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4 Avogadro’s Law—How Volume and Moles Are Related

# 4 Avogadro’s Law—How Volume and Moles Are Related

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AVOGADRO’S LAW—HOW VOLUME AND MOLES ARE RELATED

203

As the number of moles of a gas increases, its volume increases as well. Thus, dividing the

volume by the number of moles is a constant (k). The value of k is the same regardless of the

identity of the gas.

When the number of moles increases...

V

n

...the volume increases.

V

n

=

k

Thus, if the pressure and temperature of a system are held constant, increasing the number of

moles increases the volume of a gas.

Increase the

number of moles.

V=1 L

V=2 L

Volume increases.

n = 1 mol

n = 2 mol

Since dividing the volume of a gas by the number of moles is a constant, knowing the volume and

number of moles initially (V1 and n1) means we can calculate a new volume or number of moles

(V2 and n2) when one of these quantities is changed.

V1

V2

=

n1

initial conditions

n2

new conditions

To solve a problem of this sort, we follow the same three steps listed in the How To outlined in

Section 7.3A, using Avogadro’s law in step [2].

SAMPLE PROBLEM 7.6

The lungs of an average male hold 0.25 mol of air in a volume of 5.8 L. How many moles of air

do the lungs of an average female hold if the volume is 4.6 L?

ANALYSIS

This question deals with volume and number of moles, so Avogadro’s law is used to determine

a final number of moles when three quantities are known—the initial volume and number of

moles (V1 and n1), and the final volume (V2).

SOLUTION

[1]

Identify the known quantities and the desired quantity.

V1 = 5.8 L

V2 = 4.6 L

n1 = 0.25 mol

known quantities

[2]

n2 = ?

desired quantity

Write the equation and rearrange it to isolate the desired quantity, n2, on one side.

• Use Avogadro’s law. To solve for n2, we must invert the numerator and denominator on

both sides of the equation, and then multiply by V2.

smi26573_ch07.indd 203

12/3/08 11:06:39 AM

204

GASES, LIQUIDS, AND SOLIDS

V1 V2

=

n1

n2

n1

V1

Switch V and n

on both sides.

n1 V2

V1

[3]

n2

=

=

Solve for n2 by multiplying both sides by V2.

V2

n2

Solve the problem.

• Substitute the three known quantities into the equation and solve for n2.

n2

=

n1V2

=

V1

(0.25 mol)(4.6 L)

(5.8 L)

=

0.20 mol

PROBLEM 7.17

A balloon that contains 0.30 mol of helium in a volume of 6.4 L develops a leak so that its

volume decreases to 3.85 L at constant temperature and pressure. How many moles of helium

does the balloon now contain?

PROBLEM 7.18

A sample of nitrogen gas contains 5.0 mol in a volume of 3.5 L. Calculate the new volume

of the container if the pressure and temperature are kept constant but the number of moles of

nitrogen is changed to each of the following values: (a) 2.5 mol; (b) 3.65 mol; (c) 21.5 mol.

Avogadro’s law allows us to compare the amounts of any two gases by comparing their volumes.

Often amounts of gas are compared at a set of standard conditions of temperature and pressure, abbreviated as STP.

• STP conditions are: 1 atm (760 mm Hg) for pressure

273 K (0 °C) for temperature

• At STP, one mole of any gas has the same volume, 22.4 L, called the standard molar

volume.

Under STP conditions, one mole of nitrogen gas and one mole of helium gas each contain

6.02 × 1023 molecules of gas and occupy a volume of 22.4 L at 0 °C and 1 atm pressure. Since

the molar masses of nitrogen and helium are different (28.0 g for N2 compared to 4.0 g for

He), one mole of each substance has a different mass.

same volume

same number of particles

1 mol N2

1 mol He

22.4 L

6.02 × 1023 particles

28.0 g

22.4 L

6.02 × 1023 particles

4.0 g

The standard molar volume can be used to set up conversion factors that relate the volume and

number of moles of a gas at STP, as shown in the following stepwise procedure.

smi26573_ch07.indd 204

12/3/08 11:06:39 AM

AVOGADRO’S LAW—HOW VOLUME AND MOLES ARE RELATED

HOW TO

EXAMPLE

Step [1]

Step [2]

205

Convert Moles of Gas to Volume at STP

How many moles are contained in 2.0 L of N2 at standard temperature and pressure?

Identify the known quantities and the desired quantity.

2.0 L of N2

? moles of N2

original quantity

desired quantity

Write out the conversion factors.

• Set up conversion factors that relate the number of moles of a gas to volume at STP. Choose the conversion

factor that places the unwanted unit, liters, in the denominator so that the units cancel.

22.4 L

1 mol

Step [3]

1 mol

22.4 L

or

Choose this conversion

factor to cancel L.

Solve the problem.

• Multiply the original quantity by the conversion factor to obtain the desired quantity.

2.0 L

×

1 mol

22.4 L

=

0.089 mol of N2

Liters cancel.

By using the molar mass of a gas, we can determine the volume of a gas from a given number of

grams, as shown in Sample Problem 7.7.

SAMPLE PROBLEM 7.7

Burning 1 mol of propane in a gas grill adds 132.0 g of carbon dioxide (CO2) to the

atmosphere. What volume of CO2 does this correspond to at STP?

ANALYSIS

To solve this problem, we must convert the number of grams of CO2 to moles using the molar

mass. The number of moles of CO2 can then be converted to its volume using a mole–volume

conversion factor (1 mol/22.4 L).

SOLUTION

[1]

[2]

Identify the known quantities and the desired quantity.

132.0 g CO2

? L CO2

known quantity

desired quantity

Convert the number of grams of CO2 to the number of moles of CO2 using the molar mass.

molar mass

conversion factor

132.0 g CO2

×

1 mol CO2

44.0 g CO2

=

3.00 mol CO2

Grams cancel.

[3]

Convert the number of moles of CO2 to the volume of CO2 using a mole–volume

conversion factor.

mole–volume

conversion factor

3.00 mol CO2

×

22.4 L

Moles cancel.

smi26573_ch07.indd 205

1 mol

=

67.2 L CO2

12/3/08 11:06:40 AM

206

GASES, LIQUIDS, AND SOLIDS

PROBLEM 7.19

How many liters does each of the following quantities of O2 occupy at STP: (a) 4.5 mol;

(b) 0.35 mol; (c) 18.0 g?

PROBLEM 7.20

How many moles are contained in the following volumes of air at STP: (a) 1.5 L; (b) 8.5 L;

(c) 25 mL?

7.5 THE IDEAL GAS LAW

All four properties of gases—pressure, volume, temperature, and number of moles—can be combined into a single equation called the ideal gas law. The product of pressure and volume divided

by the product of moles and Kelvin temperature is a constant, called the universal gas constant

and symbolized by R.

PV

nT

=

R

universal gas constant

More often the equation is rearranged and written in the following way:

PV = nRT

For atm:

R

=

0.0821

For mm Hg:

R

=

62.4

Ideal gas law

L atm

mol K

L mm Hg

mol K

The value of the universal gas constant R depends on its units. The two most common values of R

are given using atmospheres or mm Hg for pressure, liters for volume, and kelvins for temperature.

Be careful to use the correct value of R for the pressure units in the problem you are solving.

The ideal gas law can be used to find any value—P, V, n, or T—as long as three of the quantities are

known. Solving a problem using the ideal gas law is shown in the stepwise How To procedure and

in Sample Problem 7.8. Although the ideal gas law gives exact answers only for a perfectly “ideal”

gas, it gives a good approximation for most real gases, such as the oxygen and carbon dioxide in

breathing, as well (Figure 7.6).

HOW TO

Carry Out Calculations with the Ideal Gas Law

EXAMPLE 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 °C

known quantities

Step [2]

n = ? mol

desired quantity

Convert all values to proper units and choose the value of R that contains these units.

• Convert °C to K. K = °C + 273 = 37 °C + 273 = 310. K

• Use the value of R in atm since the pressure is given in atm; that is, R = 0.0821 L · atm/mol · K.

Step [3]

Write the equation and rearrange it to isolate the desired quantity on one side.

• Use the ideal gas law and solve for n by dividing both sides by RT.

smi26573_ch07.indd 206

PV

=

nRT

PV

RT

=

n

Solve for n by dividing both sides by RT.

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THE IDEAL GAS LAW

207

How To, continued. . .

Step [4]

Solve the problem.

• Substitute the known quantities into the equation and solve for n.

n

FIGURE 7.6

=

PV

RT

=

(1.0 atm)(0.50 L)

(

0.0821

L · atm

mol · K

)

=

0.0196 rounded to 0.020 mol

(310. K)

Focus on the Human Body: The Lungs

average lung

capacity—4–6 L

trachea

average tidal

volume—0.5 L

right lung

with its

three lobes

left lung

with its

two lobes

heart

pulmonary

artery

pulmonary vein

• Humans have two lungs that contain

a vast system of air passages,

allowing gases to be exchanged

between the atmosphere and with

the bloodstream. The lungs contain

about 1,500 miles of airways that

have a total surface area about the

size of a tennis court.

• Lungs are in a sense “overbuilt,” in

that their total air volume is large

compared to the tidal volume, the

amount of air taken in or expelled

with each breath. This large reserve

explains why people can smoke for

years without noticing any significant

change in normal breathing.

• In individuals with asthma, small

airways are constricted and

inflamed, making it difficult to breathe.

alveolus

section of

alveoli

cut open

Blood in pulmonary arteries gives

up waste CO2 to the lungs so that

it can be expelled to the air.

Blood in pulmonary veins picks up

O2 in the lungs so that it can be

pumped by the heart to the body.

SAMPLE PROBLEM 7.8

ANALYSIS

smi26573_ch07.indd 207

If a person exhales 25.0 g of CO2 in an hour, what volume does this amount occupy at 1.00 atm

and 37 °C?

Use the ideal gas law to calculate V, since P and T are known and n can be determined by using

the molar mass of CO2 (44.0 g/mol).

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208

GASES, LIQUIDS, AND SOLIDS

SOLUTION

[1]

Identify the known quantities and the desired quantity.

P = 1.00 atm

T = 37 °C

25.0 g CO2

V=?L

known quantities

[2]

desired quantity

Convert all values to proper units and choose the value of R that contains these units.

• Convert °C to K. K = °C + 273 = 37 °C + 273 = 310. K

• Use the value of R with atm since the pressure is given in atm; that is,

R = 0.0821 L · atm/mol · K.

• Convert the number of grams of CO2 to the number of moles of CO2 using the molar

mass (44.0 g/mol).

molar mass

conversion factor

25.0 g CO2

×

1 mol CO2

44.0 g CO2

=

0.568 mol CO2

Grams cancel.

[3]

Write the equation and rearrange it to isolate the desired quantity, V, on one side.

• Use the ideal gas law and solve for V by dividing both sides by P.

[4]

PV

=

nRT

V

=

nRT

P

Solve for V by dividing both sides by P.

Solve the problem.

• Substitute the three known quantities into the equation and solve for V.

V

=

nRT

P

(

(0.568 mol) 0.0821

=

L · atm

mol · K

)

(310. K)

1.0 atm

=

14.5 L

PROBLEM 7.21

How many moles of oxygen (O2) are contained in a 5.0-L cylinder that has a pressure of

175 atm and a temperature of 20. °C?

PROBLEM 7.22

Determine the pressure of N2 under each of the following conditions.

a. 0.45 mol at 25 °C in 10.0 L

PROBLEM 7.23

b. 10.0 g at 20. °C in 5.0 L

Determine the volume of 8.50 g of He gas at 25 °C and 750 mm Hg.

7.6 DALTON’S LAW AND PARTIAL PRESSURES

Since the partial pressure of O2

is low at very high altitudes, most

mountain climbers use supplemental

O2 tanks above about 24,000 ft.

smi26573_ch07.indd 208

Since gas particles are very far apart compared to the size of an individual particle, gas particles

behave independently. As a result, the identity of the components of a gas mixture does not

matter, and a mixture of gases behaves like a pure gas. Each component of a gas mixture is said

to exert a pressure called its partial pressure. Dalton’s law describes the relationship between

the partial pressures of the components and the total pressure of a gas mixture.

• Dalton’s law: The total pressure (Ptotal) of a gas mixture is the sum of the partial

pressures of its component gases.

12/3/08 11:06:45 AM

DALTON’S LAW AND PARTIAL PRESSURES

209

Thus, if a mixture has three gases (A, B, and C) with partial pressures PA, PB, and PC, respectively, the total pressure of the system (Ptotal) is the sum of the three partial pressures. The partial

pressure of a component of a mixture is the same pressure that the gas would exert if it were a

pure gas.

Ptotal

total pressure

SAMPLE PROBLEM 7.9

ANALYSIS

SOLUTION

=

PA

+

PB

+

PC

partial pressures of A, B, and C

A sample of exhaled air from the lungs contains four gases with the following partial pressures:

N2 (563 mm Hg), O2 (118 mm Hg), CO2 (30 mm Hg), and H2O (50 mm Hg). What is the total

pressure of the sample?

Using Dalton’s law, the total pressure is the sum of the partial pressures.

Adding up the four partial pressures gives the total:

563 + 118 + 30 + 50 = 761 mm Hg (total pressure)

PROBLEM 7.24

CO2 was added to a cylinder containing 2.5 atm of O2 to give a total pressure of 4.0 atm of gas.

What is the partial pressure of O2 and CO2 in the final mixture?

We can also calculate the partial pressure of each gas in a mixture if two quantities are known—

[1] the total pressure and [2] the percent of each component—as shown in Sample Problem 7.10.

SAMPLE PROBLEM 7.10

ANALYSIS

Air is a mixture of 21% O2, 78% N2, and 1% argon by volume. What is the partial pressure of

each gas at sea level, where the total pressure is 760 mm Hg?

Convert each percent to a decimal by moving the decimal point two places to the left. Multiply

each decimal by the total pressure to obtain the partial pressure for each component.

SOLUTION

Fraction O2: 21% = 0.21

Fraction N2: 78% = 0.78

Fraction Ar: 1% = 0.01

PROBLEM 7.25

HEALTH NOTE

Partial pressure

0.21 × 760 mm Hg = 160 mm Hg (O2)

0.78 × 760 mm Hg = 590 mm Hg (N2)

0.01 × 760 mm Hg =

8 mm Hg (Ar)

758 rounded to 760 mm Hg

A sample of natural gas at 750 mm Hg contains 85% methane, 10% ethane, and 5% propane.

What are the partial pressures of each gas in this mixture?

The composition of the atmosphere does not change with location, even though the total atmospheric pressure decreases with increasing altitude. At high altitudes, therefore, the partial pressure of oxygen is much lower than it is at sea level, making breathing difficult. This is why mountain climbers use supplemental oxygen at altitudes above 8,000 meters.

In contrast, a hyperbaric chamber is a device that maintains air pressure two to three times higher

than normal. Hyperbaric chambers have many uses. At this higher pressure the partial pressure of

O2 is higher. For burn patients, the higher pressure of O2 increases the amount of O2 in the blood,

where it can be used by the body for reactions that fight infections.

The high pressures of a hyperbaric

chamber can be used to treat

patients fighting infections and scuba

divers suffering from the bends.

smi26573_ch07.indd 209

When a scuba diver surfaces too quickly, the N2 dissolved in the blood can form microscopic

bubbles that cause pain in joints and can occlude small blood vessels, causing organ injury. This

condition, called the bends, is treated by placing a diver in a hyperbaric chamber, where the

elevated pressure decreases the size of the N2 bubbles, which are then eliminated as N2 gas from

the lungs as the pressure is slowly decreased.

12/3/08 11:06:48 AM

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