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Chemistry II. Unit 1 Gases. The Nature of Gases. Objectives: Use kinetic-molecular theory to explain the behavior of gases. Describe how mass affects the rates of effusion and diffusion Explain how gas pressure is measured and calculate the partial pressure of a gas. - PowerPoint PPT Presentation
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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.
molarmassmolarmass
raterate
A
B
B
A
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)