chem 106, prof. t. l. heise 1 che 106: general chemistry chapter ten copyright © tyna l. heise...

Chem 106, Prof. T. L. Heise 11

CHE 106: General Chemistry

CHAPTER TEN

Copyright © Tyna L. Heise 2001

All Rights Reserved


How is matter encountered?How is matter encountered?

–NOT on the atomic or molecular scale

–As a large collection of atoms or molecules

–Recognized as solids liquids and gases

GasesGases

Chapt. 10.1


Characteristics of GasesCharacteristics of Gases

–Expands to fill container completely

–Highly compressible

–Form homogeneous mixtures regardless of identities or proportions

GasesGases

Chapt. 10.1


Why exhibit these characteristics?Why exhibit these characteristics?

–Individual particles are far apart

–Act as if they are only molecule present

GasesGases

Chapt. 10.1


Properties of Gases

Chapt. 10.2

When measuring gases, easiest properties are

1) Temperature - thermochemistry

2) Volume - solution chemistry

3) Pressure


Pressure

Chapt. 10.2

- conveys idea of force

- P = F P = pressure

A F = force

A = area

- pressure is exerted on any surface a gas contacts


Atmospheric pressureAtmospheric pressure

Gases of our atmosphere exert a force on surface of the earth due to gravity.

- force exerted is equal to mass times acceleration due to gravity

F = ma

since P = F then P = ma a = 9.8 m

A A s2

Chapt. 10.2



Chapt. 10.2

Atmospheric pressure is measured using a mercury barometer

•Tube is filled with mercury

•Small portion falls back into dish when inverted, vacuum exists above liquid in column

•Column moves up or down depending on atmospheric force on surface of mercury in dish



Chapt. 10.2

Units of Measurement:

1 atm

760 mmHg

760 torr

1.01325 x 105 Pa

101.325 kPa

Convert 13.33 kPa into atm

13.33 kPa 1 atm = 0.1316 atm

101.325 kPa


Pressures of Enclosed gasesPressures of Enclosed gases

Chapt. 10.2

Manometers

•Similar in operation to barometer

•Two types, closed tube and open tube

•Closed tube - measures pressures below atmospheric

•Difference in tube heights = pressure



Chapt. 10.2

•Open tube - measures pressures near atmospheric

•Difference in tube heights relates pressure of gas to atmospheric pressure



Chapt. 10.2



Chapt. 10.2

Sample Problem: A vessel connected to an open end

manometer is filled with gas to a pressure of 0.835 atm.

The atmospheric pressure is 755 torr.

a) which arm of manometer is higher?

b) what is the height difference?



Chapt. 10.2






a) Convert atm into torr



Chapt. 10.2







0.835 atm 760 torr =

1 atm



Chapt. 10.2







0.835 atm 760 torr = 634 torr

1 atm * which is less than atmospheric pressure



Chapt. 10.2







atmospheric pressure makes arm open to air lower and arm attached to gas container higher



Chapt. 10.2






b) Pgas = Patm + difference in heights



Chapt. 10.2






b) Pgas = Patm + difference in heights

Pgas = 634 torr

Patm = 755 torr

634 = 755 + difference



Chapt. 10.2






b) Pgas + difference in heights = Patm

Pgas = 634 torr

Patm = 755 torr

634 + X = 755 X = 121 torr


The Gas LawsThe Gas Laws

Chapt. 10.3

Four variables needed to adequately describe a gas

•Temperature (T)

•Pressure (P)

•Volume (V)

•Number of moles (n)

•Equations that express relationships between these variables are the gas laws.


Boyle’s LawBoyle’s Law

Chapt. 10.3

Boyle investigated the relationship between volume

and pressure.

As pressure increased, the volume decreased; proves an inverse relationship

PV = constant


Charles’s LawCharles’s Law

Chapt. 10.3

Charles investigated the relationship between volume

and temperature.

As temp decreased, the volume decreased; proves a direct relationship

V = constant T


Avogadro’s LawAvogadro’s Law

Chapt. 10.3

Avogadro investigated the relationship between volume

and amount of substance.

As number of molecules doubles, the volume as doubles - proves a direct relationship

Avogadro’s Hypothesis - equal volumes of gases at same temperature and pressure contain equal numbers

of molecules 22.4 L


The Ideal Gas EquationThe Ideal Gas Equation

Chapt. 10.4

Combining the previous three laws, allows for a better

mathematical look at gases.

The term R in the gas equation is called the gas constant.

The conditions of 0°C and 1 atm are referred to as standard temperature and pressure!



Chapt. 10.4

The ideal gas law allows us to isolate all variables that are to be held constant and set up proportionalities.

P1 P2

T1 =

T2

V1 V2

T1 =

T2

P1V1 = P2V2



Chapt. 10.4

Sample Exercise:

Tennis balls are usually filled with air or N2 gas to a pressure above atmospheric pressure to increase their “bounce”. If a particular tennis ball has a volume of 144 cm3 and contains 0.33 g of N2 gas, what is the pressure inside the ball at 24°C?

1) PV = nRT



Chapt. 10.4

Sample Exercise:


2) PV = nRT

P = x R = 0.08206 L-atm/mol-K

V = 144 cm3 T = 24°C

n = 0.33 g N2



Chapt. 10.4

Sample Exercise:


3) PV = nRT

P = x R = 0.08206 L-atm/mol-K

V = 144 cm3 = 144 mL = 0.144 L T = 24°C = 297 K

n = 0.33 g N2 1 mol N2 = 0.012 mol N2

28 g N2



Chapt. 10.4

Sample Exercise:


4) P = nRT = 0.012 mol N2 (0.08206 L-atm/mol-K) (297 K)

V 0.144 L



Chapt. 10.4

Sample Exercise:


2) P = nRT = 0.012 mol N2 (0.08206 L-atm/mol-K) (297 K)

V 0.144 L

P = 2.0 atm



Chapt. 10.4

Sample Exercise:

A large natural gas storage tank is arranged so that the pressure is maintained at 2.20 atm. On a cold December day when the temperature is -15°C, the volume of gas in the tank is 28,500 ft3. What is the volume of the same quantity of gas on a warm July day when the temperature is 31°C?



Chapt. 10.4

Sample Exercise:


1) V1 = V2

T1 T2



Chapt. 10.4

Sample Exercise:


1) V1 = V2 V1 = 28,500 ft3

T1 T2 T1 = -15°C

V2 = x

T2 = 31°C



Chapt. 10.4

Sample Exercise:


2) V1 = V2

T1 T2

V1 = 28,500 ft3 (12)3 in3 16.4 cm3 1 L = 8.08 x 105 L

1 ft3 1 in3 103 cm3



Chapt. 10.4

Sample Exercise:


3) V1 = V2 V1 = 8.08 x 105 L

T1 T2 T1 = -15°C = 258 K

V2 = x

T2 = 31°C = 304 K



Chapt. 10.4

Sample Exercise:


4) V1 = V2 8.08 x 105 L = X

T1 T2 258 K 304 K



Chapt. 10.4

Sample Exercise:


4) V1 = V2 8.08 x 105 L = X

T1 T2 258 K 304 K

X = 9.52 x 105 L


Further Applications of LawFurther Applications of Law

Chapt. 10.5

Ideal gas equation can be used to determine

- density of gas

- molar mass of gas

- volumes of gases formed or consumed



Chapt. 10.5

Gas Densities and Molar Mass

n = moles

V L

* multiply both sides by molar mass, M

n M = P M

V R T



Chapt. 10.5


nM = moles g

V L moles

* multiplying both sides by molar mass, M,

will give us g , so that

LDensity = PM

RT



Chapt. 10.5


this equation can algebraically be rearranged to solve for molar mass

Density = PM

RT

Molar Mass = d R T

P



Chapt. 10.4

Sample Exercise:

Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and 740.0 torr

1) M = d R T

P



Chapt. 10.4

Sample Exercise:


2) M = d R T d = 1.17 g/L

P R = 62.36 L-torr/mol-K

T = 21°C = 294 K

P = 740.0 torr



Chapt. 10.4

Sample Exercise:


3) M = d R T d = 1.17 g/L


T = 21°C = 294 K

P = 740.0 torr

= 1.17 g/L ( 62.36 L-torr/mol-K) (294 K)

740.0 torr



Chapt. 10.4

Sample Exercise:


4) M = d R T d = 1.17 g/L


T = 21°C = 294 K

P = 740.0 torr

= 1.17 g/L ( 62.36 L-torr/mol-K) (294 K)

740.0 torr

= 29.0 g/L



Chapt. 10.4

Sample Exercise:

The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere.

1) d = P M

R T



Chapt. 10.4

Sample Exercise:


2) d = P M Pressure = 1.6 atm

R T M = 28.6 g/mol

R = 0.08206 L-atm/mol-K

T = 95 K



Chapt. 10.4

Sample Exercise:


3) d = P M = 1.6 atm(28.6 g/mol)

R T 0.08206 L-atm/mol-K ( 95 K)



Chapt. 10.5

Sample Exercise:


4) d = P M = 1.6 atm(28.6 g/mol)

R T 0.08206 L-atm/mol-K ( 95 K)

= 5.9 g/L



Chapt. 10.5

Volumes of Gases in Chemical Reactions

A knowledge of gases is often important because gases are often reactants or products in chemical reactions

- coefficients in balanced equations is again going to be very important

n = PV RT

Gas Data

Moles of Gas A

Moles of Gas B

g of Gas B



Chapt. 10.5

Sample Exercise:

In the first step of the industrial process for making nitric acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the presence of a suitable catalyst. The following reaction occurs: 4NH3 + 5O2 --> 4NO + 6H20

How many liters of NH3 at 850°C and 5.00 atm are required to react with 1.00 mol of O2 in this reaction?



Chapt. 10.5

Sample Exercise:



1) No need to convert into moles of O2, we were given that information



Chapt. 10.5

Sample Exercise:



2) Convert from mol O2 into mol NH3

1.00 mol O2 4 mol NH3 = 0.800 mol NH3

5 mol O2



Chapt. 10.5

Sample Exercise:



3) Use ideal gas equation and given info to solve for final variable

PV = nRT

P = 5.00 atm n = 0.800 mol NH3 T = 850°C = 1123 K

V = X R = 0.08206 L-atm/mol-K



Chapt. 10.5

Sample Exercise:




PV = nRT V = nRT = 0.800 mol ( 0.08206) (1123 K)

P 5.00 atm



Chapt. 10.5

Sample Exercise:




PV = nRT V = nRT = 0.800 mol ( 0.08206) (1123 K)

P 5.00 atm

V = 14.7 L


Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures

Chapt. 10.6

John Dalton’s work with air allowed him to make the following observation:

total pressure of a mixture equals the sum of the pressures that each gas would exert if it were present alone



Chapt. 10.6

This equation implies that each gas behaves independently, so each gas has a unique mole quantity, and total moles equals sums of each individual amount...

nt = n1 + n2 + n3 + . . .



Chapt. 10.6

Sample Exercise:

What is the total pressure exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?

1) Solve for H2

PV = nRT



Chapt. 10.6

Sample Exercise:


2) Solve for H2 PV = nRT

P = X

V = 10.0 L

n = 2.00 g H2 1 mol H2 = + 8.00 g N2 1 mol N2 = 2 g H2 28 g N2

R = 0.08206 L-atm/mol-K

T = 273 K



Chapt. 10.6

Sample Exercise:


3) Solve for H2 PV = nRT

P = X P = nRT

V = 10.0 L V

n = 1.28 moles total = 1.28 mol (0.08206) ( 273 K) 10.0 L

R = 0.08206 L-atm/mol-K = 2.88 atm

T = 273 K



Chapt. 10.6

Sample Exercise:


4) Solve for N2

PV = nRT



Chapt. 10.6

Sample Exercise:


5) Solve for N2 PV = nRT

P = X

V = 10.0 L

n = 8.00 g N2 1 mol N2

28 g N2

R = 0.08206 L-atm/mol-K

T = 273 K



Chapt. 10.6

Sample Exercise:


6) Solve for N2 PV = nRT

P = X P = nRT

V = 10.0 L V

n = 8.00 g N2 1 mol N2 = 0.286 mol (0.08206) ( 273 K) 28 g N2 10.0 L

R = 0.08206 L-atm/mol-K = 0.641 atm

T = 273 K



Chapt. 10.6

Sample Exercise:


7) Sum individual pressures to identify total pressurePt = P(H2) + P(N2)

= 2.24 atm + 0.641 atm

= 2.88 atm



Chapt. 10.6

Partial pressures and Mole Fractions

- each gas behaves independently and so it is easy to relate the amount of a given gas to its partial pressure

- since P = nRT then P1 = n1RT and P1 = n1

V V Pt nt

Pt = ntRT

V

- the ratio n1/nt is denoted the mole fraction of gas 1

- mole fractions are unitless values expressing ratio



Chapt. 10.6

Sample Exercise:

From data gathered by Voyager 1, scientists have estimated the composition of the atmosphere of Titan, Saturn’s largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol % CH4. Calculate the partial pressure of each gas.

1) Solve for N2



Chapt. 10.6

Sample Exercise:


2) Solve for N2

Total pressure = 1220 torr

Mole fraction = 82/100 = 0.82

PN = 0.82 (1220 torr) = 1.0 x 103 torr



Chapt. 10.6

Sample Exercise:


3) Solve for Ar



Chapt. 10.6

Sample Exercise:


4) Solve for Ar


Mole fraction = 12/100 = 0.12

PAr = 0.12 (1220 torr) = 150 torr



Chapt. 10.6

Sample Exercise:


5) Solve for CH4



Chapt. 10.6

Sample Exercise:


2) Solve for CH4


Mole fraction = 6.0/100 = 0.060

PCH4 = 0.060(1220 torr) = 73 torr



Chapt. 10.6

Collecting gases over water

- when determining moles of produced gases, the best way to collect a gas sample is over water



Chapt. 10.6

Collecting gases over water

- when determining moles of produced gases, the best way to collect a gas sample is over water

- the volume of gas is measured by raising and lowering collecting container until water levels are equal inside and out…when this occurs, atmospheric pressures are equal

- total pressure inside is equal to sum of pressure of gas collected and pressure of water vapor

Ptotal = Pgas + PH2O



Chapt. 10.6

Sample exercise:

When a sample of NH4NO2 is decomposed in a test tube, 511 mL of N2 gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH4NO2 were decomposed?

1) Calculate partial pressure of N2



Chapt. 10.6

Sample exercise:



Pt = PN + Pwater

Pt = 745 torr

Pwater = 25.21 torr



Chapt. 10.6

Sample exercise:



Pt = PN + Pwater

Pt = 745 torr

Pwater = 25.21 torr

745 torr = X + 25.21 torr



Chapt. 10.6

Sample exercise:



Pt = PN + Pwater

Pt = 745 torr

Pwater = 25.21 torr

745 torr = X + 25.21 torr

PN = 720. torr



Chapt. 10.6

Sample exercise:


2) Use Ideal Gas Equation to determine moles of N2



Chapt. 10.6

Sample exercise:



PV = nRT



Chapt. 10.6

Sample exercise:



PV = nRT P = 720. Torr

V = 0.511 L

n = X

R = 62.36 L-torr/mol-K

T = 26°C = 299 K



Chapt. 10.6

Sample exercise:



PV = nRT n = PV P = 720. torr RT

V = 0.511 L = 720. torr(0.511 L)

n = X 62.36 (299 K)

R = 62.36 L-torr/mol-K

T = 26˚C = 299 K



Chapt. 10.6

Sample exercise:



PV = nRT n = PV P = 720. Torr RT

V = 0.511 L = 720. torr(0.511 L)

n = X 62.36 (299 K)

R = 62.36 L-torr/mol-K = 0.0197 mol N2

T = 26˚C = 299 K



Chapt. 10.6

Sample exercise:


3) Convert from moles of N2 to g NH4NO2



Chapt. 10.6

Sample exercise:



0.0197 mol N2 1 mol NH4NO2 64.0 g NH4NO2

1 mol N2 1 mol NH4NO2



Chapt. 10.6

Sample exercise:



0.0197 mol N2 1 mol NH4NO2 64.0 g NH4NO2

1 mol N2 1 mol NH4NO2

= 1.26 g NH4NO2


Kinetic Molecular TheoryKinetic Molecular Theory

Chapt. 10.7

Ideal gas equation describes how gases behave, but not why they do . . .

Kinetic molecular theory explains why

- gases consist of large numbers of molecules that are in continuous, random motion

- volume of the gas molecules themselves is negligible when compared to the volume of the gas as a whole

- attractive and repulsive forces between gases molecules is negligible



Chapt. 10.7

Ideal gas equation describes how gases behave, but not why they do . . .

Kinetic molecular theory explains why

- energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature is remaining constant -

i.e. collisions are elastic

- average kinetic energy is proportional to the absolute temperature



Chapt. 10.7

This theory is helpful in explaining the pressure and temperature at a molecular level:

pressure of gas is caused by the collisions of themolecules against the wallof a container

magnitude of pressureis determined by bothhow often and howforcefully the moleculesstrike the wall



Chapt. 10.7

This theory is helpful in explaining the pressure and temperature at a molecular level:

absolute temperature ofa gas is the measure ofthe average kinetic energyof its molecules

molecular motionincreases withincreasing temperature

this is on average, individual moleculeshave individual speeds



Chapt. 10.7

Applications of the Gas Laws

1. Effect of a volume increase at constant temp

- molecules must move a longer distance between collisions

- fewer collisions per unit time with wall of container

- pressure decreases



Chapt. 10.7

Applications of the Gas Laws

2. Effect of a temperature increase at constant volume

- increase in speed and u

- more collisions per unit time with wall of container

- pressure increases


Molecular Effusion and DiffusionMolecular Effusion and Diffusion

Chapt. 10.8

According to the kinetic-molecular theory, the average kinetic energy of any collection of gases,

u = 3RT R = gas constant

M T = temp

M = molar mass



Chapt. 10.8

has a specific value of u at any given temperature

-two gases at same temp have same avg. kinetic energy

-if masses are different, than the speed of particles will be different because of the inclusion of M in formula



Chapt. 10.8

The dependence of speed on mass, has several implications:

Effusion - the escape of a molecule through a tiny hole

Effusion rate is inversely proportional to the square root of molar mass

Normally rates are compared as a ratio



Chapt. 10.8

The dependence of speed on mass, has several implications:

Diffusion - spreading of one substance throughout a space or through another substance

Diffusion is also related to size of particle, however, molecule collisions make diffusion much more difficult

- average distance traveled by a molecule as it diffuses is called the mean free path, which varies with pressure


Deviations from an Ideal GasDeviations from an Ideal Gas

Chapt. 10.9

Although the ideal gas equation is useful, all real gases fail to obey the relationships to some degree

- deviation from an ideal gas occurs most at high pressure and low temperature

- to ensure as much compliance as possible to the ideal gas equation, a real gas should be considered when it is at high temperatures and low pressure

WHY?

Real gases DO - have molecular attractions

- lose energy when they collide

- have volume



Chapt. 10.9



Chapt. 10.9

Van der Waals’ Equation: takes into account volumeand molecular attraction

* constants (a) and (b) are different for each gas and must be identified using a table

P = nRT - n2a V - nb V2


Characteristics of gasesCharacteristics of gases PressurePressure Boyle’s, Charles’s, & Avogadro’s LawsBoyle’s, Charles’s, & Avogadro’s Laws Ideal gas equationIdeal gas equation Gas densities and Molar massGas densities and Molar mass Dalton’s Law of partial pressuresDalton’s Law of partial pressures Mole fractionsMole fractions Kinetic Molecular theoryKinetic Molecular theory Effusion, Diffusion,Effusion, Diffusion, u u and mean free path and mean free path Deviations from ideal gas behaviorDeviations from ideal gas behavior Van der Waals equationVan der Waals equation

Chapter Ten; ReviewChapter Ten; Review

chem 106, prof. t. l. heise 1 che 106: general chemistry chapter ten copyright © tyna l. heise...

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