friday 4/27. an introduction to gases chapter 13
TRANSCRIPT
Friday4/27
An Introduction to Gases
Chapter 13
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
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.
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.
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
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)
Pressure Calculation
What is 475 mm Hg expressed in atm?
475 mm Hg=
1 atm
760 mm Hg0.625 atm
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
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
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
Relationships
How are temperature and volume related?
How are volume and pressure related?
How are pressure and temperature related?
Reminders
Homework:Gases WS 1
Reminders:Extra Credit Due 5/11
Test Corrections due 5/1
Monday and Tuesday4/30 and 5/1
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?
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
On a Larger Scale
On a Larger Scale
Gas Laws Calculations
Get out a calculator!!!
The Gas Law
PV=nRT
P = pressure ( atm or kPa )V= volume ( L )
n= number of moles (mol)T= temperature (K)
R – The Proportionality Constant
Value depends on units
8.314L (kPa)mol (K)
0.0821L (atm)mol (K)
Or
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
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
Combined Gas Law
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iii
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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?
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
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
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
Pressure & Volume• At constant Temperature• Pressure and Volume vary inversely.
– Why? – More collisions More pressure
P1V1 = P2V2
P1V1=
n1R1T1
P2V2 n2R2T2
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
Temperature & Volume
At constant PressureVolume & Temperature vary directly.– Why?– More collisions More Volume
V1=
V2
T1 T2
P1V1=
n1R1T1
P2V2 n2R2T2
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
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
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
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?
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?
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.
Quiz Time
Have out a pencil and a calculator
Cage of Death Lab
•Determine the volume of one mole of gas at STP
•Do prelab before class
•Lab write up due
Reminders
Homework:Gases WS 2Cage of Death Pre-lab
Reminders:Extra Credit Due 5/11
Wednesday and Thursday5/2 and 5/3
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?
Cage of Death Lab
•Determine the volume of one mole of gas at STP
Gas collection tubeBalancing pressureMaking cage
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
Cage of Death Lab
4.
5.
Why is it important that the HCl and H2O don’t mix?
Not too tightNot too loose
Cage of Death Lab6.
8. (F) Today’s atmospheric pressure is…
Cage of Death Lab
9.
10.
Allow reaction to happen…
How would a bubble effect your results?
Too lowToo highJust right
Cage of Death Lab
Repeat the lab for a second trial.
CLEAN UPLiquids down drain
Solids return to container
After Lab
Done in lab? Work on Homework:Cage of Death Write UpGases WS 3
Reminders:Extra Credit due 5/11
Friday5/4
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?
Demos
How does atmospheric pressure affect gas particles?
Reminders
Homework:Study for your test
Reminders:Extra Credit Due 5/11Gases Test 5/7 or 5/8