Gases and their PropertiesGases and their Properties

Nature of GasesNature of Gases

1.1. Gas particles have mass. If you fill up a Gas particles have mass. If you fill up a basketball with air, its mass will be more than basketball with air, its mass will be more than the mass of a deflated ball. the mass of a deflated ball.

2. Gases are easy to compress: If you squeeze a 2. Gases are easy to compress: If you squeeze a gas, its volume decreasesgas, its volume decreases

3. Gases fill their containers completely: When a 3. Gases fill their containers completely: When a balloon is filled with air, the air is distributed balloon is filled with air, the air is distributed evenly. There is no place around you that is evenly. There is no place around you that is not filled with air (exception: sealed vacuum).not filled with air (exception: sealed vacuum).

Nature of Gases Cont.Nature of Gases Cont.

4.4. Different gases move through each other quite rapidly; Different gases move through each other quite rapidly; this is called diffusion. Examples gases diffusing this is called diffusion. Examples gases diffusing through air are: smelling perfume left by a person, through air are: smelling perfume left by a person, smelling a skunk nearby.smelling a skunk nearby.

5.5. Gases exert pressure: you experience air pressure Gases exert pressure: you experience air pressure when your ears pop, or the balloon inflates. Pressure when your ears pop, or the balloon inflates. Pressure = force/unit area= force/unit area

6.6. The pressure of a gas depends on its temperature: The pressure of a gas depends on its temperature: the higher the temperature, the higher the air pressure the higher the temperature, the higher the air pressure and vice versa. In the summer the tire pressure in and vice versa. In the summer the tire pressure in your car increases due to heat.your car increases due to heat.

Kinetic Molecular TheoryKinetic Molecular Theory

Ludwig BoltzmannRudolf Claussius James Maxwell

Kinetic Molecular Theory is based on six postulates:1.Gases are small particles that have a mass2.Since distances separating gas particles are large,

the volume of individual gas molecules is negligible3. Gas particles are in constant, rapid, random motion

Kinetic Molecular Theory Cont.Kinetic Molecular Theory Cont.

4.4. Collisions between gas particles or gases and Collisions between gas particles or gases and their containers are perfectly elastic (in a their containers are perfectly elastic (in a perfectly elastic collision, no energy of motion is perfectly elastic collision, no energy of motion is lost ex. bouncing ball)lost ex. bouncing ball)

5.5. The average kinetic energy of gas particles The average kinetic energy of gas particles depends only on the temperature of the gas. depends only on the temperature of the gas. The higher the temp. the greater the kinetic The higher the temp. the greater the kinetic energy (energy of motion)energy (energy of motion)

6.6. Gases exert no force on one another. The Gases exert no force on one another. The attractive forces between gas particles is attractive forces between gas particles is negligible. No hydrogen bonding. negligible. No hydrogen bonding.

Measuring Gases

In order to describe a gas, one must take into consideration four factors:

1. amount of gas2. volume 3. temperature4. pressure

Amount of gas is measured in moles of gas (n)

Volume of gas is measured in Liters = 1000 cm3

Measuring Temperature of a Gas

Temperature of a gas is usually measured in Celsius. However the gas laws are written using the Kelvin Scale.

The zero point on the Celsius scale equals the melting/freezing point of water at 1 atmosphere of pressure. The boiling point of water at 1atmosphere of pressure is set to 100 degrees on the Celsius scale.

The interval between the melting point and the boiling point of water is divided into 100 intervals, each equal to 1 degree. This interval is the same on the Kelvin scale. 1 Kelvin unit = 1o Celsius.

The zero point on the Kelvin scale is called absolute zero, the lowest possible temperature. This is the point at which all molecular motion stops. It is equal to - 273 oC.

When working with the Kelvin scale, the term "degrees" is not used. Zero degrees Celsius is written as 273 K and read as "273 Kelvins".

The zero point on the Fahrenheit scale is based on the freezing point of a mixture of salt and water and scientifically meaningless. The Fahrenheit scale is not used in science.

Conversions:Converting from Celsius to Kelvin: Example: convert 37o Celsius to Kelvin. C + 273 = K or C = K -273 37 + 273 = 310K Check out the picture to verify this answer.

To convert from Kelvin to Celsius subtract 273. Converting from Celsius to Fahrenheit: Example: convert 37o Celsius to

Fahrenheit. 9/5C + 32 = F ( 9/5 X 37 ) + 32 = 98.6oF. 9/5 = 1.8, meaning that 1 degree on the

Celsius scale = 1.8 degrees on the Fahrenheit scale. Look at the picture: a five degree interval on the Celsius scale equals a nine degree interval on the Fahrenheit scale.Remember, 37oC equals 98.6oF, because it is the body temperature.

Converting from Fahrenheit to Celsius: C = (F - 32) / 9/5 (98.6 - 32) / 9/5 = 37oC. First subtract 32, then divide by 1.8.

Instead of dividing by 1.8, you may also multiply by 5/9.

http://www.howe.k12.ok.us/~jimaskew/ptemps.htm

Calculating Pressure:

Standard Pressure is the air pressure taken at sea level. It can be measured using a variety of barometers that use different scales with different values.

The SI unit for pressure is Pascals The different units for standard

pressure are as follows: 101,325Pa = 101.325 KPa = 1atm = 760 torr = 760 mm Hg = 14.70 lbs/in2

The Standard pressures written above are all equal to one another. To convert between them, you can use their equivalencies to make conversion factors.

Image from http://www.williamsclass.com/EighthScienceWork/Atmosphere/barometerTori.jpg

How to Convert Between Pressure Scales

The standard pressures are all equal to one another. 101,325Pa = 101.325 KPa = 1 atm = 760 torr = 760 mm Hg = 14.70 lbs/in2

One can convert between two different pressures by creating a conversion factor.

For example: If a barometer reads 5 atm pressure, what is the pressure in torr?

The standard pressures indicate that 760 torr = 1 atm. By turning that equivalency into a conversion factor (760 torr/1atm or 1atm/760 torr), one can use the conversion factor to convert from atm to torr.5 atm x 760 torr =3,800 torr

1 atm

Sample Problems

What is 780 torr in mm Hg?

What is 15 atm in lbs/in2?

What is 3400 mm Hg in KPa?

What is .02 atm in Pascal?

Boyle’s Law Pressure versus Boyle’s Law Pressure versus VolumeVolume

► Robert Boyle (1627-Robert Boyle (1627-1691) was the chemist 1691) was the chemist who first proposed a law who first proposed a law to describe the behavior to describe the behavior of gases at differing of gases at differing volumes and pressures.volumes and pressures.

► Boyles law is: Boyles law is:

PP11VV11 = P = P22VV22

► As Pressure increases As Pressure increases volume must decrease volume must decrease and vice versa.and vice versa.

► Boyles law Boyles law exampleexample

Boyle’s Law is based on an Boyle’s Law is based on an Inverse RelationshipInverse Relationship

Solving Boyle’s Law Solving Boyle’s Law

► Sample problem: A Sample problem: A marshmallow has a volume marshmallow has a volume of 10mL at the initial of 10mL at the initial pressure of .89 atm. If the pressure of .89 atm. If the pressure changes to 1.02 pressure changes to 1.02 atm, what is the new atm, what is the new volume of the volume of the marshmallow? marshmallow?

► Step 1: Write out the known Step 1: Write out the known values.values.

PP11 = initial pressure = initial pressureVV11 = initial volume = initial volumePP22 = final pressure = final pressureVV22 = final volume = final volume

Step 1 continued:Step 1 continued:PP11 = 0.89 atm = 0.89 atmVV11 = 10 ml = 10 mlPP22 = 1.02 atm = 1.02 atmVV22 = x = xStep 2: Plug the values into the Step 2: Plug the values into the

formula.formula.0.89 atm x 10 ml = 1.02 atm x X0.89 atm x 10 ml = 1.02 atm x X

Step 3: Solve for X.Step 3: Solve for X.8.9 atm ml = 1.02 atm X8.9 atm ml = 1.02 atm XDivide both sides by 1.02 atm to isolate Divide both sides by 1.02 atm to isolate

X.X.8.9 atm ml8.9 atm ml = = 1.02 atm1.02 atm X X1.02 atm 1.02 atm1.02 atm 1.02 atm

8.725 ml = X8.725 ml = X

Note: Due to their inverse relationship, when Note: Due to their inverse relationship, when pressure increased, volume had to pressure increased, volume had to decreasedecrease

Example ProblemsExample Problems

► If the initial pressure of a gas is 5 atm If the initial pressure of a gas is 5 atm with a volume of 6 L, what is the new with a volume of 6 L, what is the new pressure of the gas if the volume pressure of the gas if the volume increases to 15 L?increases to 15 L?

► If the pressure of a tire is 30 psi with a If the pressure of a tire is 30 psi with a volume of 15 L, what is the pressure of volume of 15 L, what is the pressure of the tire if the volume increases to 18 the tire if the volume increases to 18 L?L?

Charles’s Law Charles’s Law ► In 1787 French physicist In 1787 French physicist

Jacques Charles studied Jacques Charles studied the effect of temperature the effect of temperature on volume when on volume when pressure is held constant pressure is held constant

► Volume and temperature Volume and temperature have a direct have a direct relationship: when relationship: when temperature temperature

iincreasesncreases,, so does volume.

Formula is:V1 = V2 T1 T2 or it can be re- written as V1T2 = V2T1

► Jacques Alexandre César CharlesJacques Alexandre César Charles (November (November 12, 1746,– April 7, 1823) was a French inventor, 12, 1746,– April 7, 1823) was a French inventor, scientist, mathematician, and balloonist.scientist, mathematician, and balloonist.

► Charles and the Charles and the Robert brothersRobert brothers launched the launched the world's first (unmanned) hydrogen-filled world's first (unmanned) hydrogen-filled balloonballoon in August 1783, then in December 1783, Charles in August 1783, then in December 1783, Charles and his co-pilot and his co-pilot Nicolas-Louis RobertNicolas-Louis Robert ascended to ascended to a height of about 1,800 feet (550 m) in a manned a height of about 1,800 feet (550 m) in a manned balloon. Their pioneering the use of hydrogen for balloon. Their pioneering the use of hydrogen for lift led to this type of balloon being named a lift led to this type of balloon being named a CharlièreCharlière

Quote from: Quote from: http://en.wikipedia.org/wiki/Jacques_Charleshttp://en.wikipedia.org/wiki/Jacques_Charles

Charles’s LawCharles’s Law

►VV11/T/T11 = V = V22/T/T22

Pressure and moles are held constant.Pressure and moles are held constant.Volume and temperature are directly Volume and temperature are directly

relatedrelatedAs temperature increases, the molecules As temperature increases, the molecules

move faster creating more pressure on move faster creating more pressure on the walls of the container. The pressure the walls of the container. The pressure is relieved by pushing the walls of the is relieved by pushing the walls of the container outward causing an increase container outward causing an increase in volume.in volume.

Volume verses Temperature Volume verses Temperature GraphGraph

►Graph of Charles Law. Volume is Graph of Charles Law. Volume is proportional to absolute temperature.proportional to absolute temperature.

How to solve Charles’ Law How to solve Charles’ Law Problems Problems

► CONVERT ALL temperature to CONVERT ALL temperature to KELVIN!!!KELVIN!!!

► Never use Celsius or Never use Celsius or Fahrenheit in the gas law Fahrenheit in the gas law formulas.formulas.

Sample Problem: A balloon has a Sample Problem: A balloon has a volume of 5L at 25volume of 5L at 25oo C. What C. What happens to the volume of the happens to the volume of the balloon if the temperature balloon if the temperature goes up to 35goes up to 35ooC? C?

Formula VFormula V11/T/T11 = V = V22/T/T22

VV11 = initial volume = initial volume

TT11 = initial temperature = initial temperature

VV22 = final volume = final volume

TT22 = final temperature = final temperature

► Step 1: Write out your know values.Step 1: Write out your know values.

VV11 = 5L = 5L

TT1 1 = 25= 25ooC +273 = 298KC +273 = 298K

VV22 = X = X

TT22 = 35 = 35ooC + 273= 308KC + 273= 308K

► Step 2: Plug the values into the Step 2: Plug the values into the formula:formula:

5L/298K = X/308K5L/298K = X/308K► Step 3: Solve for XStep 3: Solve for XCross multiply:5Lx308K = X x 298KCross multiply:5Lx308K = X x 298K1540 LK = 298K X1540 LK = 298K XDivide both sides by 298K to isolate XDivide both sides by 298K to isolate X1540 LK1540 LK = = 298K 298K XX298K 298K298K 298KX = 5.17LX = 5.17LNote: Because they are directly related, Note: Because they are directly related,

when temperature increased, so did when temperature increased, so did volumevolume

Sample Problems Sample Problems

► A gas is collected and found to fill a 2.85 L A gas is collected and found to fill a 2.85 L container at 25container at 25ooC. What will its new volume C. What will its new volume be if temperature decreases to standard be if temperature decreases to standard temperature? temperature?

► A 5.00 L sample of a gas is collected at 100 A 5.00 L sample of a gas is collected at 100 K and then allowed to expand to 20 L what K and then allowed to expand to 20 L what must the new temperature be in order to must the new temperature be in order to maintain the same pressure (as required by maintain the same pressure (as required by Charles’ Law)?Charles’ Law)?

Joseph Gay-Lussac’s LawJoseph Gay-Lussac’s Law► 1778-18501778-1850 French Physicist French Physicist

interested in the affect that interested in the affect that temperature has on pressuretemperature has on pressure

► Discovered that temperature Discovered that temperature and pressure have a direct and pressure have a direct mathematical relationship:mathematical relationship:

► Whenever temperature Whenever temperature increases, so does pressure increases, so does pressure (and vice-versa)(and vice-versa)

► His formula is as follows :His formula is as follows :

PP11/T/T11 = P = P22/T/T22

Remember: Temperature is Remember: Temperature is measured in Kelvinmeasured in Kelvin

How does Temperature affect How does Temperature affect pressure?pressure?

► If temperature is increased, the gas If temperature is increased, the gas molecules will move faster (remember: molecules will move faster (remember: kinetic molecular theory defines kinetic molecular theory defines temperature as the average speed of temperature as the average speed of molecules). The particles will collide with molecules). The particles will collide with the walls of the container more often. If the walls of the container more often. If the walls of the container are rigid the walls of the container are rigid (volume stays constant), then the (volume stays constant), then the pressure inside the container must pressure inside the container must increase.increase.

Pressure v. Temperature Pressure v. Temperature Graph Graph

Lussac’s Sample ProblemLussac’s Sample Problem

If the air inside a car has If the air inside a car has an initial temperature an initial temperature of 30of 30ooC, and a pressure C, and a pressure of 1 atm, what will the of 1 atm, what will the new pressure be if the new pressure be if the temperature increases temperature increases to 45to 45ooC?C?

PP11/T/T11 = P = P22/T/T22

PP11 = initial pressure = initial pressureTT11 = initial temperature = initial temperaturePP22 = final pressure = final pressureTT22 = final temperature = final temperature

► Step 1: Write down all your known Step 1: Write down all your known values.values.

PP11 = 1 atm = 1 atmTT11 = 30 = 30ooC + 273 = 303KC + 273 = 303KPP22 = X = XTT22 = 45 = 45ooC + 273 =318KC + 273 =318K► Step 2: Plug the values into the Step 2: Plug the values into the

formula:formula:1atm/303K = X/318K1atm/303K = X/318K► Step 3: Solve for XStep 3: Solve for XCross-multiply:Cross-multiply:318K atm = 303KX318K atm = 303KXDivide both sides by 303K to isolate X:Divide both sides by 303K to isolate X:318K atm318K atm = = 303K303KX x = 1.05 atmX x = 1.05 atm303K 303K303K 303K

Note: Because they are directly Note: Because they are directly related, when temperature increase related, when temperature increase so did pressure.so did pressure.

Sample Problems Sample Problems

►What is the pressure in a tire, if the What is the pressure in a tire, if the temperature of the tire increased from 32temperature of the tire increased from 32ooCC

to 40to 40ooC andC and its original pressure was 30 psi?its original pressure was 30 psi?

►What is the new temperature of gas in a What is the new temperature of gas in a rigid container, when pressure is increased rigid container, when pressure is increased from 1 atm to 10 atm and the original from 1 atm to 10 atm and the original temperature was 37temperature was 37ooC?C?

Combined Gas LawCombined Gas Law