Kinetic Theory of Gases

• 1. Gas particles do not attract or repel one another

• 2. Gas particles are much smaller than the distances between them

Kinetic Theory of Gases

• 3. Gas particles are in constant random motion

• 4. Gas particle collisions with other particles or their container are elastic (no kinetic energy lost)

Kinetic Theory of Gases

• 5. All gases have the same kinetic energy at a given temperature

Boyle’s Law• States that pressure and

volume are inversely proportional to one another.–As one variable increases the

other decreases

Boyle’s Law• P1V1 = P2V2

• P1V1 are the initial set of conditions

• P2V2 are the new set of conditions

Boyle’s Law• Gas in an engine cylinder is at a

pressure of 3 atm and has a volume of .65 L. The piston is then pushed down reducing the volume to .34 L, what is the new pressure in the balloon?

Boyle’s Law• P1V1 = P2V2

• (3atm)(.65L) = P2(.34L)• P2 = 5.74 atm• * As you can see the pressure

increases as the volume decreases

Boyle’s Law

Pressure

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Charles’s Law• States that as temperature

increases, volume increases

• There is a directly proportional relationship between temperature (in K) and pressure. –As one increases the other increases

Charles’s Law• V1 = V2

T1 T2

If you know three of the four values you can calculate the other

Charles’s Law• A sample of CO2 at 293oK

occupies a 4.52 L container, what volume would the container have to be reduced to in order to bring the temperature down to the freezing point of 194.5oK?

Charles’s Law• V1 = V2

T1 T2

4.52L = V2

293oK 194.5oKV1 = 3.00L

Charles’s Law

Temperature

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Gay-Lussac’s Law

• States that temperature and pressure are directly proportional to one another–As temperature increases so

does the pressure

Gay-Lussac’s Law

• P1 = P2

T1 T2

If you know three of the four values you can calculate the other

Gay-Lussac’s Law

• The pressure of a gas is raised from 4.65 atm at 30oC to 32.56 atm? What is the new temperature?

Gay-Lussac’s Law

• 4.65atm = 32.56atm

303oK T2

T2 = 2121.65 K

Gay-Lussac’s Law

Temperature

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Combined Gas Law• Shows the relationship between

pressure, temperature, and volume

• Combines Boyles, Charles’s, and Gay-Lussac’s laws into one equation

Combined Gas Law• P1V1 = P2V2

T1 T2

• Useful when dealing with many variables at one time

Combined Gas Law• A gas at 2 atm and 289oK fills a

balloon with an initial volume of .05L. If the temperature is raised to 346oK and the pressure increases to 5.6 atm, what is the new volume of the balloon?

Combined Gas Law• P1V1 = P2V2

T1 T2 • (2atm)(.05L) = (5.6atm)(V2)

289oK 346oK

V2 = .214 L

Assignment• P. 422 1-5

• P. 425 6-8

• P. 427 9-13

Avagadro’s Principle

• Equal volumes of gases that are at the same temperature and pressure have the same number of particles–1 L of He at 303o K and 1 atm and 1

L of Ne at 303o K and 1 atm contain the same number of particles

Molar Volume• The volume that a gas occupies at

0o C and at 1 atm of pressure–These conditions are known as STP

(standard temperature and pressure)

–Every gas occupies 22.4 L at STP

Molar Volume–So at STP 1 mole of any gas occupies

22.4 L–This can be used as a conversion factor

when solving stoichiometry problems22.4L1 mol

Problem• How many particles are in a

sample of gas that has a volume of 3.73L at STP?

• 3.73L x 1 mol x 6.02 x 1023 part= • 22.4 L 1 mol • 9.99 x 1022 particles

Ideal Gas Law• PV = nRT• Since we now know how to

determine the number of moles of a gas, we can use the above formula to relate four variables: Temp, Pressure, volume, and number of moles

Ideal Gas Law• The R in the formula is the

ideal gas constant.–Varies with pressure units

–See chart on p 435

How do ideal gases differ from real gases?

• At low temp or high pressure gases behave differently because of intermolecular force interaction.

• Large molecules and polar molecules also deviate from ideal behavior

Ideal gas problem• How many moles of a gas

are contained in a 4.5L container at 1235K with a pressure of 3.50 atm?

Answer• PV = nRT

• (3.5atm)(4.5L) = n(.0821L.atm/mol.K)(1235K)

• n= .15 mol

Gas Stoichiometry

• Same as stoichiometry for mass/mass or mass/volume problems.–Write what you are given first

–Cancel units until you get to the unit you need

Sample problem• What volume of H2 will be

needed to react with 2.3 L of O2 in order for complete reaction of hydrogen with oxygen?

Answer• 2.3L O2 x 2 volumes H2 = 4.6

L H2

1 volumes O2

Assignment• P. 432 25-27 odd• P. 433 29-33 odd• P. 437 41-45 odd• P. 438 45-49 odd• P. 441 57-59 odd• P. 443 61-65 odd