chapter 5 gases chapter 9/1. 1 atm = 760 mmhg t(k) = t(°c) + 273, 1 mol so 2 = 64.07 g calculate...

Chapter 5

Gases

Chapter 9/1

1 atm = 760 mmHg

T(K) = t(°C) + 273, 1 mol SO2 = 64.07 gKmolLatm 0.0821

nRT, RPV

Calculate the Volume Occupied by 637 g of SO2 (MM 64.07)

at 6.08 x 103 mmHg and –23 °C.mSO2 = 637 g, P = 6.08 x 103 mmHg, t = −23 °C,

V, L

Solution:

Solution Map:

Relationships:

Given:

Find:

P

nRT V

P, n, T, R V

L 55.2

atm 0.80

K 0520.0821mol 249.9 Kmol

Latm

P

TRnV

atm 0.08mmHg 760

atm 1mmHg 106.08 3

K 025

273C 23- (K)

T

T

g n

g 64.07

mol 1

22

2 SO mol 249.9 g 64.07

SO mol 1SO g 637

2

m = 9.988g, n = 0.250 mol, P = 1.0197 atm, T = 300. K

density, g/L

L 5530.6atm 9711.0

K 0030.082mol .2500Kmol

Latm

P

TRnV

Solution:

Calculate the Density of a Gas at 775 torr and 27 °C if 0.250 moles Weighs 9.988 g

m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C,

density, g/L

Solution Map:

Relationships:

Given:

Find:

P

TRn V

V, m d

atm 9710.1 torr760

atm 1 torr775

K 030

273C 27 (K)

T

T

P, n, T, R V

V

m d

g/L 1.65

L 553.06

g .9889

V

md

1 atm = 760 mmHg, T(K) = t(°C) + 273

KmolLatm 0.0821

nRT, RPV

V

m d

3

4

Molar Mass of a Gas

• One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law.

moles

gramsin massMassMolar

m = 0.311g, V = 0.225 L, P = 1.1658 atm, T = 328 K

Molar Mass, g/mol

mol 105447.9K 2830.0821

L .2250atm 5861.1 3

Kmol

Latm

TR

VPn atm 5861.1

mmHg 760

atm 1mmHg 886

Calculate the Molar Mass of a Gas with Mass 0.311 g that Has a Volume of 0.225 L at 55 °C and 886 mmHg.

m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,

Molar Mass, g/mol

Solution:

Solution Map:

Relationships:

Given:

Find:

TR

VP n

n, m MM

K 328

273C55 (K)

T

T

P, V, T, R n

n

m MM

g/mol 31.9 mol 109.7454

g 311.03-

n

mMM

1 atm = 760 mmHg, T(K) = t(°C) + 273

Kmol

Latm 0.0821 nRT, RPV

n

m MM

5

6

• Write a solution map:

P,V,T,R n

When using the ideal gas equation, the units of V must be L; and the units of P must be atm, or you will have to convert.

The units of T must be kelvin, K.

Information:Given: V = 0.225 L, P = 886 mmHg,

t = 55 °C, m = 0.311 gFind: molar mass, (g/mol)Equation:

PV = nRT; MM = mass/moles

Example:A sample of a gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.

Molar Mass

mass

PV = nRT

m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K,

molar mass, g/mol

What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C?m=12.0 g, V= 197 L, P=380 torr, t=127°C,

molar mass, g/mol

Solution:

Solution Map:

Relationships:

Given:

Find:

TR

VPn

n, m MM

mol 0.3K 0040.0821

L 971atm 0.50

Kmol

Latm

TR

VPn atm 50.0

torr760

atm 1 torr083

K 004

273C127 (K)

T

T

P, V, T, R n

n

mMM

g/mol 4.0 mol .03

g 2.01

n

mMM

1 atm = 760 torr, T(K) = t(°C) + 273

Kmol


n

mMM

7

8

Partial Pressure• Each gas in the mixture exerts a pressure

independent of the other gases in the mixture.

• The pressure of a component gas in a mixture is called a partial pressure.

• The sum of the partial pressures of all the gases in a mixture equals the total pressure. Dalton’s law of partial pressures. Ptotal = Pgas A + Pgas B + Pgas C +...

atm 1.00 atm 0.01 atm 0.21 atm 0.78 Ar2O2Nair PPPP

A Mixture of He, Ne, and Ar Has a Total Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg,

Determine the Partial Pressure of Ar in the Mixture.

PHe= 341 mmHg, PNe= 112 mmHg, Ptot = 558 mmHg

PAr, mmHg

Solution:

Solution Map:

Relationships:

Given:

Find:

Ptot, PHe, PNe PAr

Ptot = Pa + Pb + etc.

PAr = Ptot – (PHe + PNe)

mmHg 051

mmHg 112341558Ar

P

9

10

Finding Partial Pressure• To find the partial pressure of a

gas, multiply the total pressure of the mixture by the fractional composition of the gas.

• For example, in a gas mixture that is 80.0% He and 20.0% Ne that has a total pressure of 1.0 atm, the partial pressure of He would be:

PHe = (0.800)(1.0 atm) = 0.80 atm Fractional composition =

percentage divided by 100.

The Partial Pressure of Each Gas in a Mixture, or the Total Pressure of a Mixture, Can Be

Calculated Using the Ideal Gas Law

BAtotal

total

BAtotal

BB

AA

P PV

x R x Tn P

n n n

V

x R x Tn P

V

x R x Tn P

same theare mixture in the

everything of volumeand re temperatuthe

mixture ain B andA gasesfor

1111

atm 8959.0

L 8.7

K 9850.0821mol 0.17Kmol

Latm

XeXe

V

TRnP

XeNetotal ,KmolLatm 0.0821 PPPnRT, RPV

Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe.

Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol

PNe, atm

Solution:

Solution Map:

Relationships:

Given:

Find:

V

TRnP

Xe

Xe

nXe, V, T, R PXe

atm 2.9

atm 8950.9 atm 9.3XetotalNe

PPP

Ptot, PXe PNe

XetotalNe PPP

12

Deep Sea Divers and Partial Pressure• It is also possible to have too much O2, a condition called

oxygen toxicity.PO2 > 1.4 atm.Oxygen toxicity can lead to muscle spasms, tunnel vision, and

convulsions.

• It is also possible to have too much N2, a condition called nitrogen narcosis.Also known as rapture of the deep.

• When diving deep, the pressure of the air that divers breathe increases, so the partial pressure of the oxygen increases.At a depth of 55 m, the partial pressure of O2 is 1.4 atm.Divers that go below 50 m use a mixture of He and O2 called heliox that

contains a lower percentage of O2 than air.

13

14

Partial Pressure vs. Total Pressure

At a depth of 30 m, the total pressure of air in the diverslungs, and the partial pressure of all the gases in the air,

are quadrupled!

Gas Stoichiometry

15

16


When using the ideal gas equation, the units of V must be L; and the units of P must be atm, or you will have to convert.

The units of T must be kelvin, K.

Information:Given: 294 g KClO3

PO2 = 755 mmHg, TO2 = 305 KFind: VO2, LEquation: PV = nRTConversion Factors:

1 mole KClO3 = 122.5 g 2 mole KClO3 3 moles O2

Example:How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.

)()()( gss 23 O 3 KCl 2 KClO 2

nO2 VO2nKClO3g KClO3

3

3

KClO g 5122

KClO mol 1

. 3

2

KClO mol 2

O mol 3

P,T,R

PV=nRT

17

• Apply the solution map:Find moles of O2 made.


PO2 = 755 mmHg, TO2 = 305 KFind: VO2, LEquation: PV = nRTConversion Factors:


Solution Map: g → mol KClO3

→ mol O2 → L O2

23

2

3

33 O mol 603

KClO mol 2

O mol 3

KClO g 122.5

KClO mol 1KClO g 294 .


)()()( gss 23 O 3 KCl 2 KClO 2

18

• Apply the solution map:Convert the units.


PO2 = 755 mmHg, TO2 = 305 K, nO2 = 3.60 molesFind: VO2, LEquation: PV = nRTConversion Factors:



→ mol O2 → L O2

atm 42399.0mmHg 760

atm 1mmHg 557 P


)()()( gss 23 O 3 KCl 2 KClO 2

19

• Apply the solution map:

L 90.7

atm 42399.0

K 305 0.0821mol 60.3 Kmol

Latm

V

VP

TRn

TRnVP


PO2 = 0.99342 mmHg, TO2 = 305 K, nO2 = 3.60 molesFind: VO2, LEquation: PV = nRTConversion Factors:



→ mol O2 → L O2


)()()( gss 23 O 3 KCl 2 KClO 2

mHgO = 10.0g, P=0.750 atm, T=313 K

VO2, L

nO2 = 0.023085 mol, P = 0.750 atm, T = 313 K

VO2, L

L 791.0

atm 0.750

K3130.08206mol 850023.0Kmol

Latm

P

TRnV

Practice—What Volume of O2 at 0.750 atm and 313 K is Generated by the Thermolysis of 10.0 g of HgO?

2 HgO(s) 2 Hg(l) + O2(g), Continued

Solution:

Solution Map:

Relationships:

Given:

Find:

P

TRnV

2

2

O mol 850023.0

HgO mol 2

O mol 1

g 216.59

HgO mol 1HgO g 0.01

P, n, T, R V

1 atm = 760 mmHg, HgO = 216.59 g/mol2 mol HgO : 1 mol O2 Kmol


g HgO mol HgO mol O2

HgO mol 2

O mol 1 2

g 216.59

HgO mol 1

20

21

Calculate the Volume Occupied by 1.00 Mole of an Ideal Gas at STP.

• 1 mole of any gas at STP will occupy 22.4 L.• This volume is called the molar volume and can

be used as a conversion factor.As long as you work at STP.

1 mol 22.4 L

(1.00 atm) x V = (1.00 moles)(0.0821 )(273 K)L∙atmmol∙K

V = 22.4 L

P x V = n x R x T

22

Molar Volume

There is so muchempty spacebetween moleculesin the gas state thatthe volume of thegas is not effectedby the size of themolecules (underideal conditions).

Example —How Many Grams of H2O Form When 1.24 L H2 Reacts Completely with O2 at STP?

O2(g) + 2 H2(g) → 2 H2O(g)VH2 = 1.24 L, P = 1.00 atm, T = 273 K

massH2O, g

Solution:

Solution Map:

Relationships:

Given:

Find:

OH mol 1

g 02.18

2

OH g 998.0

OH mol 1

OH g 8.021

H mol 2

OH mol 2

H L 22.4

H mol 1H L .241

2

2

2

2

2

2

22

H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP2 mol H2O : 2 mol H2

OH mol 2

H mol 2

2

2

L 22.4

H mol 1 2

g H2OL H2 mol H2 mol H2O

2323

24


mol H2O g H2Omol H2L H2

2

2

H L 2.42

H mol 1

2

2

H mol 2

OH mol 2

Information:Given: 1.24 L H2

Find: g H2OConversion Factors:

1 mol H2O = 18.02 g2 mol H2O 2 mol H2

1 mol H2 22.4 L

Example:How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?

)()( ggg OH 2 O )(H 2 222

OH mol 1

OH g 0218

2

2.

25

• Apply the solution map:

Information:Given: 1.24 L H2

Find: g H2OConversion Factors:

1 mol H2O = 18.02 g2 mol H2O 2 mol H2

1 mol H2 22.4 LSolution Map:

L → mol H2 → mol H2O → g H2O

OH g 9880OH mol 1

OH g 18.02

H mol 2

OH mol 2

H L 22.4

H mol 1H L 1.24 2

2

2

2

2

2

22 .

Example:How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?

)()( ggg OH 2 O )(H 2 222

Practice—What Volume of O2 at STP is Generated by the Thermolysis of 10.0 g of HgO?

2 HgO(s) 2 Hg(l) + O2(g), ContinuedmHgO = 10.0 g, P = 1.00 atm, T = 273 K

VO2, L

Solution:

Solution Map:

Relationships:

Given:

Find:

g 59.216

HgO mol 1

2

2

22

O L 517.0

O mol 1

O L 2.42

HgO mol 2

O mol 1

g 216.59

HgO mol 1HgO g 0.01

HgO = 216.59 g/mol, 1 mol = 22.4 L at STP2 mol HgO : 1 mol O2

HgO mol 2

O mol 1 2

2O mol 1

L 22.4

L O2g HgO mol HgO mol O2

2626

chapter 5 gases chapter 9/1. 1 atm = 760 mmhg t(k) = t(°c) + 273, 1 mol so 2 = 64.07 g calculate...

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