Chapter 5 The Gaseous State - HCC Learning Web

Chapter 5

The Gaseous

State

Copyright © Cengage Learning. All rights reserved. 5 | 2

Contents and Concepts

Gas Laws

We will investigate the quantitative relationships that describe the behavior of gases.

1. Gas Pressure and Its Measurement

2. Empirical Gas Laws

3. The Ideal Gas Law

4. Stoichiometry Problems Involving Gas Volumes

5. 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

4

5

• Gases assume the volume and shape of their containers.

• Gases are the most compressible state of matter.

• Gases will mix evenly and completely when confined to

the same container.

• Gases have much lower densities than liquids and solids.

Physical Characteristics of Gases

NO2 gas


Gases differ from liquids and solids:

They are compressible.

Pressure, volume, temperature, and amount

are related.

Some applications

• How does a pressure cooker work?

• How is gas pressure applied in spray cans?

• How does a hot air balloon work?

• Why do we not want our tires to be full during hot summer days?

• Why do balloons deflate when left outside on a cold weather?

8

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

Pressure = Force Area

(force = mass x acceleration)

9

Manometers Used to Measure Gas Pressures

closed-tube open-tube


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

11

P a 1/V

P x V = constant

P1 x V1 = P2 x V2

Boyle’s Law

Constant temperature

Constant amount of gas

12

A sample of chlorine gas occupies a volume of 946 mL at a

pressure of 726 mmHg. What is the pressure of the gas (in

mmHg) if the volume is reduced at constant temperature to 154

mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL 154 mL

= = 4460 mmHg

P x V = constant

• 1. The pressure on a 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temperature remains constant?

• 2. A gas with a volume of 4.00 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container of the temperature remains constant?

14

Chemistry in Action:

Scuba Diving and the Gas Laws

P V

Depth (ft) Pressure

(atm)

0 1

33 2

66 3

15 As T increases V increases

Variation in Gas Volume with Temperature at Constant Pressure

16

Variation of Gas Volume with Temperature

at Constant Pressure

V a T

V = constant x T

V1/T1 = V2 /T2 T (K) = t (0C) + 273.15

Charles’ &

Gay-Lussac’s

Law

Temperature must be

in Kelvin


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.

Charles’ Law

• Volume vs. Temperature at constant pressure

• Volume is directly proportional to temperature

• Vα T @ constant P

• V=kT; V/T= k

• V1/T1 = V2/T2

• 1. If a sample of gas occupies 6.80 L at 3250 C, what will be its volume at 250C if the pressure does not change?

Note: convert Celsius to Kelvin scale by adding 273 to the given 0C.

• 2. Exactly 5.00 L of air at -50.00C is warmed to 100.00C. What is the new volume if the pressure remains constant?

20

A sample of carbon monoxide gas occupies 3.20 L at 125 0C.

At what temperature will the gas occupy a volume of 1.54 L if

the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K 3.20 L

= = 192 K

V1 /T1 = V2 /T2

T1 = 125 (0C) + 273.15 (K) = 398.15 K

Gay-Lussac’s Law

• Pressure and Temperature at constant Volume

• P is proportional to T @ constant volume

• PαT @ constant V

• P=kT; P/T=k

• P1/T1 = P2/T2

• A gas has a pressure of 6.58 kPa at 539K. What will be the pressure at 211 K if the volume does not change?

• The pressure in an automobile tire is 198 kPa at 270C. At the end of a trip on a hot sunny day, the pressure has risen to 225 kPa. What is the temperature of the air in the tire?

• Helium gas in a 2.00-L cylinder is unter 1.12 atm pressure. At 36.5 0C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas?

23

Avogadro’s Law

V a number of moles (n)

V = constant x n

V1 / n1 = V2 / n2

Constant temperature

Constant pressure

Avogadro’s Principle

• Volume is proportional to number of moles

• V α n where n is the number of moles

• More moles occupy greater volume

25

Ammonia burns in oxygen to form nitric oxide (NO) and water

vapor. How many volumes of NO are obtained from one volume

of ammonia at the same temperature and pressure?

4NH3 + 5O2 4NO + 6H2O

1 mole NH3 1 mole NO

At constant T and P

1 volume NH3 1 volume NO

Combined Gas Law

• P α T

V

P=k T

V

PV = k

T

• At 0.000C and 1.00 atm pressure a sample of gas occupies 30.0 mL. if the temperature is increased 30.00C and the gas sample is transferred to 20.0 mL container, what is the gas pressure?


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.

STP

• At STP,

• T=00C or 273 K

• P= 1atm

• At Standard temperature and pressure, molar volume is 22.4 L

• Meaning, 1 mole of any gas occupies 22.4 liters at STP

Ideal Gas Law

• Ideal gas behaves as if there is no IMF present among the molecules of gas.

• PV = R

nT

R = 1atm (22.4L)

1 mole ( 273)

R= .0821 atm-L/mol-K

30

Ideal Gas Equation

Charles’ law: V a T (at constant n and P)

Avogadro’s law: V a n (at constant P and T)

Boyle’s law: P a (at constant n and T) 1 V

V a nT

P

V = constant x = R nT

P

nT

P R is the gas constant

PV = nRT

31

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRT

P

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl

36.45 g HCl = 1.37 mol

V = 1 atm

1.37 mol x 0.0821 x 273.15 K L•atm

mol•K

V = 30.7 L

Sample problems for Ideal gas law

• If the pressure exerted by the gas at 250C in a volume of 0.044 L is 3.81 atm, how many moles of gas are present?

• Determine the celsius temperature of 2.49 moles of gas contained in a 1.00-L vessek at a pressure of 143 kPa.

33

Density (d) Calculations

d = m V

= PM RT

m is the mass of the gas in g

M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRT

P M = d is the density of the gas in g/L

34

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas?

dRT

P M = d = m

V

4.65 g

2.10 L = = 2.21

g

L

M = 2.21

g

L

1 atm

x 0.0821 x 300.15 K L•atm

mol•K

M = 54.5 g/mol

Density of gas from ideal gas equation

• D=M/V

• V=M/D

• PV=nRT

• P(Mass/D)= nRT

• Since n= mass/MW

• Then,

• P(Mass/D)=(Mass/MW)RT

• Therefore: P(MW)=DRT

• D=P(MW)

RT

• What is the density of a gas at STP that has a molar mass of 44.0 g/mol?

D=1atm(44.0g/mol)

.0821 (273K)

D= 1.96 g/L


Stoichiometry and Gas Volumes Use the ideal gas law to find moles from a given

volume, pressure, and temperature, and vice

versa.

37

Gas Stoichiometry

What is the volume of CO2 produced at 37 0C and 1.00 atm

when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

5.60 g C6H12O6 1 mol C6H12O6

180 g C6H12O6

x 6 mol CO2

1 mol C6H12O6

x = 0.187 mol CO2

V = nRT

P

0.187 mol x 0.0821 x 310.15 K L•atm

mol•K

1.00 atm = = 4.76 L


Gas Mixtures Dalton found that in a mixture of unreactive gases,

each gas acts as if it were the only gas in the

mixture as far as pressure is concerned.

39

Dalton’s Law of Partial Pressures

V and T are constant

P1 P2 Ptotal = P1 + P2

40

Consider a case in which two gases, A and B, are in a

container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nB XB =

nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT mole fraction (Xi ) = ni

nT

41

A sample of natural gas contains 8.24 moles of CH4, 0.421

moles of C2H6, and 0.116 moles of C3H8. If the total pressure

of the gases is 1.37 atm, what is the partial pressure of

propane (C3H8)?

Pi = Xi PT

Xpropane = 0.116

8.24 + 0.421 + 0.116

PT = 1.37 atm

= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

Copyright © Cengage

Learning. All rights

reserved.

5 | 42

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?

Copyright © Cengage Learning.

All rights reserved.

5 | 43

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



reserved.

5 | 44

atm0.7492N P

atm0.1532O P

atm0.03682CO P

atm0.0619OH2P

OHCOON 2222PPPPP

P = 1.00 atm



5 | 45

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

mmHg570.02N P

mmHg47.0OH2P

mmHg103.0 2O

P



5 | 46

OHCOON 2222PPPPP

570.0 mmHg

103.0 mmHg

40.0 mmHg

47.0 mmHg

P = 760.0 mmHg



reserved.

5 | 47

Mole fraction of N2

Mole fraction of H2O Mole 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

48

2KClO3 (s) 2KCl (s) + 3O2 (g)

PT = PO + PH O 2 2

Collecting a Gas over Water


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.

50

Vapor of Water and Temperature

51

?


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

= 64.06 g/mol

P = 727 mmHg

Pvapor = 21.1 mmHg

Pgas = 706 mmHg

T = 23°C = 296 K


P = 727 mmHg

Pvapor = 21.1 mmHg

Pgas = 706 mmHg

T = 23°C = 296 K P

nRTV

Molar mass NH4NO2

= 64.06 g/mol

4 2 24 2

4 2 4 2

1mol NH NO 1mol N5.68 g NH NO

64.04 g NH NO 1mol NH NO

= 0.8887 mol N2 gas


P = 727 mmHg

Pvapor = 21.1 mmHg

Pgas = 706 mmHg

T = 23°C = 296 K

n = 0.0887 mol

P

nRTV

mmHg760

atm1mmHg706

K)(296Kmol

atmL0.08206mol0.0887

V

= 2.32 L of N2

(3 significant figures)


2K

2

1mvE

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 = mass

v = velocity


Postulates of the Kinetic Theory

1. 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.

58

Kinetic theory of gases and …

• Compressibility of Gases

• Boyle’s Law

P a collision rate with wall

Collision rate a number density

Number density a 1/V

P a 1/V

• Charles’ Law


Collision rate a average kinetic energy of gas molecules

Average kinetic energy a T

P a T

59

Kinetic theory of gases and …

• Avogadro’s Law


Collision rate a number density

Number density a n

P a n

• Dalton’s Law of Partial Pressures

Molecules do not attract or repel one another

P exerted by one type of molecule is unaffected by the presence of another gas

Ptotal = SPi


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

61

The distribution of speeds

for nitrogen gas molecules

at three different temperatures

The distribution of speeds

of three different gases

at the same temperature

urms = 3RT M


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 K

CO2 molar mass =

0.04401 kg/mol


mol

kg0.04401

K296Kmol

s

mkg

8.314532

2

rmsu

25

rms 2

m1.68 10

su

2rms

m4.10 10

su

2

2

s

mkgJ

Recall

65

Gas diffusion is the gradual mixing of molecules of one gas

with molecules of another by virtue of their kinetic properties.

NH3

17 g/mol

HCl 36 g/mol

NH4Cl

r1

r2

M2 M1 =

molecular path


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

?


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

1

2.016

1

HeRate

HRate 2

70

Deviations from Ideal Behavior

1 mole of ideal gas

PV = nRT

n = PV RT

= 1.0

Repulsive Forces

Attractive Forces


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.

73

Effect of intermolecular forces on the pressure exerted by a gas.


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

75

Van der Waals equation

nonideal gas

P + (V – nb) = nRT an2 V2 ( )

}

corrected

pressure

}

corrected

volume


Use the van der Waals equation to calculate

the pressure exerted by 2.00 mol CO2 that

has a volume of 10.0 L at 25°C. Compare

this with value with the pressure obtained

from the ideal gas law.

n = 2.00 mol

V = 10.0 L

T = 25°C = 298 K

For CO2:

a = 3.658 L2 atm/mol2

b = 0.04286 L/mol


n = 2.00 mol

V = 10.0 L

T = 25°C = 298 K

L10.0

K)(298Kmol

atmL0.08206mol2.00

P

= 4.89 atm


V

nRTP

Ideal gas law:


n = 2.00 mol

V = 10.0 L

T = 25°C = 298 K

For CO2:

a = 3.658 L2 atm/mol2

b = 0.04286 L/mol

Pactual = 4.79 atm


2

2

V

an

nbV

nRTP

2

2

22

L10.0

mol

atmL3.658mol2.00

mol

L0.04286mol2.00L10.0

K298Kmol

atmL0.08206mol2.00

P

atm0.146atm4.933 P

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