Gases Chapter 13 Pg. 372-402

Goal• To learn about the behavior of gases

both on molecular and macroscopic levels.

General Characteristics of Gases

• Uniformly fills any container.

• Mixes completely with any other gas.

• Exerts pressure on its surroundings.

Force = mass X acceleration

• To understand pressure, one has to understand force.

• Weight = your mass X acceleration due to gravity

• Mr. Yoos’ forceF = 87 kg X 9.8 m/s2

F = 852.6 kg m /s2 = 850 N

Acceleration due to gravity

Pressure = Force / Area

If Mr. Yoos weighs 850 N on earth, and I am standing on a scale that is 0.5m X 0.5m or 0.25m2 , the pressure I exert is …

222

2 34000

3400

25.0

/850

m

Nr

ms

kg

m

skgm

A

FP

Let’s Say Mr. Yoos is Wearing High Heels

• Total area for the heels = 1X10-

4m2. The force is 850 N. What is the pressure?

1N/m2 = 1 Pascal or 1 Pa

Air Exerts Pressure

The “standard atmosphere” is equal to 101,325 Pa.

1 atm = 101,325 Pa1 atm = 760 torr = 760 mm Hg

Barometer

The Gas Laws

Boyle’s Law- the pressure exerted by a gas is inversely proportional to the volume the gas occupies if the temperature remains constant.

Boyle’s Law

• Example: A gas which has a pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature?

Boyle’s Law

Charles’s Law

• At constant pressure, the volume of a gas is directly proportional to the temperature (in Kelvins) of the gas.

Charles’s Law

• Example: A gas at 30.00C and 1.00 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60.00C and 1.00 atm?

Avogadro’s Law

• For a gas at constant temperature and pressure the volume is directly proportional to the number of moles of gas.

(at constant T, P)If you triple the number of moles of

gas (at constant temperature and pressure), the volume will also triple.

• Example: A 5.20 L sample at 18.00C and 2.00 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be?

The Ideal Gas Law

P in atmV in L

n in molesT in Kelvins

R =

Keep in Mind:

1. This relationship assumes that the gas behaves ideally (conditions of low pressure and high temperature). Correction factors must be added under certain conditions.

2. Keep track of dimensions! Many ideal gas law problems are best solved using DA.

3. Always list what you are given. You may be able to simplify the problem.

• Example: A sample containing 0.614 moles of a gas at 12.00C occupies a volume of 12.9 L. What pressure does the gas exert?

• Example: A sample of methane gas (CH4) at 0.848 atm and 4.0oC occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature is increased to 11.0oC?

• Example: How many moles of a gas at 104oC would occupy a volume of 6.8 L at a pressure of 270 mmHg?

Gas Stoichiometry

• Many gas law problems involve calculating the volume of a gas produced by the reaction of volume of other gases. The problem solving strategy that we have used throughout is still the same. That is, you want to relate moles of reactants to moles of products. The ideal gas law will allow you to use the following strategy:

Standard Temperature and Pressure“STP”

• P =• T = • The molar volume of an ideal gas

is _________ at STP

volume of reactants (apply the ideal gas law) moles of reactants (apply stoichiometry) moles of products (apply ideal gas law)

volume of products

• Example: A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following equation:

CO2 (s) CO2 (g)

How big will the balloon be (ie, what is the volume of the balloon), at 22.0oC and 1.04 atm, after all of the dry ice has sublimed?

• Example: 0.500 L of H2 (g) are reacted with 0.600 L of O2 (g) at STP according to the equation

2H2 (g) + O2 (g) 2H2O (g)

What volume will the H2O occupy at 1.00 atm and 350.oC?

Density and Molar Mass

PV = nRTP = nRT/V

• Example: A gas at 34.0oC and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass (MM) of the gas.

Dalton’s Law of Partial Pressures

• For a mixture of gases in a container, the total pressure is the sum of the pressures that each gas would exert if it were alone.

Because RT/V will be the same for each of the different

gases in the same container.

• Example: A volume of 2.0 L of He at 46oC, and 1.2 atm pressure, was added to a vessel that contained 4.5 L of N2 at STP. What is the total pressure and partial pressure of each gas at STP after the He is added?

The Production of Oxygen by Thermal Decomposition of

KCIO3

Mole Fraction

• The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture.

• The mole fraction of a particular component is a mixture of ideal gases is directly related to its partial pressure

Kinetic Molecular Theory of Gases

1. The volume of the individual particles of a gas can be assumed to be negligible.

2. The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas.

3. The particles are assumed to exert no forces on each other

4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

Temperature is a Measure of the Average Kinetic Energy

of a Gas

Root Mean Square Velocity

• The expression dealing with the average velocity of gas particles is called the root mean square velocity.

Where R = 8.3145 J/K mol = 8.3145 kg m2/s2 / Kmol

T = temp in Kelvins M = mass of a mole of the gas in

Kilograms

• Example: Calculate the root mean square velocity for the atoms in a sample of oxygen gas at

a. 0.0oCb. 300.oC

Effusion or Diffusion?• Diffusion- term used to describe

the mixing of gases.• Effusion- relates to the passage of

a gas through an orifice into an evacuated chamber.

Graham’s Law of Effusion

• The higher the molar mass of the gas, the slower the rate of effusion through a small orifice.

2

1

rms

rms

1

2

M

M

μ

μ

effusion of Rate

effusion of Rate

1

2

• Example: How many times faster than He would NO2 gas effuse?

With Regard to Diffusion

• The important idea is that even though gases travel very rapidly (hundreds of meters per second), their motions are in all directions, so mixing is relatively slow. The basic structure of Graham’s Law holds.

2

1

1

2

M

M

traveledDistance

traveledDistance

Chemistry in the Atmosphere

• The atmosphere is composed of 78% N2, 21% O2, 0.9% Ar, and 0.03% CO2 along with trace gases.

• The composition of the atmosphere varies as a function of distance from the earth’s surface. Heavier molecules tend to be near the surface due to gravity.

• Upper atmospheric chemistry is largely affected by UV, X-rays, and cosmic radiation emanating from space. The ozone layer is especially reactive to UV radiation.

• Manufacturing and other processes of our modern society affect the chemistry of our atmosphere. Air pollution is a direct result of such processes.

Air Pollution• Photochemical smog reactions:1. N2 (g) + O2 (g) + heat 2NO (g)

2. 2NO (g) + O2 (g) 2NO2 (g)3. NO2 (g) + radiant energy NO (g) + O (g)

4. O (g) + O2 (g) O3 (g) Ozone

Ozone causes lung and eye irritation and can be dangerous for people with asthma, emphysema, and other respiratory conditions.

Photochemical Smog

Ozone Layer

A Schematic Diagram of the Process for Scrubbing Sulfur Dioxide from

Stack Gases in Power Plants

Acid Rain

1. 2NO2 (g) + H2O (l) HNO2 (aq) + HNO3 (aq)

2. 2SO2 (g) + O2(g) 2SO3 (g)

3. SO3 (g) + H2O (l) H2SO4 (aq)

Acid Rain

Acid Rain