friday 4/27. an introduction to gases chapter 13
TRANSCRIPT
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Friday4/27
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An Introduction to Gases
Chapter 13
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Kinetic Molecular Theory
• Postulate #1– Gases consist of tiny particles (atoms or
molecules)
• Postulate #2– These particles are so small, compared
with the distances between them, that the volume (size) of the individual particles can be assumed to be negligible (zero).
– Gases are COMPRESSIBLE
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Kinetic Molecular Theory
• Postulate #3– The particles are in constant random
motion, colliding with the walls of the container. These collisions with the walls cause the pressure exerted by the gas.
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Kinetic Molecular Theory
• Postulate #4– The particles are assumed not to
attract or to repel each other.
• Postulate #5– The average kinetic energy of the gas
particles is directly proportional to the Kelvin temperature of the gas.
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Pressure
What is pressure?What is pressure?
How is it measured?How is it measured?1.1.mmHg (or Torr)mmHg (or Torr)2.2.Atmospheres (atm)Atmospheres (atm)3.3.Pascals (used in physics: Pascals (used in physics:
1 pascal = 1 newton per square meter)1 pascal = 1 newton per square meter)4. 4. psi psi
Equivalences:Equivalences:1 atm = 760 mmHg1 atm = 760 mmHg1 atm = 101,325 Pa = 101.325 kPa1 atm = 101,325 Pa = 101.325 kPa1 atm = 14.7 psi
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Pressure
Pressure of air is Pressure of air is measured with a measured with a BAROMETERBAROMETER(developed by Torricelli in (developed by Torricelli in 1643)1643)
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Pressure Calculation
What is 475 mm Hg expressed in atm?
475 mm Hg=
1 atm
760 mm Hg0.625 atm
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Dalton’s Law
“The Law of Partial Pressure”
• The total pressure of a mixture of gases is the sum of the partial pressures of the gases in the mixture.
Ptotal = PA + PB + PC
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Temperature ScalesTemperature Scales
Notice that 1 kelvin = 1 degree Celsius
Boiling point of water
Freezing point of water
CelsiusCelsius
100 ˚C100 ˚C
0 ˚C0 ˚C
100˚C100˚C
KelvinKelvin
373 K373 K
273 K273 K
100 K100 K
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Calculations Using Temperature
ALL gas calculations require temperature in Kelvin
T (K) = T(˚C) + 273.15
Body temp = 37 ˚C + 273 = 310 K
Liquid nitrogen = -196 ˚C + 273 = 77 K
ALL gas calculations require temperature in Kelvin
T (K) = T(˚C) + 273.15
Body temp = 37 ˚C + 273 = 310 K
Liquid nitrogen = -196 ˚C + 273 = 77 K
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Relationships
How are temperature and volume related?
How are volume and pressure related?
How are pressure and temperature related?
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Reminders
Homework:Gases WS 1
Reminders:Extra Credit Due 5/11
Test Corrections due 5/1
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Monday and Tuesday4/30 and 5/1
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Warm Up
When you increase the temperature in a container, do the particles of gas move faster or slower? Would this increase or decrease the pressure?
What would happen if you put a balloon in the freezer?
What would happen if you put a balloon in the oven?
Is it possible to compress a gas?
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Put a few drops of water in a can. Heat the can until the water boils. What is happening to the gas inside? Now flip the can over
into cold water. Predict what do you predict will happen?
Demo
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On a Larger Scale
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On a Larger Scale
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Gas Laws Calculations
Get out a calculator!!!
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The Gas Law
PV=nRT
P = pressure ( atm or kPa )V= volume ( L )
n= number of moles (mol)T= temperature (K)
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R – The Proportionality Constant
Value depends on units
8.314L (kPa)mol (K)
0.0821L (atm)mol (K)
Or
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The Gas Law – Problem If 7.0 moles of an ideal gas has a volume
of 12.0 L with a temperature of 300. K, what is the pressure in kPa?
P (12.0 L) =(7.0 mol)(300
K)8.31
4
L (kPa)
mol (K)
PV = nRT
P = 1454.95 kPa P = 1500 kPa
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The Gas Law – Problem If 4.00 moles of a gas has a volume of
10.0 L with a temperature of 303. K, what is the pressure in atm?
P (10.0 L)
=(4.00 mol)
(303 K)0.082
1
L (kPa)
mol (K)
PV = nRT
9.95 atm
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Combined Gas Law
fff
iii
ff
ii
TRn
TRn
VP
VP
Let’s say we have a balloon full of O2 gas AND we change some conditions. Would there be anything similar between the two gases?
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Combined Gas Law – Problem
You have 3.0 moles of a solution at 300. K and 15 atm in a 2.0 L container. If the container is heated to 350. K and the volume decreased to 1.0 L, what will the new pressure be?
P1 15 atm P2 want
V1 2.0 L V2 1.0 L
n1 3.0 moles n2 3.0 moles
R1 constant R2 constant
T1 300. K T2 350. K
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Combined Gas Law – Problemc
P1V1=
n1R1T1
P2V2 n2R2T2
P1V1=
T1
P2V2 T2
If we know that R1 = R2 and the mass is constant then
(15 atm)(2.0 L) =(300. K)
P2(1.0L) (350. K)
Replace with numbers
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Combined Gas Law – Problem
(15 atm)(2.0 L) = (300. K)
P2(1.0 L) (350. K)
(15 atm)(2.0 L)(350. K) =
P2(1.0L)(300. K)
P2 = 35 atm
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Pressure & Volume• At constant Temperature• Pressure and Volume vary inversely.
– Why? – More collisions More pressure
P1V1 = P2V2
P1V1=
n1R1T1
P2V2 n2R2T2
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P & V – Example Problem If you start with 0.500 L of a gas at 7.0 atm and you move the gas to a container with 3.5 L available, how much pressure will the gas exert?
7.0 atm (0.500 L) = P2
3.5 L
P1 (V1) = P2 (V2)
7.0 atm (0.500 L) = P2 (3.5 L)
1.0 atm = P2
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Temperature & Volume
At constant PressureVolume & Temperature vary directly.– Why?– More collisions More Volume
V1=
V2
T1 T2
P1V1=
n1R1T1
P2V2 n2R2T2
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T & V – Example Problem
If a gas is in a balloon with a volume of 12.0 L and at a temperature of 300. K, what will the volume be if you place the balloon in a freezer at 250. K?
V1=
V2
T1 T2
12.0 L=
V2
300. K250.
K
12.0 L (250. K) = V2
300. K
10.0 L = V2
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S.T.P.
• Standard Temperature and PressureStandard Temperature and Pressure
These are conditions that are universalThese are conditions that are universal
Standard Temperature: Standard Temperature:
0ºC or 273.15 K0ºC or 273.15 K
Standard Pressure: Standard Pressure:
1atm or 101.325kPa1atm or 101.325kPa
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S.T.P. – Example Problem
What is the volume What is the volume of 1 mole of a gas of 1 mole of a gas at STP?at STP?
P 1 atm
V want
n 1 mole
R 0.0821 (L)(atm)/(K)(mole)
T 273 KPV = nRT
(1atm)V = (1 mole)(0.0821 [Latm/Kmole])(273K)
V= 22.4 L
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Practice Problems
The pressure of a sample of gas is 5.00 atm and the volume is 30.0 L. If the volume is changed to 50.0 L, what is the new pressure?
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Practice Problems
A sample of gas has a volume of 50.0 L at a temperature of 300.K. What temperature would be needed for this sample to have a volume of 60.0 L if its pressure remains constant?
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Practice Problems
A 3.68g sample of a certain diatomic gas occupies a volume of 3.00 L at 1.00 atm and a temperature of 45°C. Identify this gas.
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Quiz Time
Have out a pencil and a calculator
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Cage of Death Lab
•Determine the volume of one mole of gas at STP
•Do prelab before class
•Lab write up due
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Reminders
Homework:Gases WS 2Cage of Death Pre-lab
Reminders:Extra Credit Due 5/11
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Wednesday and Thursday5/2 and 5/3
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Warm Up
A sample of gas has a volume of 90.0 L at a temperature of 303.K. What temperature would be needed for this sample to have a volume of 70.0 L if its pressure remains constant?
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Cage of Death Lab
•Determine the volume of one mole of gas at STP
Gas collection tubeBalancing pressureMaking cage
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Cage of Death Lab
2.
Why is the length of the magnesium ribbon important? Think back to stoichiometry.
3. Be careful with the HCl – 3.0 M is very corrosive
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Cage of Death Lab
4.
5.
Why is it important that the HCl and H2O don’t mix?
Not too tightNot too loose
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Cage of Death Lab6.
8. (F) Today’s atmospheric pressure is…
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Cage of Death Lab
9.
10.
Allow reaction to happen…
How would a bubble effect your results?
Too lowToo highJust right
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Cage of Death Lab
Repeat the lab for a second trial.
CLEAN UPLiquids down drain
Solids return to container
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After Lab
Done in lab? Work on Homework:Cage of Death Write UpGases WS 3
Reminders:Extra Credit due 5/11
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Friday5/4
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Warm Up
A sample of gas is in a 13.0 L container with 1.26 atm of pressure on it at 23.5 ˚C. How many moles of gas are in the sample?
If the gas in the problem above is released from its container into a 56.0 L container but the temperature remains constant, what will the new pressure be?
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Demos
How does atmospheric pressure affect gas particles?
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Reminders
Homework:Study for your test
Reminders:Extra Credit Due 5/11Gases Test 5/7 or 5/8