GENERAL CHEMISTRY

Gases and the

Kinetic Molecular

Theory

LECTURE

12

Gases CONTENTS

14-1 Properties of Gases: Gas Pressure

14-2 The Simple Gas Laws

14-3 Combining the Gas Laws: The Ideal

Gas Equation and The General Gas

Equation

14-4 Applications of the Ideal Gas

Equation

14-5 Gases in Chemical Reactions

14-6 Mixtures of Gases

14-7 Kinetic—Molecular Theory of Gases

14-8 Gas Properties Relating to the

Kinetic—Molecular Theory

14-9 Nonideal (Real) Gases

The Distinction of Gases from

Liquids and Solids

1. Gas expand to fill and assume the shape of their

containers.

The Distinction of Gases from

Liquids and Solids

1. Gas expand to fill and assume the shape of their

containers.

3. Gases are invisible – no visible particles (except Cl2,

Br2, I2)

2. Most gases have relatively low densities under normal

conditions.

The gaseous state of three halogens (group 17)

The Distinction of Gases from

Liquids and Solids

1. Gas expand to fill and assume the shape of their

containers.

5. Gases are miscible.

3. Gases are invincible – no visible particles (except Cl2,

Br2, I2)

2. Most gases have relatively low densities under normal

conditions.

4. Some gases are combustible (H2, CH4) and some are

unreactive (He, Ne)

Gas % by Volume

N2 78.09

O2 20.94

Ar 0.93

CO2 0.03

He, Ne, Kr, Xe

0.002

CH4 0.00015

H2 0.00005

The Concept of Pressure

P (Pa) = Area (m2)

Force (N) Pressure

1.00 atm, 101.325 kPa, 1.01325 bar,

760 torr, 760 mm Hg

Units of Pressure

pascual (Pa) *SI unit

standard atmosphere (atm) 1 atm = 101.325 kPa

millimeter of mercury (mmHg)

Standard Atmospheric Pressure

1 torr = 1 mmHg = atm 1

760

bar (bar) 1 bar = 1000 Pa

Physical behavior of gases

1. Pressure, P

2. Volume, V

3. Temperature, T

4. Amount, (number of moles, n)

Relationship between gas volume and

pressure – Boyle’s Law

PV = constant P a 1 V

14-2 Simple Gas Laws

*(at constant T and n)

A molecular description of Boyle’s law.

Relationship between gas volume and

temperature- Charles’ Law

V = constant x T V a T

*(at constant P and n)

A molecular description of Charles’s law.

V n V = constant x n

Relationship between gas volume and amount of

gas- Avogadro’s Law *(at constant T and P)

A molecular description of Avogadro’s law.

Standard Temperature and Pressure (STP)

•IUPAC defines standard conditions of temperature

and pressure (STP).

•At STP:

P = 1 atm = 760 torr = 760 mmHg

T = 0°C = 273.15 K

Volume of 1 mol of an ideal gas = 22.414 L = 22.4 L

(standard molar volume)

14-3 Combining the Gas Laws:

The Ideal Gas Equation

and the General Gas Equation

• Boyle’s law V 1/P

• Charles’s law V T

• Avogadro’s law V n

V nT

P

PV nT

or

The Ideal Gas Equation

R = PV

nT

PV = nRT

Applying the ideal gas equation

The General Gas Equation

R = = P2V2

n2T2

P1V1

n1T1

= P2

T2

P1

T1

The amount and volume of

the gas are constant:

Using the Gas Laws

Exercise 1

A 50.0 L cylinder contains nitrogen gas at a

pressure of 21.5 atm. The contents of the cylinder

are emptied into an evacuated tank of unknown

volume. If the final pressure in the tank is 1.55 atm,

then what is the volume of the tank?

Exercise 1 (Continued)

•Use Boyle’s Law

P1V1 = P2V2 V2 = P1V1

P2

Vtank = 644 L

V2 = 21.5 atm x 50.0 L

1.55 atm

Exercise 2

A sample of hydrogen, H2, occupies 1.00 x 102 mL

at 25.0oC and 1.00 atm. What volume would it

occupy at 50.0oC under the same pressure?

T1 = 25oC + 273 = 298 K

T2 = 50oC + 273 = 323 K

Using Charles’ Law:

= V2

T2

V1

T1

V2 = V1T2

T1 =

100 mL x 323 K

298

V2 = 108 mL

Exercise 3

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?

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

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

P1V1

n1T1

P2V2

n2T2 =

V and n remain constant

P1

T1

P2

T2 =

0.991 atm x 1 atm

760 torr = 753 torr

P2 = P1 x T2

T1 = 753 torr x

373 K

296 K

P2 = 949 torr

Exercise 4

• 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.

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

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

21oC + 273.15 = 294 K

0.885 kg x 103 g

kg

mol O2

32.00 g O2 = 27.7 mol O2 x

P = nRT

V =

27.7 mol 294 K L-atm mol-K

0.0821 x x

438 L

P = 1.53 atm

14-4 Applications of the Ideal Gas

Equation

PV = nRT and n = m

M

PV = m

M

RT

M = m

PV

RT

Molar Mass Determination

Gas Density

= m

V

PV = m

M

RT

MP

RT V

m = =

KEEP IN MIND

that gas densities are

typically much smaller than

those of liquids and solids.

Gas densities are usually

expressed in grams per liter

rather than grams per

milliliter.

Sample Problem 5 Finding the Molar Mass of a Volatile Liquid

An organic chemist isolates a colorless liquid from a petroleum sample. She places the liquid in a flask and puts the flask in a boiling water bath, which vaporizes the liquid and fills the flask with gas. She closes the flask, reweighs it, and obtains the following data:

SOLUTION:

Volume (V) of flask = 213 mL

Mass of flask + gas = 78.416 g

T = 100.0oC

Mass of empty flask = 77.834 g

P = 754 torr

m = (78.416 - 77.834) g = 0.582 g

M = m RT

VP =

0.582 g atm•L

mol•K 0.0821 373.2 K x x

0.213 L x 0.992 atm = 84.4 g/mol

PV = nRT PV = m

M

RT M = m

PV

RT

Sample Problem 6 Calculating Gas Density

A chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 (a) at STP (0oC and 1 atm) and (b) at room conditions (20.oC and 1.00 atm).

SOLUTION: PV = nRT d =

RT

M x P

d = 44.01 g/mol x 1.00 atm

atm•L

mol•K 0.0821 x 273 K

= 1.96 g/L (a)

PV = m

M

RT

Sample Problem 6 (continued) Calculating Gas Density

(b) = 1.83 g/L d = 44.01 g/mol x 1.00 atm

x 293 K atm•L

mol•K 0.0821

A chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 (a) at STP (0oC and 1 atm) and (b) at room conditions (20.oC and 1.00 atm).

d = RT

M x P PV = nRT PV =

m

M

RT

12-5 Gases in Chemical Reactions

• Ideal gas equation relates the amount of a gas to

volume, temperature and pressure

• Stoichiometric factors relate gas quantities to

quantities of other reactants or products.

Sample Problem 7 Using Gas Variables to Find Amounts of Reactants or Products

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?

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 K atm•L

mol•K 0.0821 x

1.01 atm = 18.1 L

x V = nRT/P

PV = nRT

Sample Problem 8 Using the Ideal Gas Law in a Limiting-Reactant 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: 2K(s) + Cl2(g) 2KCl(s)

n = PV

RT Cl2

5.25 L =

0.950 atm x

atm•L mol•K

0.0821 x 293 K = 0.207 mol Cl2

17.0 g K x 39.10 g K

1 mol K = 0.435 mol K

V = 5.25 L

T = 293K n = unknown

P = 0.950 atm

Cl2

Sample Problem 8 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 x 2 mol KCl

2 mol K = 0.435 mol KCl formed

0.414 mol KCl x 74.55 g KCl

mol KCl = 30.9 g KCl

continued

Cl2 is the limiting reactant.

12-6 Mixtures of Gases

• Gas laws apply to mixtures of gases.

• Partial pressure

–Each component of a gas mixture exerts a

pressure that it would exert if it were in the

container alone.

Dalton’s law of partial pressures

Ptotal = P1 + P2 + P3 + ...

Dalton’s Law of Partial Pressure

Partial Pressure

The mole fraction of a component in a mixture is described by the equation:

c1 =

n1 n1 + n2 + n3 +... =

n1 ntotal

P1= c1 x Ptotal where c1 is the mole fraction

Σcn = c1 + c2 + c3 +... = 1

Sample Problem 9 Applying Dalton’s Law of Partial Pressures

In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol N2, 17 mol 16O2, and 4.0 mol 18O2. (The isotope 18O will be measured to determine the O2 uptake.) The pressure of the mixture is 0.75 atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture.

SOLUTION:

c 18O2 = 4.0 mol 18O2

100 = 0.040

= 0.030 atm P = c x Ptotal = 0.040 x 0.75 atm 18O2

18O2

P = c x Ptotal 18O2 18O2

c =

n

ntotal

18O2 18O2

A gaseous mixture contains 3.23 g of chloroform, CHCl3, and 1.22 g

of methane, CH4. Assuming that both compounds remains as

gases, what pressure is exerted by the mixture inside a 50.0-mL

metal container at 275 oC? What pressure is contributed by CHCl3?

Sample Problem 10 Applying Dalton’s Law of Partial Pressures

Ptotal= ntotalRT

V

3.23 g CHCl3 mol CHCl3

119.35 g x = 0.027 mol CHCl3

SOLUTION:

1.22 g CH4 mol CH4

16.05 g x = 0.076 mol CH4

ntotal = 0.103 mol

= 0.103 mol

atm•L

mol•K 0.0821 548 K x x

0.050 L

Ptotal = 92.68 atm

A gaseous mixture contains 3.23 g of chloroform, CHCl3, and 1.22 g

of methane, CH4. Assuming that both compounds remains as

gases, what pressure is exerted by the mixture inside a 50.0-mL

metal container at 275 oC? What pressure is contributed by CHCl3?

Sample Problem 10 (continued) Applying Dalton’s Law of Partial Pressures

n CHCl3 = 0.027 mol

n CH4 = 0.076 mol ntotal = 0.103 mol

Ptotal = 92.68 atm

PCHCl3= cCHCl3 x Ptotal SOLUTION:

cCHCl3 = 0.027 mol

0.103 mol = 0.262

PCHCl3= 0.0262 x 92.68 atm

PCHCl3= 24.29 atm

Collecting a gas over water

Ptot = Pbar = Pgas + PH2O

Pgas = Pbar - PH2O

Sample Problem 11 Calculating the Amount of Gas Collected over Water

Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reacts with water:

CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)

For a sample of acetylene that is collected over water, total gas pressure (adjusted to barometric pressure) is 738 torr and the volume is 523 mL. At the temperature of the gas (23oC), the vapor pressure of water is 21 torr. How many grams of acetylene are collected?

0.943 atm 0.523 L x n

C2H2 =

atm•L

mol•K 0.0821 x 296 K

= 0.0203 mol

0.0203 mol x 26.04 g C2H2

mol C2H2 = 0.529 g C2H2

SOLUTION:

P C2H2 = 738 torr – 21 torr = 717 torr 717 torr x

atm

760 torr

= 0.943 atm

grams ? PV = nRT n= PV

RT moles grams

Visualizing Molecular Motion

12-7 Kinetic Molecular Theory of Gases

• Particles are point masses in constant,

random, straight line motion.

• Particles are separated by great

distances.

• Collisions are rapid and elastic.

• No force between particles.

• Total energy remains constant.

Distribution of molecular speeds at three temperatures.

Kinetic Energy and Temperature

NAm = Mw:

Ek = m x u 2 1

2

Ek = mass x speed2 1

2

Ek = T 3

2

R

NA

m u 2 = T 3

2

R

NA

1

2

u 2 = T 3

2

R

NAm

1

2

u 2 = 3RT

Mw M

3RTu rms

M

3RTu rms

Distribution of Molecular Speeds – the effect

of mass and temperature

14-8 Gas Properties Relating to the

Kinetic-Molecular Theory

A. Diffusion

–Migration of molecules of

different substances due to

random molecular motion.

B. Effusion

–Escape of gas molecules

from their container through

a tiny pinhole.

Graham’s Law

rate of effusion of A

rate of effusion of B

(urms )A

(urms )B

3RT/MA

3RT/MB

MB

MA

rateA

rateB =

√Mw B

√Mw A

Applying Graham’s Law of Effusion

A mixture of helium (He) and methane (CH4) is placed in an effusion apparatus. Calculate the ratio of their effusion rates (He: CH4).

SOLUTION:

M of CH4 = 16.04 g/mol M of He = 4.003 g/mol

CH4

He rate

rate = √ 16.04

4.003 = 2.002 /

Sample Problem 12

A sample of hydrogen, H2, was found to effuse through a pinhole 5.2 times as rapidly as the same volume of unknown gas (at the same temperature and pressure). What is the molecular weight of the unknown gas?

Applying Graham’s Law of Effusion Sample Problem 13

rateH2

rateunk =

√Munk

√MH2

5.2 =

√Munk

√2 g/mol

2

27.04 =

Munk

2 g/mol

Munk = 54.08 g/mol

SOLUTION:

12-9 Real Gases: Deviation from

Ideal Behavior

Gas particles have finite volume

Vreal gas > Videal gas

Presence of attractive force between

gas molecules

Preal gas < Pideal gas

Compressibility factor

PV/nRT = 1 for ideal gas.

Deviations for real gases.

PV/nRT > 1 - molecular volume is significant.

PV/nRT < 1 – intermolecular forces of attraction.

Deviation from Ideal Behavior

The behaviour of real gases – compressibility

factor as a function of pressure at 0°C

Gases tend to behave ideally at high temperatures and low pressure.

Gases tend to behave nonideally at low temperature and high pressure.

Behavior of Gases

van der Waals Equation

P + n2a

V2 V – nb = nRT

a - intermolecular attraction

b - volume of gas molecules

adjusts

pressure up

adjusts

volume down

A 1.74 g sample of a compound that contains only carbon and hydrogen contains 1.44 g of carbon and 0.300 g of hydrogen. At STP 101 mL of the gas has a mass of 0.262 gram. What is its molecular formula?

QUIZ

104252

52

HC)H(C :formulaMolecular

229

58.1

g/mol 1.58L 0.101atm 00.1

K 273K mol

atm L0.0821)(262.0

PV

RT g=MW

RTMW

g PV

nRT PV

29 = masswith HC

5.20.120

0.2971

0.120

0.120

H mol 297.0H g 1.01

H mol 1H g 0.300 = atoms H mol ?

C mol 120.0C g 0.12

C mol 1C g 1.44 = atoms C mol ?

g

HC

Answer

3 pts

4 pts

1 pt

Summary

• Simple gas laws and the Ideal gas equation

• Determining MW and densities of gases using the ideal gas equation

• Chemical reactions involving gases

• Mixtures of gases

• Kinetic molecular theory

• Rate of effusion

• Ideal gas vs. Real gases