CHAPTER 11 Gases - Mr. Cowmeadow's Chemistry Classes ...

MODERN CHEMISTRY CHAPTER 11 SOLUTIONS MANUALCopyright © by Holt, Rinehart and Winston. All rights reserved.

121

C H A P T E R 1 1

GasesPractice, p. 365

1. Given: P = 1.75 atm a. 1.75 atm × ⎯101.

a3t2m5 kPa⎯ = 177 kPa

b. 1.75 atm × ⎯760

amtmm Hg⎯ = 1330 mm Hg

A-1. Given: P = 745.8 mm Hg a. 745.8 mm × ⎯7160

atmmm

⎯ = 0.9813 atm

b. 0.9813 atm × ⎯7160

attmorr

⎯ = 745.8 torr

c. 0.9813 atm × ⎯101.

a3t2m5 kPa⎯ = 99.43 kPa

ATE, Additional Sample Problem, p. 365

2. Given: P = 72.7 atm

Unknown: P in Pa72.7 atm × × = 7.37 × 106 Pa

1000 Pa⎯1 kPa

101.325 kPa⎯⎯

1 atm

1. Given: Volume abun-dance in air of:N2 = 78.08%O2 = 20.95%Ar = 0.934%, and CO2 = 0.035%

Unknown: Partial pres-sures of eachgas at Ptotal= 760. mm

N2 = 760 mm Hg × 0.7808 = 593.4 mm Hg

O2 = 760 mm Hg × 0.2095 = 159.2 mm Hg

Ar = 760 mm Hg × 0.00934 = 7.10 mm Hg

CO2 = 760 mm Hg × 0.00035 = 0.27 mm Hg

Additional Example Problem, p. 366


B-1. Given: T = 27.0°C

PT =743.3 mm Hg

Unknown: PNe

PT = PNe + PH2O

PH2O at 27.0°C = 26.7 mm Hg (from Appendix A-8)

PNe = PT − PH2O = 743.3 − 26.7 = 716.6 mm Hg

1. Given: T = 20.0°C

PH2 = 742.5 torr

PH2O at 20°C = 17.5 torr

Unknown: PT

PT = PH2 + PH2O = 742.5 + 17.5 = 760.0 torr

Practice, p. 367

Practice, p. 372

1. Given: V1 = 752 mL

T1 = 25.0 °C = 298 K

T2 = 100 °C = 373 K

Unknown: V2

⎯VT1

1⎯ = ⎯VT2

2⎯

V2 = ⎯V

T1T

1

2⎯ = ⎯752 m

2L98

×K373 K

⎯ = 941 mL


122

C-1. Given: V1 = 450 mL

P1 = 1

P2 = 15

Unknown: V2

ATE, Additional Sample Problems, p. 370

C-2. Given: V1 = of helium =125 mL

P1 = 0.974 atm

P2 = 1.000 atm

Unknown: V2

P1V1 = P2V2

V2 = ⎯P

P1V

2

1⎯ = = 122 mL He(0.974 atm)(125 mL)⎯⎯⎯

1.000 atm

2. Given: V1 = 375 mL

T1 = 0.0°C = 273 K

V2 = 500 mL

Unknown: T2

⎯VT1

1⎯ = ⎯VT2

2⎯

T2 = ⎯T

V1V

1

2⎯ = ⎯(273

3K7)5(5

m00

LmL)

⎯ = 364 K

364 K − 273 = 91°C

Practice, p. 370

1. Given: V1 = 500 mL He

P1 = 1 atm

P2 = 0.5 atm

Unknown: V2

P1V1 = P2V2

V2 = ⎯P

P1V

2

1⎯ = = 1000 mL He(1 atm)(500 mL He)⎯⎯⎯

0.5 atm

5. Given: T = 23.0°C

PT = Patm= 785 mm Hg

Unknown: PN2

PT = PN2 + PH2O

PH2O at 23°C = 21.1 mm Hg (from Appendix A-8)

PN2 = PT − PH2O = 785 − 21 = 764 mm Hg

4. a. Given: P =151.98 kPa

b. Given: P = 456 torr

151.98 kPa × ⎯101

1.3

a2t5m

kPa⎯ = 1.4999 atm

456 torr × ⎯7160

attmorr

⎯ = 0.600 atm

Section Review, p. 367

P1V1 = P2V2

P2 = ⎯P

P1V

2

1⎯ = ⎯(1)(45

105

mL)⎯ = 30.0 mL


123

D-1. Given: V1 = 5.5 L

T1 = 25°C = 298 K

T2 = 100°C = 373 K

Unknown: V2

⎯VT1

1⎯ = ⎯TV

2

2⎯

V2 = ⎯V

T1T

1

2⎯ = ⎯(5.5

2L9)8(3

K73 K)⎯ = 6.9 L


Practice, p. 374

1. Given: T1 = 120°C = 393 K

P1 = 1.07 atm

T2 = 205°C = 478 K

Unknown: P2

⎯TP1

1⎯ = ⎯

TP2

2⎯

P2 = ⎯PT1T

1

2⎯ = = 1.30 atm(1.07 atm)(478 K)⎯⎯⎯

393 K

3. Given: P1 = 1.20 atm

T1 = 22°C = 295 K

P2 = 2.00 atm

Unknown: T2

⎯TP1

1⎯ = ⎯

TP

2

2⎯

T2 = ⎯T

P1P

1

2⎯ = = 492 K

492 K − 273 = 219°C

(295 K)(2.00 atm)⎯⎯⎯

1.20 atm

2. Given: T1 = 122°C = 395 K

T2 = 205°C = 478 K

P1 = 1.07 atm

Unknown: P2

P2 = ⎯PT1T

1

2⎯ = ⎯1.07 a

3t9m5

×K

478 K⎯ = 1.29 atm


E-1. Given: T1 = 20°C= 293 K

P1 = 1.0 atm

T2 = 500°C= 773 K

Unknown: P2

⎯TP1

1⎯ = ⎯

TP

2

2⎯

P2 = ⎯PT1T

1

2⎯ = ⎯(1.0 a

2tm93

)(K773 K)⎯ = 2.6 atm

1. Given: V1 = 27.5 mL

T1 = 22.0°C = 295 K

P1 = 0.974 atm

T2 = 15.0°C = 288 K

P2 = 0.993 atm

Unknown: V2

⎯P

T1V

1

1⎯ = ⎯PT2V

2

2⎯

V2 = ⎯P

P1V

2T1T

1

2⎯ = = 26.3 mL(0.974 atm)(27.5 mL)(288 K)⎯⎯⎯⎯

(0.993 atm)(295 K)

Practice, p. 375

2. Given: V1 = 700 mL

T1 = 0°C = 273 K

P1 = 1.00 atm

V2 = 200 mL

T2 = 30.0°C = 303 K

Unknown: P2 (in Pa)

⎯P

T1V

1

1⎯ = ⎯PT2V

2

2⎯

P2 = ⎯PT1V

1V1T

2

2⎯ = × ⎯101.

a3t2m5 kPa⎯ = 394 kPa

= 3.94 × 105 Pa

(1.00 atm)(700 mL)(303 K)⎯⎯⎯⎯

(273 K)(200 mL)


124

F-1. Given: T1 = 27.0°C = 300 K

P1 = 0.200 atm

V1 = 80.0 mL

T2 = 0°C = 273 K

P2 = 1.00 atm

Unknown: V2

⎯P

T1V

1

1⎯ = ⎯PT2V

2

2⎯

V2 = ⎯P

P1V

2T1T

1

2⎯ = = 14.6 mL(0.200 atm)(80.0 mL)(273 K)⎯⎯⎯⎯

(1.00 atm)(300 K)

ATE, Additional Sample Problems, p. 375

F-2. Given: V1 = 75 mL

T1 = 0°C = 273 K

P1 = 1.00 atm

T2 = 17°C = 290 K

P2 = 0.97 atm

Unknown: V2

⎯P

T1V

1

1⎯ = ⎯PT2V

2

2⎯

V2 = ⎯P

P1V

2T1T

1

2⎯ = = 82 mL(1.00 atm)(75 mL)(290 K)⎯⎯⎯

(0.97 atm)(273 K)


2. Given: V1 = 200.0 mL

P1 = 0.960 atm

V2 = 50.0 mL

Unknown: P2

P1V1 = P2V2

P2 = ⎯PV1V

2

1⎯ = = 3.84 atm(0.960 atm)(200.0 mL)⎯⎯⎯

50.0 mL

3. Given: V1 = 1.55 L

T1 = 27.0°C = 300. K

T2 = −100.0°C = 173. K

(Pressure is constant)

Unknown: V2

⎯VT1

1⎯ = ⎯VT2

2⎯

V2 = ⎯V

T1T

1

2⎯ = ⎯1.55

3L00

×K173 K⎯ = 0.894 atm


125

4. Given: V1 = 2.0 m3

T1 = 100.0 K

P1 = 100.0 kPa

T2 = 400.0 K

P2 = 200.0 kPa

Unknown: V2

⎯PT1V

1

1⎯ = ⎯PT2V

2

2⎯

V2 = ⎯P

T1V

1P1T

2

2⎯ =

V2 = 4.0 m3

100.0 kPa × 2.0 m3 × 400.0 K⎯⎯⎯⎯

100.0 K × 200.0 kPa

1. a. Given: 3O2(g) →2O3(g),

24 O2molecules

Unknown: number of O3molecules

24 molecules O2 × ⎯23

mm

ooll

OO

3

2⎯ = 16 molecules O3

Additional Example Problems, p. 380

b. Given: 12 L O2

Unknown: L O3

12 L O2 × ⎯23

mm

ooll

OO

3

2⎯ = 8 L O3

2. Given: 2 Cl2 (g) + 7 O2(g) → 2 Cl2O7

35 L O2 × ⎯27

mm

oollCO

l

2

2⎯ = 10. L Cl2

2. Given: V = 14.1 L H2at STP

Unknown: n: number ofmoles of H2at STP

n = ⎯22.

V4 L

H/2

mol⎯ = ⎯

221.44.

L1/Lmol

⎯ = 0.629 mol H2

1. Given: n = 7.08 mol N2at STP

Unknown: V of N2 at STP

V = 7.08 mol N2 × ⎯22

m.4ol

L⎯ = 159 L N2

Practice, p. 381

G-1. Given: n = 0.0580 molNO at STP

Unknown: V of NO atSTP

V of NO = 0.0580 mol NO × ⎯22

m.4ol

L⎯ = 1.30 L



126

2. Given: n = 0.0035 molCH4 at STP

Unknown: V of CH4 inmL at STP

V = (0.0035 mol)�⎯22m.4ol

L⎯��⎯1000

LmL⎯� = 78 mL CH4


1. Given: n = 0.325 mol H2

V = 4.08 L

T = 35°C = 308 K

Unknown: P in atm

P = nRT/V = = 2.01 atm(0.325 mol)�⎯0.08

m21

olL• K

• atm⎯�(308 K)

⎯⎯⎯⎯4.08 L

Practice, p. 385

2. Given: V = 8.77 L

n = 1.45 mol

T = 20°C = 293 K

Unknown: P in atm

P = nRT/V = = 3.98 atm

(1.45 mol)�⎯0.08m21

olL• K

• atm⎯�(293 K)

⎯⎯⎯⎯8.77 L

1. Given: V = 4.55 L O2

Unknown: V of H2 gas

2H2(g) + O2(g) → 2H2O(g)

V = (4.55 L O2)�⎯21 LL

HO2

2⎯� = 9.10 L H2

Practice, p. 382

2. Given: V of CO = 0.626 L

Unknown: V of O2 gas

2O2 + 4CO → 4CO2

V = (0.626 L CO)�⎯42LL

CO

O2⎯� = 0.313 L O2

3. Given: V = 708 L NO2

Unknown: V of NO gasproduced

3NO2(g) + H2O(l) → 2HNO3(l ) + NO(g)

V = (708 L NO2)�⎯31LL

NN

OO

2⎯� = 236 L NO

H-1. Given: V = 3.14 L XeF6

Unknown: V of XeV of F

Xe(g) + 3F2(g) → XeF6(g)

V of Xe = (3.14 L XeF6)�⎯11LLXXeFe

6⎯� = 3.14 L Xe

V of F = (3.14 L XeF6) ⎯(1

(3L

LX

FeF

2)

6)⎯ = 9.42 L F2



127

7. Given: 2N2O (g) →2N2 (g) + O2 (g)

2.22 L N2O

Unknown: Volume N2and O2,density ofmixed prod-ucts at STP

2.22 L N2O × ⎯22mmoollNN

2

2

O⎯ = 2.22 L N2

2.22 L N2O × ⎯21mmoollNO

2

2

O⎯ = 1.11 L O2

At STP, 1 mol = 22.4 L

For N2, density = ⎯218.

m02

olg

⎯ × ⎯212m.4

oLl

⎯ = 1.25 g/L

For O2, density = ⎯312.

m00

olg

⎯ × ⎯212m.4

oLl

⎯ = 1.43 g/L

density of mixed products = ⎯23

⎯ × 1.25 g/L + ⎯13

⎯ × 1.43 g/L

density = 1.31 g/L

6. Given: V = 22.9 L

n = 14.0 mol

T = 12°C = 285 K

Unknown: P in atm

PV = nRT

P = ⎯nR

VT

⎯ =

P = 14.3 atm

14.0 mol × 0.0821 ⎯Lm

•

ola

•

tmk

⎯ × 285 K

⎯⎯⎯⎯22.9 L

5. Given: V = 4.44 L

T = 22.55°C = 295.55 K

15.4 g O2

Unknown: P

PV = nRT

P = ⎯nR

VT

⎯ =

P = 2.63 atm

15.4 g O2 × 0.0821 ⎯Lm

•

ola

•

tmk

⎯ × 295.55 K⎯⎯⎯⎯⎯

�⎯312.m00

olgOO

2

2⎯� × 4.44 L

I-1. Given: V = 2.07 L He

n = 2.88 mol

T = 22°C =295 K

Unknown: P of He in atm

P = ⎯nR

VT

⎯

= = 33.7 atm He

(2.88 mol)�⎯0.08m21

olL• K

• atm⎯�(295 K)

⎯⎯⎯⎯2.07 L



128

1. Given: VO2 = 420 m/s at25°C (298 K)

Unknown: VHe at 298 K

⎯VVO

H

2

e⎯ = ��VHe = �� VO2 = �⎯

342.0.000a� amm

uu

⎯� × 420 m/s = 1200 m/sMO2⎯MHe

MO2⎯MHe

Additional Example Problems, p. 386

2. Given: VH2 = 1.81 ×103 m/s

VX = 312 m/s

Unknown: molar mass of X, MX

⎯V

V

H

X

22

2

⎯ = ⎯MM

H

X

2

⎯

MX = ⎯VH2

2

V

×

X2MH2⎯ =

MX = 70.3 g/mol

(1.84 × 103 m/s)2 (2.02 g/mol)⎯⎯⎯⎯

(312 m/s)2

2. Given: H2 rate of effusion = 9 timesthat of unknowngas

Unknown: M of unknowngas (X)

=

�M�X� = � �(�M�H�2�)

= � � (�2� g�/m�o�l�) = 12.7

MX ≈ 160 g/mol

9⎯1

rate of effusion of H2⎯⎯⎯rate of effusion of X

√MX�⎯√MH2�

rate of effusion of H2⎯⎯⎯rate of effusion of X

Practice, p. 388

1. Given: identities of 2 gases, CO2and HCl

Unknown: relative ratesof effusion

= = = 0.9�3�6�.5� g/mol⎯⎯�4�4� g/mol

√MHCl�⎯√MCO2�

rate of effusion of CO2⎯⎯⎯rate of effusion of HCl

3. Given: rate of effusion ofNe = 400 m/s

Unknown: rate of effusionof butane,C4H10, atsame temperature

=

rate of effusion of butane = (rate of effusion of neon)� �= (400 m/s)� � = 235 m/s

�2�0� g/mol⎯⎯�5�8� g/mol

�M�N�e�⎯⎯�M�C�4H�10

�

�M�C�4H�10�

⎯⎯�M�N�e�

rate of effusion of neon⎯⎯⎯

rate of effusion of butane

J-1. Given: N2 rate of effusion = 1.7times that of other gas

Unknown: (1) M of other gas (X)(2) identity ofother gas

=

�M�X� = � �(�M�N�2�)

= �⎯11.7⎯�(�2�8� g�/m�o�l�) = 8.995

(1) MX = 81 g/mol

(2) Krypton (average atomic mass of Kr = 83.8)

rate of effusion of N2⎯⎯⎯ rate of effusion of X

�M�X�⎯�M�N�2

�rate of effusion of N2⎯⎯⎯rate of effusion of X



129

3. Given: rate of effusion ofa gas = 1.6 timesthat of CO2

Unknown: M of unknowngas (X)

=

�M�X� = (�M�C�O�2�)� �

= (�4�4� g/mol)�⎯11.6⎯� = 4.14

MX = 4.142 = 17 g/mol

rate of effusion CO2⎯⎯⎯rate of effusion X

�M�C�O�2�

⎯⎯�M�X�

rate of effusion of X⎯⎯⎯rate of effusion of CO2


5. Given: T = 25°C = 298 K

Unknown: molecular vel-ocities of H2O,He, HCl, BrF,NO2

=

(Molecular velocities of 2 different gases are inversely proportional to thesquare root of their molar masses.)

MH2O = 18 g/mol

MHe = 4 g/mol

MHCl = 36.5 g/mol

MBrF = 98.8 g/mol

MNO2 = 46 g/mol

Rates of effusion: BrF < NO2 < HCl < H2O < He

�MHe�⎯�MH2O�

rate of effusion of H2O⎯⎯⎯rate of effusion of He

4. Given:

= ⎯116⎯

Unknown: ⎯MM

A

B⎯

rate of diffusion of A⎯⎯⎯rate of diffusion of B

= ⎯116⎯

⎯MM

B

A⎯ = ⎯

2516

⎯

�M�B�⎯�M�A�

6. Given: VX = ⎯12

⎯ VO2, where

X is either HBr orHI

Unknown: molar massand identity of X

⎯V

V

O

X

22

2

⎯ = = ⎯MM

O

X

2

⎯

MX = 4 × MO2 = 4 × 32.00 g/mol = 128.0 g/mol

X is HI

VO22

⎯

�⎯12

⎯ VO2�2


130

18. a. Given: P1 = 350. torrV1 = 200. mLP2 = 700. torr

Unknown: V2

b. Given: V1 = 2.4 × 105 LP2 = 180 mm HgV2 = 1.8 × 103 L

Unknown: P1

P1V1 = P2V2

V2 = ⎯P

P1V

2

1⎯ = = 100 mL

P1V1 = P2V2

P1 = ⎯PV2V

1

2⎯ = = 1.4 mm Hg(180 mm Hg)(1.8 × 103 L)⎯⎯⎯

2.4 × 105 L

(350 torr)(200 mL)⎯⎯⎯

700 torr

11. Given: T = 35.0°CPT = 742.0 torr

Unknown: Pgas

PH2O at 35°C = 42.2 mm

PT = PH2O + Pgas

Pgas = PT − PH2O = 742.0 − 42.2 = 699.8 torr

10. Given: PCO2= 0.285 torr

PN2= 593.525 torr

PT = 1 atm =760 torr

Unknown: PO2

PT = PCO2+ PN2

+ PO2

PO2= PT − (PCO2

+ PN2)

= 760 torr − (593.525 + 0.285) = 166.190 torr

9. a. Given: P = 125 mm

Unknown: P in atm

b. Given: P = 3.20 atm

Unknown: P in Pa

c. Given: P = 5.38 kPa

Unknown: P in mm Hg

P = (125 mm)�⎯7160

atmmm

⎯� = 0.164 atm

P = (3.20 atm)�⎯101.a3t2m5 kPa⎯� = 324.24 kPa = 3.24 × 105 Pa

P = (5.38 kPa)�⎯1011.3

a2t5m

kPa⎯��⎯760

amtmm Hg⎯� = 40.4 mm Hg

P = (1.25 atm)�⎯76a0tmtorr⎯� = 950 torr

P = (2.48 × 10−3 atm)�⎯76a0tmtorr⎯� = 1.88 torr

P = (4.75 × 104 atm)�⎯76a0tmtorr⎯� = 3.61 × 107 torr

P = (7.60 × 106 atm)�⎯76a0tmtorr⎯� = 5.78 × 109 torr

8. a. Given: P = 1.25 atm

Unknown: P in torr

b. Given: P = 2.48 ×10−3 atm

c. Given: P = 4.75 ×104 atm

d. Given: P = 7.60 ×106 atm

Chapter Review

7. Given: V2 = (1.40)(V1)

Unknown: P2

P2 = ⎯PV1V

2

1⎯ = ⎯(760

1m.4

m0

)(1)⎯ = 543 mm


131

19. a. Given: V1 = 80.0 mLT1 = 27°C

= 300 KT2 = 77°C

= 350 K

Unknown: V2

b. Given: V1 = 125 LV2 = 85.0 LT2 = 127°C

= 400 K

Unknown: T1

c. Given: T1 = −33°C = 240 K

V2 = 54.0 mLT2 = 160°C

= 433 K

Unknown: V1

⎯VT1

1⎯ = ⎯TV

2

2⎯

V2 = = = 93.3 mL

=

T1 = = = 588 K = 315°C

=

V1 = = = 29.9 mL(54.0 mL)(2.40 K)⎯⎯⎯

433 KV2T1⎯

T2

V2⎯T2

V1⎯T1

(125 L)(400 K)⎯⎯

85.0 LV1T2⎯

V2

V2⎯T2

V1⎯T1

(80.0 mL)(350 K)⎯⎯

300 KV1T2⎯

T1

20. Given: V1 = 140.0 mLT1 = 67°C = 340 KV2 = 50.0 mL

Unknown: T2

=

T2 = = = 121 K = −152°C(50.0 mL)(340 K)⎯⎯

140.0 mLV2T1⎯

V1

V2⎯T2

V1⎯T1

21. Given: V1 = 240. mLP1 = 0.428 atmP2 = 0.724 atm

Unknown: V2

P1V1 = P2V2

V2 = ⎯P

P1V

2

1⎯ = = 142 mL(0.428 atm)(240 mL)⎯⎯⎯

(0.724 atm)

24. Given: T1 = −73°C = 200 KP1 = 1P2 = 2

Unknown = T2

T2 = = = 400 K = 127°C(2)(200)⎯

1P2T1⎯

P1

22. Given: T1 = 47°C = 320 KP1 = 0.329 atmT2 = 77°C = 350 K

Unknown: P2

⎯TP1

1⎯ = ⎯

TP

2

2⎯

P2 = = = 0.360 atm(0.329 atm)(350 K)⎯⎯⎯

320 KP1T2⎯T1

23. Given: T1 = 47°C = 320 KP1 = 1.03 atmV1 = 2.20 LT2 = 107°C = 380 KP2 = 0.789 atm

Unknown: V2

=

V2 = = = 3.41 L(1.03 atm)(2.20 L)(380 K)⎯⎯⎯

(0.789 atm)(320 K)P1V1T2⎯

P2T1

P2V2⎯T2

P1V1⎯T1

25. Given: V1 = 155 cm3

P1 = 22.5 kPaV2 = 90.0 cm3

Unknown: P2

P1V1 = P2V2

P2 = ⎯PV1V

2

1⎯ = = 38.8 kPa(22.5 kPa)(155 cm3)⎯⎯⎯

(90.0 cm3)


132

26. Given: V1 = 450.0 mL

Unknown: V2

a. P2 = 2P1

b. P2 = ⎯14

⎯P1

P1V1 = P2V2

V2 = = = 225.0 mL

V2 = = 1800 mL(1)(450.0 mL)⎯⎯

⎯14

⎯

(1)(450.0 mL)⎯⎯

2P1V1⎯

P2

27. Given: V1 = 1.00 × 106 mLP1 = 575 mm HgP2 = 1.25 atm

Unknown: V2

P1V1 = P2V2, V2 = ⎯PP1V

2

1⎯

P1 = (575 mm)�⎯7160

atmmm

⎯� = 0.756 atm

V2 = = 6.05 × 105 mL(0.756 atm)(1.00 × 106 mL)⎯⎯⎯⎯

1.25 atm

28. Given: T1 = 27°C = 300 KP1 = 0.625 atmP2 = 1.125 atm

Unknown: T2

⎯TP1

1⎯ = ⎯

TP

2

2⎯

T2 = = = 540 K = 267°C(1.125 atm)(300 K)⎯⎯⎯

0.625 atmP2T1⎯

P1

29. Given: V1 = 1.75 LT1 = −23°C = 250 KP1 = 150 kPaV2 = 1.30 LP2 = 210 kPa

Unknown: T2

=

T2 = = = 260 K = −13°C(210 kPa)(1.30 L)(250 K)⎯⎯⎯

(150 kPa)(1.75 L)P2V2T1⎯

P1V1

P2V2⎯T2

P1V1⎯T1

30. Given: P1 = 7.75 × 104 PaT1 = 17°C = 290 KV1 = 850.cm3

V2 = 720. cm3

P2 = 8.10 × 104 Pa

Unknown: T2

=

T2 = = = 257 K = −16°C(8.10 × 104 Pa)(720 cm3)(290 K)⎯⎯⎯⎯

(7.75 × 104 Pa)(850 cm3)

P2V2T1⎯P1V1

P2V2⎯T2

P1V1⎯T1

31. Given: V1 = 250 LT1 = 22°C = 295 KP1 = 0.974 atmT2 = −52°C = 221 KP2 = 0.750 atm

Unknown: V2

=

V2 = = = 243 L(0.974 atm)(250 L)(221 K)⎯⎯⎯

(0.750 atm)(295 K)P1V1T2⎯

P2T1

P2V2⎯T2

P1V1⎯T1

32. Given: V1 = 250 LT1 = 295 KP1 = 0.974 atmV2 = 400 LP2 = 0.475 atm

Unknown: T2

=

T2 = = = 230 K = −43°C(0.475 atm)(400 L)(295 K)⎯⎯⎯

(0.974 atm)(250 L)P2V2T1⎯

P1V1

P2V2⎯T2

P1V1⎯T1


133

33. Given: V1 = 5.05 m3

P1 = 20.°C = 293 K

P1 = 9.95 × 104 PaT2 = 0°C = 273 KP2 = 1.01 325 ×105 Pa

Unknown: V2 in m3 perday

=

V2 = =

= 4.62 × 10−4 m3

�⎯15 bmreinaths⎯��⎯6h

0omur

in⎯��⎯24

dhaoyurs⎯� = 21 600 breaths/day

�⎯21 600da

byreaths⎯��⎯4.62

b×re

1a0t

−

h

4 m3⎯� = 9.98 m3/day

(9.95 × 104 Pa)(5.05 m3)(273 K)⎯⎯⎯⎯

(1.01 325 × 105 Pa)(293 K)

P1V1T2⎯P2T1

P2V2⎯T2

P1V1⎯T1

40. Given: V = 5.00 L O2

n = 1.08 × 1023

molecules

a. Unknown: number ofmoleculesin 5.00 LH2

b. Unknown: number ofmoleculesin 5.00 LCO2

c. Unknown: number ofmoleculesin 10.00 LNH3

� ��⎯1LLHO

2

2⎯�(5.00 L H2) = 1.08 × 1023 molecules H2

� ��⎯1LLCO

O

2

2⎯�(5.00 L CO2) = 1.08 × 1023 molecules CO2

� ��⎯L1 LNH

O2

3⎯�(10.00 L NH3) = 2.16 × 1023 molecules NH3

1.08 × 1023 molecules⎯⎯⎯

5.00 L O2

1.08 × 1023 molecules⎯⎯⎯

5.00 L O2

1.08 × 1023 molecules⎯⎯⎯

5.00 L O2

41. a. Unknown: number ofmoles in22.4 L N2at STP

b. Unknown: number ofmoles in5.60 L Cl2at STP

c. Unknown: number ofmoles in0.125 L Neat STP

d. Unknown: number ofmoles in70.0 mLNH3 atSTP

V = M × ⎯22

m.4ol

L⎯

n = ⎯22.4

VL/mol⎯ = ⎯

222.42.

L4/Lmol

⎯ = 1.00 mol N2

n = ⎯22.4

VL/mol⎯ = ⎯

225.4.6

L0/Lmol

⎯ = 0.250 mol Cl2

n = ⎯22.4

VL/mol⎯ = ⎯

220..412

L5/mL

ol⎯ = 5.58 × 10−3 mol Ne

n = ⎯22.4

VL/mol⎯ = �⎯22

7.04.0L

m/m

Lol

⎯��⎯1000L

mL⎯� = 3.13 × 10−3 mol NH3


134

42. a. Unknown: m in g of11.2 L H2at STP

b. Unknown: m in g of2.80 L CO2at STP

c. Unknown: m in g of15.0 mLSO2 at STP

d. Unknown: m in g of3.40 cm3 F2at STP

m = (11.2 L)�⎯212m.4

oLl

⎯��⎯2.01m4ogl

H2⎯� = 1.01 g H2

m = (2.80 L)�⎯212m.4

oLl

⎯��⎯44mg

oClO2⎯� = 5.50 g CO2

m = (15.0 mL)�⎯1000L

mL⎯��⎯212m

.4oLl

⎯��⎯64mg

oSlO2⎯� = 0.0429 g SO2

m = (3.40 cm3)�⎯cm

m

L3⎯��⎯1000

LmL⎯��⎯212m

.4oLl

⎯��⎯38m

golF2⎯� = 5.77 × 10−3 g F2

43. a. Unknown: V in L of8.00 g O2 atSTP

b. Unknown: V in L of3.50 g COat STP

c. Unknown: V in L of0.017 g H2Sat STP

d. Unknown: V in L of 2.25 ×105 kg NH3at STP

V = (8.00 g O2)�⎯22m.4ol

L⎯��⎯32

mgoOl

2⎯� = 5.60 L O2

V = (3.50 g CO)�⎯22m.4ol

L⎯��⎯28

mgoClO

⎯� = 2.80 L CO

V = (0.0170 g H2S)�⎯22m.4ol

L⎯��⎯34

mg

oHl

2S⎯� = 0.0112 L H2S

V = (2.25 × 105 kg)�⎯10k0g0 g⎯��⎯22

m.4ol

L⎯��⎯17

mg

oNlH3

⎯� = 2.96 × 108 L NH3

44. Given: V = 75.0 L CO2

a. Unknown: number ofL C2H2required

b. Unknown: number ofL H2O produced

c. Unknown: number of L O2required

2C2H2 + 5O2 → 4CO2 + 2H2O

(75.0 L CO2)�⎯24LL

CC

2

OH

2

2⎯� = 37.5 L C2H2

(75.0 L CO2)�⎯24 LL

HCO

2O

2⎯� = 37.5 L H2O

(75.0 L CO2)�⎯45LL

COO2

2⎯� = 93.8 L O2


135

45. Given: CuO(s) + H2(g) →Cu(s) + H2O(g)

V = 5.60 L H2 atSTP

a. Unknown: n of H2

b. Unknown: n of Cu produced

c. Unknown: number ofgrams Cuproduced

n = ⎯RPV

T⎯ = = 0.250 mol H2

n of Cu = (0.250 mol H2)�⎯11 mm

ooll

CH

u

2⎯� = 0.250 mol Cu

(0.250 mol)�⎯63.m5

oglCu

⎯� = 15.9 g Cu

(1.0 atm)(5.60 L)⎯⎯⎯

�⎯0.08m21

olL• K

• atm⎯�(273 K)

46. Given: P = 0.961 atmV = 29.0 L CH4T = 20°C =293 K

Unknown: (a) V of CO2

(b) V of H2O

CH4 + 2O2 → CO2 + 2H2O

a. V = (29.0 L CH4)�⎯11 LL

CC

HO2

4⎯� = 29.0 L CO2

b. V = (29.0 L CH4)�⎯21 LL

HCH

2O

4⎯� = 58.0 L H2O vapor

47. Given: air = 20.9% O2 byvolume.

a. V = 25.0 L C8H18

Unknown: V of airneeded forcombustion

b. Unknown: (1) V CO2produced(2) V H2Oproduced

2C8H18 + 25O2 → 16CO2 + 18H2O

V = (25.0 L C8H18)�⎯22L5

CL

8

OH

2

18⎯� = 312.5 L O2

V air needed = �⎯2100.09

LL

aOir

2⎯�(312.5 L O2) = 1.50 × 103 L air

(1) V = (25.0 L C8H18)�⎯216L

LC8

CHO

1

2

8⎯� = 200. L CO2

(2) V = (25.0 L C8H18)�⎯218L

LC

H

8H2O

18⎯� = 225 L H2O vapor

48. Given: V = 4.50 × 102 mLCO

V = 825 mL H2

a. Unknown: reactantpresent inexcess

b. Unknown: amount ofCO remain-ing afterreaction

c. Unknown: volume ofCH3OHproduced

CO(g) + 2H2(g) → CH3OH(g)

(4.50 × 102 mL CO)�⎯12 mm

LL

CH

O2⎯� = 900 mL H2 needed

(825 mL H2)�⎯12 mm

LL

CH

O

2⎯� = 413 mL CO needed

CO is present in excess. (4.50 × 102 mL > 413 mL)

450 mL − 413 mL = 37 mL CO

(825 mL H2)�⎯1 m2Lm

CLHH3O

2

H⎯� = 413 mL CH3OH


136

49. a. Given: V = 2.50 L HF

n = 1.35 mol

T = 320. K

Unknown: P in atm

b. Given: V = 4.75 L NO2

n = 0.86 mol

T = 300. K

Unknown: P

c. Given: V = 7.50 ×104 mL CO2

n = 2.15 mol

T = 57°C = 330 K

Unknown: P

P = ⎯nR

VT

⎯ = = 14.2 atm

P = ⎯nR

VT

⎯ = = 4.5 atm

(7.50 × 104 mL)�⎯1000L

mL⎯� = 75 L

P = ⎯nR

VT

⎯ = = 0.777 atm

(2.15 mol)�⎯0.08m21

olL• K

• atm⎯�(330. K)

⎯⎯⎯⎯75 L

(0.86 mol)�⎯0.08m21

olL• K

• atm⎯�(300. K)

⎯⎯⎯⎯4.75 L

(1.35 mol)�⎯0.08m21

olL• K

• atm⎯�(320. K)

⎯⎯⎯⎯2.50 L

50. a. Given: n = 2.00 mol H2

T = 300. K

P = 1.25 atm

Unknown: V in L

b. Given: n = 0.425 molNH3

T = 37°C =310 K

P = 0.724 atm

Unknown: V in L

c. Given: m = 4.00 g O2

T = 57°C =330 K

P = 0.888 atm

Unknown: V in L

V = ⎯nR

PT

⎯ = = 39.4 L H2

V = ⎯nR

PT

⎯ = = 14.9 L NH3

V = ⎯nR

PT

⎯ = = 3.81 L O2

(4.00 g)�⎯m3o2l O

g2⎯��⎯0.08

m21

olL• K

• atm⎯�(330 K)

⎯⎯⎯⎯⎯⎯⎯⎯0.888 atm

(0.425 mol)�⎯0.08m21

olL• K

• atm⎯�(310 K)

⎯⎯⎯⎯0.724 atm

(2.00 mol)�⎯0.08m21

olL• K

• atm⎯�(300 K)

⎯⎯⎯⎯1.25 atm

51. a. Given: V = 1.25 L

T = 250. K

P = 1.06 atm

Unknown: n

b. Given: V = 0.80 L

T = 27°C =300 K

P = 0.925 atm

Unknown: n

n = ⎯RPV

T⎯ = = 0.0646 mol

n = ⎯RPV

T⎯ = = 0.030 mol

(0.925 atm)(0.80 L)⎯⎯⎯

�⎯0.08m21

olL• K

• atm⎯�(300 K)

(1.06 atm)(1.25 L)⎯⎯⎯

�⎯0.08m21

olL• K

• atm⎯�(250. K)


137

56. a. Unknown: relativerate of effu-sion of H2and N2

b. Unknown: relativerate of effu-sion of F2and Cl2

= = = 3.72

= = = 1.37�71 g/m�ol�⎯⎯�38 g/m�ol�

�MCl2�⎯�MF2

�rate of effusion of F2⎯⎯⎯rate of effusion of Cl2

�28.00� g/mol�⎯⎯�2.02 g�/mol�

�MN2�

⎯�MH2

�rate of effusion of H2⎯⎯⎯rate of effusion of N2

57. Unknown: relative aver-age velocity ofH2 and Ne

⎯vveelloocciittyy

ooff

NH

e2⎯ = = = 3.16

�20.17�9 g/m�ol�⎯⎯

�2.02 g�/mol��MNe�⎯�MH2

�

58. Given: velocity of Cl2 mol-ecules = 324 m/s

Unknown: velocity of SO2molecules

=�MSO2�

⎯�MCl2�

velocity of Cl2⎯⎯velocity of SO2

velocity of SO2 = (velocity of Cl2)� �= (324 m/s)� � = 341 m/s

�71 g/m�ol�⎯⎯�64 g/m�ol�

�MCl2�⎯�MSO2�

52. a. Given: V = 5.60 L O2

P = 1.75 atm

T = 250. K

Unknown: m in g

b. Given: V = 3.50 L NH3

P = 0.921 atm

T = 27°C =300 K

Unknown: m

c. Given: V = 0.125 L SO2P = 0.822 atm

T = −5°C =268 K

Unknown: m

n = ⎯RPV

T⎯

n = = 0.478 mol O2

m = (0.478 mol)�⎯32mgoOl

2⎯� = 15.3 g O2

n = ⎯RPV

T⎯ = = 0.131 mol NH3

m = (0.131 mol)�⎯17mg

oNlH3⎯� = 2.23 g NH3

n = ⎯RPV

T⎯ = = 4.67 × 10−3 mol SO2

m = (0.0047 mol)�⎯64.0m7

oglSO2⎯� = 0.299 g SO2

(0.822 atm)(0.125 L)⎯⎯⎯

�⎯0.08m21

olL• K

• atm⎯�(268 K)

(0.921 atm)(3.50 L)⎯⎯⎯

�⎯0.08m21

olL• K

• atm⎯�(300 K)

(1.75 atm)(5.60 L)⎯⎯⎯

�⎯0.08m21

olL• K

• atm⎯�(250 K)

c. Given: V = 0.750 L

T = 50°C =223 K

P = 0.921 atm

Unknown: n

n = ⎯RPV

T⎯ = = 0.0377 mol

(0.921 atm)(0.75 L)⎯⎯⎯

�⎯0.08m21

olL• K

• atm⎯�(223 K)


138

64. Given: V1 = 4.00 LT1 = 304 KP1 = 755 mmV2 = 4.08 LP2 = 728 mm

Unknown: T2

⎯P

T1V

1

1⎯ = ⎯PT2V

2

2⎯

T2 = = = 299 K(728 mm)(4.08 L)(304 K)⎯⎯⎯

(755 mm)(4.00 L)P2V2T1⎯

P1V1

65. Given: P1 = 4.62 atmV1 = 2.33 LV2 = 1.03 L

Unknown: P2 (in torr)

P1V1 = P2V2

P2 = = = 10.45 atm

(10.45 atm)�⎯76a0tmtorr⎯� = 7940 torr

(4.62 atm)(2.33 L)⎯⎯⎯

1.03 LP1V1⎯V2

66. Given: V2 = 2.00 × 107 LP2 = 20.0 atmP1 = 1.0 atm

Unknown: V1

P1V1 = P2V2

V1 = = = 4.00 × 108 L(20.0 atm)(2.00 × 107 L)⎯⎯⎯

1.0 atmP2V2⎯

P1

61. Given: V1 = 295 mLT1 = 36°C = 309 KT2 = 55°C = 328 K

Unknown: V2

⎯VT1

1⎯ = ⎯TV

2

2⎯

V2 = ⎯VT1

1⎯ = = 313 mL(295 mL)(328 K)⎯⎯

(309 K)

62. Given: V1 = 638 mLP1 = 0.893 atmT1 = 12°C = 285 KV2 = 881 mLT2 = 18°C = 291 K

Unknown: P2

⎯P

T1V

1

1⎯ = ⎯PT2V

2

2⎯

P2 = = = 0.660 atm(0.893 atm)(638 mL)(291 K)⎯⎯⎯⎯

(285 K)(881 mL)P1V1T2⎯T1V2

63. Given: T1 = 84°C = 357 KP1 = 0.503 atmP2 = 1.20 atm

Unknown: T2 in °C

⎯TP1

1⎯ = ⎯

TP

2

2⎯

T2 = = = 852 K = 579°C(1.20 atm)(357 K)⎯⎯⎯

(0.503 atm)P2T1⎯

P1

60. Given: V1 = 2.30 LT1 = 311 KT2 = 295 K

Unknown: V2

⎯VT1

1⎯ = ⎯TV

2

1⎯

V2 = ⎯V

T1T

1

2⎯ = = 2.18 L(2.30 L)(295 K)⎯⎯

(311 K)

59. Given: PT = 6.11 atmPA = 1.68 atmPB = 3.89 atm

Unknown: PC

PT = PA + PB + PC

PC = PT − (PA + PB) = 6.11 − (1.68 + 3.89) = 0.54 atm


139

67. Given: effusion rate ofgas X = 0.850times effusion rate of NO2

Unknown: MX

=

�MX� = � �(�MNO2�)

= �⎯01.8.0500⎯�(�46 g/m�ol�) = 8.0 g/mol

MX = (8.0)2 = 64 g/mol

effusion rate of NO2⎯⎯⎯effusion rate of X

�MNO2�

⎯�MX�

effusion rate of X⎯⎯⎯effusion rate of NO2

68. Given: V = 265 mL =0.265 L Cl2 at STP

Unknown: m of Cl2

n = (0.265 L Cl2)�⎯212m.4

oLl

⎯��⎯70.9m0

oglCl2⎯� = 0.839 g Cl2

69. Given: n = 3.11 mol CO2

P = 0.820 atm

T = 39°C = 312 K

Unknown: V in L

PV = nRT

V = ⎯nR

PT

⎯ = = 97.2 L

(3.11 mol CO2)�⎯0.08m21

olL• K

• atm⎯�(312 K)

⎯⎯⎯⎯⎯⎯⎯⎯0.820 atm

70. Unknown:effusion rate of CO⎯⎯⎯effusion rate of SO3

= = = 1.7�80 g/m�ol�⎯⎯�28 g/m�ol�

�MSO3�

⎯�MCO�

effusion rate of CO⎯⎯⎯effusion rate of SO3

71. Given: m = 0.993 g

V = 0.570 L

T = 281 K

P = 1.44 atm

Unknown: M

M = ⎯mPRVT

⎯ = = 27.9 g/mol

(0.993 g)�⎯0.08m21

olL• K

• atm⎯�(281 K)

⎯⎯⎯⎯⎯⎯⎯⎯(1.44 atm)(0.570 L)

72. Given: V = 1000. cm3 =1000. mL3 = 1 L

T = 32°C = 305 K

P = (752 mm Hg)

�⎯760amtmm Hg⎯� =

0.99 atm

Unknown: n

PV = nRT

n = ⎯RPV

T⎯ = = 0.0395 mol He

(0.99 atm)(1 L)⎯⎯⎯

�⎯0.08m21

olL• K

• atm⎯�(305 K)

73. Given: T = 16°C = 289 K

P = 0.982 atm

M = 7.40 g

V = 3.96 L

Unknown: V at STP; M

⎯P

T1V

1

1⎯ = ⎯P

T2V

2

2⎯

V2 = ⎯P

P1V

2T1T

1

2⎯

= = 3.67 L

M = ⎯mPRVT

⎯ = = 45.1 g/mol

(7.40 g)�⎯0.08m21

olL• K

• atm⎯�(289 K)

⎯⎯⎯⎯⎯⎯⎯⎯(0.982 atm)(3.96 L)

(0.982 atm)(3.96 L)(273 K)⎯⎯⎯⎯

(1.00 atm)(289 K)


140

5. Given: VX = 2VCH4

MCH4 = 16.05 g/mol

Unknown: MX

⎯V

VA

B

2

2⎯ = ⎯MM

B

A⎯

MX = ⎯M

(

C

2

H

V4

C

V

H

C

4)

H242

⎯ = ⎯16

4.05⎯ = 4.01 g/mol ≈ 4 g/mol

6. Given: N2(g) + 3H2(g) →2NH3(g)

Molar volume N2= molar volume H2

3 L N2

3 L H2

Unknown: L N2 remainingafter reaction

3 L H2 �⎯31 LL

HN2

2⎯� = 1 L N2 reacted

3 L N2 available − 1 L N2 reacted = 2 L N2 remaining

4. Given: T1 = 100.0°C = 373 K

T2 = 300.0°C = 573 K

P1 = 3.0 atm

Unknown: P2

⎯TP1

1⎯ = ⎯

TP2

2⎯

P2 = ⎯P

T1T

1

2⎯ = ⎯

3.0 at3m73

×K573 K

⎯ = 4.6 atm

3. Given: n = 0.500 mol

V = 10.0 L

T = 20.°C = 293 K

Unknown: P

PV = nRT

P = ⎯nR

VT

⎯ =

P = 120 kPa

0.500 mol × 8.314 ⎯mL •

okl •

PKa

⎯ × 293 K

⎯⎯⎯⎯10.0 L

2. Given: V1 = 150 mL

P1 = 0.923 atm

P2 = 0.987 atm

Unknown: V2

V2 = ⎯P

P1V

2

1⎯ = = 140 mL0.923 atm × 150 mL⎯⎯⎯

0.987 atm

Standardized Test Prep, p. 397

2. Given: V1 = 785 mL = 0.785 L

P1 = 0.879 atm

P2 = 0.994 atm

Unknown: V2

P1V1 = P2V2

V2 = ⎯P

P1V

2

1⎯ = = 694 mL0.879 atm × 0.785 L⎯⎯⎯

0.994 atm

Math Tutor, p. 396

CHAPTER 11 Gases - Mr. Cowmeadow's Chemistry Classes ...

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