Chapter 12

Gases

Properties of GasesCompressibilityMassVolume (fill container)Exert PressureDiffuse through other gasesLow Density

Kinetic Molecular Theory (KMT)1. All gases are matter (have mass and take up

space).2. These particles are in constant, rapid and

random motion.3. All collisions are perfectly elastic. (no energy is

lost in the collision)4. The force of gas particle collisions on the walls of

the container creates pressure; however, gas particles do not exert force on each other.

5. At a given temperature, all gas particles have the same amount of kinetic energy (temperature).

6. The distance between gas particles is very large.

Gas Pressure

Gas pressure results from the force of colliding particles on a given area.

The metric unit of force is the newton (similar to the ounce in the English system).

A newton of force acting on a square meter is called a pascal.

Normal atmospheric pressure at sea level is 101.3 kPa.

Measuring GasesIn order to describe a gas, we use four variables

to make predictions about the behavior of the particles.1. Amount of a gas (n) expressed in moles2. Amount of space the gas takes up, volume (V),

as measured in liters. *note: this is the volume of the container

3. Measure of kinetic energy, temperature (T), in degrees Kelvin

1. *note: K = C + 273

4. Measure of the force exerted by the gas particles on the walls of the container, pressure (P)

Kinetic Energy and Temperature

Temperature is an indication of kinetic energy.

Fra

ctio

n of

Par

ticl

es

Speed of Particles

T1

T2 T2 > T1

Kelvin scale

The Kelvin Temperature Scale

(Absolute Zero Scale)

-300 -200 -100 0 100 oCelcius scaleKin

etic

Ene

rgy

of M

olec

ules

0

-273oC

0 73 173 273 373

K = oC + 273

PressurePressure is the measure of the amount of

force of an object per unit area.Newton / m2 = Pascal

Atmospheric Pressure is the amount of pressure exerted by the entire atmosphere on the surface of the Earth.A barometer is an instrument that measures

atmospheric pressureA manometer is an instrument that measures

the pressure of an enclosed gas

Mercury Barometer

Manometer

STP

Pressure and temperature changes affect gas volume.

To compare experimental results, scientist convert their results to standard temperature and pressure (STP)

Standard Conditions:

Standard Temperature = 0oC (273 K)

Standard Pressure = 101.3 kPa (1atm) {760 torr}

Pressure UnitsKilopascals (kPa)

Atmospheres (atm)1 atm = 101.3 kPa

Torricellis (torr)

1 torr = 1 mm Hg 760 torr = 101.3 kPa(other units are listed on page 420)

Pressure Units

Atmosphere (atm)Millimeter Hg (mmHg)

Pascal (Pa)Pounds per square inch

(psi)Kilopascal (kPa)

1 atm = Standard Pressure

1 atm = 760 mmHg1 atm = 101,325 Pa1 atm = 14.7 psi1 atm = 101.3 kPa

ConversionsCovert using Dimensional AnalysisInformation you have x conversion factor

(unknown/known)Convert 665 mmHg to kPa

665 mmHg x 101.3 kPa = 88.6 kPa 760 mmHg

Classwork: page 421 #1-4Homework: page 422 #1-12

Boyle’s Law

Pressure and Volume are inversely proportional at a given temperature and number of molecules.

VP = k (a constant value)

Pressure Effects on Volume

V 1/P

Inverse Proportionality

Boyle’s Law continuedV1P1 = k1

and V2P2 = k2

since k1 = k2 (that’s why its called a constant)

V1P1 = V2P2

This is the mathematical expression for Boyle’s Law

A Boyle’s Law Problem

A gas in a 242 cm3 container exerts a pressure of 87.6 kPa.

What volume would the gas occupy at standard atmospheric pressure?

Identify Variables

V1 =

V2 =

P1 =

P2 =

242 cm3

?

87.6 kPa

101.3 kPa

Two Methods of Solving

1. Algebraic:

V1P1 = V2P2

V2 = V1P1

P2

209 cm3

V2 = 242 cm3 x 87.6 kPa101.3 kPa

V2 =

Another Method2. Logic and reason:

The new volume (V2) depends upon the old volume (V1) multiplied by a pressure change.

V2 = 242 cm3 x _______

Since the pressure is increasing, the volume should decrease.

Multiply by a fraction less than 1.

87.6

101.3

Charles’s LawAt a given pressure and number of

molecules, Volume and Temperature are directly proportional.

V = kT

Two Temperature Scales

-300 -200 -100 0 100 oCelcius scale

0 73 173 273 373 Kelvin scale

0

Vol

ume

y = mx + b

V = kT + bV = kT + 0

V = kT

b

b

Charles’s Law Continued

V1 = k1

T1

and V2 = k2

T2

since k1 = k2

V1 = V2

T1 T2This is the mathematical expression for Charles’s Law

A Charles’s Law Problem

A 225 cm3 volume of gas is collected at 58oC. What volume would the gas occupy at standard temperature?

Identify Variables

V1 =

V2 =

T1 =

T2 =

225 cm3

?

58oC + 273 = 331 K

0oC + 273 = 273 K

Two Methods of Solving

1. Algebraic:

V1 / T1 = V2 / T2

V2 = V1T2

T1

V2 = 225 cm3 x 273 K 331 K

V2 = 186 cm3

The Other Method

2. Logic and reason:

The new volume (V2) depends upon the old volume (V1) multiplied by a temperature change.

V2 = 225 cm3 x _______

Since the temperature is decreasing, the volume should decrease.

Multiply by a fraction less than 1.

273

331

Gay-Lussac’s Law

P1 = k1

T1

and

P2 = k2

T2

since k1 = k2

P1 = P2

T1 T2

This is the mathematical expression for Gay-Lussac’s Law

The Combined Gas LawIn most laboratory situations, both

temperature and pressure change.

The Combined Gas Law is used to calculate the affects of these changes.

A Combination of Gas Laws

Boyle’s Law:V 1/P

Charles’s Law:V T

V T / P

To change a proportionality to an equality… multiply by a proportionality constant.

V = kT PVP = k T

V1P1 = V2P2

T1 T2

Gay-Lussac’s Law: P T

V1P1 = V2P2

T1 T2 Solve for each variable

V2 = V1P1T2 / T1P2

T2 = V2P2T1 / V1P1

P2 = V1P1T2 / T1V2

Sample Problem

What volume would 955 cm3 of a gas measured at 58 oC and 108.0 kPa occupy at 76 oC and 123.0 kPa?

V1 =

V2 =

T1 =

T2 =

P1 =

P2 =

955 cm3

?

331 K

349 K

108.0 kPa

123.0 kPa

Method #1V1P1 / T1 = V2P2 / T2

V2 = V1P1T2 / (T1P2)

V2 = (955 cm3) (108.0 kPa) (349 K)

(331 K) (123.0 kPa)

V2 = 884 cm3

Method #2

V2 = 955 cm3 x --------- x ---------

temperature increasesvolume increases

V2 = 884 cm3

349

331

pressure increases

volume decreases

123

108

Gas Law Variables

P

T

V

When T is constant P VWhen P is constant T V

When V is constant T P

Dalton’s Law of Partial Pressures

Gases are often collected by water displacement.

These gases contain water vapor.To find the pressure of the dry gas alone,

Dalton’s Law of Partial Pressures is used.

The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each gas in the mixture.

Dalton’s Law of Partial Pressures

Gases are often collected by water displacement.

These gases contain water vapor.To find the pressure of the dry gas alone,

Dalton’s Law of Partial Pressures is used.

The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each gas in the mixture.

The Mathematical Expression of Dalton’s Law

PT = P1 + P2 + …. Pn

Example:

Gas A = 1.0 atmGas B = 1.5 atmGas C = 0.5 atm

PT =3.0 atm

The Pressure of a “Dry” Gas When a gas is collected over water,

the total pressure is the result of the dry gas plus the water vapor pressure.

PT = Pgas + Pvapor

To determine the pressure of the dry gas alone, subtract the water vapor pressure from the total pressure. ( P. 401 )

Pgas = PT - Pvapor

Diffusion

If a bottle of ammonia were opened at the front of the room, the odor could soon be detected at the back of the room.

Diffusion is the random movement of particles through another substance.

Effusion is gas under pressure escaping through a small opening.

Diffusion

Kinetic Energy

The equation for finding the kinetic energy of a particle:

K.E. = ½ mv2

m = mass of the particlev = velocity (speed) of the particle

Kinetic Energy

K.E. = ½ mv2

#@*~!!*~##~@!&

Graham’s LawIf gases A and B are at the same

temperature…..

____ ____ K.E.A = K.E.B

½ mAv2A = ½ mBv2

B

mAv2A = mBv2

B

Collecting terms…

v2A = mB

v2B mA

vA = mB

vB mA

This is the mathematical expression for Graham’s Law

The Meaning of Graham’s Law

“The ratio of the velocities of two gases is equal to the square root of the inverse ratios of their masses.”

mAv2A = mBv2

B

if: mA > mB

then: vB > vA

(The lightest gas always travels faster)See practice problems on page 438

A Graham’s Law Experiment

HCl(g) + NH3(g) NH4Cl(s)

HCl NH3

v = d/t

vA = dA/tA

vB = dB/tB

tA = tB

vA / vB = dA / dB

(experimental)

vA / vB = √mB / mA

(accepted)

% error:

| exp. – accepted |accepted

x 100

Ideal Gases

The gas laws we have studied only apply to ideal gases.

There are no ideal gases.

Properties of Ideal Gases Ideal gases have:

1. Point mass

(Particles have mass but occupy no volume)

2. No mutual attractions

(No matter how close or how slow the

particles are, they do not attract each other)

Ideal Gas Laws Work

Real gas particles occupy negligible volume compared to the total volume occupied by the gas…

(unless under very high pressures)

Real Gases

Ideal Gas Laws Work

•Real gas particles occupy negligible volume compared to the total volume occupied by the gas…

(unless under very high pressures)

•Real gas particles have such weak intermolecular attraction forces that they do not affectively attract each other…

(unless at very low temperatures)

Real Gas LawsIdeal gas laws work for real gases except at

very high pressures or at very low temperatures.

Why not use real gas laws that work all the time?

Ideal gas law: PV = KT

Solve for VReal gas law: [(P + n2a)/V2] [V – nb] = KT

Solve for V

The Real Gas Law Solution

V = -(P + n2a) (P + n2a)2 – 4(KT)(P +n2a)(nb) 2KT

Manometer Problems

A Closed Arm Manometer

165 torr x 101.3 kPa 760 torr= 22.0 kPa

= ? kPa

Manometer Problems

An Open Arm Manometer

The gas exerts 120 torr more pressure than the atmosphere.

Gas pressure =

872 torr

872 torr x 101.3 kPa = 760 torr116 kPa