Chapter 5 The Gaseous State. Copyright © Houghton Mifflin Company. All rights reserved.5 | 2...

Chapter 5The Gaseous

State

Copyright © Houghton Mifflin Company. All rights reserved. 5 | 2

Contents and Concepts

Gas LawsWe will investigate the quantitative relationships that describe the behavior of gases.

1. Gas Pressure and Its Measurement2. Empirical Gas Laws3. The Ideal Gas Law4. Stoichiometry Problems Involving Gas Volumes5. Gas Mixtures; Law of Partial Pressures


Kinetic-Molecular Theory

This section will develop a model of gases as molecules in constant random motion.

6. Kinetic Theory of Gases

7. Molecular Speeds; Diffusion and Effusion

8. Real Gases


Gases are compressible.

Gas volume depends on temperature and pressure.

Pressure, volume, temperature, and amount of a gas are related by the ideal gas law:

PV = nRT


Pressure, P

The force exerted per unit area

It can be given by two equations:

The SI unit for pressure is the pascal, Pa.

(pascal)Pasm

kgm

s

m

m

kg223

dgh P A

FP

(pascal)Pasm

kg

ms

mkg

22

2

Pressure exerted by a column of fluid


Volume: ( =height, =area)

Mass: ( =density)

Force:

( =acceleration due to gravity)

Pressure: / ( ) /

V hA h A

m Vd d

F ma Vdg hAdg

g

P F A hAdg A hdg



A barometer is a device for measuring the pressure of the atmosphere.

A manometer is a device for measuring the pressure of a gas or liquid in a vessel.


The water column would be higher because its density is less by a factor equal to the density of mercury to the density of water.

OHOHHgHg 22hgdhgd

gdhP

OH

HgHgOH

2

2

d

dhh


Empirical Gas Laws

All gases behave quite simply with respect to temperature, pressure, volume, and molar amount. By holding two of these physical properties constant, it becomes possible to show a simple relationship between the other two properties.

The studies leading to the empirical gas laws occurred from the mid-17th century to the mid-19th century.


Boyle’s Law

The volume of a sample of gas at constant temperature varies inversely with the applied pressure.

The mathematical relationship:

In equation form:

PV

1

ffii

constant

VPVP

PV


Figure A shows the plot of V versus P for 1.000 g O2 at 0°C. This plot is nonlinear.

Figure B shows the plot of (1/V) versus P for 1.000 g O2 at 0°C. This plot is linear, illustrating the inverse relationship.


At one atmosphere the volume of the gas is 100 mL. When pressure is doubled, the volume is halved to 50 mL. When pressure is tripled, the volume decreases to one-third, 33 mL.


When a 1.00-g sample of O2 gas at 0C is placed in a containerat a pressure of 0.50 atm, it occupies a volume of 1.40 L.

When the pressure on the O2 is doubled to 1.0 atm, the volume is reduced to 0.70 L, half the original volume.

?


A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant?

Vi = 38.7 mLPi = 751 mmHgTi = 21°C

Vf = ?Pf = 359 mmHgTf = 21°C

PV

PV = PV VPi i

i i f f ff


Vi = 38.7 mLPi = 751 mmHgTi = 21°C

Vf = ?Pf = 359 mmHgTf = 21°C

mmHg)(359

mmHg)mL)(751(38.7f V

= 81.0 mL(3 significant figures)

f

iif P

VPV


A graph of V versus T is linear. Note that all lines cross zero volume at the same temperature, -273.15°C.


The temperature -273.15°C is called absolute zero. It is the temperature at which the volume of a gas is hypothetically zero.

This is the basis of the absolute temperature scale, the Kelvin scale (K).

When working with gas laws (and almost anything related to thermochemistry), the first thing you should do is convert the temperature to Kelvin


Charles’s Law

The volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K).


In equation form:

TV

f

f

i

i

constant

T

V

T

VT

V


A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume.

As the air inside warms, the balloon expands to its orginial size.


A 1.0-g sample of O2 at a temperature of 100 K and a pressure of 1.0 atm occupies a volume of 0.26 L.

When the absolute temperature of the sample is raised to 200 K, the volume of the O2 is doubled to 0.52 L.

?


You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your calculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many milliliters of gas would you obtain at 27°C?

Vi = 79.4 mLPi = 760 mmHgTi = 0°C = 273 K

Vf = ?Pf = 760 mmHgTf = 27°C = 300. K

i f

i f

V V T V= V

T T Tf i

fi


Vi = 79.4 mLPi = 760 mmHgTi = 0°C = 273 K

Vf = ?Pf = 760 mmHgTf = 27°C = 300. K

K)(273

mL)K)(79.4(300.f V

= 87.3 mL(3 significant figures)

i

iff T

VTV


Combined Gas Law

The volume of a sample of gas at constant pressure is inversely proportional to the pressure and directly proportional to the absolute temperature.


In equation form:

P

TV

f

ff

i

ii

constant

T

VP

T

VPT

PV

?


Divers working from a North Sea drilling platform experience pressure of 5.0 × 101 atm at a depth of 5.0 × 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C?

Vi = 5.0 LPi = 5.0 × 101 atmTi = 4°C = 277 K

Vf = ?Pf = 1.0 atmTf = 11°C = 284 K


Vi = 5.0 LPi = 5.0 × 101 atmTi = 4°C = 277 K

Vf = ?Pf = 1.0 atmTf = 11°C = 284. K

fi

iiff PT

VPTV

atm)K)(1.0(277

L)atm)(5.010xK)(5.0(284 1

f V

= 2.6 x 102 L(2 significant figures)


Avogadro’s Law

Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules.


Standard Temperature and Pressure (STP)

The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure.

The molar volume, Vm, of a gas at STP is 22.4 L/mol.

The volume of the yellow box is 22.4 L. To its left is a basketball.

Ideal Gas LawP1V1=P2V2 or

PV=constant (at constant n and T)

V1/T1=V2/T2 or

V/T=constant (at constant n and P)

Avogadro’s Law V1/V2=n1/n2 or

V/n=constant (ant constant P and T)

PV/nT=constant=R

PV=nRT



Ideal Gas Law

The ideal gas law is given by the equation

PV=nRT

The molar gas constant, R, is the constant of proportionality that relates the molar volume of a gas to T/P.

?


A 50.0-L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder?

V = 50.0 LP = 17.1 atmT = 23°C = 296 K

RT

PVn

K)(296Kmol

atmL0.08206

L)atm)(50.0(17.1

n

mass = 986 g(3 significant figures)mol

g28.01mol35.20mass


mnM PMn P m dRTPV = nRT or d = = M

V RT V V RT Pm

mor

Gas Density and Molar Mass

Using the ideal gas law, it is possible to calculate the moles in 1 L at a given temperature and pressure. The number of moles can then be converted to density (grams per liter).

To find molar mass, find the moles of gas, and then find the ratio of mass to moles.

?


What is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm?

Mm = 16.04 g/molP = 3.50 atmT = 125°C = 398 K

RT

PMd m

K)(398Kmol

atmL0.08206

atm))(3.50mol

g(16.04

dfigures)tsignifican(3L

g1.72d

?


A 500.0-mL flask containing a sample of octane (a component of gasoline) is placed in a boiling water bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0°C. The mass of the vapor required to fill the flask is 1.57 g. What is the molar mass of octane? (Note: The empirical formula of octane is C4H9.) What is the molecular formula of octane?


P = 634 mmHg = 0.8342 atm

P

dRTM m

atm)(0.8342

K368.2Kmol

atmL0.08206

L

g3.140

m

M

figures)tsignifican(3mol

g114mM

d = 1.57 g/0.5000 L = 3.140 g/L

T = 95.0°C = 368.2 K


2

mol

g57

mol

g114

n

Molecular formula: C8H18

Molar mass = 114 g/molEmpirical formula: C4H9

Empirical formula molar mass = 57 g/mol


Stoichiometry and Gas Volumes

Use the ideal gas law to find moles from a given volume, pressure, and temperature, and vice versa.

?


When a 2.0-L bottle of concentrated HCl was spilled, 1.2 kg of CaCO3 was required to neutralize the spill. What volume of CO2 was released by the neutralization at 735 mmHg and 20.°C?


First, write the balanced chemical equation:

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

3

2

3

33

3

CaCOmol1

COmol1

CaCOg100.09

CaCOmol1CaCOg10x1.2

Moles of CO2 produced = 11.99 mol

Second, calculate the moles of CO2 produced:

Molar mass of CaCO3 = 100.09 g/mol


n = 12.0 molP = 735 mmHg = 0.967 atmT = 20°C = 293 K

P

nRTV

atm)(0.967

K)(293Kmol

atmL0.08206mol12.0

V

= 3.0 × 102 L(2 significant figures)


Gas Mixtures

Dalton found that in a mixture of unreactive gases each gas acts as if it were the only the only gas in the mixture as far as pressure is concerned.


Originally (left), flask A contains He at 152 mmHg and flask B contains O2 at 608 mmHg. Flask A is then filled with oil forcing the He into flask B (right). The new pressure in flask B is 760 mmHg


Partial Pressure

The pressure exerted by a particular gas in a mixture

Dalton’s Law of Partial Pressures

The sum of the partial pressures of all the different gases in a mixture is equal to the total pressure of the mixture:

P = PA + PB + PC + . . .

?


A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample?


mL10

L1mL100.0

K308Kmol

atmL0.08206

Ng28.01

Nmol1Ng0.0830

3

2

22

N2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

Og32.00

Omol1Og0.0194

3

2

22

O2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

COg44.01

COmol1COg0.00640

3

2

22

CO2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

OHg18.01

OHmol1OHg0.00441

3

2

22

OH2P

atm0.749

atm0.153

atm0.0368

atm0.0619


atm0.7492N P

atm0.1532O P

atm0.03682CO P

atm0.0619OH2P

OHCOON 2222PPPPP

P = 1.00 atm

?


The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570.0 mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air?

mmHg40.02CO P

Hg103.02O P

mmHg570.02N P

mmHg47.0OH2P


OHCOON 2222PPPPP

570.0 mmHg103.0 mmHg

40.0 mmHg47.0 mmHg

P = 760.0 mmHg


Mole fraction of N2

Mole fraction of H2OMole fraction of CO2

Mole fraction of O2

mmHg760.0

mmHg47.0

mmHg760.0

mmHg40.0

mmHg760.0

mmHg103.0

mmHg760.0

mmHg570.0

Mole fraction N2 = 0.7500

Mole fraction O2 = 0.1355

Mole fraction CO2 = 0.0526

Mole fraction O2 = 0.0618


Collecting Gas Over Water

Gases are often collected over water. The result is a mixture of the gas and water vapor.

The total pressure is equal to the sum of the gas pressure and the vapor pressure of water.

The partial pressure of water depends only on temperature and is known (Table 5.6).

The pressure of the gas can then be found using Dalton’s law of partial pressures


The reaction of Zn(s) with HCl(aq) produces hydrogen gas according to the following reaction:

Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

The next slide illustrates the apparatus used to collect the hydrogen. The result is a mixture of hydrogen and water vapor.



5.6)Table(See

mmHg16.5PC,19At

mmHg769

OH2

P

mmHg16.5mmHg7692

22

22

H

OHH

OHH

P

PPP

PPP

mmHg752.52H P

places)decimal(no

mmHg7532H P

?


You prepare nitrogen gas by heating ammonium nitrite:

NH4NO2(s) N2(g) + 2H2O(l)

If you collected the nitrogen over water at 23°C and 727 mmHg, how many liters of gas would you obtain from 5.68 g NH4NO2?

Molar mass NH4NO2

= 64.05 g/mol

P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 K


P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 K P

nRTV

Molar mass NH4NO2

= 64.04 g/mol

3

2

3

33 CaCOmol1

COmol1

CaCOg64.04

CaComol1CaCOg5.68

= 0.8869 mol CO2 gas


P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 Kn = 0.8869 mol

P

nRTV

mmHg760

atm1mmHg706

K)(296Kmol

atmL0.08206mol0.0887

V

= 2.32 L of CO2

(3 significant figures)

Kinetic-Molecular Theory (Kinetic Theory)

A theory, developed by physicists, that is based on the assumption that a gas consists of molecules in constant random motion.

Kinetic energy is related to the mass and velocity:

m = massv = velocity

Program demonstrating kinetic-molecular theory:http://people.chem.byu.edu/rbshirts/research/boltzmann_3d


2K 2

1mvE


Postulates of the Kinetic Theory1. Gases are composed of molecules whose

sizes are negligible.2. Molecules move randomly in straight lines in all

directions and at various speeds.3. The forces of attraction or repulsion between

two molecules (intermolecular forces) in a gas are very weak or negligible, except when the molecules collide.

4. When molecules collide with each other, the collisions are elastic.

5. The average kinetic energy of a molecule is proportional to the absolute temperature.


An elastic collision is one in which no kinetic energy is lost. The collision on the left causes the ball on the right to swing the same height as the ball on the left had initially, with essentially no loss of kinetic energy.


P

Each of the gas laws can be derived from the postulates.

For the ideal gas law:

frequency of collision x average force


muNV

uP

1

The average force depends on the mass of the molecules, m, and its average speed, u; it depends on momentum, mu.

The frequency of collision is proportional to the average speed, u, and the number of molecules, N, and inversely proportional to the volume, V.


Rearranging this relationship gives

The average kinetic energy of a molecule of mass m and average speed u is 1/2mu2.

Thus PV is proportional to the average kinetic energy of the molecule.

2NmuPV


However, the average kinetic energy is also proportional to the absolute temperature and the number of molecules, N, is proportional to moles of molecules. We now have

Inserting the proportionality constant, R, gives

nTPV

nRTPV


Molecular Speeds

According to kinetic theory, molecular speeds vary over a wide range of values. The distribution depends on temperature, so it increases as the temperature increases.

Root-Mean Square (rms) Molecular Speed, u

A type of average molecular speed, equal to the speed of a molecule that has the average molecular kinetic energy

mM

RTu

3


When using the equation

R = 8.3145 J/(mol K)

T must be in kelvins

Mm must be in kg/mol

?


What is the rms speed of carbon dioxide molecules in a container at 23°C?

mM

RTu

3rms

T = 23°C = 296 KCO2 molar mass =

0.04401 kg/mol


mol

kg0.04401

K296Kmol

s

mkg

8.314532

2

rmsu

2

25

rmss

m1.68x10u

s

mx104.10 2

rms u

2

2

s

mkgJ

Recall

?


Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor?

Hydrogen will diffuse more quickly by a factor of 1.4.

2.016

4.002

4.002

12.016

1

HeRate

HRate 2



a. He will reach the end first because it has a smaller molar mass.

b. Open the valves at two different times, allowing Ar more time by a factor equal to the square root of the ratio of molar masses of Ar to He, or approximately 3.16 times longer.


Maxwell predicted the distributions of molecular speeds at various temperatures. The graph shows 0°C and 500°C.

3/ 22 2( ) 4 exp( / 2 )

2

Mf u u Mu RT

RT


Diffusion

The process whereby a gas spreads out through another gas to occupy the space uniformly

Below NH3 diffuses through air. The indicator paper tracks its progress.


Effusion

The process by which a gas flows through a small hole in a container. A pinprick in a balloon is one example of effusion.


Graham’s Law of Effusion

At constant temperature and pressure, the rate of effusion of gas molecules through a particular hole is inversely proportional to the square root of the molecular mass of the gas.

mM

1moleculesofeffusionofrate


Real Gases

At high pressure the relationship between pressure and volume does not follow Boyle’s law. This is illustrated on the graph below.


At high pressure, some of the assumptions of the kinetic theory no longer hold true:

1. At high pressure, the volume of the gas molecule (Postulate 1) is not negligible.

2. At high pressure, the intermolecular forces (Postulate 3) are not negligible.


Van der Waals Equation

An equation that is similar to the ideal gas law, but which includes two constants, a and b, to account for deviations from ideal behavior

The term V becomes (V – nb).

The term P becomes (P + n2a/V2).

Values for a and b are found in Table 5.7

nRTnbVV

anP

2

2


Other Resources

Visit the student website at http://www.college.hmco.com/pic/ebbing9e

http://www.college.hmco.com/pic/ebbing9e

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