5-1

Gases

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5-2

An Overview of the Physical States of Matter

The Distinction of Gases from Liquids and Solids

1. Gas volume changes greatly with pressure.

2. Gas volume changes greatly with temperature.

3. Gases have relatively low viscosity.

4. Most gases have relatively low densities under normal conditions.

5. Gases are miscible.

5-3

Figure 5.1 The three states of matter.

5-4

Effect of atmospheric pressure on objects

at Earth’s surface.

5-5

A mercury barometer.

5-6

5-7

Converting Units of Pressure

PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects

the CO2 released in an evacuated flask. The CO2 pressure is

291.4 mmHg. Calculate the CO2 pressure in torrs,

atmospheres, and kilopascals.

SOLUTION:

PLAN: Construct conversion factors to find the other units of pressure.

291.4 mmHg x 1 torr

1 mmHg

= 291.4 torr

291.4 torr x1 atm

760 torr

= 0.3834 atm

0.3834 atm x 101.325 kPa

1 atm

= 38.85 kPa

5-8

Figure 5.4 The relationship between the volume

and pressure of a gas.

Boyle’s law

5-9

Figure 5.13 A molecular description of Boyle’s law.

5-10

Figure 5.5

The relationship between the volume and temperature of a gas.

Charles’s law

5-11

Figure 5.15 A molecular description of Charles’s law.

5-12

Boyle’s law At constant temperature, the volume occupied by a

fixed amount of gas is inversely proportional to the applied

(external) pressure.

n and T are fixed

Charles’s law At constant pressure, the volume of a fixed amount

of gas is directly proportional to its absolute (Kelvin) temperature.

V a TP and n are fixed

V

T= constant

V = constant x T

PV = constant

constant

P

V =V a1

P

P1V1 = P2V2

V1

T1

V2

T2

=

5-13

Amontons’s law At constant volume, the pressure exerted by a fixed

amount of gas is directly proportional to its absolute temperature.

P a TV and n are fixed

P

T= constant

P = constant x T

Combined gas law V aT

PV = constant x

T

P

PV

T= constant

P1

T1

P2

T2

=

P1V1

T1

P2V2

T2

=

5-14

An experiment to study the relationship between the

volume and amount of a gas.

5-15

The volume of 1 mol of an ideal gas compared with some

familiar objects.

5-16

THE IDEAL GAS LAW

PV = nRT

R = PV

nT=

1 atm x 22.414 L

1 mol x 273.15 K=

0.0821 atm•L

mol•K

Figure 5.9

R is the universal gas constant

3 significant figures

5-17

Applying the Volume-Pressure Relationship

PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies

24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases

the pressure on the trapped air to 2.64 atm. Assuming constant

temperature, what is the new volume of air (in L)?

PLAN:

5-18

Applying the Volume-Pressure Relationship

SOLUTION:

P1 = 1.12 atm P2 = 2.64 atm

V1 = 24.8 cm3 (convert to L) V2 = unknown

n and T are constant and

24.8 cm3 1 mL

1 cm3

L

103 mL

= 0.0248 L

P1V1

n1T1

P2V2

n2T2

=

P1V1 = P2V2

P1

P2

V2 = =1.12 atm x 0.0248 L

2.46 atm= 0.0105 L

x x

V1 x

5-19

Applying the Pressure-Temperature Relationship

PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that

opens if the internal pressure exceeds 1.00 x 103 torr. It is filled

with methane at 23oC and 0.991 atm and placed in boiling water

at exactly 100oC. Will the safety valve open?

PLAN:

5-20

Applying the Pressure-Temperature Relationship

SOLUTION:

P1 = 0.991 atm (convert to torr) P2 = unknown

T1 = 23oC (convert to K) T2 = 100oC (convert to K)

P1V1

n1T1

P2V2

n2T2

=

P1

T1

P2

T2

=

0.991 atm x

1 atm

760 torr = 753 torr

P2 = P1 x T2

T1

= 753 torr x373 K

296 K= 949 torr

V and n remain constant and

5-21

Solving for an Unknown Gas Variable at Fixed

Conditions

PROBLEM: A steel tank has a volume of 438 L and is filled with 0.885 kg of

O2. Calculate the pressure of O2 at 21oC.

PLAN:

SOLUTION:

V, T and mass, which can be converted to moles (n), are given. We

use the ideal gas law to find P.

V = 438 L T = 21oC (convert to K)

n = 0.885 kg O2 (convert to mol) P = unknown

21oC + 273.15 = 294 K0.885 kg x 103 g

kg

mol O2

32.00 g O2

= 27.7 mol O2

P = nRT

V=

27.7 mol 294 Katm•L

mol•K0.0821x x

438 L

= 1.53 atm

x

5-22

Figure 5.11

Summary of the stoichiometric relationships among the

amount (mol, n) of gaseous reactant or product and the gas

variables pressure (P), volume (V), and temperature (T).

5-23

Using Gas Variables to Find Amounts of

Reactants or Products

PROBLEM: Copper reacts with oxygen impurities in the ethylene used to

produce polyethylene. The copper is regenerated when hot H2

reduces the copper(II) oxide, forming the pure metal and H2O.

What volume of H2 at 765 torr and 225oC is needed to reduce

35.5 g of copper(II) oxide?

PLAN: Since this problem requires stoichiometry and the gas laws, we

have to write a balanced equation, use the moles of Cu to

calculate moles of H2 and then volume of H2 gas.

5-24

Using Gas Variables to Find Amounts of

Reactants or Products

SOLUTION: CuO(s) + H2(g) Cu(s) + H2O(g)

35.5 g CuO x mol CuO

79.55 g CuO

1 mol H2

1 mol Cu= 0.446 mol H2

0.446 mol H2 x

498 Katm•L

mol•K0.0821 x

1.01 atm

= 18.1 L

x

5-25

Using the Ideal Gas Law in a Limiting-Reactant Problem

PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)]

to form ionic metal halides. What mass of potassium chloride forms

when 5.25 L of chlorine gas at 0.950 atm and 293 K reacts with 17.0 g

of potassium?

SOLUTION:

PLAN: After writing the balanced equation, we use the ideal gas law to find the

number of moles of reactants, the limiting reactant and moles of product.

2K(s) + Cl2(g) 2KCl(s)

n = PV

RTCl2

5.25 L= 0.950 atm x

atm•Lmol•K

0.0821 x 293 K

= 0.207 mol Cl2

17.0 g K x

39.10 g K

mol K= 0.435 mol K

V = 5.25 L

T = 293K n = unknown

P = 0.950 atm

Cl2

5-26

Using the Ideal Gas Law in a Limiting-Reactant Problem

SOLUTION:

0.207 mol Cl2 x 2 mol KCl

1 mol Cl2

= 0.414 mol

KCl formed

0.435 mol K x2 mol KCl

2 mol K= 0.435 mol

KCl formed

0.414 mol KCl x74.55 g KCl

mol KCl= 30.9 g KCl

Cl2 is the limiting reactant.

5-27

Sample Problems

5-28

A sample of chlorine gas is confined in a 5.0-L

container at 328 torr and 37°C. How many moles

of gas are in the sample?

5-29

Plan: Given the volume, pressure, and temperature of a gas, the number of moles of the

gas can be calculated using the ideal gas equation, PV = nRT.

The gas constant, R = 0.0821 L•atm/mol•K, gives pressure in atmospheres and

temperature in Kelvin. The given pressure in torr must be converted to atmospheres and

the temperature converted to Kelvin.

Solution: PV = nRT or n = PV/RT

P = = 0.43158 atm; V = 5.0 L; T = 37°C + 273 = 310 K 1 atm

328 torr760 torr

(0.43158 atm)(5.0 L)

L•atm0.0821 (310 K)

mol •K

PVn =

RT

= 0.08479 = 0.085 mol chlorine

5-30

What is the effect of the following on the volume

of 1 mol of an ideal gas?

(a) The pressure is tripled (at constant T).

(b) The absolute temperature is increased by a

factor of 3.0 (at constant P).

5-31

Plan: Use the relationship

Solution:

a) As the pressure on a gas increases, the molecules move closer together, decreasing

the volume. When the pressure is tripled, the volume decreases to one third of the

original volume at constant temperature (Boyle’s Law).

V2 = ⅓V1

b) As the temperature of a gas increases, the gas molecules gain kinetic energy. With

higher energy, the gas molecules collide with the walls of the container with greater

force, which increases the size (volume) of the container. If the temperature is

increased by a factor of 3.0 (at constant pressure) then the volume will increase by a

factor of 3.0 (Charles’s Law).

V2 = 3V1

1 1 2 2

1 2

= PV P V

T T

1 1 2 1 12

2 1 1

( )( )(1) = =

(3 )(1)

V PT V PV

P T P

1 1 2 1 12

2 1 1

( )(1)(3 ) = =

(1)( )

V PT V TV

P T T

5-32

How many grams of phosphorus react with 35.5

L of O2 at STP to form tetraphosphorus

decaoxide?

P4(s) 5O2(g) + P4O10(s)

5-33

Plan: We can find the moles of oxygen from the standard molar volume of gases (1 L

of gas occupies 22.4 L at STP) and use the stoichiometric ratio from the balanced

equation to determine the moles of phosphorus that will react with the oxygen.

Solution:

P4(s) + 5 O2(g) P4O10(s)

Mass P4 =

= 39.2655 = 39.3 g P4

2 4 42

2 2 4

1 molO 1 mol P 123.88 g P35.5 LO

22.4 L O 5 mol O 1 mol P

5-34

How many grams of phosphine (PH3) can form

when 37.5 g of phosphorus and 83.0 L of

hydrogen gas react at STP?

P4(s) + H2(g) PH3(g) [unbalanced]

5-35

Plan: To find the mass of PH3, write the balanced equation and find the number of moles

of PH3 produced by each reactant. The smaller number of moles of product indicates the

limiting reagent. Solve for moles of H2 using the ideal gas equation.

Solution: P4(s) + 6 H2(g) 4 PH3(g)

From the ideal gas equation:

Moles hydrogen = 3.705357 mol H2 (unrounded)

PH3 from P4 =

= 1.21085 mol PH3

PH3 from H2 =

= 2.470238 mol PH3

P4 is the limiting reactant.

Mass PH3 =

= 41.15676 = 41.2 g PH3

344

4 4

4 mol PH1 mol P37.5 g P

123.88 g P 1 mol P

32

2

4 mol PH3.705357 mol H

6 mol H

3 344

4 4 3

4 mol PH 33.99 g PH1 mol P37.5 g P

123.88 g P 1 mol P 1 mol PH