1.60 atm

9.100 (a) We use molar mass to calculate the number of moles of CO2 (44.01 g/mol).

Moles CO2 = 150.0 g 1 mol44.01 g

× = 3.41 mol CO2

(b) We use molar mass to calculate the number of moles of O2 (32.00 g/mol).

Moles O2 = 24.0 g 1 mol32.00 g

× = 0.750 mol O2

(c) To calculate the partial pressure of CO2 we use the ideal gas law (PV = nRT). We know the number of moles of CO2 (part (a)) and temperature (50.0°C), but the volume is missing. To calculate the volume, we take advantage of a concept from Dalton’s law of partial pressures: each gas occupies the total volume of the container. Because we know the total pressure, the number of moles, and the temperature, we can calculate the total volume.

Ptotal = 4.25 atm

ntotal = 3.41 mol + 0.750 mol = 4.16 mol

T = 25.0°C + 273.15 = 298.2 K

Vtotal = ?

totaltotal

4.16 molnRTV =P

=( ) L atm0.08206 ⋅

mol K⋅298.2 K

( )4.25 atm

= 23.9 L

Using the total volume, we can calculate the partial pressure of CO2 using the moles of CO2.

22

COCO

total

3.41 moln RT

P =V

=( ) L0.08206 atm

mol⋅

K⋅323.2 K

( )23.9 L

= 3.48 atm

(d) To calculate the partial pressure of O2, we can do exactly as we did in (c) or we can use Dalton’s law of partial pressures. Either method gives the same result. According to Dalton's law of partial pressures the total pressure is the sum of the pressure of O2and CO2. Mathematically, we can express Dalton's law of partial pressures as:

2 2total CO OP = P + P

2 2O total COP = P P− = 4.25 atm − 3.48 atm = 0.77 atm

9−35

9.101 Given gas density (g/L), we can calculate molar mass (g/mol) using the molar volume of a gas (L/mol). Under STP conditions, the molar volume of a gas is 22.414 L/mol. Multiplying density (g/L) by molar volume (L/mol) converts density units to molar mass units:

(a) MM = 1.785 g1 L

22.414 L×

1 mol = 40.01 g/mol

(b) MM = 1.340 g1 L

22.414 L×

1 mol = 30.03 g/mol

(c) MM = 2.052 g1 L

22.414 L×

1 mol = 45.99 g/mol

(d) MM = 0.905 g1 L

22.414 L×

1 mol = 20.3 g/mol

(e) MM = 0.714 g1 L

22.414 L×

1 mol = 16.0 g/mol

9.102 Given gas density (g/L), we can calculate molar mass (g/mol) using the molar volume of a gas (L/mol).

We begin by calculating the volume occupied by 1.00 mol of the gas at 27.0°C and 745 torr:


× = 0.980 atm


1 molV =

( ) L atm0.08206 ⋅mol K⋅

300.2 K

( )0.980 atm

= 25.1 L

This volume is the volume occupied by 1.00 mol of gas, so it is the molar volume which can be written as 25.1 L/mol. Multiplying density (g/L) by molar volume (L/mol) converts density units to molar mass units:

(a) MM = 2.436 g1 L

25.1 L×

1 mol = 61.2 g/mol

(b) MM = 0.842 g1 L

25.1 L×

1 mol = 21.2 g/mol

(c) MM = 1.325 g1 L

25.1 L×

1 mol = 33.3 g/mol

(d) MM = 3.450 g1 L

25.1 L×

1 mol = 86.7 g/mol

(e) MM = 1.706 g1 L

25.1 L×

1 mol = 42.9 g/mol

9.103 The five postulates of the kinetic-molecular theory are:

1) Gases are composed of small, widely separated particles. 2) Gas particles behave independently of each other.

9−36

3) Gas particles move in straight lines. 4) Gas pressure results from the force exerted by the particles in the container. This force is the sum of

the forces exerted by the particles as they bounce off the container walls. 5) The average kinetic energy of gas particles depends only on the absolute temperature.

The key to understanding the postulates 1−3 of the kinetic-molecular theory is recognizing that particles (atoms or molecules) in the gas phase are not strongly attracted to each other. Particles that are strongly attracted tend to be liquids or solids at room temperature. In addition, because they are not attracted to each other, the particles move through space on straight paths. Postulate 4 also is easy to understand. As particles strike the container walls, they exert a force on those walls. This is analogous to the force you feel when you walk into a wall (if you try this, make sure no one is looking). One person walking into a wall exerts only a small force, but if the whole class ran into the same wall at the same time, they could do some damage! The forces of individual particles are additive. The fifth postulate is most difficult to understand. When a gas particle absorbs energy, its only options are to store it as potential energy or use it as kinetic energy. Kinetic energy is energy of motion. Particles travel faster and vibrate faster when they have more energy.

9.104 Kinetic energy of gas particles is directly related to pressure. As the kinetic energy of the particles

increases the pressure exerted by the particles increases. 9.105 As the temperature of a gas sample increases the kinetic energy of the particles increases. Recall that

kinetic energy is the energy of motion. This means that increasing the temperature results in an increase in particle velocity. Faster-moving particles strike the walls of a container with greater force than slower-moving particles, so the pressure is higher. In addition, when the particles are moving faster, the frequency of their collisions with the container walls also increases.

9.106 As the temperature decreases, the volume of the balloon decreases because the helium atoms have less

forceful collisions with the balloon wall. At the macroscopic level, the balloon at lower temperature should be smaller. At the microscopic level, at the higher temperature the helium atoms are moving faster and are less dense (image A). At the lower temperature, the atoms are moving more slowly and are more dense (image B).

A B

9.107 The pressure of a gas depends on the force and frequency of collisions in a given area. The density of gas

particles is inversely proportional to volume. If the volume decreases, the density of the gas sample increases. Because the density of gas particles increases with decreasing volume, we can conclude that the frequency of collisions in a given area also increases with decreasing volume. This means that the pressure increases.

9.108 The pressure of a gas depends on the force and frequency of collisions in a given area. If we introduce

more gas particles into a container without allowing the container to expand, the density of gas particles increases. This increase results in more frequent particle collisions, which in turn creates a greater force on the container wall. As long as the temperature remains constant, the energy of each collision remains constant as well.

9−37

9.109 Because the gases are all at the same temperature, their kinetic energies are the same. The equation for kinetic energy, KE, is: KE = ½ mv2. This means that if two objects have the same kinetic energy, the heavier object will move more slowly than the lighter object. For a gas sample (with many particles), the equation is described in terms of the average kinetic energy: KEave = ½ m(vave)2. This equation shows us that, at a given temperature, gas particles with the largest molar mass will move the slowest. Of the substances listed, CO2 has the largest molar mass, so it has the lowest velocity. H2 has the smallest molar mass, so it has the highest velocity.

(lowest velocity) CO2 < CH4 < He < H2 (highest velocity)

9.110 For gases, average kinetic energy is directly proportional to temperature. At any given temperature, particles with larger molar masses move more slowly, on average, than particles with smaller molar masses, but their average kinetic energies are equal. This means that CO2, which has the largest molar mass, has the same kinetic energy as H2, even though the lighter particle moves faster. All the gases have the same average kinetic energy because they are at the same temperature.

9.111 Gas particles with smaller molar masses have higher average velocities than gas particles with larger molar

masses, when at the same temperature. As a result, gases with smaller molar masses diffuse faster than gases with larger molar masses.

9.112 As the velocity of gas particles increases, they collide more frequently with the walls of a container.

Effusion rates are greater when the frequency of collisions is high. At any temperature, heavier gas particles move more slowly than lighter particles, so the rate of effusion for heavier particles will be lower than that for lighter particles.

9.113 The He atoms are lighter than Ne atoms, so the He atoms will have a higher average velocity, and therefore

effuse faster than the Ne atoms. 9.114 Lighter particles move more rapidly than heavier particles; therefore, lighter particles effuse at a higher rate

than heavier particles. Argon, with its lower molar mass, will effuse out of the container faster than Xe. 9.115 The volumes of gases that react are directly proportional to the stoichiometric coefficients in the balanced

chemical equation for the reaction. If the products are at the same temperature and pressure as the reactants, the gaseous reaction product volumes are also proportional to the stoichiometric coefficients of the chemical equation. For the given reaction, this means that 2 L H2 react with 1 L O2.

2H2(g) + O2(g) → 2H2O(g)

We can calculate the volume of H2 required to react with 12 L of O2 using this problem solving map:

2 2volume ratioVolume O Volume H→

Volume of O2 = 212 L O 2

2

2 L H1 L O

× = 24 L H2

9.116 The volumes of gases that react are directly proportional to the stoichiometric coefficients in the balanced chemical equation for the reaction. If the products are at the same temperature and pressure as the reactants, the gaseous reaction product volumes are also proportional to the stoichiometric coefficients of the chemical equation. For the given reaction, this means that 3 L H2 reacts with 1 L N2.

3H2(g) + N2(g) → 2NH3(g)

We can calculate the volume of nitrogen required to react with 9 L of hydrogen using the following problem solving map:

2 2volume ratioVolume H Volume of N→

9−38

Volume of N2 = 29 L H 2

2

1 L N3 L H

× = 3 L N2

9.117 When we measure the volumes of gases in chemical reactions under the same pressure and temperature

conditions, the volume ratios are equivalent to the mole ratios from the balanced chemical equation. For the reaction of hexane with oxygen, we can say that 2 L of C6H14 react with 19 L of O2 to produce 12 L of CO2 and 14 L of H2O.

Volume CO2 = 6 148.00 L C H 2

6 14

12 L CO2 L C H

× = 48.0 L CO2

Volume O2 = 6 148.00 L C H 2

6 14

19 L O2 L C H

× = 76.0 L O2

9.118 When we measure the volumes of gases in chemical reactions under the same pressure and temperature conditions, the volume ratios are equivalent to the mole ratios from the balanced chemical equation. For the reaction of ammonia with oxygen, we can write 4 L NH3 form 4 L NO.

Volume NO = 41250 L NH4

4 L NO4 L NH

× = 1250 L NO

To calculate the volume of gas at STP we use the combined gas law.

1 1 22

1 2= ×

PV TVT P

Temperature in kelvins = 325°C + 273.15 = 598 K

P1 V1 T1 P2 V2 T2 4.25 atm 1250 L 325°C 1 ? 273.15 K

2(4.25 atm

=V )(1250 L)598 K

273.15 K×

1 atm = 2.43 × 103 L

9.119 Use the ideal gas law to calculate the number of moles of N2O produced in the reaction. Then use this number of moles of N2O to calculate the number of moles, and mass, of NH4NO3 (80.05 g/mol) required for the reaction.

2 2 4 3 4 3mole ratio Volume N O mol N O mol NH NO g NH NO→ → →PV = nRT MM


× = 3.75 atm

Temperature in K = 42°C + 273.15 = 315.2 K

(3.75 atm=

PVn =RT

)(145 L )L0.08206 atm⋅mol K⋅

315.2 K

( ) = 21.0 mol N2O

Mass NH4NO3 = 221.0 mol N O 4 31 mol NH NO×

21 mol N O4 3

4 3

80.05 g NH NO1 mol NH NO

× = 1.68 × 103 g NH4NO3

9−39

9.120 Use the ideal gas law to calculate the number of moles of NO produced in the reaction. Then use this number of moles of NO to calculate the number of moles, and mass, of Cu (63.55 g/mol) required for the reaction.

mole ratio Volume NO mol NO mol Cu g Cu→ → →PV = nRT MM


× = 0.954 atm


(0.954 atm=

PVn =RT

)(15 L )L0.08206 atm⋅mol K⋅

293.2 K

( ) = 0.595 mol NO

Mass Cu = 0.595 mol NO3 mol Cu

×2 mol NO

63.55 g Cu1 mol Cu

× = 56.7 g Cu

9.121 At higher altitudes, the pressure on the outside of the balloon is lower than the pressure the balloon

experiences on the ground. The balloon expands or contracts so that the external pressure and the internal pressure are the same.

9.122 It is easiest to think of a gas in a flexible container. As the temperature of the gas increases, the kinetic

energy of the particles increases. This kinetic energy increase results in an increase in the total force of the gas particles on the container walls, causing the container to expand. As the container expands, the gas density decreases. This density decrease results in a decrease in the frequency of particle collisions that continues until the force exerted by the particles on the inside of the container is equal to the force exerted by gas particles on the outside of the container. If we heat a gas that is not inside a container, the same process happens, except that the gas particles are simply pushing other gas particles out of the way.

9.123 The density of a gas decreases when it is heated and its volume is allowed to expand, or if the gas sample

size in a fixed volume container is reduced. A hot air balloon actually has an opening in the bottom. As the air in the balloon is heated, the gas expands and the “extra” air leaves from the opening at the bottom of the balloon. Since gas molecules are escaping and the volume is remaining fairly constant, the air in the hot air balloon becomes less dense than the surrounding air, allowing the balloon to float.

9.124 The water in the tank exerts pressure on the bubbles in the tank. Because the water pressure is greater at

the bottom of the tank than at the top, the bubbles expand as they rise to the surface. 9.125 We are looking at a change of state (i.e. “Assume the pressure at the surface is 760 torr and changes to 150

torr.”). Therefore, we can assume that this is a combined gas law problem, and organize the data into a table as shown below:

P1 V1 T1 P2 V2 T2 760 torr nc ? 150 torr nc 218 K

*nc = no change

Next, we write the combined gas law, cancel the variables that do not change, and solve for T1:

Combined gas law: 1 1P V 2 2

1=

P VT 2T

21 1

2×

TT = PP

= 760 torr 218 K150 torr

× = 1.10 × 103 K

9−40

9.126 Although the gas molecules move at high velocity, they do not travel in straight lines. They bump into other gas molecules, and so are constantly changing direction which adds distance between an object and our noses.

9.127 The table below summarizes the changes on the macroscopic and microscopic levels.

What will happen if… Macroscopic view Microscopic view

temperature increases and pressure remains constant balloon gets bigger

molecules move more rapidly molecules collide more forcefully collision frequency increases molecules are further apart

temperature decreases and pressure remains constant balloon gets smaller

molecules move more slowly molecules collide less energetically collision frequency decreases molecules are closer together

10,000 feet (temperature decreases and pressure decreases)

balloon probably gets a little bigger

molecules move more slowly molecules collide less energetically collision frequency decreases

9.128 Use the ideal gas law to calculate the number of moles N2 that form when the bag fills. Then use this

number of moles of N2 to calculate the number of moles, and mass, of NaN3 (65.02 g/mol) required to produce the N2.

2 2 3 3mole ratio Volume N mol N mol NaN g NaN→ → →PV = nRT MM


× = 1.50 atm


(1.50 atm=

PVn =RT

)(2.5 L )L0.08206 atm⋅mol K⋅

298.2 K

( ) = 0.153 mol N2

Mass NaN3 = 20.153 mol N 32 mol NaN×

23 mol N3

3

65.02 g NaN 1 mol NaN

× = 6.63 g NaN3

9.129 To calculate the volume of O2, we must first determine the number of moles of O2 produced. We can

calculate the number of moles from the mass of HgO that reacted, the molar mass of HgO (216.6 g/mol), and the balanced chemical equation.

2 2 mole ratiog HgO mol HgO mol O Volume O→ → →MM PV = nRT

Moles O2 = 27.0 g HgO1 mol HgO

×216.6 g HgO

21 mol O2 mol HgO

× = 0.0623 mol O2

Temperature in kelvins = 50.0°C + 273.15 = 323.2 K

2O0.0623 mol

=n RT

V =P

( ) L atm0.08206 ⋅mol K⋅

323.2 K

( )0.947 atm

= 1.75 L O2

9−41

9.130 First, we use the ideal gas law to calculate the number of moles of argon in the sample:

Pressure of Ar in atm = 765 torr 1 atm760 torr

× = 1.01 atm


(1.01 atm=

PVn =RT

)(3.00 L )L0.08206 atm⋅mol K⋅

300.2 K

( ) = 0.123 mol Ar

Next, we use the ideal gas law to calculate the volume of argon after bubbling it through water. Note: We use Dalton’s law of partial pressures to determine the pressure of the argon alone:

total Ar waterP P P= + Ar total waterP = P P− = 778 torr − 13 torr = 765 torr

Pressure of Ar in atm = 765 torr 1 atm760 torr

× = 1.01 atm


0.123 mol=

nRTV =P

( ) L atm0.08206 ⋅mol K⋅

288.2 K

( )1.01 atm

= 2.88 L

9.131 If we assume that the reaction takes place at 745 torr and 25.0°C, we can use the stoichiometric coefficients

to calculate the volume of the butene combusted:

Volume C4H8 = 212.0 L O 4 8

2

1 L C H6 L O

× = 2.00 L C4H8

Then, we can use the combined gas law to calculate the volume occupied by this amount of butene at 188°C and 2.50 atm:

P1 V1 T1 P2 V2 T2 745 torr 2.00 L 25.0°C 2.50 atm ? 188°C

Pressure of C4H8 in atm = 745 torr 1 atm760 torr

× = 0.980 atm

Convert temperatures to kelvins:

T1 = 25.0°C + 273.15 = 298.2 K

T2 = 188°C + 273.15 = 461 K

Solve the combined gas law for V2:

1 1 22

1 2= ×

PV TVT P

= (0.980 atm )(2.00 L)298.2 K

461 K×

2.50 atm = 1.21 L

9−42

9.132 We can use the molar masses of N2 (28.02 g/mol) and Ar (39.95 g/mol) to calculate the number of moles of each in the mixture.

Moles N2 = 72.0 g 1 mol28.02 g

× = 2.57 mol N2

Moles Ar = 66.0 g 1 mol39.95 g

× = 1.65 mol Ar

We can either calculate the total pressure due to the total number of moles of gas in the sample, or we can calculate the partial pressures of each gas separately, and then add them to find the total pressure (as shown below):


total

2.57 mol2

2

NN

n RTP =

V=( ) L0.08206 atm

mol⋅

K⋅318.2 K

( )2.00 L

= 33.6 atm

total

1.65 molAr

Arn RTP =V

=( ) L0.08206 atm

mol⋅

K⋅318.2 K

( )2.00 L

= 21.5 atm

The total pressure is: PT = 33.6 atm + 21.5 atm = 55.1 atm.

9.133 First, we use the balanced chemical equation and the molar mass of CuO (79.55 g/mol) to calculate the

number of moles of H2 required to react with 85.0 g of CuO. Then we can use the ideal gas law to calculate the volume of H2 gas.

2 2 CuO Mole ratio PV nRTMass CuO mol CuO mol H Volume H=→ → →MM

Moles H2 = 85.0 g CuO1 mol CuO

×79.55 g CuO

21 mol H1 mol CuO

× = 1.07 mol H2



× = 0.950 atm

1.07 molnRTV =

P=( ) L atm0.08206 ⋅

mol K⋅300.2 K

( )0.950 atm

= 27.7 L

9.134 The “Before” image shows the molecules at higher pressure. Because the density of the particles is higher,

and they appear to be traveling at the same velocity, the pressure must be higher. The “After” image shows the particles at the same temperature as in the “Before” image (they appear to have the same kinetic energy) but at a lower pressure. This can happen only if the volume of the container increases (i.e. Boyle’s law: as pressure decreases, volume increases) or if molecules are effusing from a hole in the container. In either case, there would be a reduction in density without a change in kinetic energy.

9−43

9.135 Because the gas volumes are all measured at the same conditions, we can use the mole ratios as volume ratios:

(a) Volume H2S = 25.00 L H 2

2

1 L H S1 L H

× = 5.00 L

(b) Volume NH3 = 25.00 L H 3

2

2 L NH3 L H

× = 3.33 L

(c) Volume C6H12 = 25.00 L H 6 12

2

1 L C H3 L H

× = 1.67 L

9.136 From the mass of CaCO3 and its molar mass (100.09 g/mol), we can calculate the number of moles of CaCO3 that decompose each day. Then we can use the balanced chemical equation to determine the number of moles of CO2 formed, and the ideal gas law to determine the volume of this CO2 at the stated conditions.

3 3 3 2 21000 g 1 kg mole ratiokg CaCO g CaCO mol CaCO mol CO L CO=→ → → →MM PV = nRT

Moles CO2 = 531.00 10 kg CaCO× 31000 g CaCO×

3kg CaCO31 mol CaCO

×3100.09 g CaCO

2

3

1 mol CO1 mol CaCO

× = 9.99 × 105 mol CO2

T = 25.0°C + 273.15 = 298.2 K


× = 0.967 atm

59.99 10 mol×

=nRTV =

P

( ) L atm0.08206 ⋅mol K⋅

298.2 K

( )0.967 atm

= 2.53 × 107 L

9.137 From Chapter 5, we learned that, when heated, calcium carbonate decomposes to form calcium oxide and

carbon dioxide gas:

CaCO3(s) → CaO(s) + CO2(g) Reaction 1

The reaction of CaO with carbon is:

2CaO(s) + 5C(s) → 2CaC2(s) + CO2(g) Reaction 2

The equation given in this problem is:

CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g) Reaction 3

If we determine the number of moles of CaCO3 (limestone) that react, then we can calculate the number of moles of C2H2 that form using the stoichiometric coefficients from reactions 1-3.

3 2 2 2mole ratio mole ratio mole ratiomol CaCO mol CaO mol CaC mol C Hreaction 1 reaction 2 reaction 3→ → →

Moles CaCO3 = 35.00 g CaCO 3

3

1 mol CaCO100.09 g CaCO

× = 0.0500 mol CaCO3

Moles C2H2 = 30.0500 mol CaCO1 mol CaO

×31 mol CaCO

22 mol CaC×

2 mol CaO2 2

2

1 mol C H1 mol CaC

× = 0.0500 mol C2H2

9−44

Finally, we use the ideal gas law to calculate the temperature of the gas:

0.750 atmPVT = =nR

( ) 1.25 L( )0.0500 mol( ) L0.08206 atm⋅

mol K

⋅

= 228 K (−45°C)

9.138 From the mass of KClO3 (122.55 g/mol) we can calculate the number of moles of O2 formed. Then we can

calculate the volume of O2 produced at 735.0 torr and 70.0°C. Because the O2 is collected over water, we recognize that the pressure of the collected gas mixture will include the vapor pressure of water (Table 9.2).

3 3 2 2Dalton 's lawof partial p ressure

mole ratioMass KClO mol KClO mol O L OMM PV=nRT→ → →

Moles O2 = 313.5 g KClO 31 mol KClO×

3122.55 g KClO2

3

3 mol O2 mol KClO

× = 0.165 mol O2

The total pressure of the gas mixture (O2 plus H2O) is 735 torr, and the partial pressure of water vapor at 70.0°C is 233.7 torr. This means that the O2 pressure at this temperature is:

2OP = 735 torr – 233.7 torr = 501 torr


× = 0.659 atm

T = 70.0°C + 273.15 = 343.2 K

0.165 mol=

nRTV =P

( ) L atm0.08206 ⋅mol K⋅

343.2 K

( )0.659 atm

= 7.05 L

9.139 (a) Assuming the temperature and pressure are kept constant, the number of moles of a gas and its volume are directly proportional to each other. Increasing the number of moles of gas increases the volume occupied by the gas.

(b) Pressure and temperature must be held constant for this relationship to be true. We can use the ideal gas law to show this:

knRTV = nP

= ×

where k is constant as long as T and P are constant. This means volume is directly proportional to the number of moles of gas at constant T and P.

9.140 (a) Air is a mixture of gases. To calculate the average molar mass of air, we calculate the volume occupied

by 1 mole of air. Then we can determine the mass of the air in that sample volume using the density of air under the specified conditions. Because the mass we calculate represents the mass of 1 mol of air, it is the molar mass of air. Because 760.0 torr = 1 atm, we can simply substitute this pressure into the calculation.

T = 25.0°C + 273.15 = 298.2 K

1 mol=

nRTV =P

( ) L atm0.08206 ⋅mol K⋅

298.2 K

( )1.000 atm

= 24.47 L

9−45

Mass of gas = 24.47 L 1.186 g1 L

× = 29.02 g

The molar mass of air is 29.02 g/mol (b) The molar masses of O2 and N2 are 32.00 g/mol and 28.02 g/mol, respectively. Assuming that air

contains only O2 and N2, we can calculate the percentage of each as follows:

Mass difference O2 – N2 = 32.00 g/mol – 28.03 g/mol = 3.98 g/mol

Mass difference Air – N2 = 1.00 g/mol

Percentage of N2 = 3.98 g / mol 1.00 g / mol 100%3.98 g / mol

−× = 74.9% N2

The numerator of the percentage represents how much closer the average mass is to the mass of nitrogen. The denominator represents the total difference in the two masses. The percentage of O2 is 25.1%. Alternatively, we can use the fractional abundances to calculate the average mass:

Average mass = 2 2 2 2N N O O× + ×F MM F MM

where MM represents the molar mass of each substance and F represents the fractional abundance of each substance (what we are trying to find). Both the fractions are unknown, so we need a second equation:

Sum of fractional abundances = 2 2N O+F F = 1.00

The sum the fractional abundances is 1.00 (i.e. 100%). We can solve the two equations and calculate the percentages of N2 and O2.

9.141 We can determine the molar mass of the liquid by dividing the mass of the liquid by the number of moles

of liquid. Mass is given, but we need to determine the number of moles from the information given. We are given the pressure, temperature, and volume of the vaporized liquid, so we can use the ideal gas law to calculate the number of moles of vapor (which is the same as the number of moles of liquid).

Volume in liters = 127 mL 1 L1000 mL

× = 0.127 L


× = 0.909 atm

Temperature in kelvins = 98°C + 273.15 = 371 K

0.909 atm

= =PVnRT

( ) 0.127 L( )L0.08206 atm⋅mol K⋅

371 K

( ) = 3.79 × 10−3 mol

Molar mass of liquid = 30.495 g

3.79 10 mol−× = 131 g/mol

9−46

9.142 If you make a close inspection of the data, they appear to be linear. There are actually three data points in the question. The implied data point is that the air pressure is 760 mm at an altitude of 0 ft. If you graph the data and extrapolate to 30,000 ft you find that the air pressure will be very low; about 50 torr. Under these circumstances, the partial pressure of oxygen would be too low to survive.

9.143 If we calculate the total number of moles of gas produced by the reaction, we can calculate their total

volume.

3 5 2 3 3 5 2 3mole ratiog C H (ONO ) mol C H (ONO ) mol gas Volume gas→ → →MM PV = nRT

The molar mass of nitroglycerine is 227.1 g/mol. Note: We are not concerned with the volume produced by any one gas, so we can group the stoichiometric coefficients of the gases:

1 mol C3H5(ONO2)3 = 12 + 10 + 6 + 1 = 29 mol gas

Moles gas = 3 5 2 31.00 g C H (ONO ) 3 5 2 31 mol C H (ONO )×

3 5 2 3227.1 g C H (ONO ) 3 5 2 3

29 mol gas4 mol C H (ONO )

× = 0.0319 mol gas

Temperature in K = 275°C + 273.15 = 548 K

0.0319 mol=

nRTV =P

( ) L atm0.08206 ⋅mol K⋅

548 K

( )2.00 atm

= 0.718 L

9.144 The water pressure decreases as the mass of the water above the bubble decreases. Like a balloon, an air bubble changes volume so that the pressure inside the bubble is the same as outside the bubble. As pressure on the outside of the bubble decreases, the air bubble volume increases to allow for decrease in air pressure inside the bubble.

9.145 (a) Average kinetic energy does not change when volume is changed because it is constant at a given

temperature. (b) Average molecular velocity does not change when volume is changed because for a given gas, average

velocity only changes when kinetic energy changes: KEave = m(vave)2. (c) Pressure increases as the gas particles become more crowded with resulting greater collisions per unit

area. 9.146 (a) Average kinetic energy decreases with temperature. (b) Average molecular velocity decreases because kinetic energy decreases: KEave = m(vave)2. (c) Pressure decreases because gas particles are colliding less frequently and with less force. 9.147 (a) Average kinetic energy decreases with temperature. (b) Average molecular velocity decreases because kinetic energy decreases. (c) Volume decreases to maintain constant pressure as gas particle velocity decreases.

9−47

9.148 To calculate moles of each gas we rearrange the ideal gas law PV=nRT to solve for moles: PV

nRT

= . To

calculate mass, we multiply moles by the molar mass of the gas. (a) Before substituting P, V and T into the formula, we check to see if each has the correct units.

Temperature is given in kelvin units and volume is given in liters so no conversions are needed for T and V. However pressure units must be converted from units of torr to atm:

1 atm 389.0 torr 0.5120 atm760 torr

P = × =

Substituting into the rearranged gas law gives:

0.512 atmPVnRT

= =20.0 L×

L0.08206 atm⋅mol K⋅

300.0 K×0.416 mol

= Ne

Mass of neon = 0.416 mol Ne1

20.18 g Ne

mol Ne× 8.39 g Ne=

(b) The calculated number of moles of hydrogen is the same as that for neon because identity of the element is irrelevant in the ideal gas law calculation. The mass is different because hydrogen has a lower molar mass.

Moles of hydrogen = 0.416 mol H2

Mass of hydrogen = 20.416 mol H 2

2

2.016 g H

1 mol H× 20.839 g H=

(c) Moles of carbon dioxide = 0.416 mol CO2

Mass of carbon dioxide = 20.416 mol CO 2

2

44.01 g CO

1 mol CO× 218.3 g CO=

9.149 Under the identical conditions of temperature and pressure, the ratio of gas densities is the same as the ratio of their molar masses because equal volumes of gases contain the same number of gas particles (Avogadro’s hypothesis). Therefore, to find a gas that has 8.0 times the density we must identify a gas with a molar mass that is 8.0 times the molar mass of H2:

Molar mass of unknown gas = molar mass H2 ´ 8.0 = 2.016 g/mol ´ 8.0 = 16 g/mol

The molar mass of O2 is 32.00 g/mol, the molar mass of CH4 is 16.04 g/mol, and the molar mass of Ne is 20.18 g/mol. The molar mass of CH4 is the best match, so of the three possibilities, it is most likely the unknown gas.

9.150 (a) The air pressure is lower at higher altitudes, so the air inside the sealed bottle is at a lower pressure

than after the plane descends. The higher pressure at the Earth’s surface pushes in on the bottle much more than the air inside the bottle, causing the bottle to decrease volume by crushing. This is an application of Boyle’s law.

(b) This is another application of Boyle’s law. The lower atmospheric pressure at higher altitudes causes the pressure inside the bag to also be lower, causing its volume to be greater. Once the volume of the bag can no longer increase, the pressure inside the bag increases to greater than the external pressure.

(c) As the temperature in the trunk and in the soccer ball increases, volume and pressure increase due to increased air particle collisions. If the soccer balls are old and worn in places, the air pressure can be great enough to break the seams.

(d) As the air in a hot air balloon is heated up, the air molecules increase in velocity and spread out (Charles’ Law) as some escape through the large hole in the bottom of the balloon. The remaining air molecules become much less crowded, and the mass per unit volume decreases, decreasing the density.

9.151 The pressure of a gas can be increased by reducing the volume, increasing the temperature, or adding more

gas to the container. Another way would be to increase the pressure of the surrounding air if the gas is in an elastic container.

9−48

9.152 (a) The partial pressure of water vapor at 50.0°C is 92.5 torr.

(b) The molecular-level image shows there are 5 times as many O2 molecules as water molecules, so the partial pressure of O2 is 5 times that of H2O (Dalton’s law of partial pressures):

2OP = 92.5 torr ´ 5 = 462 torr

(c) The total pressure is the sum of the partial pressures: Ptotal =

2OP + 2H OP = 462.5 torr + 92.5 torr = 555 torr

(d) We know the partial pressure of O2, temperature, and volume so we can solve for moles of O2 using PV=nRT after converting pressure to atm units, temperature to kelvin units, and volume to mL units:

0.60855 atmPVnRT

= = 0.1200 L×


323.15 K×0.002754 mol

=

(e) To determine moles of KClO3, we use mole ratio obtained from the coefficients in the balanced equation as the conversion factor:

3 2Moles of KClO 0.00276 mol O= 3

2

2 moles of KClO3 mol O

× 30.00184 mol KClO=

9.153 (a) 91 atmPVnRT

= =1.0 L×


740 K×1.5 mol

=

(b) Mass of carbon dioxide = 21.5 mol CO 2

2

44.01 g CO

1 mol CO× 266 g CO=

(c) To calculate density we divide mass by volume. Mass was calculated in (b). The volume used to determine moles in (a) was 1.0 L so this is the volume we must use to calculate density:

mass 66 gDensity 66 g/Lvolume 1.0 L

= = =

Density in units of g/mL 66 gL

=1 L

× 0.066 g/mL1000 mL

=

CONCEPT REVIEW

9.154 Answer: C; Pressure is the force applied per unit area. When considering gas particles, it is the pressure they exert on the walls of a container divided by the surface area of the container.

A. The densities of gases are less than the densities of liquids and solids. B. Gases consist of particles that are relatively far apart compared to liquids and solids. D. Warm air is less dense than cold air. E. When the temperature of a gas increases, the density of the gas decreases.

9.155 Answer: B; To compare the magnitudes of the pressure, they should be converted to a common unit. The

value 12.1 lb/in2 (A) is equal to 0.823 atm. The pressure 34.1 inches Hg (C), converted to atmospheres is 0.0449 atm. In part D, 760 torr is equal to 1.0 atm. The pressure 64,000 Pa (E) converted to atmospheres is 0.63 atm. The order of increasing pressure is C <E <A <D <B.

9.156 Answer: E; According to Charles’s law, volume will decrease as temperature decreases at constant

pressure.

A. The balloon will expand because the low heat will slow the molecules down. B. Temperature has no effect on the behavior of gases. Volume is directly proportional to temperature (on

the Kelvin scale).

9−49

C. Molecules do not change in size when temperature changes. D. If temperature increases, the balloon expands because the molecules move faster and occupy a larger

volume. 9.157 Answer: A; When volume decreases, atoms will be closer together.

B. With a volume decrease a given volume of gas will contain more atoms. C. Atoms do not get larger with changes in volume, pressure, or temperature. D. This image does not show a change in volume of the piston. With no volume change at constant

temperature, there wouldn’t be a change in the number of atoms. Also, atom sizes do not change. E. This image does not show a change in volume of the piston. If this were corrected, the image will still

be wrong because atom sizes do not change. 9.158 Answer: C; This is a statement of Boyle’s law.

A. The volume of a gas is inversely proportional to pressure at constant temperature. B. The volume of a gas is directly proportional to temperature at constant pressure. D. Equal volumes of different gases will contain the same moles of gas at a given temperature and

pressure. E. The volume of a gas is directly proportional to the moles of gas at constant temperature and pressure.

9.159 Answer: C; The moles of helium can be calculated by rearranging the ideal gas law: PVn = RT

, where R is

the ideal gas constant, L atm0.08206mol K

R = ⋅⋅

. Following the units of the constant, volume must be expressed

in liters, pressure in atmospheres, and temperature in kelvins.

A. This mathematical operation has the incorrect rearrangement of the ideal gas law to solve for moles of gas. Also, temperature is not expressed in kelvins.

B. This expression has the correct relationship to solve for moles but does not show the conversion of pressure and torr to atm. Also, the temperature is not expressed in kelvins.

D. This expression has the correct conversion for torr to atom and shows temperature in kelvins. However, the ideal gas law is incorrectly rearranged to solve for moles.

E. This mathematical operation has the correct relationship to solve for moles but has incorrect values of pressure, volume, and temperature.

9.160 Answer: D; Substituting the moles of gas, temperature, and pressure into the ideal gas law:

1.25 mol

= nRTV = P

( ) L atm0.08206 ⋅mol K⋅

50.0 273.15 K

+

( )

978 torr 1 atm ×760 torr

= 25.8 L

A. The volume decreases by a factor of 0.84. B. The initial volume can be calculated using the ideal gas law:

( ) ( )L atm1.25 mol 0.08206 25.0 273.15 K

mol K = = 30.8 L1 atm755 torr

760 torr

nRTV = P

⋅ + ⋅ ×

C. The volume is expected to decrease because the pressure increases by a greater factor than the temperature increase.

E. The volume of one mol of any gas at standard temperature and pressure is 22.4 L.

9−50

9.161 Answer: E; According to the ideal gas law, equal volumes of different gases contain the same number of moles of gas. Substituting volume, temperature, and pressure into the ideal gas law:

( )( )

( )

1.12 atm 2.50 L = = 0.114 mol

L atm0.08206 25.0 273.15 Kmol K

PVn = RT ⋅ + ⋅

The molar mass can then be used to calculate the grams of each gas. Since density is mass/volume, the gas with highest molar mass will have the greatest density.

A. 2

2

N 2

28.02 g0.114 mol N 1 molDensity = = 3.21 g/L N

2.50 L

×

B. 2

2

O 2

32.00 g0.114 mol O 1 molDensity = = 3.66 g/L O

2.50 L

×

C. 2

2

CO 2

44.01 g0.114 mol CO 1 molDensity = = 5.04 g/L CO

2.50 L

×

D. He

4.003 g0.114 mol He 1 molDensity = = 0.458 g/L He

2.50 L

×

E. 2

2

SO 2

64.06 g0.114 mol SO 1 molDensity = = 7.33 g/L SO

2.50 L

×

9.162 Answer: C; The average kinetic energy of gas particles depends only on the absolute temperature, not the

identity of the gas.

A. The He and O2 containers contain the same number of gas molecules, but have different masses. B. There are equal moles of He and O2. D. The He molecules on average are moving faster than the O2 molecules. E. The volume of the O2 container has the same volume as the He container.

9.163 Answer: B; Under STP conditions the gas with the greatest number of moles will occupy the greatest

volume. Using the molar mass of each gas, we can calculate the moles and then use molar volume (24.414 L/1 mol) to determine the volume occupied by each gas.

A. 2NVolume = 46.2 g 2

1 mol N ×28.02 g

24.414 L 1 mol

× 2 = 37.0 L N

B. 2HVolume = 24.3 g 2

1 mol H ×2.016 g

24.414 L 1 mol

× 2 = 270 L H

C. 2COVolume = 54.1 g 2

1 mol CO ×44.01 g

24.414 L 1 mol

× 2 = 27.6 L CO

D. ArVolume = 39.4 g 1 mol Ar ×39.95 g

24.414 L 1 mol

× = 22.1 L Ar

E. 3NHVolume = 21.5 g 3

1 mol NH ×17.03 g

24.414 L 1 mol

× 3 = 28.3 L NH

The order of increasing volume is D < C < E < A < B 9.164 Answer: E; Gases are composed of small and widely separated particles.

A. The average kinetic energy of gas particles depends only on the absolute temperature of the gas.

9−51

B. The average velocity of gas particles increases when temperature increases. C. The volume occupied by gas particles is very small compared to the total volume occupied by the gas. D. A gas particle moves in rapid, straight-line motion until it collides with another particle or the

container. When collisions do occur, they are perfectly elastic. 9.165 Answer: B; The gas with the lowest molar mass is expected to have the greatest velocity. The order of

increasing average velocity is C < A < D < E < B 9.166 Answer: B; According to Gay-Lussac’s law of combining volumes, gases combine in simple whole-number

volume proportions with the volume ratios or reactant and products the same as the mole ratios as long as conditions of temperature and pressure remain constant. The volume of CO2 produced can be calculated from the following relationship:

2CO 4 10Volume = 5.00 L C H 2

4 10

8 L CO

2 L C H× 2 = 20.0 L CO

A. The volume of O2 required to react with butane can be calculated in a similar way:

2O 4 10Volume = 5.00 L C H 2

4 10

13 L O

2 L C H× 2 = 32.5 L O

The volume of O2 required to completely react with the butane is 32.5 L. C. The ratio of volumes of reactants to products is 15:18. The total volume of gases is expected to

increase. D. The volume of H2O produced can be calculated from the mole ratios indicated by the balanced

equation:

2H O 4 10Volume = 5.00 L C H 2

4 10

10 L H O

2 L C H× 2 = 25.0 L H O

The volume of H2O produced is 25.0 L. E. The final volume of gases produced can be calculated from the mole ratios indicated by the balanced

equation:

products 4 10Volume = 5.00 L C H 2 2

4 10

18 L products (CO +H O)

2 L C H× = 45.0 L products

The final volume of gases produced is 45.0 L. 9.167 Answer: C; From Table 9.2 the vapor pressure of water at 29°C is 30.0 torr. If oxygen gas is collected over

water at a total pressure of 765 torr, the partial pressure of oxygen is 735 torr:

2total O waterP = P + P

2O total water = 765 torr 30.0 torr = 735 torrP = P P− −

A. 3mol O = 7.5 g KClO231 mol KClO

×3122.55 g KClO

2

3

3 mol O

2 mol KClO× 2 0.092 mol O=

0.092 mol of oxygen is expected to be produced. B. The partial pressure of water at this temperature is 30 torr. D. 7.5 g of KClO3 reacts.

E. 3mol KClO = 7.5 g KClO33

31 mol KClO

122.55 g KClO× 0.061 mol KClO3=

0.061 mol of KClO3 reacts.

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