Chemistry IIUnit 1 Gases

The Nature of GasesObjectives:1. Use kinetic-molecular theory to explain the

behavior of gases.2. Describe how mass affects the rates of

effusion and diffusion3. Explain how gas pressure is measured and

calculate the partial pressure of a gas

Properties of Gasestake the shape of their containerlow densityCompressibleMixtures are homogeneousFluids (flow)

Diffusion and effusionDiffusion: movement of one material through

anotherEffusion: a gas escapes through a tiny opening

Graham’s law of effusion

Lighter particles effuse faster than heavier particles.

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raterate

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Gas pressureResults from collisions of gas particles with an

object.In empty space there are no particles, there is

no pressure and (vacuum.)Atmospheric pressure (air pressure): due to

atoms and molecules in air. Barometer: used to measure atmospheric pressure.

Units for measuring pressure: Pascal (Pa) Standard atmosphere (atm) Millimeters of mercury (mmHg)

1 atm = 760 mmHg = 101.3 kPa1kpa = 1000 paStandard pressure: 1 atm

Factors affecting gas pressureAmount of gas Volume Temperature

Standard temperature : 0C (273K)

Converting between units of pressure1. A pressure gauge records a pressure of 450

kPa. What is the measurement expressed in atmospheres and millimeters of mercury?

For converting to atm:450 kpa x 1 atm = 4.4 atm 101.3kPaFor converting to mmHg:

450kPa x 760 mmHg = 3.4 x 103 mmHg 101.3 kPa

2. What pressure in kilopascals and in atmospheres, does a gas exert at 385 mmHg?

51.3 kPa, 0.507 atm

3. The pressure on the top of Mount Everest is 33.7 kPa. Is that pressure greater or less than 0.25atm?

33.7 kPa is greater than 0.25 atm

Reaction_to_Air_Pressure_Below_Sea_Level.asf

Dalton’s Law of Partial PressuresThe total pressure of a mixture of gases is

equal to the sum of the pressures of all the gases in the mixture.

Ptotal= P1 + P2 + P3 + … Pn

CW p 405 #1-3 p409 #4-7

Gas LawsObjectives1. Describe the relationships among the

temperature, pressure, and volume of a gas2. Use the gas laws to solve problems

Boyle’s Law : Pressure and VolumeStates that for a given mass of gas at

constant temperature, the volume of a gas varies inversely with pressure.

If pressure increases, volume decreases; if pressure decreases, volume increases.

Volume could be in liters (L), mL or cm3

1L=1000 mL 1 cm3= 1 mL

P1 x V1 = P2 x V2

P: pressure 1: initial condition V: volume 2: final condition

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Using Boyle’s Law1. A balloon with 30.0L of helium at 103kPa

rises to an altitude where the pressure is only 25.0kPa. What is the volume of the helium (at constant temperature)?

P1 x V1 = P2 x V2

P1= 103 kPa V1= 30.0 LP2= 25.0 kPa V2=?

V2 = P1V1

P2

= (103 kPa x 30.0L) 25.0 kPa = 123.6 L

2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure of the container now (at constant temperature)?

P1 x V1 = P2 x V2

P1= 205 kPa V1= 4.00 LP2= ? V2=12.0 L

P2 = P1V1

V2

= (205 kPa x 4.00L) 12.0L = 68.3 kPa

Classwork: pg 443 #1-3

Charles’s Law: Temperature and VolumeStates that the temperature of an enclosed

gas varies directly with the volume at constant pressure.

As temperature increases, volume increases.

V1 = V2T1 T2V1: initial volume V2: final volumeT1: initial temperature T2: final

temperatureTemperature has to be in Kelvin scale.K =C + 273

Volume and Temperature

As a gas is heated, it expands.This causes the density of thegas to decrease.

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Using Charles’s LawEx.1 A balloon inflated in a room at 24C has a volume of

4.00L . The balloon is then heated to a temperature of 58C. What is the new volume ?

T1= 24C T2=58C = 24C +273 = 58C +273 = 297K = 331 KV1= 4.00 L V2= ?

Since temperature increases, you expect the volume to increase.

V1 = V2T1 T2

V1T2 = V2 T1(4.00Lx 331K) = V2 297K4.56 L = V2

Using Charles’s LawEx.2 A sample of SO2 gas has a volume of 1.16L at a

temperature of 23C. At what temperature (in C) will the gas have a volume of 1.25L?

T1= 23C = 296K T2=?V1= 1.16 L V2= 1.25

Since volume increases, you expect the temperature to increase.

V1 = V2T1 T2

T2 = V2 x T1 V1T2 = 1.25L x 296

K 1.16 LT2 = 319 K -273 = 46.0C

Classwork: p 446 #4-7

Combined Gas LawDescribes the relationship among the

pressure, temperature and volume, when the amount of gas is constant.

P1V1 = P2V2 T1 T2Standard temperature and pressure (STP):

0C, 1 atm (101.3 kPa)Useful conversions:1L =1000 mL ; 1mL =1cm3 ; 1dm3 = 1 L

Using the combined gas law:1. The volume of a gas filled balloon is 30.0L at

313K and 153 kPa. What would the volume be at standard temperature and pressure (STP)?

Classwork: p 450 #11,12 p 984 #8,9

T1= 313 K T2=0 C=273 K (at STP)

P1= 153 kPa P2 = 101.3 kPa (at STP)

V1= 30.0L V2= ? P1V1 = P2V2 T1 T2

P1V1T2 = V2 P2T1

39.5 L = V2

(153kPa x30.0L x 273K) = V2 (101. 3 kPa x 313 K)