05 gases 6.pdf

(1)

CHAPTER 11: GASES

Solids

Liquids

Gases

particles that are

close together in a

fixed arrangement

particles that are

close together but

mobile

particles that are

far

apart

definite volume

an indefinite volume

(same volume as

container)

Maintains shape

indefinite shape

(same shape as

container)

indefinite shape

(same shape as

container)

particles that move

very slowly

particles that move

slowly

particles that move

very fast

Gases are compressible

when we apply a force

Gases are compressible

because there is so

much empty space

between gas particles.

Applying a force pushes

the particles into the

empty space,

compressing the gas.

Liquids are not very compressible because

there is little empty space between particles.

How do Gases Differ From Liquids and Solids?

1)

A sample of gas assumes both the shape and volume of the

container.

2)

Gases are compressible.

3)

The densities of gases are much smaller than those of liquids

and solids and are highly variable depending on temperature

and pressure.

4)

Gases form homogeneous mixtures (solutions) with one

The Kinetic Molecular Theory Helps us

Understand the Behavior of Gases

Gas Laws

describe

HOW

gases behave. The

kinetic molecular

theory (KMT)

describes

WHY

gases behave as they do.

1. A gas consists of small particles

that move

randomly

and

rapidly

.

2. Gas particles are in constant

(2)

The

Kinetic Molecular Theory (KMT)

Describes

WHY

Gases Behave As They Do

3. A gas is composed of particles that

are separated by large distances. The

size of gas particles is so small

compared to the space between the

particles that the volume occupied by

a gas molecule is negligible. Most of

the volume of a gas is empty space.

Gases have low densities (d = m/V):

Density H2O(l) at 25°C = 0.997 g/mL

Density H2O(g) at 25°C = 0.000737 g/mL

The water molecules are 1353 times further apart in the gaseous state!

The

Kinetic Molecular Theory (KMT)

Describes

WHY

4. Gas particles exert

no

attractive or repulsive

forces

on one another.

Gas particles are far

apart and fill a

container of any size

and shape.

He atom He atom

Repulsive forces Attractive forces

The

Kinetic Molecular Theory (KMT)

Describes

WHY

5. When gas particles

collide

with each other, they

rebound

and travel in new

directions. Collisions of

the gas particles are

elastic. Thus, the total

kinetic energy of the gas

particles is constant at a

constant temperature.

Elastic

collision Inelastic collision

k

E



T

The

Kinetic Molecular Theory (KMT)

Describes

WHY

6. The average

kinetic energy

of a collection of gas

particles is directly

proportional

to

the

Kelvin

temperature

of the gas. The kinetic energy of

gas particles increases with increasing

temperature.

KE

( )

avg

=

3

2

RT

The

Kinetic Molecular Theory

(

KMT

) Describes

WHY

Gases are compressible because molecules in the gas phase are

separated by large distances (assumption 1).

Pressure is the result of the

collisions of gas molecules with

the walls of their container

(assumption 2).

Decreasing volume increases

the frequency of collisions.

Pressure increases as collision

frequency increases.

Molecular (Root-Mean-Square) Speed,

u

rms

The total kinetic energy of a mole of gas is equal to:

R

= 8.314 J/K mol

The average kinetic energy of one molecule is:

For one mole of gas:

RT

3 2

mu2

1

2 u2

m is the mass

is the mean square speed

A

N_1mu2 _ 3RT

2 2 rearrange and Take the square root

(m x NA = M)



M

rms

RT

(3)

Molecular (Root-Mean-Square) Speed, u

rms

The

root-mean-square (rms) speed (

u

rms

)

is the speed of a molecule

with the average kinetic energy in a gas sample.

u

rms is directly proportional to temperature



M

rms

RT

u 3

Molecular (Root-Mean-Square) Speed, u

rms

The

root-mean-square (rms) speed (

u

rms

)

is the speed of a molecule

with the average kinetic energy in a gas sample.

u

_rms

is inversely proportional to the square root of M.



M

rms

RT

u 3

Molecular (Root-Mean-Square) Speed, urms

When two gases are at the same temperature, it is possible

to compare the the u

rms

values of the different gases.

( ) ( )

rms rms

u u 

1 2

2 1

M M

Worked Example 11.1

Strategy Use equation and the molar masses of He and CO2 to

determine the ratio of their root-mean-square speeds. When solving a problem such as this, it is generally best to label the lighter of the two molecules as molecule 1 and the heavier as molecule 2. This ensures that the result will be greater than 1, which is relatively easy to interpret.

Determine how much faster a helium atom moves, on average, than a carbon dioxide molecule at the same temperature.

Solution The molar masses of He and CO2 are 4.003 and 44.02 g/mol,

respectively.

On average, He atoms move 3.316 times as fast as CO2 molecules at the same

temperature.

( ) ( )

rms rms

u u 

1 2

2 1

M M

44.02 g 1 mol _{= 3.316} 4.003 g

1 mol

urms(He)

urms(CO2) =

Think About It Remember that the relationship between molar

mass and molecular speed is reciprocal. A CO2 molecule has

approximately 10 times the mass of an He atom. Therefore, we should expect an He atom, on average, to be moving approximately √10 times (~3.2 times) as fast as a CO2 molecule.

Diffusion and Effusion

Diffusion

is the mixing of gases as the result of random motion and

frequent collisions.

Effusion

is the escape of gas molecules from a container to a region

of vacuum.

Lighter Gases Diffuse and Effuse More Rapidly

Than Heavier Gases

Graham’s law

states that the rate of diffusion or effusion of a

gas is inversely proportional to the square root of its molar mass.

 1

M

(4)

Problems Using

u

rms

Equation

Determine the molar mass and identity of a diatomic gas

that moves 4.67 times as fast as CO2.

Solution:

Step 1:

Use the equation below to determine the molar mass of the

unknown gas:

u

rms

(unknown gas) = 4.67 x

u

rms

(CO

2

)

( ) ( )

rms rms

u u 

1 2

2 1

M M





( )

u

u   M

rms rms 2

unknown _4.67 44.02 g/mol

CO unknown

M = 2.018 g/mol The gas must be H2.

When Gas Particles Collide with the

Walls of a Container, they Exert a

Pressure

(

P

) is the

force

(

F

) exerted per unit

area

(

A

).

Pressure =

Force

=

F

Area

A

The SI unit of force is the newton (N), where 1 N = 1 kg•m/s2

The SI unit of pressure is the pascal (Pa),

defined as 1 newton per square meter. 1 Pa = 1 N/m2

Atmospheric

Pressure

The pressure exerted by a column of air

from the top of the atmosphere to the

surface of the earth is about

1 atmosphere at sea level.

Atmospheric Pressure

A

barometer

is a device used to

measure atmospheric pressure.

Atmospheric pressure is expressed in

terms of the height of the barometer’s

mercury column, usually in

millimeters of mercury

(mm Hg).

1 mm Hg = 1 torr

1 atm

=

760 mmHg

=

760 torr

= 14.7 psi = 101.325 kPa

A Straw Works Like a

Barometer

When a straw is put into a

glass of orange soda, the

pressure inside and outside of

the straw are the same

, so the

liquid levels inside and outside

of the straw are the same

.

When a person sucks on the

straw, the pressure inside the

straw is lowered

. The greater

pressure on the surface of the

liquid outside of the straw

pushes the liquid up the straw.

Limitations of a Straw

Even if you formed a perfect vacuum,

atmospheric pressure could only

push orange soda to a total height of

about

10 m

(about

34 feet

).

This is because a column of water

10 m high exerts the same pressure

(14.7 lb/in

2

) as the gas molecules in

(5)

Measurement of Pressure

A

manometer

is a device used to measure pressures other than

atmospheric pressure.

Blood Pressure

A

sphygmomanometer

is a device

used to measure blood pressure.

Blood pressure is expressed in

millimeters of mercury (mm Hg)

above

atmospheric pressure.

1 atm = 760 mm Hg

120

80

=

120

+

760

mm Hg

80

+

760

mm Hg

=

880

mm Hg

840

mm Hg

=

1

.

16

atm

1

.

11

atm

Worked Example 11.2

Strategy Use P = hdg to calculate pressure. Remember that the height must be expressed in meters and density must be expressed in kg/m3_.

Calculate the pressure exerted by a column of mercury 70.0 cm high. Express the pressure in pascals, in atmospheres, and in bars. The density of mercury is 13.5951 g/cm3_.

Solution

h = 70.0 cm × = 0.700 m

d =

g = 9.80665 m/s2 1 m 100 cm

13.5951 g

cm3 × _{1000 g}1 kg × 100 cm _{1 m}

3

= 1.35951×104_kg/m3

Worked Example 11.2 (cont.)

Solution

pressure = 0.700 m × × = 9.33×104_kg/m∙s2

9.33×104_Pa

9.33×104_Pa× = 0.921 atm

0.921 atm × = 0.933 bar 1.35951×104_kg

m3 9.80665 m _s2

1 atm 101,325 Pa

1.01325 bar 1 atm

Think About It Make sure your units cancel properly in this type of problem. Common errors include forgetting to express height in meters and density in kg/m3_{. You can avoid these errors by becoming familiar with the value of}

atmospheric pressure in the various units. A column of mercury slightly less than 760 mm is equivalent to slightly less than 101,325 Pa, slightly less than 1 atm, and slightly less than 1 bar.

Inverse

Relationship versus

Direct

Relationship

When a change in one property causes a change in another

property, the two properties are related.

If the changes occur in opposite directions, the properties

have an

inverse

relationship

. If one quantity

increases

, the

other quantity

decreases

.

A

direct relationship

is one in which the related properties

increase or decrease together.

Pressure and Volume: Boyle

’

s Law

(a) (b) (c)

P (mmHg) 760 1520 2280

V (mL) 100 50 33

V

P



1

(6)

Pressure and Volume: Boyle

’

s Law

For a fixed amount of gas at constant temperature, the

pressure

and

volume

of the gas are

inversely

related.

When one quantity

increases

, the other

decreases

.

V

µ

1

P

1

´

V

1

=

P

2

´

V

2

or

If the

volume

of a

cylinder of gas is

halved

,

the

pressure

of the

gas inside the

cylinder

doubles

.

Boyle’s Law and Breathing:

Inhalation

During

inhalation

,

•

The

rib cage expands

and

the diaphragm lowers.

•

This

increases the volume

of the lungs.

•

Increasing the volume causes

the

pressure to decrease

.

•

Air is drawn into the lungs

to equalize the pressure.

Boyle’s Law and Breathing:

Exhalation

During

exhalation

,

•

The

rib cage contracts

and

the diaphragm is raised.

•

This

decreases the volume

of the lungs.

•

Decreasing the volume

causes the

pressure to

increase.

•

Air is expelled out of the

lungs

to equalize the

pressure.

Qualitative Boyle’s Law

A sample of helium gas in a balloon has a volume of 10. L

at a pressure of 0.90 atm. At 1.40 atm (

T

constant), is the

new volume represented by A, B, or C?

Qualitative Boyle’s Law

At a higher pressure (

T

constant), the new volume is

represented by the smaller balloon.

Using Boyle’s Law

to Calculate a New Gas Volume or Pressure

Calculate the volume of a sample of gas at 5.75 atm if it

occupies 5.14 L at 2.49 atm. (Assume constant

temperature.)

Solution:

Step 1:

Use the relationship below to solve for V

2

:

P

1

V

1

= P

2

V

2





2.49 atm 5.14 L



2.26 L

5.75 atm

P V

V

P

1 1 2

(7)

Worked Example 11.3

Strategy Use P1V1 = P2V2 to solve for V2.

If a skin diver takes a breath at the surface, filling his lungs with 5.82 L of air, what volume will the air in his lungs occupy when he dives to a depth where the pressure of 1.92 atm? (Assume constant temperature and that the pressure at the surface is exactly 1 atm.)

Solution P1 = 1.00 atm, V1 = 5.82 L, and P2 = 1.92 atm.

V2 = = = 3.03 L P1×_P V1

2

1.00 atm×5.82 L 1.92 atm

Think About It At higher pressure, the volume should be smaller. Therefore, the answer makes sense.

Boyle

’

s Law Calculations:

P

1

V

1

=

P

2

V

2

Freon-12, CCl

2

F

2

, is used in refrigeration systems. What is the new

volume (L) of an 8.0 L sample of Freon gas after its pressure is changed

from 550 mmHg to 200 mmHg at constant

T

?

A sample of oxygen gas has a volume of 12.0 L at 600 mmHg. What is

the new pressure when the volume changes to 36.0 L?

Temperature and Volume: Charles

’

s Law

Heating at constant pressure: volume increases Cooling at constant pressure:

volume decreases

Temperature and Volume: Charles

’

s Law

For a fixed amount of gas

at constant pressure

, the

volume

of

the gas is

proportional

to its

Kelvin

temperature

.

If one quantity

increases

, the other

increases

as well.

V

µ

T

V

1

T

1

=

V

2

T

2

or

Temperature and Volume: Charles

’

s Law

If the temperature of the cylinder is

doubled

,

the volume of the gas inside the cylinder

doubles

.

Hot Air is Less Dense Than Cool Air Causing a

Hot Air Balloon to Lift OFF

Less dense

inside balloon.

(8)

Charles’s Law can be used to explain

how wind currents form at the beach

A sample of gas originally occupies 29.1 L at 0.0

Charles’ Law Calculations

°

C. What is its

new volume when it is heated to 15.0

°

C? (Assume constant

pressure.)

Solution:

Step 1:

Use the relationship below to solve for V

2

: (Remember that

temperatures must be expressed in kelvin.

V

T



T

1 2

29.1 L 288.15 K

30.7 L

273.15 K

V

T

V

T





1 2



2 1

Charles

’

s Law Calculations:

A balloon has a volume of 785 mL at 21

°

C. If the

balloon is placed in an ice water bath (ie., the temperature

drops to 0

°

C), what is the new volume of the balloon (

P

constant)?

A sample of oxygen gas has a volume of 420 mL at a

temperature of 18

°

C. At what temperature, in

°

C, will

the volume of the oxygen be 640 mL (

P

and

n

constant)?

V

1

T

1

=

V

2

T

2

Temperature and Pressure: Gay-Lussac

’

s Law

Heating at constant volume: pressure increases Cooling at constant volume:

pressure decreases

Temperature and Pressure: Gay-Lussac

’

s Law

For a fixed amount of gas at constant volume,

the

pressure

of a gas is

proportional

to its

Kelvin temperature

.

If one quantity

increases

, the other

increases

as well.

P

µ

T

P

1

T

1

=

P

2

T

2

or

Increasing

the

temperature

increases

the

kinetic energy

of the gas particles,

causing

the

pressure

exerted by the particles

to increase

.

Gay-Lussac

’

s Law Calculations

The air in a tank has a pressure of 640 mm Hg at 23

°

C.

When placed in sunlight, the temperature rose to 48

°

C.

What was the pressure in the tank?

P

1

T

1

=

P

2

T

2

P

2

=

P

1

´

T

2

T

1

=

640

( )

321

296

=

694

mm

Hg

P

1

T

1

=

(9)

Gay-Lussac

’

s Law Calculations

A gas has a pressure of 2.0 atm at 18

°

C. What is the new

pressure when the temperature is 62

°

C?

(

V

and

n

constant)

A gas has a pressure of 645 mmHg at 128

°

C. what is the

temperature in

°

C if the pressure increases to 824 mmHg (

n

and

V

remain constant)?

P

1

T

1

=

P

2

T

2

Vapor Pressure and

Boiling Point

The

vapor pressure of

water

•

Is the pressure above

water at equilibrium in a

closed container

•

At the boiling point

is

equal to the external

(atmospheric)

pressure

Vapor Pressure and

Boiling Point

Vapor Pressure

•

Is the pressure of gas

molecules above the surface

of a liquid

•

At the boiling point

is

equal to the external

(atmospheric) pressure



Boiling Point of Water

•

Depends on the vapor

pressure

•

Is lower at higher altitudes

•

Is increased by using an

autoclave to increase

external pressure



bp in

Denver

PV

µ

T

P

1

V

1

T

1

=

P

2

V

2

T

2

or

Combined Gas Law

All

three gas laws

can be

combined

into one equation.

This equation is used for determining the

effect of

changing two factors

on the third factor.

Worked Example 11.6

Strategy In this case, because there is a fixed amount of gas, we use P1V1/T1 =

P2V2/T2. The only value we don’t know is V2. Temperatures must be expressed in

kelvins. We can use any units of pressure, as long as we are consistent. If a child releases a 6.25-L helium balloon in the parking lot of an amusement park where the temperature is 28.50°C and the air pressure is 757.2 mmHg, what will the volume of the balloon be when it has risen to an altitude where the temperature is -34.35°C and the air pressure is 366.4 mmHg?

Think About It Note that the solution is essentially multiplying the original volume by the ratio of P1 and P2, and by the ratio of T2 to T1. The effect of

decreasing external pressure is to increase the balloon volume. The effect of decreasing temperature is to decrease the volume. In this case, the effect of

Solution T1 = 301.65 K, T2 = 238.80 K.

V2 = = = 10.2 L P_P1T2V1

2T1

(10)

Combined Gas Law Calculations

A sample of carbon dioxide gas has a volume of 15.2 L at a

pressure of 1.35 atm and a temperature of 33

°

C. Determine its

volume at a temperature of 35

°

C and a pressure of 3.50 atm.

P

1

V

1

T

1

=

P

2

V

2

T

2

V

2

=

P

1

V

1

T

1

æ

è

ç

ö

ø

÷

T

2

P

2

æ

è

ç

ö

ø

÷ =

1.35

´

15.2

306

æ

è

ç

ö

ø

÷

308

3.50

æ

è

ç

ö

ø

÷ =

5.90

L

Combined Gas Law Calculations

A sample of helium gas has a volume of 0.180 L, a pressure of

0.800 atm, and a temperature of 29

°

C. At what temperature

in

°

C will the helium have a volume of 90.0 mL and a

pressure of 3.20 atm?

A gas has a volume of 675 mL at 35

°

C and 0.850 atm

pressure. What is the volume in mL of the gas at -95

°

C and a

pressure of 802 mmHg?

Volume and Moles: Avogadro

’

s Law

The presence of additional molecules

causes an increase in pressure.

’

s Law

When the pressure and temperature are held constant, the

volume

of a gas is

proportional

to the

number

of

moles

present.

If one quantity

increases

, the other

increases

as well.

V

µ

n

V

1

n

1

=

V

2

n

2

or

’

s Law

When you blow up a balloon, its volume increases because you are adding more air molecules.

If the balloon has a hole in it, air leaks out, causing its volume to

decrease.

V

µ

n

V

1

n

1

=

V

2

n

2

or

Worked Example 11.5

Strategy Apply Avogadro’s law to determine the volume of a gaseous product.

If we combine 3.0 L of NO and 1.5 L of O2, and they react according to the

balanced equation 2NO(g) + O2(g) → 2NO2(g), what volume of NO2 will be

produced? (Assume that the reactants and products are all at the same temperature and pressure.)

Solution Because volume is proportional to the number of moles, the balanced equation determines in what volume ratio the reactants combine and the ratio of product volume to reactant volume. The amounts of reactants given are stoichiometric amounts.

According to the balanced equation, the volume of NO2 formed will be equal to

the volume of NO that reacts. Therefore, 3.0 L of NO2 will form.

(11)

Avogadro

’

s Law Calculations

What volume in liters of water vapor will be produced when 34 L of

H

2

and 17 L of O

2

react according to the equation below:

2H

2

(

g

) + O

2

(

g

) → 2H

2

O(

g

)

Assume constant pressure and temperature.

Solution:

Step 1:

Because volume is proportional to the number of moles, the

balanced equation determines in what volume ratio the reactants

combine and the ratio of product volume to reactant volume. The

amounts of reactants given are stoichiometeric amounts.

34 L of H

2

O will form.

Avogadro

’

s Law Calculations

V

1

n

1

=

V

2

n

2

A 2.00 mol sample of a gas is known to occupy 30.0 L of

volume. What is the volume occupied by 1.50 mol?

V

2

=

V

1

n

1

æ

è

ç

ö

ø

÷

´

n

2

=

30.0

2.00

æ

è

ç

ö

ø

÷

(

1.50

)

=

22.5

L

V

1

n

1

=

V

2

n

2

Avogadro

’

s Law Calculations

If 0.75 mole of helium gas occupies a volume of 1.5 L,

what volume will 1.2 moles of helium occupy at the same

temperature and pressure?

V

1

n

1

=

V

2

n

2

Combining all the previous gas laws into one law gives you:

The Ideal Gas Law:

PV

=

nRT

PV

=

nRT

R

=

PV

nT

=

(22.4

L

)

×

(1.00

atm

)

(1.00

mole

)

×

(273

K

)

=

0.0821

L

×

atm

mole

×

K

R

is a constant, called the

ideal gas constant

.

1 mole

of any gas occupies

22.4 L

at STP (

273 K

and

1 atm

)

PV = nRT

Ideal Gas Law Calculations:

PV

=

nRT

How many moles of gas are contained

in a typical human breath that takes in

0.50 L of air at 1.0 atm pressure and 37

°

C?

Step [1]

Identify the

known quantities

and the

desired quantity

.

P

= 1.0 atm

V

= 0.50 L

T

= 37

o

C

(12)

Ideal Gas Law Calculations:

PV

=

nRT

Step [2]

Convert all values to proper units and

the value of

choose

R

that contains these units

.

•

The pressure is given in atm, so use the following

R

value:

R

= 0.0821 L • atm

mol • K

•

Temperature is given in

o

C, but must be in K

:

K =

o

C + 273

K = 37

o

C + 273

K = 310 K

PV

=

nRT

Step [3]

Write the equation and rearrange it to isolate

the desired quantity on one side.

PV

=

nRT

Solve for

n

by dividing both sides by

RT

.

PV

RT

=

n

Step [4]

Solve the problem.

PV

RT

n

=

(1.0 atm) (0.50 L)

L • atm

mol • K

(310 K)

0.0821

0.020 mol

Answer

=

PV

=

nRT

Calculate the volume, in liters, of 0.100 mol of O

2

gas at

0

°

C and 2.00 atm pressure.

PV

=

nRT

V

=

nRT

P

=

0.100

mol

(

)

(

0.0821

Latm

/

Kmol

)

(

273

K

)

2.00

atm

=

1.12

L

Ideal Gas Law Calculations:

PV

=

nRT

Dinitrogen monoxide (N

2

O), aka laughing gas, is used by

dentists as an anesthetic. If a 20.0 L tank of laughing gas

contains 2.86 moles of N

2

O at 23

°

C, what is the pressure in

mmHg inside the tank?

A cylinder contains 5.0 L of O

2

at 20.0

°

C and 0.85 atm.

How many grams of oxygen are in the cylinder?

Worked Example 11.7

Strategy Convert the temperature in °C to kelvins, and use the ideal gas equation to solve for the unknown volume.

Calculate the volume of a mole of ideal gas at room temperature (25°C) and 1 atm.

Think About It With the pressure held constant, we should expect the volume to increase with increased temperature. Room temperature is higher than the standard temperature for gases (0°C), so the molar volume at room temperature (25°C) should be higher than the molar volume at 0°C–and it is.

Solution The data given are n = 1 mol, T = 298.15 K, and P = 1.00 atm. Because the pressure is expressed in atmospheres, we use R = 0.08206 L∙atm/K∙mol in order to solve for volume in liters.

V = = = 24.5 L nRT _P (1 mol)(0.08206 L∙atm/K∙mol)(298.15 K) _{1 atm}

Molar Volume

Calculate the volume, in liters, of 1.000 mol of an ideal gas at

0.00

°

C and 1.000 atm pressure.

PV

=

nRT

V

=

nRT

P

=

1.000

(

)

(

0.08206

)

(

273.15

)

1.000

=

22.41

L

The volume of an ideal gas at STP is

22.41 L

,

known as the

molar volume

.

STP

0°C and 1 atm

(13)

Standard Temperature and Pressure, STP

The volume of gases can be compared at STP

when they have

•

The same temperature

Standard Temperature (

T

) = 0°C or 273 K

•

The same pressure

Standard Pressure (

P

) = 1 atm or 760 mmHg

1 mole

of any gas at STP =

22.4 L

Molar Volume:

Same volume (

22.4 L

),

Same # of particles (

1 mol

)

6.02 x 10

23

particles

How to Convert Moles of Gas to Volume at STP

1 mol gas

(at STP)

= 22.4 L

How many moles are contained in 2.0 L of N

2

at

standard temperature and pressure.

Step [1]

Identify the

quantity

.

known quantities

and the

desired

2.0 L of N

2

original quantity

desired quantity

? moles of N

2

Step [2]

Write out the conversion factors.

22.4 L

1 mol

or

22.4 L

1 mol

Choose this one to cancel L

Step [3]

Set up and solve the problem.

2.0 L x 1 mol

22.4 L =

0.089 mol N

2

Liters cancel

Answer

1 mol gas

(at STP)

= 22.4 L

What is the volume occupied by 2.75 moles of N

2

gas at STP?

What is the volume at STP of 4.00 g of CH

4

?

How many grams of He are present in 8.00 L of He at STP?

Which Formula Do I Use?

Law

Variables

Constants

Boyle

’

s

P

,

V

n, T

Charles

V

,

T

n, P

Gay-Lussac

P

,

T

n, V

Combined

P

,

V

,

T

n

Avogadro

V

,

n

P, T

(14)

Ideal Gas Law

If you can remember this one law, you can derive

(work out) all the others.

PV

=

nRT

Gather all the variables (the things that change)

on the left side of the equation.

Put all the others (constants) on the right side of

the equation.

Derivation of Boyle’s Law

PV

=

nRT

P and V are on the left side and all the others are on the right side.

P

1

´

V

1

=

P

2

´

V

2

Boyle’s law

PV

=

nRT

Constant

PV

T

=

nRT

T

Derivation of Charles’s Law

PV

=

nRT

Divide both sides by T…

Gather V and T on the left side and all the others are on the right side.

Cancel T on the right side

PV

T

=

nR

Divide both sides by P…

PV

PT

=

nR

P

Cancel P on the left side

V

T

=

nR

P

Derivation of Charles’s Law

Charles’s law

V

1

T

1

=

V

2

T

2

Constant

V

T

=

nR

P

PV

T

=

nRT

T

Derivation of Gay-Lussac’s Law

PV

=

nRT

Gather P and T on the left side and all the others are on the right side.

PV

T

=

nR

Divide both sides by V…

PV

VT

=

nR

V

Cancel V on the left side

P

T

=

nR

V

Gay-Lussac’s law

P

1

T

1

=

P

2

T

2

P

T

=

nR

V

Constant

(15)

PV

T

=

nRT

T

Derivation of Combined Gas Law

PV

=

nRT

Gather P, V, and T on the left side and all the others are on the right side.

PV

T

=

nR

PV

T

=

nR

Constant

Combined Gas law

P

1

V

1

T

1

=

P

2

V

2

T

2

PV

n

=

nRT

n

Derivation of Avogadro’s Law

PV

=

nRT

Divide both sides by n…

Gather V and n on the left side and all the others are on the right side.

Cancel n on the right side

PV

n

=

RT

Divide both sides by P…

PV

Pn

=

RT

P

Cancel P on the left side

V

n

=

RT

P

Avogadro’s law

V

1

n

1

=

V

2

n

2

V

n

=

RT

P

Constant

Derivation of Avogadro’s Law

Calculate the Density of a Gas

Using the Ideal Gas Law

Using algebraic manipulation, it is possible to solve for variables

other than those that appear explicitly in the ideal gas equation.

PV = nRT

n

P

V



RT

n

P

V

RT

 



M

P

d

RT



M

d_M is the density (in g/L) _{is the molar mass (in g/mol)}

g

V

=

P

(

mol

.

wt

.)

RT

Calculate the Density of a Gas

Using the Ideal Gas Law

What pressure would be required for helium at 25

°

C to have the

same density as carbon dioxide at 25

°

C and 1.00 atm?

P

d

RT



M

Solution:

Step 1: Use the equation below to calculate the density of CO2 at

25°C and 1 atm.













d = 1.00 atm 44.01 g/mol = 1.7966 g/L

Calculate the Density of a Gas

Using the Ideal Gas Law

What pressure would be required for helium at 25

°

C to have the

same density as carbon dioxide at 25

°

C and 1 atm?

Solution:

Step 2: Use the density of CO2 found in step 1 to calculate the

pressure for He at 25°C.

 





P 4.003 g/mol





1.7966 g/L =

0.08206 Latm/mol K 298.15 K

(16)

Worked Example 11.8

Strategy Use d = PM/RT to solve for density. Because the pressure is expressed in atm, we should use R = 0.08206 L∙atm/K∙mol. Remember the express temperature in kelvins.

Carbon dioxide is effective in fire extinguishers partly because its density is greater than that of air, so CO2 can smother the flames by depriving them of

oxygen. (Air has a density of approximately 1.2 g/L at room temperature and 1 atm.) Calculate the density of CO2 at room temperature (25°C) and 1.0 atm.

Think About It The calculated density of CO2 is greater than that of air under

the same conditions (as expected). Although it may seem tedious, it is a good idea to write units for each and every entry in a problem such as this. Unit cancellation is very useful for detecting errors in your reasoning or your solution setup.

Solution The molar mass of CO2 is 44.01 g/mol.

d = = PM _RT _{0.08206 L∙atm/K∙mol)(298.15 K) = 24.5 L}(1 atm)(44.01 g/mol)

Gas Densities Calculation

At 100

°

C and 720 mm Hg, what is the density of

carbon dioxide?

d

=

720

mmHg

760

mmHg

atm

æ

è

ç

ö

ø

÷

44.01

g

mol

(

)

0.08206

L

×

atm

mol

×

K

(

)

(

373

K

)

=

1.36

g

L

Molar Mass of a Gas Calculation

(

mol

.

wt

.)

=

dRT

P

upon rearranging…

Assuming ideal behavior, what is the molecular weight of a

gas with a density of 2.50 g/L at 98

°

C and 0.974 atm?

(

mol

.

wt

.)

=

2.50

g L

(

)

0.08206

L×atm mol×K

(

)

(

371

K

)

0.974

atm

=

78.1

g

mol

=

78.1

u

d

=

P

(

mol

.

wt

.)

RT

Real Gases

The

van der Waals equation

is useful for gases that do not behave

ideally.





an

P

V nb

nRT

V



















2

Experimentally measured pressure

corrected

pressure term volume term corrected Container volume

Real Gases

The

van der Waals equation

is useful for gases that do not behave

ideally.





an

P

V nb

nRT

V



















2

Real Gases

Calculate the pressure exerted by 0.35 mole of oxygen gas in a

volume of 6.50 L at 32

°

C using (a) the ideal gas equation and(b)

the van der Waals equation.

Solution:

Step 1:

Use the ideal gas equation to calculate the pressure of O

2

.

PV = nRT



0.35 mol 0.08206 Latm/mol K 305.15 K





_{1.3 atm} 6.50 L

nRT P

V

(17)

Real Gases

volume of 6.50 L at 32

°

C using (a) the ideal gas equation and(b)

Solution:

Step 2:

Use table 11.6 to find the values of

a

and

b

for O

2

.

Real Gases

volume of 6.50 L at 32

°

C using (a) the ideal gas equation and (b)

Solution

Step 3:

Use the van der Waals equation to calculate

P

.

P

= 1.3 atm





an

P

V nb

nRT

V



















2

         

2 2 0.35 mol 0.08206 Latm/mol K 305.15 K 1.36atmL /mol 0.35 mol

6.50 L 0.35 mol 0.0318 L/mol 6.50 L

( )

nRT an P

V nb V

 _   



 

 

2 2

Gas Mixtures

When two or more gases are placed in a container, each gas behaves

as though it occupies the container alone.

1.00 mole of N

2

in a 5.00 L container at 0

°

C exerts a pressure of

4.48 atm.

Addition of 1.00 mole of O

2

in the same container exerts an

additional 4.48 atm of pressure.

The total pressure of the mixture is the sum of the

partial pressures

(P

i

)

:

P

total

=

P

_N2

+

P

_O2

= 4.48 atm + 4.48 atm = 8.96 atm

2 N

(1mol)(0.08206 L atm/K mol)(273.15 K) 4.48 atm 5.00 L

P    

2 O

(1mol)(0.08206 L atm/K mol)(273.15 K) 4.48 atm 5.00 L

P    

Dalton

’

s Law of Partial Pressures

The

total pressure

(

P

_total

) of a gas mixture is

the

sum

of the

partial pressures

of its component gases.

P

Total

=

P

1

+

P

2

+

P

3

+

...

The

partial pressure

of a gas is the

pressure the gas would

exert if it were by itself

in the container.

Dalton

’

s Law of Partial Pressures

The total pressure depends on the total number of gas

particles and not on the types of particles.

0.5 mole O

2

0.3 mole He

0.2 mole Ar

1.0 mole

1.0 mole N

2

0.4 mole O

2

0.6 mole He

1.0 mole

1.0 atm _{1.0 atm} _{1.0 atm}

Gas Mixtures

Dalton

’

s Law Calculations

A sample of exhaled air contains four gases with

the following partial pressures:

N

2

(

563 mm Hg

),

O

2

(

118 mm Hg

),

CO

2

(

30 mm Hg

), and

H

2

O

(

50 mm

Hg

). What is the total pressure of the sample?

P

_total

=

P

_N2

+

P

_O2

+

P

_CO2

+

P

_H2O

P

_total

=

563

+

118

+

30

+

50

P

_total

= 761 mm Hg

(18)

Dalton

’

s Law Calculations

The total pressure exerted by a mixture of O

2

, N

2

, and

He gases is 1.50 atm. What is the partial pressure, in

atmospheres, of the O

2

, given that the partial pressures

of the N

2

and He are 0.75 and 0.33 atm, respectively.

P

O

₂

=

P

Total

-

P

N

₂

-

P

He

=

1.50

-

0.75

-

0.33

=

0.42

atm

P

Total

=

P

1

+

P

2

+

P

3

+

...

P

Total

=

P

1

+

P

2

+

P

3

+

...

Dalton

’

s Law Calculations

A scuba tank contains O

2

with a pressure of 0.450 atm and He

at 855 mmHg. What is the total pressure in mmHg in the tank?

For a deep descent, a scuba diver uses a mixture of helium and

oxygen with a total pressure of 8.00 atm. If the oxygen has a

partial pressure of 1280 mmHg,

what is the partial pressure of the helium?

P

Total

=

P

1

+

P

2

+

P

3

+

...

In the

lungs

, O

2

enters the blood,

while CO

2

from the blood is released.

In the

tissues

, O

2

enters the cells,

which releases CO

2

into the blood.

Dalton

’

s Law Calculations

Determine the partial pressures and the total pressure in a 2.50-L

vessel containing the following mixture of gases at 15.8

°

C:

0.0194 mol He, 0.0411 mol H

2

, and 0.169 mol Ne.

Solution:

Step 1:

Since each gas behaves independently, calculate the partial

pressure of each using the ideal gas equation:







PHe 

0.0194 mol He 0.08206 Latm/mol K 288.95 K _{0.184 atm} 2.50 L







P2 

2 H

0.0411 mol H 0.08206 Latm/mol K 288.95 K 0.390 atm 2.50 L







PNe 

0.169 mol Ne 0.08206 Latm/mol K 288.95 K 1.60 atm 2.50 L

Dalton

’

s Law Calculations

Determine the partial pressures and the total pressure in a 2.50-L

vessel containing the following mixture of gases at 15.8

°

C:

0.0194 mol He, 0.0411 mol H

2

, and 0.169 mol Ne.

Solution:

Step 2:

Use the equation below to calculate total pressure.

P

total

= 0.184 atm + 0.390 atm + 1.60 atm = 2.17 atm

total

=

i

P



P

Dalton’s Law of Partial Pressures

P

Total

=

P

1

+

P

2

+

P

3

+

...

P

1

=

n

1

RT

V

æ

è

ç

ö

ø

÷

P

Total

=

(

n

1

+

n

2

+

n

3

+

...

)

RT

V

æ

è

ç

ö

ø

÷ =

n

Total

RT

V

æ

è

ç

ö

ø

÷

P

2

=

n

2

RT

V

æ

è

ç

ö

ø

÷

(19)

Dalton’s Law Calculation

What is the total pressure when the two gases are completely mixed?

P

Total

=

(

n

H

₂

+

n

CO

₂

)

RT

V

æ

è

ç

ö

ø

÷

P

Total

=

æ

è

ç

2.016

1.00

+

44.01

88.0

ö

ø

÷

(

0.08206

3.00

)

( )

300

æ

è

ç

ö

ø

÷ =

20.5

atm

1.00 L 1.00 g H2 27.0°C

2.00 L 88.0 g CO2 27.0°C

Mole Fractions in Gas Mixtures

Each gas in a mixture behaves independently.

P

1

=

n

1

RT

V

æ

è

ç

ö

ø

÷

P

t

=

n

t

RT

V

æ

è

ç

ö

ø

÷

P

1

P

t

=

n

1

RT

V

æ

è

ç

ö

ø

÷

n

t

RT

V

æ

è

ç

ö

ø

÷

P

1

P

t

=

n

1

n

t

PV

=

nRT

P

1

P

t

=

n

1

n

t

upon rearranging

P

1

=

n

1

n

t

æ

è

ç

ö

ø

÷

P

t

n

1

n

t

æ

è

ç

ö

ø

÷ =

C

1

Mole fraction

The relative amounts of the components in a gas mixture can be

specified using

mole fractions.

There are three things to remember about mole fractions:

1)

The mole fraction of a mixture component is always less than 1.

2)

The sum of mole fractions for all components of a mixture is

always 1.

3)

Mole fractions are dimensionless.

i

i total

=

n



Χi is the mole fraction.

niis the moles of a certain component

ntotalis the total number of moles.

Mole Fraction Calculation #1

What is the mole fraction of H

2

in a gaseous mixture containing

1.0 g H

2

, 8.0 g O

2

, and 16 g CH

4

?

n

H

2

n

t

æ

è

ç

ö

ø

÷ =

C

H

₂

C

H2

=

1.0

2.016

æ

è

ç

ö

ø

÷

1.0

2.016

æ

è

ç

ö

ø

÷ +

8.0

32.00

æ

è

ç

ö

ø

÷ +

16

16.04

æ

è

ç

ö

ø

÷

æ

è

ç

ö

ø

÷

=

0.28

Mole Fraction Calculation #2

On a humid day in summer, the mole fraction of gaseous H

2

O

(water vapor) in the air at 25

°

C can be as high as 0.0287.

Assuming a total pressure of 0.977 atm, what is the partial

pressure in atmospheres of H

2

O in the air?

P

H

₂

O

=

n

H

2

O

n

t

æ

è

ç

ö

ø

÷

P

t

n

H

2

O

n

t

æ

è

ç

ö

ø

÷ =

C

H

₂

O

(20)

Worked Example 11.13

Strategy Use the ideal gas equation to calculate the total number of moles in the cylinder. Subtract moles of N2 from the total to determine moles of NO. Divide

moles NO by total moles to get mole fraction.

In 1999, the FDA approved the use of nitric oxide (NO) to treat and prevent lung disease, which occurs commonly in premature infants. The nitric oxide used in this therapy is supplied to hospitals in the form of a N2/NO mixture. Calculate the

mole fraction of NO in a 10.00-L gas cylinder at room temperature (25°C) that contains 6.022 mol N2 and in which the total pressure is 14.75 atm.

Solution The temperature is 298.15 K.

total moles = =

mol NO = total moles – N2 = 6.029 – 6.022 = 0.007 mol NO

χNO = = = 0.001

(14.75 atm)(10.00 L)

(0.08206 L∙atm/K∙mol)(298.15 K) = 6.029 mol

PV RT

nNO

ntotal

0.007 mol NO 6.029 mol

Think About It To check your work, determine χ_N2 by subtracting χNO from 1. Using each mole fraction and the total pressure,

calculate the partial pressure of each component using χi = Pi/Ptotal

and verify that they sum to the total pressure.

Stoichiometry Flowchart

grams of known moles of known moles of unknown grams of unknown

liters of known liters of unknown

Molarity of

solution Molarity of solution

P, V, T of known P, V, T of unknown

Reactions with Gaseous Reactants and Products

What mass (in grams) of Na

2

O

2

is necessary to consume 1.00 L of

CO

2

at STP?

2Na

2

O

2

(

s

) + 2CO

2

(

g

) → 2Na

2

CO

3

(

s

) + O

2

(

g

)

Solution:

Step 1:

Convert 1.00 L of CO

2

at STP to moles using the ideal gas

equation.

PV = nRT

(1 atm)(1.00 L) = n(0.08206 Latm/mol K)(273.15 K)

n = 0.04461 moles CO2

Reactions with Gaseous Reactants and Products

What mass (in grams) of Na

2

O

2

is necessary to consume 1.00 L of

CO

2

at STP?

2Na

2

O

2

(

s

) + 2CO

2

(

g

) → 2Na

2

CO

3

(

s

) + O

2

(

g

)

Solution:

Step 2:

Determine the stoichiometric amount of Na

2

O

2

.

   

  _ _

 

2 2 2

2

2 mol Na O 77.98 g

0.04461 moles CO × × = 3.48 g

2 mol CO mol

If 12.8 g of aluminum reacts with HCl, how many liters of

H

2

would be formed at 715 mm Hg and 19°C?

2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g) known unknown

V = ?

T = 19 + 273 = 292 K

P = 715 mm Hg

n = 0.712 mol

PV

=

nRT

12.8 g Al 26.98 gmol

æ

è ç ç

ö

ø ÷ ÷ 32

æ è

ç ö_ø÷ = 0.712 mol H2

Stoichiometry Calculation

If 12.8 g of aluminum reacts with HCl, how many liters of

H

2

would be formed at 715 mm Hg and 19

°

C?

2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g) known unknown

V

=

nRT

P

=

0.712

(

)

(

0.08206

)

( )

292

715

760

æ

è

ç

ö

ø

÷

(21)

Strategy Convert the given mass of Na2O2 to moles, use the balanced equation

to determine the stoichiometric amount of CO2, and then use the ideal gas

equation to convert moles of CO2 to liters.

Sodium peroxide (Na2O2) is used to remove carbon dioxide from (and add

oxygen to) the air supply in spacecrafts. It works by reacting with CO2 in the air

to produce sodium carbonate (Na2CO3) and O2.

2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g)

What volume (in liters) of CO2 (at STP) will react with a kilogram of Na2O2?

Worked Example 11.14 (cont.)

Solution The molar mass of Na2O2 is 77.98 g/mol (1 kg = 1000 g). (Treat the

specified mass of NaO2 as an exact number.)

1000 g Na2O2×

12.82 mol Na2O2×

V_CO2 =

= 12.82 mol Na2O2

1 mol Na2O2

77.98 g Na2O2

2 mol CO2

2 mol Na2O2 = 12.82 mol Na2O2

(12.82 mol CO2)(0.08206 L∙atm/K∙mol)(273.15 K)

1 atm = 287.4 L CO2

Think About It The answer seems like an enormous volume of CO2. If you

check the cancellation of units carefully in ideal gas equation problems, however, with practice you will develop a sense of whether such a calculated volume is reasonable.

Reactions with Gaseous Reactants and Products

Although there are no empirical gas laws that focus on the

relationship between

n

and

P

, it is possible to rearrange the ideal

gas equation to find the relationship.

PV = nRT

The change in pressure in a reaction vessel can be used to

determine how many moles of gaseous reactant are consumed:

V

n P

RT

 

V

n

P

RT

   

rearrangement

Strategy Use Δn = ΔP×(V/RT) to determine Δn, the number of moles CO2

consumed.

Another air-purification method for enclosed spaces involves the use of “scrubbers” containing aqueous lithium hydroxide, which react with carbon dioxide to produce lithium carbonate and water:

2LiOH(aq) + CO2(g) → Li2CO3(s) + H2O(l)

Consider the air supply in a submarine with a total volume of 2.5×105_{L. The}

pressure of 0.9970 atm, and the temperature 25°C. If the pressure in the submarine drops to 0.9891 atm as the result of carbon dioxide being consumed by an aqueous lithium hydroxide scrubber, how many moles of CO2 are consumed.

Solution ΔP = 0.9970 atm – 0.9891 atm = 7.9×10-3_atm,_V_{= 2.5}×105_{L, and}

T = 298.15 K.

Δn_CO2 = 7.9×10-3_atm× _{(0.08206 L∙atm/K∙mol)}2.5×105 L _×_{(298.15 K) = 81 moles CO}2

consumed

Think About It Careful cancellation of units is essential. Note that this amount of CO2 corresponds to 162 moles or 3.9 kg of LiOH.

(It’s a good idea to verify this yourself.

Reactions with Gaseous Reactants and Products

The volume of gas produced by a chemical reaction can be measured using an apparatus like the one shown below.

When gas is collected over water in this manner, the total pressure is the sum of two partial pressures:

Ptotal = Pcollected gas + PH2O

(22)

Collecting Gases Over Water

Zn + 2HCl ZnCl2 + H2

P

Total

=

P

gas

+

P

H

₂

O

Collecting Gases Over Water Calculation

Exactly 100 mL of oxygen is collected over water at

23

°

C and 800 torr.

Calculate the volume of the dry oxygen at 0

°

C and

760 torr.

P

Total

=

P

O

₂

+

P

H

₂

O

P

O

₂

=

P

Total

-

P

H

₂

O

=

800

-

21.07

=

779

torr

Exactly 100 mL of oxygen is collected over water at 23

°

C and

800 torr.

°

C and 760 torr.

V

1

= 100 mL

T

1

= 23 + 273 = 296 K

P

1

= 779 torr

V

2

= ?

T

2

= 0 + 273 = 273 K

P

2

= 760 torr

Combined Gas law

P

1

V

1

T

1

=

P

2

V

2

T

2

Exactly 100 mL of oxygen is collected over water at 23

°

C and

800 torr.

°

C and 760 torr.

P

1

V

1

T

1

=

P

2

V

2

T

2

V

2

=

P

1

V

1

T

1

æ

è

ç

ö

ø

÷

T

2

P

2

æ

è

ç

ö

ø

÷ =

779

´

100

296

æ

è

ç

ö

ø

÷

273

760

æ

è

ç

ö

ø

÷ =

94.5

mL

Calculate the mass of O

2

produced by the decomposition of KClO

3

when 821 mL of O

2

is collected over water at 30.0

°

C and 1.015

atm.

Solution:

Step 1:

Use Table 11.5 to determine the vapor pressure of water at

30.0

°

C.

Calculate the mass of O

2

3

when 821 mL of O

2

°

C and 1.015

atm.

Solution:

Step 2:

Convert the vapor pressure of water at 30.0

°

C to atm and

then use Dalton’s law to calculate the partial pressure of O

2

.

P

total

=

P

_O2

+

P

_H2O

P

_O2

=

P

total

–

P

_H2O

P

_O2

= 1.015 atm – 0.041842 atm = 0.973158 atm

P

2

H O @ 30.0 C

1 atm

(23)

Calculate the mass of O

2

3

when 821 mL of O

2

°

C and 1.015

atm.

Solution:

Step 3:

Convert to moles of O

2

using the ideal gas equation and then

find mass.

PV = nRT













PV n

RT

   2

0.973158 atm 0.821 L _{0.032117 moles O}

0.08206 Latm/mol K 303.15 K

 

_ _

2 32.00 g 2

0.032117 moles O 1.03 g O