Pressure Force acting on a unit area of a surface.
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Transcript of Pressure Force acting on a unit area of a surface.
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PressureForce acting on a unit area of a surface
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Standard Atmospheric PressureBarometric Pressure
• Air has weight and therefore exerts pressure.
• A standard Atmosphere (1atm) is the average pressure at sea level.
• What happens to air pressure as you increase your elevation?
• How do we measure pressure?
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Kinetic Molecular Theory
• Particles in an ideal gas…– have elastic collisions. – are in constant, random, straight-line motion.– don’t attract or repel each other.– have an avg. KE directly related to Kelvin temperature.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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Characteristics of Ideal Gases
Gases expand to fill any container.
Gases are fluids (like liquids).
Gases have very low densities.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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Characteristics of Gases
• Gases can be compressed.
• Gases undergo diffusion.– random motion
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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Properties of Gases
V = volume of the gas (liters, L)
T = temperature (Kelvin, K)
P = pressure (atmospheres, atm)
n = amount (moles, mol)
Gas properties can be modeled using math.Model depends on:
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Pressure - Temperature - Volume Relationship
P T V P T V
Gay-Lussac’s P T
Charles V T
P T
V
P T
V P T V P T V
Boyle’s P 1V ___
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Boyle’s LawP1V1 = P2V2
(Temperature is held constant)
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• As the pressure on a gas increases
• As the pressure on a gas increases -
the volume decreases
• Pressure and volume are inversely related
1 atm
4 Liters
2 atm
2 Liters
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Pressure vs. Volume for a Fixed Amount of Gas
(Constant Temperature)
0 100 200 300 400 500
Pressure Volume PV
(Kpa) (mL)
100 500 50,000
150 333 49,950
200 250 50,000
250 200 50,000
300 166 49,800
350 143 50,500
400 125 50,000
450 110 49,500
Vol
ume
(mL)
100
200
300
400
500
600
Pressure (KPa)
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Boyle’s Law Illustrated
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 404
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Pressure Units
• Atmosphere• Feet of water
• mm Hg• cm Hg• Torr• Barr
• mbarr• kPa• Pa
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PV Calculation (Boyle’s Law)
A quantity of gas has a volume of 120 dm3 when confined under a pressure of 93.3 kPa at a temperature of 20 oC. At what pressure will the volume of the gas be 30 dm3 at20 oC?
P1 x V1 = P2 x V2
(93.3 kPa) x (120 dm3) = (P2) x (30 dm3)
P2 = 373.2 kPa
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Volume and Pressure
Two-liter flask
The molecules arecloser together; thedensity is doubled.
The average molecules hits the wall twice as often. The total number of impacts with the wall is doubled and the pressure is doubled.
One-liter flask
Bailar, Jr, Moeller, Kleinberg, Guss, Castellion, Metz, Chemistry, 1984, page 101
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Charles' Law
This means, for example, that Volume goes up as Temperature goes up.
Jacques Charles(1746 - 1823)
Isolated boron and studied gases.Balloonist.
A hot air balloon is a good example of Charles's law.
VV and TT are directly related.
T1 T2
V1 V2=
(Pressure is held constant)
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• Raising the temperature of a gas increases the pressure if the volume is held constant.
• The molecules hit the walls harder.
• The only way to increase the temperature at constant pressure is to increase the volume.
Temperature
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VT Calculation (Charles’ Law)
At constant pressure, the volume of a gas is increased from150 dm3 to 300 dm3 by heating it. If the original temperatureof the gas was 20 oC, what will its final temperature be (oC)?
T1 = 20 oC + 273 = 293 KT2 = X KV1 = 150 dm3
V2 = 300 dm3
150 dm3
293 K= 300 dm3
T2
T2 = 586 K
oC = 586 K - 273
T2 = 313 oC
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Temperature and the Pressure of a Gas
High in the mountains, Richard checked the pressure of his car tires and observed that they has 202.5 kPa of pressure. That morning, the temperature was -19 oC. Richard then drove all day, traveling through the desert in the afternoon. The temperature of the tires increased to 75 oC because of the hot roads. What was the new tire pressure? Assume the volume remained constant. What is the percent increase in pressure?
P1 = 202.5 kPa P2 = X kPaT1 = -19 oC + 273 = 254 KT2 = 75 oC + 273 = 348 K
202.5 kPa 254 K
= P2
348 K
P2 = 277 kPa
% increase = 277 kPa - 202.5 kPa x 100 % 202.5 kPa
or 37% increase
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The pressure and absolute temperature (K) of a gas are directly related – at constant mass & volume
P
T
Gay-Lussac’s Law
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Temperature (K)
Pressure(torr)
P/T(torr/K)
248 691.6 2.79
273 760.0 2.78
298 828.4 2.78
373 1,041.2 2.79
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The Combined Gas Law
When measured at STP, a quantity of gas has a volume of 500 dm3. What volume will it occupy at 0 oC and 93.3 kPa?
P1 = 101.3 kPaT1 = 273 KV1 = 500 dm3
P2 = 93.3 kPaT2 = 0 oC + 273 = 273 KV2 = X dm3
(101.3 kPa) x (500 dm3) = (93.3 kPa) x (V2)
273 K 273 K
V2 = 542.9 dm3
1 1 2 2
1 2
PV PV
T T
(101.3) x (500) = (93.3) x (V2)
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Standard Temperature and Pressure
• 0o C
• 1 atm = 760 mmHg = 760 torr = 101.3 kPa
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Collecting Gasses over a liquid (such as water)
• Some water vapor will be mixed with the collected gas and will exert pressure.
• The partial pressure of the water vapor for each temperature is constant.
• So, take the pressure of the gas and subtract the partial pressure of water at the given temperature.