208

CHAPTER 6 GASES

PRACTICE EXAMPLES 1A (E) The pressure measured by each liquid must be the same. They are related

through P = g h d Thus, we have the following g hDEG dDEG = g hHg dHg. The g’s cancel; we substitute known values: 9.25 mDEG ×1.118 g/cm3

DEG = hHg× 13.6 g/cm3

Hg

3

Hg 3

1.118g/cm9.25m 0.760m Hg, = 0.760m Hg =760.mmHg

13.6g/cmh P

1B (E) The solution is found through the expression relating density and height: TEG TEG Hg Hgh d h d

We substitute known values and solve for triethylene glycol’s density: 9.14 mTEG ×dTEG = 757 mmHg ×13.6 g/cm3

Hg. Using unit conversions, we get

dTEG= 3 30.757 m13.6 g/cm 1.13 g/cm

9.14 m

2A (E) We know that Pgas = Pbar + ΔP with Pbar = 748.2 mmHg. We are told that ΔP

= 7.8 mmHg. Thus, Pgas = 748.2 mmHg + 7.8 mmHg = 756.0 mmHg. 2B (M) The difference in pressure between the two levels must be the same, just

expressed in different units. Hence, this problem is almost a repetition of Practice Example 6-1. hHg=748.2 mmHg – 739.6 mmHg=8.6 mmHg. Again we have g hg dg=g hHg dHg. This becomes hg × 1.26 g/cm3 glycerol = 8.6 mmHg ×13.6 g/cm3 Hg

hg 3mmHg

g cm Hg

g cm glycerolmm glycerol 8 6

136

12693

3

.. /

. /

3A (M) A = r2 (here r = ½(2.60 cm × 1 m

100 cm) = 0.0130 m)

0.0130 m)2 = 5.31 × 10-4 m2 F = m × g = (1.000 kg)(9.81 m s-2) = 9.81 kg m s-2 = 9.81 N

P = A

F = -4 2

9.81 N

5.31 10 m= 18475 N m-2 or 1.85 × 104 Pa

P (torr) = 1.85 × 104 Pa × = 139 torr

Chapter 6: Gases

209

3B (M) Final pressure = 100 mb. 101, 325 Pa

100 mb 1013.25 mb

= 1.000 × 104 Pa

The area of the cylinder is unchanged from that in Example 6-3, (1.32 × 10-3 m2).

P = F

A = 1.000 × 104 Pa =

-3 2

F

1.32×10 m

Solving for F, we find F = 13.2 (Pa)m2 = 13.2 (N m-2)m2 = 13.2 N F = m × g = 13.2 kg m s-2 = -2m 9.81 m s

total mass = mass of cylinder + mass added weight = m = g

F =

-2

-2

13.2 kg m s

9.81 m s= 1.35 kg

An additional 350 grams must be added to the top of the 1.000 kg (1000 g) red cylinder to increase the pressure to 100 mb. It is not necessary to add a mass with the same cross sectional area. The pressure will only be exerted over the area that is the base of the cylinder on the surface beneath it.

4A (M) The ideal gas equation is solved for volume. Conversions are made within the equation.

33

33

1 mol NH 0.08206L atm20.2g NH 25 273 K

17.03g NH mol K24.4 L NH

1atm752 mmHg

760 mmHg

nRTV

P

4B (E) The amount of Cl g2b g is 0.193 mol Cl2 and the pressure is 0.980 atm (0.993

barr × (1 atm/1.01325 barr) = 0.980 atm). This information is substituted into the ideal gas equation after it has been solved for temperature.

1 1

0.980atm 7.50 L464K

0.193mol 0.08206Latm mol K

PVT

nR

5A (E) The ideal gas equation is solved for amount and the quantities are substituted.

10.5atm 5.00 L2.11mol He

0.08206 L atm30.0 273.15 K

mol K

PVn

RT

5B (M)

n = PV

RT=

-7 33

-1 -1

1 atm 1000L6.67 10 Pa 3.45 m

101325 Pa 1m(0.08206 L atm K mol ) (25 273.15)K

= 9.28 × 10-10 moles of N2

molecules of N2 = 9.28 × 10-10 mol N2 × 23

2

2

6.022 10 molecules of N

1 mole N

molecules of N2 = 5.59 × 1014 molecules N2

Chapter 6: Gases

210

6A (E) The general gas equation is solved for volume, after the constant amount in moles is cancelled. Temperatures are converted to kelvin.

VV PT

P T21 1 2

2 1

100211

.

.mL 2.14atm 37.8 + 273.2 K

1.02atm 36.2 + 273.2 KmL

b gb g

6B (M) The flask has a volume of 1.00 L and initially contains O g2 b g at STP. The mass

of O g2b g that must be released is obtained from the difference in the amount of

O g2b g at the two temperatures, 273 K and 373 K. We also could compute the masses

separately and subtract them. We note that 1.00 bar is 0.987 atm.

2 2 2STP O O O100 C

21 1

2

1 1mass released = n n

273K 373K 273K 373K

0.987 atm 1.00 L 1 1 32.00 g= = 0.378 g O

0.08206 L atm mol K 273 K 373 K 1 mol O

PV PV PVM M M

R R R

7A (M) The volume of the vessel is 0.09841 L. We substitute other values into the

expression for molar mass.

0.08206L atm40.4868g 40.1305g 22.4 273.2 K

mol K 86.4g/mol1atm

772mmHg 0.09841L760mmHg

mRTM

PV

7B (M) The gas’s molar mass is its mass (1.27 g) divided by the amount of the gas

in moles. The amount can be determined from the ideal gas equation.

nPV

RT

FHG

IKJ

737 107

0 0820625 273

0 0424

mm Hg1 atm

760mm HgL

Latmmol K

Kmol gas

.

..

b g

1.27 g= = 30.0 g/mol

0.0424 molM

This answer is in good agreement with the molar mass of NO, 30.006 g/mol. 8A (M) The molar mass of He is 4.003 g/mol. This is substituted into the expression for density.

dMP

RT

4 003 0 987

0 08206 2980162

1. .

..

g mol atm

L atm mol K Kg / L

-1 -1

When compared to the density of air under the same conditions (1.16 g/L, based on the “average molar mass of air”=28.8g/mol) the density of He is only about one seventh as much. Thus, helium is less dense (“lighter”) than air.

Chapter 6: Gases

211

8B (M) The suggested solution is a simple one; we merely need to solve for mass of gas from density and its moles from the ideal gas law.

m(gas) D V 1.00 g/L 1.00 L 1.00 g

1

1atm745mmHg 1.00 L

760mmHgPVn 0.0312 mol

RT 0.08206L atm K 382 K

Therefore, the molar mass of the gas is as follows: g 1.00 g

M 32.0 g/molmol 0.0312 mol

The molecular weight suggests that the gas is O2. 9A (M) The balanced equation is 2 2 33 2NaN s Na l N (g)( ) ( )

2 2

1atm776mmHg 20.0L

760mmHgmoles N 0.821mol N

0.08206Latm(30.0 273.2) K

mol K

P V

R T

Now, solve the stoichiometry problem.

3 33 2 3

2 3

2 mol NaN 65.01 g NaNmass NaN = 0.821 mol N = 35.6 g NaN

3 mol N 1 mol NaN

9B (M) Here we are not dealing with gaseous reactants; the law of combining volumes

cannot be used. From the ideal gas equation we determine the amount of N g2b g per

liter under the specified conditions. Then we determine the amount of Na(l) produced simultaneously, and finally the mass of that Na(l).

1 atm

Pressure: 1.0 barr 0.987 atm1.01325 barr

2

0.987 atm 1.000L 2mol Na 22.99g Namass of Na(l) 0.619g Na(l)

0.08206Latm 3mol N 1mol Na(25 273) Kmol K

10A (E) The law of combining volumes permits us to use stoichiometric coefficients for volume

ratios.

O volume L NO g L O

L NO L O g2

22= 1.00

5

4= 1.25b g b g

Chapter 6: Gases

212

10B (E)The first task is to balance the chemical equation. There must be three moles of hydrogen for every mole of nitrogen in both products (because of the formula of NH3 )

and reactants: N g H g NH g2 2 3+ 3 2b g b g b g . The volumes of gaseous reactants and

products are related by their stoichiometric coefficients, as long as all gases are at the same temperature and pressure.

volume NH g L H g L NH g

L H g L NH3 2

3

23= 225

2

3= 150.b g b g b g

b g

11A (M) We can work easily with the ideal gas equation, with a the new temperature of

T = 55+ 273 = 328b g K K . The amount of Ne added is readily computed.

nNe g Ne1 mol Ne

20.18g Nemol Ne 12 5 0 619. .

Pn RT

V

total

mol0.08206Latm

mol KK

Latm

( . . )

.

175 0 619 328

5013

11B (E) The total volume initially is 2.0 + 8.0 = 10.0L L L. These two mixed ideal gases

then obey the general gas equation as if they were one gas.

PPV T

V T21 1 2

2 1

100

2 055

.

..

atm 10.0 L 298 K

L 273Katm

12A (M) The partial pressures are proportional to the mole fractions.

P

nP

P P P

H OH O

tottot

2

2 22

CO tot H O 2

2

2

2 2

n

mol H O

mol CO mol H Oatm = 0.0348atm H O(g)

atm atm = 2.47 atmCO g

0 00278

0197 0 002782 50

2 50 0 0348

.

. ..

. . ( )

12B (M) Expression (6.17) indicates that, in a mixture of gases, the mole percent equals

the volume percent, which in turn equals the pressure percent. Thus, we can apply these volume percents—converted to fractions by dividing by 100—directly to the total pressure.

2

2

N pressure = 0.7808 748 mmHg = 584 mmHg,

O pressure = 0.2095 748 mmHg = 157 mmHg,

2CO pressure = 0.00036 748 mmHg = 0.27 mmHg,

Ar pressure = 0.0093 748 mmHg = 7.0 mmHg

13A (M) First compute the moles of H g2 b g , then use stoichiometry to convert to moles of HCl.

2

1atm(755 25.2) torr 0.0355L

760 mmHg 6mol HCIamount HCI = 0.00278mol HCI

0.08206 L atm 3mol H(26 273) Kmol K

Chapter 6: Gases

213

13B (M) The volume occupied by the O g2 b g at its partial pressure is the same as the volume

occupied by the mixed gases: water vapor and O g2 b g . The partial pressure of O g2b g is

found by difference. O pressure total pressure mmHg H O pressure mmHg2 2= 749.2 23.8 = 725.4 b g

2 21

1atm725.4 mm Hg 0.395 L

760 mm HgP Vmol O 0.0154 mol O

R T 0.08206 L atm K 298 K

2 22 2 2

2 2

2 mol Ag O 231.74 g Ag Omass Ag O = 0.0154 mol O 7.14 g Ag O

1 mol O 1 mol Ag O

Mass% Ag2O = 7.14/8.07 × 100 = 88.4% The volume of the dry gas in turn is determined as follows:

1 10.0154 mol 0.08206 L atm mol K 298 K0.382 L

749.2 / 760 atmV

14A (M) The gas with the smaller molar mass, NH3 at 17.0 g/mol, has the greater root-mean-square speed

2 2 1 1

rms 1

3 3 8.3145kg m s mol K 298K661m/s

0.0170kg mol

RTu

M

14B (D)

bullet speed = 2180mi 1h 5280ft 12in. 2.54cm 1m

974.5m/s1h 3600s 1mi 1ft 1in. 100 cm

Solve the rms-speed equation (6.20) for temperature by first squaring both sides.

2

rms

3RTu

M

2 3

2

rms 22

2

974.5m 2.016 10 kg1s 1mol H

76.75K8.3145kg m3

3s mol K

u MT

R

We expected the temperature to be lower than 298 K. Note that the speed of the bullet is about half the speed of a H2 molecule at 298 K. To halve the speed of a molecule, its temperature must be divided by four.

15A (M) The only difference is the gas’s molar mass. 2.2 10 4

2 mol N effuses through the orifice in 105 s.

2

2

24

2

? mol O 28.014 g/mol0.9357

2.2 10 mol N 31.999 g/molN

O

M

M

moles 4 42 2O = 0.9357 2.2 10 = 2.1 10 mol O

Chapter 6: Gases

214

15B (M) Rates of effusion are related by the square root of the ratio of the molar masses

of the two gases. H2 , effuses faster (by virtue of being lighter), and thus requires a shorter time for the same amount of gas to effuse.

2

2 2

2

2 2H N

2 2

2.016 g H /mol Htime = time 105 s 28.2 s

28.014 g N /mol NH

N

M

M

16A (M) Effusion times are related as the square root of the molar mass. It requires 87.3 s for Kr to effuse.

unk unk

Kr

unknown time 131.3 ssubstitute in values 1.50

Kr time 87.3 s 83.80 g/mol

M M

M

Munk g / mol g / mol 1.504 83.80 = 1.90 102 2b g

16B (M) This problem is solved in virtually the same manner as Practice Example 18B. The lighter

gas is ethane, with a molar mass of 30.07 g/mol.

C H time = Kr time C H

Kr)s

30.07 g C H / mol C H

83.80 g Kr / mol Krs2

2 2 6 2 66

6 87 3 52 3 M

M

( )

(. .

17A (D) Because one mole of gas is being considered, the value of n a2 is numerically the same as the value of a , and the value of nb is numerically the same as the value of b .

2

2

2 2

2

2 2

0.083145 L barr1.00 mol 273 K

3.66 L barr molmol K(2.00 0.0427) L (2.00 L)

11.59 barr 0.915 barr

= 10.68 barr CO g compared with 10.03 barr for Cl g

nRT n a

PV nb V

ideal

0.083145 L barr1.00 mol 273 KnRT mol KP 11.35 ~ 11.4 barr

V 2.00 L

Cl g2b g shows a greater deviation from ideal gas behavior than does CO g2 b g .

17B (M) Because one mole of gas is being considered, the value of n a2 is numerically the same as the value of a , and the value of nb is numerically the same as the value of b .

2 2

2 2

0.083145 L barr1.00 mol 273 K 1.47 L barrmol K 11.58 atm 0.368 atm

(2.00 0.0395) L (2.00 L)

= 11.2 barr CO g

nRT n a

PV nb V

compared to 10.03 barr for Cl g2b g , 11.2 barr for CO, and 11.35 barr for CO g2 b g . Thus, Cl g2b g displays the greatest deviation from ideality, 11.4 barr.

Chapter 6: Gases

215

INTEGRATIVE EXERCISE

A. (M) First, convert the available data to easier units. 101.3×103 Pa × (1 atm/1.013×105 Pa) = 1 atm, and 25 °C = 298 K. Then, assume that we have a 1 L container of this gas, and determine how many moles of gas are present:

1

PV (1.00 atm)(1.00 L)n = = = 0.0409 mol.

RT 0.08206 L atm K 298 K

Knowing the density of the gas (1.637 g/L) and its volume (1 L) gives us the mass of 1 L of gas, or 1.637 g. Therefore, the molar mass of this gas is: (1.637 g / 0.0409 mol) = 40.03 g/mol Now, determine the number of moles of C and H to ascertain the empirical formula:

22

2 2

22

2 2

1 mol CO 1 mol Cmol C: 1.687 g CO 0.0383 mol

44.01 g CO 1 mol CO

1 mol H O 2 mol Hmol H: 0.4605 g H O 0.05111 mol

18.02 g H O 1 mol H O

Dividing by the smallest value (0.0383 mol C), we get a H:C ratio of 1.33:1, or 4:3. Therefore, the empirical formula is C3H4, which has a molar mass of 40.07, which is essentially the same as the molar mass calculated. Therefore, the actual formula is also C3H4. Below are three possible Lewis structures:

HC C CH3

C C C

H

H

H

H

C

HC CH2

H

Chapter 6: Gases

216

B. (D) First, let us determine the amount of each element in the compound:

3 22

2 2

3 22

2 2

1 mol CO 1 mol Cmol C: 151.2 10 g CO 0.003435 mol C

44.01 g CO 1 mol CO

12.01 g Cg C: 0.003435 mol C = 0.04126 g C

1 mol C1 mol H O 2 mol H

mol H: 69.62 10 g H O 0.18.02 g H O 1 mol H O

3

12

007727 mol H

1.01 g Hg H: 0.007727 mol H = 0.007804 g H

1 mol H

PV 1 atm 9.62 10 L 2 mol Nmol N: = 0.0008589 mol N

RT 1 mol N0.08206 L atm K (273 K)

14.01 g Ng N: 0.0008589 mol N = 0.

1 mol N

01203 g N

Therefore, the mass of O is determined by subtracting the sum of the above masses from the mass of the compound: g O: 0.1023 (0.04126 0.007804 0.01203) 0.04121 g O

1 mol Omol O: 0.04121 g O = 0.002576 mol O

16.0 g O

To determine the empirical formula, all of the calculated moles above should be divided by the smallest value. Doing so will give the following ratios: C: 0.003435/0.0008589 = 4 H: 0.007727/0.0008589 = 9 N: 1 O: 0.002576/0.0008589 = 3 The empirical formula is C4H9NO3, and has a molar mass of 119.14 g/mol. To determine the actual formula, we have to calculate its molecular mass. We know the density at a given volume and therefore we need to find out the number of moles. Before that, we should convert the experimental conditions to more convenient units. T = 127 °C + 273 = 400 K, and P = 748 mm Hg/760 mm Hg = 0.9842 atm.

1

PV (0.9842 atm) 1 Ln = = = 0.02999 mol

RT 0.08205 L atm K (400 K)

MM = g/mol = 3.57 g/0.02999 = 119 g/mol

Therefore, the empirical and molecular formulas are the same, being C4H9NO3.

Chapter 6: Gases

217

EXERCISES

Pressure and Its Measurement 1. (E)

(a) P 7361

0 968 mmHg atm

760 mmHg atm.

(b) 1 atm

P 0.776 bar 0.766 atm1.01325 bar

(c) P 8921

117 torr atm

760 torr atm.

(d) P =1000 Pa 1 atm

225 kPa 2.22 atm1 kPa 101,325 Pa

2. (E)

(a) h 0 984760

748. atm mmHg

1 atm mmHg (b)

h 928 torr = 928 mmHg

(c) 22

2 2

12 in 2.54 cm 10 mmH O 1 mmHg 1 mHg142 ft H O 3.18 mHg

1 ft 1 in 1 cm H O 13.6 mmH O 1000 mmHgh

3. (E)

bnz bnz Hg Hg

3

bnz 3

We use:

0.760 m Hg 13.6 g/cm Hg0.970 atm 11.4 m benzene

1 atm 0.879 g/cm benzene

h d h d

h

4. (E)

4 4gly gly CCl CCl

34

gly 4 3

We use:

1.59 g/cm CCl3.02 m CCl 3.81 m glycerol

1.26 g/cm glycerol

h d h d

h

5. (E) P = Pbar – h1 = 740 mm Hg – 30 mm (h1) = 710 mm Hg 6. (E) P = Pbar + h1 = 740 mm Hg + 30 mm (h1) = 770 mm Hg

Chapter 6: Gases

218

7. F = m × g and 1 atm = 101325 Pa = 101325 kg m-1 s-2 = P = F

A =

-2

2

9.81 m s

1 m

m

mass (per m2) = -1 -2 2

-2

101325 kg m s 1 m

9.81 m s

= 10329 kg

(Note:1 m2 = (100 cm)2 = 10,000 cm2)

P (kg cm-2) = m

A=

2

10329 kg

10,000 cm= 1.03 kg cm-2

8. (E) Start by noting that 1 atm is defined as 101325 Pa. Pascals is the unit name used to describe

SI units of force (F) per unit area, and is expressed as N/m2. Since F = ma, the MKS units for pressure are kg·m·s-2·m-2, or kg·m-1·s-2. The Imperial unit of measuring force is pounds-force, or lbf. Considering acceleration due to gravity, an object weighing 1 N has a mass of (1/9.8 m/s2) 0.102 kg. We now can convert the MKS units to psi as follows:

2 2

2 2 2

N 0.102 kg 1 lb 1 m (2.54 cm)101325 = 14.7 psi

m 1 N 0.453 kg (100 cm) 1 in

The Simple Gas Laws

9. (E)

(a) V = 26.7762

385= 52.8 L

mmHg

mmHg L

(b) V = 26.7762

368= 7 27 L

mmHg

atm 760 mmHg

1 atm

L.

.

10. (E) Apply Charles’s law: . 26 273 299 KiV kT T

(a) 3371 K273 98 371 K = 886 mL 1.10 10 mL

299 KT V

(b) 2253 K273 20 253 K = 886 mL 7.50 10 mL

299 KT V

11. (E) Charles’ Law states that V1/T1 = V2/T2. Therefore,

22

3.0 L 1.50 L, and T = 225 K

450 K T

12. (E) 0.725 L

105 kPa 33.8 kPa (0.334 atm)2.25 L

if i

f

VP P

V

Chapter 6: Gases

219

13. (E) P PV

Vi ff

i

F

HGIKJ 721

187550 6 mmHg

35.8 L L

35.8L

1 atm

760 mm H O atm

2

.

14. (E) We let P represent barometric pressure, and solve the Boyle’s Law expression

below for P.

342.0 mL ( 85 mmHg) 37.7 mL 42.0 37.7 3.2 10P P P P

323.2 10

7.4 10 mmHg42.0 37.7

P

15. (E) Combining Boyle’s and Charles’ Law, we get the following expression:

1 1 2 2

1 2

P V P V=

T T

Therefore,

3 3

2 2 12 3 3

1 1

0.340 atm 5.00 10 m 300 KP V TT = = = 255 K

P V 1.000 atm 2.00 10 m

16. (E) First, convert temperatures from Celsius to the Kelvin scale.

22 C + 273.15 = 295.15 K -22 C = (-22 + 273.15) = 251.15 K

Volume decrease is proportional to the ratio of the temperatures in Kelvin.

volume contraction = 251.15 K

0.851295.15 K

(volume will be about 85% of its original volume).

17. (E) STP: P = 1 barr and T = 273.15. P (1 barr) = 0.9869 atm

Ar

0.9869 atm 0.0750 L 39.948 g Armass = = 0.132 g Ar

L atm 1 mol Ar0.08206 273.15 KK mol

18. (M) STP: P = 1 barr (0.9869 atm) and T = 273.15 K (note: one mole of gas at STP occupies 22.698

L).

22

2

1 mol Cl Latm250.0 g Cl 0.08206 273.15K

70.906 g K molnRTV= = = 80.08 L Cl

P 0.9869atm

Alternatively 22 2

1 mol Cl 22.698 L at STP= = 250.0 g Cl = 80.03 L Cl

70.906 g 1 molmV n V

Chapter 6: Gases

220

19. (M)

(a) Conversion pathway approach:

3 3 3

3

3

1 mol PH 34.0 g PH 1000 mg PH1 L mass = 27.6 mL

1000 mL 22.698 L STP 1 mol PH 1 g

= 41.3 mg PH

Stepwise approach:

3

3

3

3

3

3 3

3 3

1 L27.6 mL

1000 mL

1 mol PH

22.698 L STP

34.0 g PH

1 mol PH

1000 mg PH

1 g

0.0276 L

0.0276 L 0.001216 mol PH

0.001216 mol PH 0.0413 g PH

0.0413 g PH 41.3 mg PH

(b)

23

3 33

203

6.022 10 moleculesnumber of molecules of PH = 0.001216 mol PH

1 mol PH

number of molecules of PH = 7.32 10 molecules

20. (E)

(a)

617 7

23

2

1 mol Rn 222 g Rn 10 gmass = 5.0 10 atoms = 8.30 10 mol

1 mol 1 g6.022 10 atoms

mass = 1.8 10 g Rn g

(b) 6

7 22.698 L 10 L volume = 8.30 10 mol = 19 L Rn g

1 mol 1 L

21. (M) At the higher elevation of the mountains, the atmospheric pressure is lower than

at the beach. However, the bag is virtually leak proof; no gas escapes. Thus, the gas inside the bag expands in the lower pressure until the bag is filled to near bursting. (It would have been difficult to predict this result. The temperature in the mountains is usually lower than at the beach. The lower temperature would decrease the pressure of the gas.)

Chapter 6: Gases

221

22. (M) Based on densities, 1 = 13.6 2 mHg mH O . For 30 m of water,

22

1 mHg30 mH O = 2.2 mHg

13.6 mH O water

1000 mm 1 atm2.2 mHg = 2.9 atm

1 m 760 mmHgP

To this we add the pressure of the atmosphere above the water:

total 2.9 atm 1.0 atm 3.9 atmP .

When the diver rises to the surface, she rises to a pressure of 1.0 atm. Since the pressure is about one fourth of the pressure below the surface, the gas in her lungs attempts to expand to four times the volume of her lungs. It is quite likely that her lungs would burst.

General Gas Equation

23. (E) Because the number of moles of gas does not change, P V

TnR

P V

Ti i

i

f f

f

is

obtained from the ideal gas equation. This expression can be rearranged as follows.

VV P T

P Tfi i f

f i

4 25 748 2732 268

742 2732 2564 30

. ( . . )

( . . ).

L mmHg K

mmHg K L

24. (E) We first compute the pressure at 25 C as a result of the additional gas. Of course, the gas pressure increases proportionally to the increase in the mass of gas, because the mass is proportional to the number of moles of gas.

P =12.5

10.0762 = 953

g

g mmHg mmHg

Now we compute the pressure resulting from increasing the temperature.

P =62 + 273

25+ 273953 = 1.07 10 3b g

b g K

K mmHg mmHg

25. (E) Volume and pressure are constant. Hence niTi = PV

R= nfTf

f

i

n

n = i

f

T

T=

(21 273.15) K

(210 273.15) K

= 0.609 (60.9 % of the gas remains)

Hence, 39.1% of the gas must be released. Mass of gas released = 12.5 g ×39.1

100= 4.89 g

26. (M) First determine the mass of O2 in the cylinder under the final conditions.

mass O atm L g / mol

L atmmol K

K g O2 2

n M

PV

RTM

115 34 0 32 0

0 08206 22 273517

. . .

. ( ).

mass of O to be released g g g O2 2= 305 51.7 = 253

Chapter 6: Gases

222

Ideal Gas Equation 27. (M) Assume that the CO g2 b g behaves ideally and use the ideal gas law: PV nRT=

2

4

1 mol CO L atm89.2 g 0.08206 (37 273.2) K

1000 mL44.01 g mol K5.32 10 mL

1 atm 1 L737 mmHg760 mmHg

nRTV

P

28. (M) PnRT

V

FHG

IKJ

358 46 2732

12 82 29

. ( . )

..

g O1 mol O

32.00 g O0.08206 L atm

mol K K

L atm

22

2

29. (E) 11.2 atm 18.5 L 83.80 g/mol

mass 702 g KrL atm

0.08206 (28.2+273.2) Kmol K

PVn M M

RT

30. (M)

Mol He = 7.41 g He × (1 mol He/4.003 g He) = 1.85 mol

3

3 3

3.50 atm 72.8 L1.68 10 K

L atm1.85 mol 0.08206

mol K

( C) = (1.68 10 273) 1.41 10 C

PVT

nR

t

31. (M) 9 15gas 23

1 mol gasn 5.0 10 molecules gas 8.3 10 mol gas

6.022 10 molecules gas

We next determine the pressure that the gas exerts at 25 C in a cubic meter

15

113

33

L atm8.3 10 mol gas 0.08206 298.15 K

101,325 PaK molP = = 2.1 10 Pa1 atm1 L10 dm

1 m1 m 1 dm

32. (M)

2

2

2CO 2 2

2

22

CO 3

2

1 mol COn 1242 g CO = 28.20 mol CO T = 273.15 + (-25) = 248.15 K

44.010 g CO

0.250 m 1000 LV = = 3.1416 1.75 m = 85.9 L

2 1 m

L atm 101,325 Pa28.20 mol CO 0.08206 248.15 K

K molP =

r h

51 atm = 6.77 10 Pa

85.9 L

Chapter 6: Gases

223

33. (E) The basic ideal law relationship applies here. Molar volume is the amount of volume that one mole of a gas occupies. If PV=nRT, then molar volume is V/n, and the relationship rearranges to:

m

V R T = V =

n P

(a) Vm = (298 K·0.08206 L·atm·K-1)/1.00 atm = 24.4 L·mol-1 (b) Patm = 748 mmHg / 760 mmHg = 0.978 atm Vm = (373 K·0.08206 L·atm·K-1)/0.978 atm = 31.3 L·mol-1 34. (E) The molar volume formula given above can be rearranged to solve for T. T = (P/R)·Vm = (2.5 atm/0.08205 L·atm·K-1)×22.4 L·mol-1 = 682 K Determining Molar Mass 35. (M) Use the ideal gas law to determine the amount in moles of the given quantity of gas.

-1

L atm0.418 g 0.08206 339.5 K

mol K1 atm

743 mmHg 0.115 L760 mmHg

= 104 g molmRT

PVM

Alternatively 1 atm 1 L

743 mmHg 115 mL760 mmHg 1000 mL

0.00404 mol gasL atm

0.08206 (273.2 66.3) Kmol K

PVn

RT

M =

0.418

0.00404= 103

g

mol g / mol

36. (M) Use the ideal gas law to determine the amount in moles in 1 L of gas.

-1

1 atm725 mmHg 1 L

mass760 mmHg0.028 mol gas =

L atm0.08206 415 K

mol K

0.841 g01 = = 30.0 g mol

moles 0.02801 mol

PVn

RTM

37. (M) First we determine the empirical formula for the sulfur fluoride. Assume a 100 g sample of SxFy.

1 mol S 1 mol Fmoles S = 29.6 g S = 0.923 mol S moles F = 70.4 g F = 3.706 mol F

32.064 g S 18.9984 g F

Dividing the number of moles of each element by 0.923 moles gives the empirical formula SF4. To find the molar mass we use the relationship:

-1 -1 -1-1dRT 4.5 g L 0.08206 L atm K mol 293 K

Molar mass = = = 108 g molP 1.0 atm

-1

4 4-1

molecular formula mass 108 g molThus, molecular formula empirical formula = SF = SF

empirical formula mass 108.06 g mol

Chapter 6: Gases

224

38. (M) We first determine the molar mass of the gas.

T P= 24.3+ 273.2 = 297.5 = 7421

760= 0.976 K mmHg

atm

mmHg atm

1 1mRT 2.650 g 0.08206 L atm mol K 297.5 K

M= 155 g/mol1 LPV

0.976 atm 428 mL1000 mL

Then we determine the empirical formula of the gas, based on a 100.0-g sample. 1 mol C

mol C = 15.5 g C = 1.29 mol C 0.649 1.99 mol C12.01 g C

1 mol Clmol Cl = 23.0 g Cl = 0.649 mol Cl 0.649 1.00 mol Cl

35.45 g Cl

1 mol Fmol F = 61.5 g F = 3.24 mol F 0.649 4.99 mol F

19.00 g F

Thus, the empirical formula is C ClF2 5 , which has a molar mass of 154.5 g/mol. This is the same as the experimentally determined molar mass. Hence, the molecular formula is C ClF2 5 .

39. (M)

(a)

L atm0.231 g 0.08206 (23 273) K

mol K 55.8 g/mol1 atm 1 L

749 mmHg 102 mL760 mmHg 1000 mL

mRTM

PV

(b) The formula contains 4 atoms of carbon. (5 atoms of carbon gives a molar mass of at least 60―too high―and 3 C atoms gives a molar mass of 36―too low to be made up by adding H’s.) To produce a molar mass of 56 with 4 carbons requires the inclusion of 8 atoms of H in the formula of the compound. Thus the formula is C H4 8 .

40. (M) First, we obtain the mass of acetylene, and then acetylene’s molar mass. mass of acetylene g g g= 56.2445 56.1035 = 0.1410 acetylene

L atm0.1410 g 0.08206 (20.02 273.15) K

mol K 26.04 g/mol1 atm 1 L

749.3 mmHg 132.10 mL760 mmHg 1000 mL

mRTM

PV

The formula contains 2 atoms of carbon (3 atoms of carbon gives a molar mass of at least 36-too high and 1C atom gives a molar mass of 12-too low to be made up by adding H’s.) To produce a molar mass of 26 with 2 carbons requires the inclusion of 2 atoms of H in the formula. Thus the formula is C2H2.

Chapter 6: Gases

225

Gas Densities

41. (E)

3

L atm1.80 g/L 0.08206 (32 273) K

760 mmHgmol K28.0 g/mol 1 atm

1.21 10 mmHg

MP dRTd P

RT M

P

22

2

28.0 g N1 LMolar volume N 15.56 L/mol

1.8 g 1 mol N

42. (E) MdRT

M

2 56 0 08206 22 8 273 2

75662 5

. . ( . . ).

g / L L atmmol K

K

mmHg1 atm

760 mmHg

g / mol

62.5 g 1 L

Molar volume 24.4 L/molmol 2.56 g

43. (M)

(a) 28.96 g/mol 1.00 atm

1.18 g/L airL atm

0.08206 (273 25)Kmol K

MPd

RT

(b) dMP

RT

44 0 100

0 08206 273 25180

. .

. ( ).

g / mol CO atmL atmmol K

K g / L CO2

2

Since this density is greater than that of air, the balloon will not rise in air when filled with CO2 at 25C ; instead, it will sink!

44. (E) dMP

RTT

MP

RT

becomes

g / mol atmL atmmol K

g / LK = 181 C

44 0 100

0 08206 118454

. .

. .

45. (E)

L atm2.64 g/L 0.08206 (310+273)K

mol K becomes 124 g/mol1 atm

775 mmHg760 mmHg

MP dRTd M

RT P

Since the atomic mass of phosphorus is 31.0, the formula of phosphorus molecules in the vapor must be P4 . 4 31.0 = 124 atoms / moleculeb g

Chapter 6: Gases

226

46. (M) We first determine the molar mass of the gas, then its empirical formula. These two pieces of information are combined to obtain the molecular formula of the gas.

L atm2.33 g/L 0.08206 296 K

mol K 57.7 g/mol1 atm

746 mmHg760 mmHg

dRTM

P

1 mol Cmol C = 82.7 g C = 6.89 mol C 6.89 1.00 mol C

12.0 g C

1 mol Hmol H = 17.3 g H = 17.1 mol H 6.89 2.48 mol H

1.01 g H

Multiply both of these mole numbers by 2 to obtain the empirical formula, C H2 5 , which has an empirical molar mass of 29.0 g/mol. Since the molar mass (calculated as 57.7 g/mol above) is twice the empirical molar mass, twice the empirical formula is the molecular formula, namely, C H4 10 .

Gases in Chemical Reactions

47. (E) Balanced equation: 3 8 2 2 2C H g + 5 O g 3 CO g + 4 H O l

Use the law of combining volumes. O volume L C H L O

L C H L O2 3 8

2

3 82= 75.6

5

1= 378

48. (E) Each mole of gas occupies 22.7 L at STP.

2 22

2

2

3 mol H (g) 22.7 L H (g) at STP1 mol AlH STP volume 1.000 g Al

26.98 g Al 2 mol Al(s) 1 mol H

= 1.262 L H g

49. (M) Determine the moles of SO g2b g produced and then use the ideal gas equation.

Conversion pathway approach:

6 22

62

1 mol SO3.28 kg S 1000 g S 1 mol Smol SO = 1.2 10 kg coal

100.00 kg coal 1 kg S 32.1 g S 1 mol S

1.23 10 mol SO

62

72

L atm1.23 10 mol SO 0.08206 296 K

mol K 3.1 10 L SO1 atm

738 mmHg760 mmHg

nRTV

P

V

Chapter 6: Gases

227

Stepwise approach:

6 4

4 7

7 6

6 622

3.28 kg S1.2 10 kg coal 3.94 10 kg S

100.00 kg coal

1000 g S3.94 10 kg S 3.94 10 g S

1 kg S

1 mol S3.94 10 g S 1.23 10 mol S

32.1 g S

1 mol SO1.23 10 mol S = 1.23 10 mol SO

1 mol S

6

27

2

L atm1.23 10 mol SO 0.08206 296 K

mol K 3.1 10 L SO1 atm

738 mmHg760 mmHg

nRTV

P

50. (M) Determine first the amount of CO g2 b g that can be removed. Then use the ideal gas law.

22 2

1 mol CO1000 g 1 mol LiOHmol CO = 1.00 kg LiOH = 20.9 mol CO

1 kg 23.95 g LiOH 2 mol LiOH

VnRT

P

20 9 25 9 273 2

7511

519. ( . . ) mol 0.08206

L atmmol K

K

mmHg atm

760 mmHg

L CO (g)2

51. (M) Determine the moles of O2, and then the mass of KClO3 that produced this amount of O2.

2 2

1 atm 1 L738 mmHg 119 mL

760 mmHg 1000 mLmol O 0.00476 mol O

L atm0.08206 (22.4 273.2)K

mol K

mass KClO mol O mol KClO

mol O

g KClO

mol KClO g KClO3 2

3

2

3

33= 0.00476

2

3

122.6

1= 0.389

% KClO g KClO

g sample KClO3

33=

0 389

357100% = 10.9%

.

.

52. (M) Determine the moles and volume of O2 liberated. 2 2 1 +2 2 2 2 H O aq H O O gb g b g b g

2 2 2 2 22

2 2 2 2

2

0.0300 g H O 1 mol H O 1 mol O1.01 gmol O = 10.0 mL soln

1 mL 1 g soln 34.0 g H O 2 mol H O

= 0.00446 mol O

Chapter 6: Gases

228

2

2

L atm0.00446 mol O 0.08206 (22 273) K

1000 mLmol K 109 mL O1 atm 1 L752 mmHg

760 mmHg

V

53. (M) First we need to find the number of moles of CO(g)

Reaction is 3 7 32 3 8 2 CO g H g C H g H O la f a f a f a f

2

2

CO

2

HH (required) 2

1 atm28.5 L 760 torrPV 760 torn = = = 1.27 moles CO

L atmRT 0.08206 273.15 KK mol

7 mol H L atm1.27 mol CO 0.08206 299 Kn RT 3 mol CO K molV = = = 73.7 LH

1 atmP 751 mmHg760 mmHg

54. (M) (a) In this case, H2 is the limiting reagent.

33 2 3

2

2 L NHVolume NH = 313 L H = 209 L NH

3 L H

(b) Moles of NH3 (@ 315 C and 5.25 atm)

1 132 3

2

2 mol NH5.25 atm 313 L3.41 10 mol H = 2.27 10 mol NH

L atm 3 mol H0.08206 (315 273)Kmol K

13

23

L atm2.27 10 mol NH 0.08206 298 K

mol K(@25 C, 727mmHg) = 5.80 10 L NH1 atm

727 mmHg760 mmHg

V

Mixtures of Gases 55. (M) Determine the total amount of gas; then use the ideal gas law, assuming that the

gases behave ideally. 1 mol Ne 1 mol Ar

moles gas = 15.2 g Ne + 34.8 g Ar20.18 g Ne 39.95 g Ar

= 0.753 mol Ne + 0.871 mol Ar = 1.624 mol gas

VnRT

P= =

1624 26 7 2732

7.15= 5.59

. ( . . ) mol 0.08206L atmmol K

K

atm L gas

Chapter 6: Gases

229

56. (M) 2 24. L H (g)2 at STP is 20.100 mol H (g). After 0.10 mol He is added, the

container holds 0.20 mol gas. At STP, pressure is in barr; 1 barr is 0.987 atm L atm

0.20 mol 0.08206 (273 100)Kmol K 6.2 L gas

0.987 atm

nRTV

P

57. (M) The two pressures are related, as are the number of moles of N g2 b g to the total

number of moles of gas.

2 2

28.2 atm 53.7 Lmoles N 61.7 mol N

L atm0.08206 (26 273) K

mol K

PV

RT

2

75.0 atmtotal moles of gas = 61.7 mol N = 164 mol gas

28.2 atm

32

20.18 g Nemass Ne = 164 mol total 61.7 mol N = 2.06 10 g Ne

1 mol Ne

58. (M) Solve a Boyle’s law problem for each gas and add the resulting partial pressures.

2H He

2.35L 3.17 L762 mmHg 324mmHg 728 mmHg 418 mmHg

5.52 L 5.52 LP P

2total H He 324 mmHg + 418 mmHg 742 mmHgP P P

59. (M) Initial pressure of the cylinder

P = nRT

V=

-1 -122

2

1 mol O(1.60 g O )(0.08206 L atm K mol )(273.15 K)

31.998 g O

2.24 L

= 0.500 atm

We need to quadruple the pressure from 0.500 atm to 2.00 atm.

The mass of O2 needs to quadruple as well from 1.60 g 6.40 g or add 4.80 g O2

(this answer eliminates answer (a) and (b) as being correct).

One could also increase the pressure by adding the same number of another gas (e.g. He) mass of He = nHe × MMHe

(Note: moles of O2 needed = 4.80 g × 2

2

1 mol O

31.998 g O = 0.150 moles = 0.150 moles of He)

mass of He = 0.150 moles × 4.0026 g He

1 mol He= 0.600 g He ((d) is correct, add 0.600 g of He)

Chapter 6: Gases

230

60. (M) (a) First determine the moles of each gas, then the total moles, and finally the total pressure.

22 2

2

2

1 mol H 1 mol Hemoles H = 4.0 g H moles He = 10.0 g He

2.02 g H 4.00 g He

= 2.0 mol H = 2.50 mol He

L atm(2.0 2.50) mol 0.08206 273 K

mol K 23 atm4.3 L

nRTP

V

(b) P PH2

He2atm

2.0 mol H

4.5 mol total atm atm atm = 13 atm 23 10 23 10

61. (M)

(a)

6 6

6 6ben

1 mol C H L atm0.728 g 0.08206 (35 273) K

78.11 g C H mol K 760 mmHg

2.00 L 1 atm= 89.5 mmHg

nRTP

V

Ptotal = 89.5 mmHg C H g mmHg Ar g mmHg6 6 + 752 = 842b g b g

(b) P Pbenzene Ar mmHg mmHg 89 5 752.

62. (D) (a) The %CO2 in ordinary air is 0.036%, while from the data of this problem, the

%CO2 in expired air is 3.8%. P

P

{

{ }

.

..

CO expired air}

CO ordinary air

CO

COCO (expired air to ordinary air)2

2

2

22

38%

0 036%11 102

(b/c) Density should be related to average molar mass. We expect the average molar

mass of air to be between the molar masses of its two principal constituents, N2 (28.0 g/mol) and O2 (32.0 g/mol). The average molar mass of normal air is approximately 28.9 g/mol. Expired air would be made more dense by the presence of more CO2 (44.0 g/mol) and less dense by the presence of more H O2 (18.0 g/mol). The change might be minimal. In fact, it is, as the following calculation shows.

2 2exp.air 2 2

2 2

2 22 2

2 2

28.013 g N 31.999 g O0.742 mol N + 0.152 mol O

1 mol N 1 mol O

44.01 g CO 18.02 g H O+ 0.038 mol CO + 0.059 mol H O

1 mol CO 1 mol H O

39.9 g Ar+ 0.009 mol Ar

1 mol Ar

M

= 28.7 g/mol of expired air

Chapter 6: Gases

231

Since the average molar mass of expired air is less than the average molar mass of ordinary air, expired air is less dense than ordinary air. Calculating the densities:

(28.7 g/mol)(1.00 atm)d(expired air) = 1.13 g/L

(0.08206 L atm/K mol)(310 K)

d(ordinary air) 1.14 g/L

63. (E) 1.00 g H2 0.50 mol H2 1.00 g He 0.25 mol He Adding 1.00 g of He to a vessel that only contains 1.00 g of H2 results in the number

of moles of gas being increased by 50%. Situation (b) best represents the resulting mixture, as the volume has increased by 50%

64. (E) The answer is (b), because the volume at 275 K with 1.5 times as many atoms is roughly 4/3

larger than the drawing. 65. (M) In this problem, you don’t need to explicitly solve for moles of gas, since you are looking at

the relationship between pressure and volume.

2

2

4.0 atm 1.0 LPV 4.0mol O = = =

RT RT RT2.0 atm 2.0 LPV 4.0

mol N = = = RT RT RT

total mol. of gas = 8.0/RT

Therefore,

nRT 8.0 RTP = = = 4.0 atm

V RT 2.0

66. (M) Like the above problem, you also don’t need to explicitly solve for moles of each gas to get

the total moles, because you are ultimately looking at pressure/volume relationships:

0.75 atm 1.0 LPV 0.75mol He = = =

RT RT RT0.45 atm 2.5 LPV 1.125

mol Xe = = = RT RT RT

1.20 atm 1.0 LPV 1.2mol Ar = = =

RT RT RTtotal mol. of gas = 3.075/RT

The total volume after the opening of the valves is the sum of the volumes of the flasks (VFlask)

and the tubes (VTube). Therefore,

Chapter 6: Gases

232

Flask Tube Tube

nRT 3.075 RTP = = = 0.675 atm

V V RT 4.5 V

Solving for VTube, we get a volume of 0.055 L.

Collecting Gases over Liquids 67. (M) The pressure of the liberated H g2 b g is 744 23.8 = 720.mmHg mmHg mmHg

2

2

3 mol H1 mol Al L atm1.65 g Al 0.08206 (273 25)K

26.98 g 2 mol Al mol K2.37 L H (g)

1 atm720. mmHg

760 mmHg

nRTV

P

This is the total volume of both gases, each with a different partial pressure. 68. (D) (a) The total pressure is the sum of the partial pressures of O g2b g and the vapor

pressure of water.

2 2 2

2

total O H O O

O

756 mmHg = 19 mmHg

(756 19) mmHg = 737 mmHg

P P P P

P

(b) The volume percent is equal to the pressure percent.

2 2 2O O2

total total

737 mmHg of O

756 mm Hg total%O (g) by volume 100% 100% 100% 97.5%

V P

V P

(c) Determine the mass of O g2b g collected by multiplying the amount of O2

collected, in moles, by the molar mass of O g2b g .

2

2O 2

2

1 atm 1 L737. mmHg 89.3 mL

760 mmHg 1000 mL 32.00 g Omass 0.115 g O

L atm 1 mol O0.08206 (21.3 273.2)Kmol K

PV

RT

69. (M) We first determine the pressure of the gas collected. This would be its “dry gas”

pressure and, when added to 22.4 mmHg, gives the barometric pressure.

PnRT

V

FHG

IKJ

146

10 08206 297

116

760729

. .

.

g mol O

32.0 g OL atmmol K

K

L

mmHg

1 atm mmHg

2

2

barometric pressure mm Hg mmHg mmHg= 729 + 22.4 = 751

Chapter 6: Gases

233

70. (M) We first determine the “dry gas” pressure of helium. This pressure, subtracted from the barometric pressure of 738.6 mmHg, gives the vapor pressure of hexane at 25C .

PnRT

V

FHG

IKJ

1072

10 08206 298 2

8 446

760589 7

. . .

..

g mol He

4.003 g HeL atmmol K

K

L

mmHg

1 atm mmHg

vapor pressure of hexane = 738.6 589.7 = 148.9 mmHg 71. (M) The first step is to balance the equation:

2NaClO3

2NaCl + 3O2 The pressure of O2 is determined by subtracting the known vapor pressure of water at the given

temperature from the measured total pressure. PO2 = PTOT – PH2O = 734 torr – 21.07 torr = 713 torr Patm = 713 mmHg / 760 mmHg = 0.938 atm

2 1

3

32 3

2 3

3

0.938 atm 0.0572 LPVmol O = = = 0.00221 mol

RT 0.08206 L atm K (296 K)

Mass of NaClO is then determined as follows:

2 mol NaClO 106.44 g0.00221 mol O = 0.1568 g NaClO

3 mol O 1 mol NaClO

%NaClO

0.1568 g = 100 17.9%

0.8765 g

72. (M) The work for this problem is nearly identical to the above problem.

2KClO3

2KCl + 3O2 The pressure of O2 is determined by subtracting the known vapor pressure of water at the given

temperature from the measured total pressure. PO2 = PTOT – PH2O = 323 torr – 25.22 torr = 298 torr

Patm = 298 mmHg / 760 mmHg = 0.392 atm

Chapter 6: Gases

234

2 1

3

32 3

2 3

3

0.392 atm 0.229 LPVmol O = = = 0.003657 mol

RT 0.08206 L atm K (299 K)

Mass of KClO is then determined as follows:

2 mol KClO 122.54 g0.003657 mol O = 0.299 g KClO

3 mol O 1 mol KClO

0.%KClO =

299 g 100 72%

0.415 g

Kinetic-Molecular Theory 73. (M) Recall that 1 J = kg m2 s-2

2 -2

rms 32

2

J 1 kg m s3 8.3145 303 K

mol K 1 J3326 m/s

70.91 10 kg Cl

1 mol Cl

RTu

M

74. (M)

1

1 11 1

22

3

2 = Square both sides and solve for . = 4 273 K = 1092 K273 K3

RTT TM T TTRT

M

Alternatively, recall that 1 J = kg m2 s-2

uRT

Mrms m/s 3

184 103. Solve this equation for temperature with urms doubled.

2 3 3 23rms

2 -2

2.016 10 kg/mol (2 1.84 10 m /s)1.09 10 K

3 J 1 kg m s3 8.3145

mol K 1 J

MuT

R

75. (M)

2 2rms

J3 8.3145 298K

3 mol K 0.00783 kg / mol = 7.83 g/mol. or 7.83 .( ) mi 1 hr 5280 ft 12 in. 1 m

2180hr 3600 sec 1 mi 1 ft 39.37 in.

RTM u

u

76. (E) A noble gas with molecules having urms at 25C greater than that of a rifle bullet will

have a molar mass less than 7.8 g/mol. Helium is the only possibility. A noble gas with a slower urms will have a molar mass greater than 7.8 g/mol; any one of the other noble gases will have a slower urms .

77. (E) We equate the two expressions for root mean square speed, cancel the common

factors, and solve for the temperature of Ne. Note that the units of molar masses do not

Chapter 6: Gases

235

have to be in kg/mol in this calculation; they simply must be expressed in the same units.

3 3 300

4 003

3

2018

RT

M

R R T

K Ne

. . Square both sides: Ne300 K

4.003 20.18

T

Solve for TNe: 3

Ne

20.18300 K 1.51 10 K

4.003T

78. (E) um , the modal speed, is the speed that occurs most often, 55 mi/h

38 + 44 + 45 + 48 + 50 + 55 + 55 + 57 + 58 + 60average speed = = 51.0 mi/h =

10u

2 2 2 2 2 2 2 2 2 2

rms

38 + 44 + 45 + 48 + 50 + 55 + 55 + 57 + 58 + 60 26472 miles = = 51.5

10 10 hu

79. (D) The greatest pitfall of this type of problem is using improper units. Therefore, convert everything to SI units.

MM O2 = 32.0 g/mol = 32.0×10-3 kg/mol

326

2 23

32.0 10 kg 1 molMass of O molecule: = 5.314 10 kg

mol 6.022 10 molec.

R =8.3145 J·mol-1·K-1, or kg·m2/(s2·mol·K) Now, we must determine the urms first to determine kinetic energy:

3

3RT 3 8.3145 298 482 m/s

M 32.0 10rmsu

Kinetic energy of an O2 molecule is as follows:

2 26 2 211 1 m = (5.314 10 kg)(482 m/s) = 6.17 10 J/molecule

2 2k rmse u

80. (D) To calculate the total kinetic energy for a certain quantity of gas molecules, we must first

calculate the (average) kinetic energy for one molecule and then apply it to the bulk.

MM N2 = 28.0 g/mol = 28.0×10-3 kg/mol 3

262 23

28.0 10 kg 1 molMass of N molecule: = 4.650 10 kg

mol 6.022 10 molec.

R =8.3145 J·mol-1·K-1, or kg·m2/(s2·mol·K) Now determine the urms first to determine kinetic energy:

3

3RT 3 8.3145 298 515 m/s

M 28.0 10rmsu

Kinetic energy of an N2 molecule is as follows:

Chapter 6: Gases

236

2 26 2 211 1 m = (4.650 10 kg)(515 m/s) = 6.17 10 J/molecule

2 2k rmse u

Therefore, the kinetic energy of 155 g of N2 is:

23 2142

22 2

1 mol N 6.022 10 molec. 6.17 10 J: 155 g N = 4.11 10 J

14.0 g N 1 mol N 1 molec.ke

Diffusion and Effusion of Gases 81. (M)

rate (NO

rate (N O

(N O

(NO

mol NO

mol N O/2

2

2

2

2

2

)

)

)

)

.

..

/

.

M

M

x t

t

44 02

46 010 9781

0 00484

mol NO mol mol NO2 2= 0.00484 0.9781= 0.00473 82. (M)

2 2

2

rate (N ) mol (N )/38 s 64s (unknown)1.68

rate (unknown) mol (unknown)/64s 38s ( )

M

M N

2 22(unknown)=(1.68) (N ) (1.68) (28.01g/mol )=79g/m olM M

83. (M)

(a) rate (N

rate (O

(O

(N2

2

2

2

)

)

)

)

.

..

M

M

32 00

28 01107 (b) rate (H O

rate (D O

(D O

(H O2

2

2

2

)

)

)

)

.

..

M

M

20 0

18 02105

(c) rate ( CO

rate ( CO

( CO

( CO

142

122

122

142

)

)

)

)

.

..

M

M

44 0

46 00 978 (d) rate ( UF

rate ( UF

( UF

( UF

2356

2386

2386

2356

)

)

)

).

M

M

352

3491004

84.

(M)

2

2

rate of effusion O

rate of effusion SO = 2

2

M(SO )

M(O ) =

64

32 = 2 = 1.4

O2 will effuse at 1.4 times the rate of effusion of SO2. The situation depicted in (c) best represents the distribution of the molecules. If 5 molecules of SO2 effuse, we would expect that 1.4 times as many O2 molecules effuse over the same period of time

(1.4 × 5 = 7 molecules of O2). This is depicted in situation (c).

85. (M) For ideal gases, the effusion rate is inversely proportional to their molecular mass. As such,

the rate of effusion of a known gas can be determined if the rate of effusion for another gas is known:

He

Ne

MMrate of effusion of Ne

rate of effusion of He MM

Chapter 6: Gases

237

Since effusion is loosely defined as movement of a fixed number of atoms per unit time, and since in this problem we are looking at the time it takes for the same number of moles of both Ne and He to effuse, the above equation can be rearranged as follows:

He He

Ne Ne

time MM =

time MM

4.00 =

22 h 20.18

Ne He

Ne He

mol mol

time time

x

Solving for x,

= 22 4.00 / 20.18 = 9.80 hx

86. (M) The same reasoning discussed above works in this problem:

Hg Hg

Rn Rn

Rn

Rn

time MM =

time MM

1 200.59 =

1.082 MM

Solving for MM gives a value of 235 g/mol.

HgRn

Rn Hg

molmol

time time

Nonideal Gases

87. (M) For 2

2 22 vdw 2

Cl (g), 6.49L atm and 0.0562L. = nRT n a

n a nb PV nb V

At 0ºC, Pvdw = 9.90 atm and Pideal = 11.2 atm, off by 1.3 atm or + 13%

(a) ideal

0.08206Latm1.00mol 373K

mol KAt 100 C 15.3atm

2.00L

nRTP

V

2

vdw 2

ideal

0.08206Latm1.00mol

mol K 6.49L atm0.0422 atm 1.62atm

2.00 0.0562 L 2.00L

=0.0422 373K 1.62 14.1atm is off by 1.2 atm or +8.5%

T

P T

P

(b) At 200 C

1.00mol0.08206Latm

mol KK

2.00Latmideal

PnRT

V

473

19 4.

vdw 2 3 5 ideal=0.0422 473K 1.62 18.3 atm is off by 1.0 atm or +5.5%P P

(c) At 400 C

1.00mol0.08206Latm

mol KK

2.00Latmideal

PnRT

V

673

27 6.

vdw ideal=0.0422 673K 1.62 26.8atm is off by 0.8 atm or +3.0%P P

Chapter 6: Gases

238

88. (M) (a) V = 100.0 L,

2 2 2

2 2

2

0.083145 L barr1.50 mol 298K 1.50 7.857 L barrmol K

(100.0 1.50 0.0879)L (100.0 L)

= 0.3721 0.0018 barr SO g 0.3704 atm

vdw

nRT n aP

V nb V

ideal

0.083145 L barr1.50 mol 298K

mol K 0.372 barr100.0L

P

The two pressures are almost equal. (b) V = 50.0 L, Pvdw = 0.739 barr; Pideal = 0.744 barr.

Here the two pressures agree within a few percent.

(c) V = 20.0 L, Pvdw = 1.82 barr; Pideal = 1.86 barr

(d) V = 10.0 L, Pvdw = 3.60 barr; Pideal = 3.72 barr

89. (E) The van der Waals parameter b is defined as the excluded volume per mole, or the volume that is taken up by 1 mole of gas once converted to a liquid.

From Table 6-5, bHe = 0.0238 L/mol. Therefore, the volume of a single He atom is: Conversion pathway approach:

31237 3

23 3

1 10 pm0.0238 L 1 mol He 1 m = 3.95 10 pm / He atom

mol He 6.022 10 atoms 1000 L 1 m

Step-wise approach:

2623

326 29 3

312

29 3 7 33

0.0238 L 1 mol He 3.95 10 L/atom

mol He 6.022 10 atoms

1 m 3.95 10 L/atom = 3.95 10 m /atom

1000 L

1 10 pm3.95 10 m /atom = 3.95 10 pm / He atom

1 m

V = (4/3) π r3. Rearranging to solve for r gives 3r= 3V 4π . Solving for r gives an atomic

radius of 211.3 pm.

Chapter 6: Gases

239

90. (M) (a) The process is very similar to the above example. From Table 6-5, bCH4 = 0.0431 L/mol. Therefore, the volume of a single CH4 molecule is:

3123

7 34423 3

4

1 10 pm1 mol CH0.0431 L 1 m = 7.16 10 pm / CH molec.

mol CH 6.022 10 molec. 1000 L 1 m

V = (4/3) π r3. Solving for r gives a molecular radius of 258 pm. This value is greater than r=228 pm, because our calculated value is based on b, which is the volume taken up by a mole of gas that has been condensed into a liquid. (b) Compression ratio is given as follows:

3 1

3 13

1 1

1 mL 1 L 16.043 gV = 2.43 10 L mol

66.02 g 1000 mL mol

PV (100 bar)(2.43 10 L mol )Z= = 9.0 10

RT 0.083145 bar L mol K 325 K

m

m

Integrative and Advanced Exercises

91. (E) A millimeter of mercury is 1/760 of an atmosphere of pressure at sea level. The density of mercury (13.5951 g/cm3) needs to be included in this definition and, because the density of mercury, like that of other liquids, varies somewhat with temperature, the pressure will vary if the density varies. The acceleration due to gravity (9.80665 m/s2) needs to be included because pressure is a force per unit area, not just a mass per unit area, and this force depends on the acceleration applied to the given mass, i.e., force = mass × acceleration.

92. (E) Of course, we express the temperatures in Kelvin:

K. 3 C 270– and K, 23 C 250– K, 73 C 200– K, 173 C 100- K, 273 C 0

(A) 3

3 110.0 cm0.0250 cm K

400 K

Vk

T

3 1 30.0250 cm K 273 K = 6.83 cmV

31313

31313

cm0.08K3Kcm0.02503cm0.58K23Kcm0.0250

cm8.1K73Kcm0.02503cm4.33K173Kcm0.0250

VV

VV

Chapter 6: Gases

240

(B)

313133

cm7.31K273Kcm0.0500Kcm0500.0K400

cm0.20 Vk

T

V

3131

3131

cm2.0K3K3cm0.0500cm2.123K K30.0500cm

cm7.3K73K3cm0.0500cm65.8173KK3cm0.0500

VV

VV

(C)

313133

cm3.27K273Kcm0.100Kcm100.0K400

cm0.40 Vk

T

V

313313

313313

cm3.0K3Kcm0.100cm3.223K Kcm100.0

cm3.7K73Kcm100.0cm3.17173KKcm100.0

VV

VV

As expected, in all three cases the volume of each gas goes to zero at 0 K.

93. (M) Initial sketch shows 1 mole of gas at STP with a volume of ~22.4 L.

Velocity of particles = vi. Note that velocity increases as T (i.e., quadrupling temperature,

results in a doubling of the velocity of the particles). a) Pressure drops to 1/3 of its original value (1 atm 0.333 atm). The number of

moles and temperature are constant. Volume of the gas should increase from 22.4 L 67 L while the velocity remains unchanged at vi.

b) Pressure stays constant at its original value (1 atm torr). The number of moles is

unchanged, while the temperature drops to half of its value (273 K 137 K). Volume of the gas should decrease from 22.4 L 11.2 L while the velocity

drops to about 71% to 0.71vi( T 0.5 T ).

c) Pressure drops to 1/2 of its original value (1 atm 0.5 atm). The number of

moles remains constant, while the temperature is increased to twice it original value (273 K 546 K). Volume of the gas should quadruple, increasing the volume from 22.4 L 90 L. Velocity of the molecules increases by 41% to

1.41vi( T 2 T ).

d) Pressure increases to 2.25 times its original value (2.25 atm) owing to a

temperature increase of 50% (273 K 410 K) and a 50% increase in the number of particles (1 mole 1.5 moles). The volume of the gas should remain unchanged at 22.4 L, while the velocity increases by 22% to

1.22vi( T 1.5 T ).

Chapter 6: Gases

241

n = 1 mol P = 1 atm V = 22.4 L T = 273 K

v = vi

(initial)

n = 1 mol P = 0.333 atm V = 11.2 L T = 273 K

v = vi

(a)

n = 1 mol P = 1 atm V = 11.2 L T = 137 K v = 0.71vi

(b)

n = 1 mol P = 0.500 atm V = 11.2 L T = 273 K

v = vi

(c)

n = 1 mol P = 2.25 atm V = 22.4 L T = 273 K

v = vi

(d)

94. (M) We know that the sum of the moles of gas in the two bulbs is 1.00 moles, and that both bulbs

have the same volume, and are at the same pressure because they are connected. Therefore,

1 1 2 2

1 2

2 1

1 2

n T =n T

n T 350 K= = 1.556

n T 225 K

n =1.556 n

Therefore, n2(1.556) + n2 = 1.00. Solving the equation gives (Flask 2) n2=0.391, and (Flask 1) n1 = 0.609.

95. (M) First, calculate the number of moles of gas:

1

PV (1.00 atm)(6.30 L)n= = 0.2377 mol

RT 0.082058 L atm K (323 K)

Molecular mass (MM) then can be calculated: MM = (10.00 g)/(0.2377 mol) = 42.069 g/mol Now we must determine the mole ratio of C to H: 85.6 g C × (1 mol C/12.01 g C) = 7.13 mol 14.4 g H × (1 mol H/1.01 g H) = 14.3 mol

Chapter 6: Gases

242

Now, divide both by the smallest number: C:C mole ratio = 1 H:C mole ratio = 2 Therefore, the empirical formula is CH2, with a formula unit molar mass of ~14.03 g/mol. To determine molecular formula, divide MM by formula unit MM: 42.069/14.03 = 3.00. Therefore, the molecular formula is C3H6.

96. (M) First we determine the molar mass of the hydrocarbon.

L atm0.7178g 0.08206 (65.0 273.2) KmRT mol KM 52.0 g/mol

0.00987 atmPV99.2 kPa 0.3907 L

1 kPa

Now determine the empirical formula. A hydrocarbon contains just hydrogen and carbon. 1 mol CO 1 mol C2 amount C 2.4267g CO 0.05514 mol C 0.05514 1.000 mol C

2 44.01 g CO 1 mol CO2 2

1 mol H O 2 mol H2 amount H 0.4967g H O 0.05513 mol H 0.05514 1.000 molH2 18.02 g H O 1 mol H O

2 2

The empirical formula is CH. This gives an empirical molar mass of 13 g/mol, almost precisely one-fourth of the experimental molar mass. The molecular formula is therefore C4H4.

97. (M) Stepwise approach: Note that three moles of gas are produced for each mole of NH4NO3 that decomposes.

4 34 3

4 3 4 3

-1 -1

1 mol NH NO 3 moles of gasamount of gas = 3.05 g NH NO × × = 0.114 mol gas

80.04 g NH NO 1 mol NH NO

T = 250 + 273 = 523 K

nRT 0.114 mol × 0.08206 L atm mol K × 523 KP = = = 2.25 atm

V 2.18 L

Conversion pathway:

-1 -14 34 3

4 3 4 3

nRTP =

V1 mol NH NO 3 moles of gas

3.05 g NH NO × × × 0.08206 L atm mol K × 523 K80.04 g NH NO 1 mol NH NO

P = 2.18 L

= 2.25 atm

Chapter 6: Gases

243

98. (M) First, let’s convert the given units to those easier used: P = 101 kPa × (1 barr/101 kPa) × (1 atm/1.01 barr) = 0.9901 atm T = 819 °C + 273 K = 1092 K

4 24 2 4 2

4 24 2

1

1 mol NH NOmol NH NO 128 g NH NO = 1.998 mol

64.052 g

3 mol gasmol gas = 1.998 mol NH NO = 5.994

1 mol NH NO

5.994 mol 0.082058 L atm K 1092 KnRTV = = = 542 L

P 0.9901atm

99. (M)

(a) nH2 = 1.00 g/2.02 g/mol = 0.495 mol H2; nO2= 8.60 g/32.0 g/mol = 0.269 mol O2

2 2

1 1

O

0.769 mol 0.08206 L atm mol K 298 K P P 12.5 atm

1.50 0Ltotal

total H

n RTP

V

2 2

2 2 2

The limiting reagent in the production of water is H (0.495 mol) with O (0.269 mol) in excess.

2 H (g) O (g) 2 H O(l)

(initial) 0.495 mol

2 2

2

O O

1 1

O

0.269 mol 0

after reaction ~0 0.022 mol 0.495 mol

so P P

0.022 mol 0.08206 L atm mol K 29 P

total H

total

P

n RT

V

8 K 0.359 atm

1.50 L23.8 mm Hg

0.359 atm 0.39 atm760 mm Hg/atmtotalP

100. (M)

(a) 2 22 2

2

3.8 mL CO 1 mL O4.0 L air 1000 mL mL O /min 76 mL O /min

1 min 1 L 100 mL air 2 mL CO

(b) We first determine the amount of O2 produced per minute, then the rate of consumption of Na2O2.

21 1

2 2 2 2 22 2

2 2 2

1 atm735 mmHg 0.967 atm T 25 273 298 K

760 mmHg

0.967 atm 0.076 L0.0030 mol O

0.08206 L atm mol K 298 K

0.0030 mol O 2 mol Na O 77.98 g Na O 60 minrate 28 g Na O /h

1 min 1 mol O 1 mol Na O 1 h

P

PVn

RT

Chapter 6: Gases

244

101. (E) Determine relative numbers of moles, and the mole fractions of the 3 gases in the 100.0 g gaseous mixture.

22 2 2

2

22 2 2

2

1 mol Namount N 46.5 g N 1.66 mol N 2.86 mol 0.580

28.01 g N

1 mol Neamount Ne 12.7 g Ne 0.629 mol Ne 2.86 mol 0.220

20.18 g Ne

1 mol Clamount Cl 40.8 g Cl 0.575 mol Cl 2.86 mol 0.201

70.91 g Cl

tota

2 2l amount 1.66 mol N 0.629 mol Ne 0.575 mol Cl 2.86 mol

Since the total pressure of the mixture is 1 atm, the partial pressure of each gas is numerically very close to its mole fraction. Thus, the partial pressure of Cl2 is 0.201 atm or 153 mmHg

102. (M) Let us first determine the molar mass of this mixture.

1 1 1

0.518 g L 0.08206 L atm mol K 298 K13.4 g/mol

1 atm721 mmHg

760 mmHg

dRTM

P

Then we let x be the mole fraction of He in the mixture.

13.4 g/mol 4.003 g/mol (1.000 ) 32.00 g/mol 4.003 32.00 32.00

32.00 13.40.664

32.00 4.003

x x x x

x

Thus, one mole of the mixture contains 0.664 mol He. We determine the mass of that He and then the % He by mass in the mixture.

4.003 g He 2.66 g Hemass He 0.664 mol He 2.66 g He %He 100% 19.9% He

1 mol He 13.4 g mixture

103. (M) The volume percents in a mixture of gases also equal mole percents, which can be converted to mole fractions by dividing by 100.

2 22 2

2 2

22

2

28.013 g N 31.999 g O0.7808 mol N 0.2095 mol O

air 1 mol N 1 mol O

44.010 g CO 39.948 g Ar0.00036 mol CO 0.0093 mol Ar =28.96 g/mol

1 mol CO 1 mol Ar

M

Chapter 6: Gases

245

104. (M) (a)

2 2TOT O N OP = P +P = 154 torr + 612 torr = 766 torr

We know that 2 2O O TOTP = χ P . Therefore,

2Oχ = 154/766 = 0.20

To determine the wt% of O2 and N2O, we have to use the mole fractions and determine the mass of each species:

22 2

2

22 2

2

2

2

32.0 g Omass O : 0.20 mol O = 6.432 g

1 mol O

44.0 g N Omass N O: 0.80 mol N O = 35.20 g

1 mol N O

6.432wt% O = 100 = 15.4%

6.432 35.20wt% N O =100 15.4 84.6%

(b) Apparent molar mass of the mixture: (32.00 g/mol)(154 torr/766 torr) + (44.02 g/mol)(612 torr/766 torr) = 41.6 g/mol

105. (M) The amount of N2(g) plus the amount of He(g) equals the total amount of gas. We use this

equality, and substitute with the ideal gas law, letting V symbolize the volume of cylinder B.

2 2 2N He total N N He He total totaln + n = n becomes P V + P V = P V

since all temperatures are equal. Substitution gives (8.35 atm × 48.2 L) + (9.50 atm × V) = 8.71 atm (48.2 L + V)

L2371.850.9

402420 8.71 atm L 420 9.50 atm L 402

VVV

106. (M) To construct a closed-end monometer, fill the tube with mercury, making sure to

remove any air bubbles. Inverting the tube results in a column of mercury with a vacuum in the space directly above the column.

When the container is evacuated, the heights of both mercury columns are equal. As

the gas pressure in the container increases, the heights of the mercury columns will change. Because the system is closed, Pbar is not needed. One drawback of this system is that gas pressure much larger than one atmosphere cannot easily be handled

vacuum

difference in height of columns of mercury is the pressure of the gas

vacuum

pressure equals difference in heights of the two columns of mercury

gas

Chapter 6: Gases

246

unless a very long column is used. (> 1 meter). Otherwise, the column of mercury would be pushed up to the closed end of the monometer at a critical pressure, and remain unchanged with increasing pressure.

An open-ended monometer, would be better suited to measuring higher pressure than

lower pressure. A gas pressure of one atmosphere results in the columns of mercury being equal. A sidearm of 760 mm pressure would allow one to measure pressures up to 2 atmospheres.

107. (D)

(a) Mav = g g g g g

mol mol mol mol mol

8.0 23.2 17.7 1.1 50.544.01 + 28.01 + 2.016 + 16.043 + 28.01

100 100 100 100 100

Mav = 24.56 g/mol density = P M

R T=

1 atm g763 mmHg× 24.56

760 mmHg mol g=1.015

LL atm0.08206 296 K

K mol

(b) Pco = Ptotal × V %

100 % = 763 mmHg ×

23.2 %

100 % = 177 mmHg or 0.233 atm

(c) CO(g) + ½ O2(g) CO2(g) H2(g)+ ½ O2(g) H2O(g) CH4(g)+ 2 O2(g) CO2(g) + 2 H2O(g) Use the fact that volume is directly proportional to moles when the pressure and

temperature are constant. 1000 L of producer gas contains: 232 L CO which requires 116 L O2 177 L H2 which requires 88.5 L O2 11 L CH4 which requires 22 L O2 226.5 L O2 (Note: air is 20.95 % O2 by volume)

Thus, the reaction requires 3226.5L=1.08×10 L

0.2095.

108. (M) First, balance the equation: C20H32O2 + 27 O2 → 20 CO2 + 16 H2O

20 32 2 22 20 32 2

20 32 2 20 32 2

2

1 mol C H O 27 mol Omol O needed: 2000 g C H O

304.52 g C H O 1 mol C H O

= 177.33 mol O

Using the ideal gas law, we can determine the volume of 177.33 mol of O2:

1

2 2

177.33 mol 0.082058 L atm K 298 KnRTvol O = = = 4336.30 L O

P 1.00 atm

Chapter 6: Gases

247

To determine the volume of air needed, we note that O2 represents 20.9% of air by volume: x(0.205) = 4336.30 L. Solving for x gives 20698 L, or 2.070×104 L.

109. (M) First recognize that 3 moles of gas are produced from every 2 moles of water, and compute the number of moles of gas produced. Then determine the partial pressure these gases would exert.

22

2 2

1 1

1 mol H O 3 mol gasamount of gas 1.32 g H O 0.110 mol gas

18.02 g H O 2 mol H O

0.110 mol 0.08206 L atm mol K 303 K 760 mmHg717 mmHg

2.90 L 1 atm

nRTP

V

Then the vapor pressure (partial pressure) of water is determined by difference.

Pwater = 748 mmHg - 717 mmHg = 31 mmHg 110. (D) First, balance the equations:

Fe + 2 HCl → 2 FeCl2 + H2 2 Al + 6 HCl → 2 AlCl3 + 3 H2 Then, we calculate the partial pressure of O2:

2 2O TOT H OP =P - P =841 - 16.5 = 824.5 torr

2 1

1.0849 atm 0.159 Lmol H = = 0.007192 mol

0.082058 L atm K 292 K

Now, we will try to express the calculated moles of H2 using the mole relationship between the metals and H2:

2 22

1 mol H 3 mol Hmol H = mol + mol = 0.007198 mol

1 mol Fe 2 mol Al

1 mol Fe 1 mol Al 1.0 mass + 1.5 0.1924 mass

55.85 g Fe 26.98 g Al

0.007192 0.017905 mass 0.01070 0.055597 mass

Fe Al

Fe Fe

Fe

Fe

Solving for massFe yields 0.09307 g of Fe.

%Fe = (0.09307/0.1924) ×100 = 48.4%

Chapter 6: Gases

248

111. (M) The total pressure of the mixture of O2 and H2O is 737 mmHg, and the partial pressure of H2O is 25.2 mmHg.

(a) The percent of water vapor by volume equals its percent pressure.

2 2

25.2 mmHg% H O 100% 3.42% H O by volume

737 mmHg

(b) The % water vapor by number of molecules equals its percent pressure, 3.42% by number.

(c) One mole of the combined gases contains 0.0342 mol H2O and 0.9658 mol O2.

2 22 2

2 2

2 2

22 2

18.02 g H O 31.999 g Omolar mass = 0.0342 mol H O× + 0.9658 mol O ×

1 mol H O 1 mol O

=0.616 g H O + 30.90 g O = 31.52 g

0.616 g H O% H O= ×100% = 1.95 % H O by mass

31.52 g

112. (D)

(a) 1 mol of the mixture at STP occupies a volume of 22.414 L. It contains 0.79 mol He and 0.21 mol O2 .

STP densitymass

volume

mol He g He

mol He mol O

g O mol O

L g / L

22

2= =

0 794 0031

0 2132 01

22.414= 0.44

..

..

25C is a temperature higher than STP. This condition increases the 1.00-L volume containing 0.44 g of the mixture at STP. We calculate the expanded volume with the combined gas law.

Vfinal L K

K L final density

g

L g / L 1.00

25+ 273.2

273.2= 1.09 =

0.44

1.09= 0.40

b g

We determine the apparent molar masses of each mixture by multiplying the mole fraction (numerically equal to the volume fraction) of each gas by its molar mass, and then summing these products for all gases in the mixture.

2 2air

2 2

28.01 g N 32.00 g O 39.95 g Ar0.78084 + 0.20946 + 0.00934

1 mol N 1 mol O 1 mol ArM

= 21.87 + 6.703 + 0.373 = 28.952 2 g N g O g Ar g / mol air

2mix 2

2

4.003 g He 32.00 g O 9.9 g mixture0.79 + 0.21 = 3.2 g He + 6.7 g O =

1 mol He 1 mol O molM

Chapter 6: Gases

249

(b) In order to prepare two gases with the same density, the volume of the gas of smaller molar mass must be smaller by a factor equal to the ratio of the molar masses. According to Boyle’s law, this means that the pressure on the less dense gas must be larger by a factor equal to a ratio of molar masses.

Pmix atm atm 28.95

9.91.00 = 2.9

113. (M) First, determine the moles of Cl2 and NaClO. Then, determine the limiting reagent

2 1

4.66 atm 1.0 Lmol Cl = = 0.2007 mol

0.082058 L atm K 283 K

mol NaClO = (0.750 L)(2.00 M) = 1.50 mol

22

4 mol NaClO0.2007 mol Cl 0.8028 mol NaClO needed

1 mol Cl

Therefore, NaClO is the excess reagent. Now, we must calculate the theoretical yield of ClO2:

2 22 2

2 2

2 mol ClO 67.45 g ClO0.2007 mol Cl 27.07 g ClO

1 mol Cl 1 mol ClO

25.9%yield = 100 95.7%

27.07

114. (M) First, determine the moles of Na2S2O3, then use the chemical equations given to determine

the moles of O3 in the mixture:

2 2 3

3 33 2 2 3 3

2 2 3 3

mol Na S O = 0.0262 L 0.1359 M = 0.003561 mol

1 mol I 1 mol Omol O : 0.003561 mol Na S O 0.001780 mol O

2 mol Na S O 1 mol I

Using the ideal gas law, we can determine the moles of gas:

3

1

4

0.993 atm 53.2 Lmoles of gas = = 2.2123 mol

0.082058 L atm K 291 K

0.001780 mol = 8.046 10

2.2123 mol totalO

115. (D) First compute the pressure of the water vapor at 30.1°C.

2

2

2H O

1 mol H O 0.08206 L atm0.1052 g (30.1 273.2) K

18.015 g H O mol K 760 mmHgP 13.72 mmHg

8.050 L 1 atm

Chapter 6: Gases

250

The water vapor is kept in the 8.050-L container, which means that its pressure is proportional to the absolute temperature in the container. Thus, for each of the six temperatures, we need to calculate two numbers: (1) the pressure due to this water (because gas pressure varies with temperature), and (2) 80% of the vapor pressure. The temperature we are seeking is where the two numbers agree.

K 273.2) 1.30(

K 273.2)(TmmHg 72.13)(

water

TP

For example,mmHg 3.13

K 273.2) 1.30(

K 273.2)(20. mmHg 13.72C)20(

P

T 20. °C 19. °C 18. °C 17. °C 16. °C 15. °C Pwater, mmHg 13.3 13.2 13.2 13.1 13.1 13.0 80.0% v.p.,

mmHg 14.0 13.2 12.4 11.6 10.9 10.2

At approximately 19°C, the relative humidity of the air will be 80.0%.

116. (D)

Pgas

(mmHg) 1/A 1/V

(cm-1) 810.8 0.0358 739.8 0.0328 896.8 0.0394 996.8 0.0437 1123 0.0493 1278 0.0562 1494 0.0658 1796 0.0787 2216 0.0980 2986 0.1316

Factors that would affect the slope of this straight line are related to deviations real gases exhibit from ideality. At higher pressures, real gases tend to interact more, exerting forces of attraction and repulsion that Boyle’s Law does not take into account.

117. (D) (a) T = 10 C = 283 K M = 28.96 g mol-1 (from question 99) or 0.02896 kg mol-1

R = 2

kg m8.314472

s K mol g = 9.80665

2

m

s h = 12in 2.54cm 1m

14494ft× × × 4417.8m1ft 1in 100cm

2

2

o

kg m0.02896 9.80665 4417.8m

mol s

kg m2.303 8.314472 283K

s K mol 0.2314Mgh

2.303RT 760 10 = 446 mmHgP = P × 10 760 × 10mmHg

Plot of Pressure versus A-1

Equation of the line

P = 22670(A-1) + 2.5377

0

500

1000

1500

2000

2500

3000

3500

0 0.05 0.1 0.15A-1(cm-1)

Pre

ssu

re (

mm

Hg

)

Chapter 6: Gases

251

(b) h = 12in 2.54cm 1m

900ft× × × 274.3m1ft 1in 100cm

2

2

o o o

kg m0.02896 9.80665 274.3m

mol s

kg m2.303 8.314472 283K

s K mol 29 1P = P 10 = 0.967455P = P smaller

30 30or

118. (M)

(a) rmsu is determined as follows:

1 1

rms 3

3 8.3145 J mol K 298 K3RTu 482 m/s

M 32.00 10 kg

(b)

3/22 2

u

3/2 3 232 3

u

MF 4 u exp Mu / 2RT

2 RT

32.00 10 (498)32.00 10F 4 (498) exp 1.92 10

2 (8.3145)(298) 2(8.3145)(298)

119. (D) Potential energy of an object is highest when the kinetic energy of the object is zero and the

object has attained its maximum height. Therefore, we must determine the kinetic energy. But first, we have to determine the velocity of the N2 molecule.

1 1

rms 3

2 23k rms

3 2k p

3 8.3145 J mol K 300 K3RTu 517 m/s

M 28.00 10 kg

1 1e mu 28.00 10 kg 517 m/s 3742 J

2 2

e e m g h 28.00 10 kg 9.8 m/s h = 3742 J

Solving for h, the altitude reached by an N2 molecule is 13637 m or 13.6 km. 120. (D)

Chapter 6: Gases

252

121. (D) (a) First multiply out the left-hand side of the equation.

2

32

2

2

)(V

abn

V

anPnbPVnRTnbV

V

anP

Now multiply the entire equation through by V2, and collect all terms on the right-hand side.

32232 0 abnV an PnbV PV nRTV

Finally, divide the entire equation by P and collect terms with the same power of V, to obtain:

00

3223

P

abnV

P

anV

P

bPRTnV

(b)

22 2

2

1 mol CO185 g CO 4.20 mol CO

44.0 g COn

3 2

2 2 2 3 2 2

3 2

1 1(0.08206 L atm mol K 286 K) (0.0429 L/mol 12.5 atm)0 4.20 mol

12.5 atm

(4.20 mol) 3.61 L atm mol (4.20 mol) 3.61 L atm mol 0.0429 L/mol

12.5 atm 12.5 atm

8.07 5.09 0.918

V V

V

V V V

We can solve this equation by the method of successive approximations. As a first value, we

use the volume obtained from the ideal gas equation:

L7.89atm12.5

K286KmolatmL0.08206mol4.20 11

P

nRTV

Chapter 6: Gases

253

V = 7.89 L (7.89)3 - 8.06 (7.89)2 + (5.07 × 7.89) - 0.909 = 28.5 >0 Try V = 7.00 L (7.00)3 - 8.06 (7.00)2 + (5.07 × 7.00) - 0.909 = -17.4 <0 Try V = 7.40 L (7.40)3 - 8.06 (7.40)2 + (5.07 × 7.40) - 0.909 = 0.47 >0 Try V = 7.38 L (7.38)3 - 8.06 (7.38)2 + (5.07 × 7.38) - 0.909 = -0.53 <0 Try V = 7.39 L (7.39)3 - 8.06 (7.39)2 + (5.07 × 7.39) - 0.909 = -0.03 < 0

The volume of CO2 is very close to 7.39 L. A second way is to simply disregard the last term and to solve the equation 0 = V3 - 8.06 V2 + 5.07 V This equation simplifies to the following quadratic equation: 0 = V2 - 8.06 V + 5.07, which is solved with the quadratic formula.

L37.7

2

74.14

2

68.606.8

2

07.54)06.8(06.8

2

V

The other root, 0.69 L does not appear to be reasonable, due to its small size.

122. (D)

(a) The van der Waals gas equation is used to determine the actual pressure.

2

2

22

22

n aP V nb nRT

V

1 1.3821 0.083145 280nRT n aP 96.5 bar = 9.65 MPa

V nb V 0.2168 1 0.0319 0.2168

9.65-10%error = 100 3.5%

10

(b) The volume using the ideal gas law is given as follows: V = nRT/P = (1 mol)(8.3145 kPa·L·K-1·mol-1)(280 K)/(10×103 kPa) = 0.233 L %error = (0.233 – 0.2168)/0.2168 × 100 = 7.47%

123. (M) (a)

3 -1 3 -1

2 6 -2

3 -1 6 -2

23 -1 3 -1

V = 500 cm mol B = -21.89 cm molB CPV = RT 1+ +

T= 273 K C = 1230 cm molV V

L L atm -21.89 cm mol 1230 cm molP 0.500 = 0.08206 273 K 1+ +

mol K mol 500 cm mol 500 cm mol

LP 0.500

m

L atm

22.4 0.961L atm mol= 22.40 0.961 P= = 43.1 atm

Lol mol0.500

mol

Chapter 6: Gases

254

(b) The result is consistent with Figure 6-20. In Figure 6-20, the compressibility factor is slightly below 1 at pressures between 0 and 400 atm. At a pressure of ~50 atm, the compressibility factor is just slightly below 1. Using the data provided, a pressure of 43.1 atm has a compressibility factor of 0.961, which is slightly below 1. The data is consistent with the information provided in Fig 6-20.

124. (M)

2

32 1 1

1atm752 mm Hg 21.07 mm Hg 0.9618 atm

760 mm Hg

0.9618 atm 0.202 Lmol H 7.99 10 mol

0.08206 L atm mol K 296 K

HP

PV

RT

2 2 2

2

H H H

-3 2H 2

n = n (from Al) + n (from Mg) Let = mass of Al in g, therefore 0.156 = mass Mg in grams

3 mol H 1 mol H1 mol Al 1 mol Mgn = 7.99 10 mol H = (0.156 )

26.98 g Al 2 mol Al 24.305 g Mg

x x

x x

2

-32

1 mol Mg

7.99 10 mol H = 0.0556 + 0.006418 0.04114

0.001572 = 0.01446 = 0.1087 g Al

0.1087 g Al mass % Al = 100% = 69.7 % Al

0.156 g mixture

mass % Mg = 100% 69.7 % = 30.3 % Mg

x x

x x

FEATURE PROBLEMS 125. (M) Boyle’s Law relates P and V, i.e., P × V = constant. If V is proportional to the

value of A, and Pgas = Pbar + PHg (i.e. the pressure of the gas equals the sum of the barometric pressure and the pressure exerted by the mercury column), then a comparison of individual A × P products should show a consistent result or a constant.

A (cm)

Pbar (atm)

PHg (mmHg)

Pgas (mmHg)

A × Pgas

(cm mmHg) 27.9 739.8 71 810.8 22621 30.5 739.8 0 739.8 22564 25.4 739.8 157 896.8 22779 22.9 739.8 257 996.8 22827 20.3 739.8 383 1123 22793 17.8 739.8 538 1278 22745 15.2 739.8 754 1494 22706 12.7 739.8 1056 1796 22807 10.2 739.8 1476 2216 22601 7.6 739.8 2246 2986 22692

Chapter 6: Gases

255

Since consistent A × Pgas results are observed, these data conform reasonably well (within experimental uncertainty) to Boyle’s Law.

126. (E) Nitryl Fluoride 65.01 u (49.4

100) = 32.1 u of X

Nitrosyl Fluoride 49.01 u (32.7

100) = 16.0 u of X

Thionyl Fluoride 86.07 u (18.6

100) = 16.0 u of X

Sulfuryl Fluoride 102.07 u (31.4

100) = 32.0 u of X

The atomic mass of X is 16 u which corresponds to the element oxygen. The number of atoms of X (oxygen) in each compound is given below: Nitryl Fluoride = 2 atoms of O Nitrosyl Fluoride = 1 atom of O Thionyl Fluoride = 1 atom of O Sulfuryl Fluoride = 2 atoms of O

127. (M) (a) The N g2 b g extracted from liquid air has some Ar(g) mixed in. Only O g2b g was

removed from liquid air in the oxygen-related experiments. (b) Because of the presence of Ar(g) [39.95 g/mol], the N g2 b g [28.01 g/mol]

from liquid air will have a greater density than N g2 b g from nitrogen

compounds. (c) Magnesium will react with molecular nitrogen

2 3 23 Mg s + N g Mg N s but not with Ar. Thus, magnesium

reacts with all the nitrogen in the mixture, but leaves the relatively inert Ar(g) unreacted.

(d) The “nitrogen” remaining after oxygen is extracted from each mole of air (Rayleigh’s mixture) contains 0.78084 + 0.00934 = 0.79018 mol and has the mass calculated below.

2

2

mass of gaseous mixture = 0.78084 28.013 g/mol N + 0.00934 39.948 g/mol Ar

mass of gaseous mixture = 21.874 g N + 0.373 g Ar = 22.247 g mixture.

Then, the molar mass of the mixture can be computed: 22.247 g mixture / 0.79018 = 28.154 mol g/mol. Since the STP molar volume of an ideal gas is 22.414 L, we can compute the two densities.

d N g / mol

L / mol g / mol2 =

28.013

22.414= 1.2498b g 28.154 g/mol

mixture = = 1.2561 g/mol22.414 L/mol

d

These densities differ by 0.50%.

Chapter 6: Gases

256

128. (M) (a) First convert pressures from mmHg to atm:

Density (g/L) Pressure (atm) Density/Pressure (g/L.atm) 1.428962 1.0000 1.428962 1.4290 1.071485 0.75000 1.428647 1.4286 0.714154 0.50000 1.428308 1.4283 0.356985 0.25000 1.42794 1.4279 average = 1.4285 g/L.atm

(b) 2OM =

dRT

P

2OM = 1.4285 g/L atm 0.082058 L atm/mol K 273.15 K

2OM = 32.0175 g/mol Thus, the atomic mass of O2 =

2OM / 2 = 16.0086

This compares favorably with the value of 15.9994 given in the front of the textbook.

129. (D) Total mass = mass of payload + mass of balloon + mass of H2

Use ideal gas law to calculate mass of H2: PV = nRT = mass(m)

Molar mass(M)RT

m = PVM

RT=

3 3 -33

3 3 3

12in 2.54cm 1×10 L g1atm 120ft × × × 2.016

1cm mol1ft 1in

Latm0.08206 273K

K mol

= 306 g

Total mass = 1200 g + 1700 g + 306 g 3200 g

We know at the maximum height, the balloon will be 25 ft in diameter. Need to find out what mass of air is displaced. We need to make one assumption – the volume percent of air is unchanged with altitude. Hence we use an apparent molar mass for air of 29 g mol-1 (question 99). Using the data provided, we find the altitude at which the balloon displaces 3200 g of air.

Note: balloon radius = 25

2 = 12.5 ft. volume =

4

3()r3 =

4

3(3.1416)(12.5)3 = 8181 ft3

Convert to liters:

3 3 -33

3 3 3

12in 2.54cm 1×10 L8181ft × × × = 231,660L

1cm1ft 1in

At 10 km: m = PVM

RT=

2 1atm g2.7×10 mb× (231,660 L) 29

1013.25 mb mol

L atm0.08206 223K

K mol

= 97,827 g

At 20 km: m = PVM

RT=

1 1atm g5.5×10 mb× (231,660 L) 29

1013.25 mb mol

L atm0.08206 217K

K mol

= 20,478 g

Chapter 6: Gases

257

At 30 km: m = PVM

RT=

1 1atm g1.2×10 mb× (231,660 L) 29

1013.25 mb mol

L atm0.08206 230K

K mol

= 4,215g

At 40 km: m = PVM

RT=

0 1atm g2.9×10 mb× (231,660 L) 29

1013.25 mb mol

L atm0.08206 250K

K mol

= 937 g

The lifting power of the balloon will allow it to rise to an altitude of just over 30 km.

SELF-ASSESSMENT EXERCISES 130. (E)

(a) atm: Pressure exerted by the atmosphere at sea level per unit area, or 760 mm Hg (b) STP: Standard temperature and pressure, defined as a pressure of 1 barr at 273 K (c) R: Gas constant, which is the ratio between the product of the molar volume of a gas and

pressure, and temperature. (d) Partial pressure: The ratio between the pressure of a gas in a container and the total pressure

of the container (e) urms: Root mean squared velocity of gas molecules in a sample

131. (E)

(a) Absolute zero: the lowest theoretical temperature, and the temperature at which all molecular/atomic vibrations cease.

(b) Collection of gas over water: The process of isolating a gas (which does not react with water) generated by a reaction by bubbling it through a bottle and measuring its volume by the displacement of water.

(c) Effusion of a gas: the escape of gas molecules from their container through a tiny orifice or pinhole.

(d) Law of combining volumes: the volume ratio of gases consumed and generated is the same as their mole ratio, provided that temperature and pressure are kept constant throughout.

132. (E)

(a) A barometer measures atmospheric pressure, whereas a manometer measures the pressure of a generated gas in a closed vessel.

(b) Celsius temperature uses the freezing and boiling points of water to generate a temperature scale, whereas the Kelvin scale uses the temperature below which no molecular vibrations can occur as the zero point.

(c) The ideal gas equation states the relationship between pressure, volume, moles, and temperature in an ideal gas. The general gas equation uses the ideal gas equation to set up a linear relationship between P, V, n, and T (P1V1/n1T1 = P2V2/n2T2).

Chapter 6: Gases

258

(d) An ideal gas is one in which the gas molecules themselves don’t occupy a volume, and that their interaction with each other is negligible. A real gas takes into account molecular volume of the gas and intermolecular/interatomic interactions.

133. (E) The answer is (d). The following shows the values for each:

(a) P = g·h·d = (9.8 m/s2)(0.75 m)(13600 kg/m3) = 99960 Pa (b) Just for a rough approximation, we assume the density of air to be the same as that of

nitrogen (this underestimates it a bit, but is close enough). The density of N2 is 0.02802 g/22.7 L = 1.234 g/L = 1.234 kg/m3. P = (9.8 m/s2)(16093 m)(1.234 kg/m3) = 194616 Pa.

(c) P = g·h·d = (9.8 m/s2)(5.0 m)(1590 kg/m3) = 77910 Pa (d) P = nRT/V = (10.00 g H2 × 1 mol/2.02 g)(0.083145 L·barr·K-1)(273 K)/(22.7 L) = 4.95 barr

= 495032 Pa 134. (E) The answer is (c), because the temperature decreases from 100 °C (373 K) to 200 K. 135. (M) P1/T1 = P2/T2. To calculate T2, rearrange the formula:

T2 = 2.0 barr / (1.0 barr × 273 K) = 546 K 136. (M) The answer is (d).

1 1 2 2

1 2

2

P V P V=

T T

1 atm 22.4 L 298 KV = 16.3 L

273 K 1.5 atm

137. (M) The answer is (b). Since the same number of moles of all ideal gases occupy the same

volume, density is driven by the molar mass of the gas. Therefore, Kr has the highest density because it has a molar mass of 83 g/mol.

138. (E) The answer is (a). Using the formula urms = (3RT/M)1/2, increases T by a factor of 2 increase

urms by a factor of 2 . 139. (M)

(a) False. They both have the same kinetic energy (b) True. All else being equal, the heavier molecule is slower. (c) False. The formula PV=nRT can be used to confirm this. The answer is 24.4 L (d) True. There is ~1.0 mole of each gas present. All else being equal, the same number of

moles of any ideal gas occupies the same volume. (e) False. Total pressure is the sum of the partial pressures. So long as there is nothing else but

H2 and O2, the total pressure is equal to the sum of the individual partial pressures.

Chapter 6: Gases

259

140. (E) The answer is (c). Partial pressures are additive, so:

2 2

2

2

TOT H O O

TOT H OO

P =P +P

P P 751 21P (atm)= 0.96 atm

760 mm Hg 760

141. (M) The answer is (a). First, determine the # moles of NH3 using the ideal gas law relationship:

n = PV/RT = (0.500 atm × 4.48 L) / (0.08206 L·atm K-1 × 273 K) = 0.100 mol If 1.0 mole of a substance has 6.022×1023 atoms, 0.100 moles has 6.022×1022 atoms.

142. (M) The answer is (b). Since PV = nRT, the number of moles of O2 needed to satisfy the

conditions of the problem is:

n = (2.00 atm × 2.24 L) / (0.08206 L·atm K-1 × 273 K) = 0.200 moles

The amount of O2 available = 22 2

2

1 mol O1.60 g O 0.050 mol O

32.0 g O

To have an additional 0.150 mol in the system, we would add 0.6 g of He to the container.

143. (E) The volumes of both gases were measured at the same temperature and pressure. Therefore, the proportionality constant between volume and moles for both gases is the same (that is, volume can essentially replace moles in the following calculations):

22

3 L CO25.0 L H = 10.7 L CO needed

7 L H

So, all of the H2 and 10.7 L of CO are consumed, and 1.3 L of CO remain.

144. (M) The answer is (a), that is, the partial pressure of H2 in the container is less than SO2. The

reasoning can be derived from the kinetic molecular theory of gases. Gas molecules with smaller molar masses travel faster, and as such can escape faster from an orifice. This is expressed in Graham’s Law in section 6.8.

145. (M) The answer is (c). Gases behave more ideally at high temperatures and low pressures. 146. (M)

(a) He or Ne: Ne has higher a and b values. (b) CH4 or C3H8: C3H8 has higher a and b values (c) H2 or Cl2: Cl2 has higher a and b values

Chapter 6: Gases

260

147. (D) We know that pressure is force per unit area, that is:

F m gP =

A A

Using the fact that area 2 2A=πr =π(D/2) and that mass m=d V , and that volume of a cylindrical tube is V=A h (where h is the height of the liquid in the tube), we can express the pressure formula as follows:

d A h gF d V gP = = =d h g

A A A

Therefore,

2

d V gd h g, and,

Ad d

h=A π D/2

As we can see, the height, h is inversely proportional to D. That is, the larger the diameter of the tube, the shorter the height of the liquid.

148. (D) First, convert the given information to more useful units. The pressure (752 torr) is

equivalent to 0.9895 atm (752 mm/760 mm Hg), and the temperature is 298 K. Then, use the ideal gas relationship to determine how many moles of gas are present, assuming 1 L of gas:

1

PV = nRT

0.9895 atm 1 LPVn = 0.04046 mol

RT 0.08206 L atm K 298 K

From the problem, we know the mass of the gas per 1 L, as expressed in the density. Mass of 1 L of this gas is 2.35 g. The molar mass of this substance is therefore: MM = g/mol = 2.35 g/0.04046 mol. = 58.082 g/mol Now, let’s calculate the empirical formula for the hydrocarbon gas and see how it compares to the molar mass:

1 mol Cmol C = 82.7 g C = 6.89 mol

12.01 g C

1 mol Hmol H = 17.3 g H = 17.1 mol

1.01 g H

H:C ratio = 17.1 mol / 6.89 mol = 2.5

The ratio between H and C is 2.5:1 or 5:2, making the empirical formula C2H5. The mass of this formula unit is 29.07 g/mol. Comparing to the calculated molar mass of the gas, it is smaller by a factor of 2. Therefore, the gas in question in C4H10 or butane.

Chapter 6: Gases

261

149. (E) N2 comprises 78.084% of atmosphere, oxygen 20.946%, argon 0.934%, and CO2 0.0379%. To graphically show the scale of this difference, divide all values by the smallest one (CO2). Therefore, for every single mark representing CO2, we need 2060 marks for N2, 553 marks for O2, and 25 for Ar.

150. (M) To construct a concept map, one must first start with the most general concepts. These

concepts are not defined by or in terms of other concepts discussed in those sections. In this chapter, pressure is the overarching concept (6-1). The topics that fall under the rubric of pressure are liquid and gas pressure, and measuring pressure. Simple gas laws are derived from the concept of pressure (6-2). Simple gas laws include Boyle’s Law, Charles’ Law and Avogadro’s Law. These laws combine to form another subtopic, the Ideal Gas Law and the General Gas Equation. Take a look at the subsection headings and problems for more refining of the general and specific concepts.