KMT and Gas Laws States of Matter, Kinetic Molecular Theory, Diffusion, Properties of Gases, and Gas...
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Transcript of KMT and Gas Laws States of Matter, Kinetic Molecular Theory, Diffusion, Properties of Gases, and Gas...
KMT and Gas Laws
States of Matter, Kinetic Molecular Theory, Diffusion,
Properties of Gases, and
Gas Laws
Standards4. The kinetic molecular theory describes the motion of atoms and molecules and explains the properties of
gases. As a basis for understanding this concept: a. Students know the random motion of molecules and their collisions with a surface create the observable
pressure on that surface.
4. b. Students know the random motion of molecules explains the diffusion of gases.
4. c. Students know how to apply the gas laws to relations between the pressure, temperature, and volume of any amount of an ideal gas or any mixture of ideal gases.
4. d. Students know the values and meanings of standard temperature and pressure (STP).
4. e. Students know how to convert between the Celsius and Kelvin temperature scales.
4. f. Students know there is no temperature lower than 0 Kelvin.
4. g.* Students know the kinetic theory of gases relates the absolute temperature of a gas to the average kinetic energy of its molecules or atoms.
4. h.* Students know how to solve problems by using the ideal gas law in the form PV = nRT.
4. i. * Students know how to apply Dalton’s law of partial pressures to describe the composition of gases and Graham’s law to predict diffusion of gases.
States of Matter
Solid
Liquid
Gas
Plasma
Freezing
MeltingDep
ositi
onSu
blim
ation
CondensationBoiling
Deionization
Ionization
KMT
• The path of any individual molecule could best be described as random.
KMT – Kinetic Molecular Theory
Molecular Motion• The state of matter depends
on how much energy (motion) the molecules, atoms, or ions have.
• The state of matter also depends on how attracted the atoms, molecules, or ions are to each other.
Diffusion• Diffusion – when a substance spreads out in a
gas or liquid.
• Examples: 1. Perfume eventually reaching the far side of a
room. 2. Kool-Aid dissolving into water.
Temperature (T)• Kinetic energy is the energy of motion.
• Temperature is defined as a measure of the average kinetic energy of the atoms or molecules.
Temperature (T)
• There are two scales and an absolute unit. (degrees Fahrenheit, degrees Celsius, and Kelvin)
Temperature Scales
Water Boils
Human Body
Water Freezes
212°F
98.7°F
32°F
100°C
37°C
0°C
373K
310K
273K
Temperature ScalesSurface of Sun
Room Temp.
Absolute Zero
9,941°F
70°F
-460°F
5,505°C
21°C
-273°C
5,778K
294K
0K
Converting Temperatures• Fahrenheit Celsius
°C = (°F – 32)×(5/9)
• Celsius Fahrenheit°F = °C ×(9/5) + 32
• Celsius KelvinK = °C + 273.15
Absolute Zero• At Zero Kelvin (0 K or –273.15 °C), atoms
and molecules stop moving.
• There is no temperature lower than absolute zero (0 K).
Volume (V)• How much space is occupied by a fluid.
Liquid Gas • Usually gases are measured in Liters (L)
Pressure (P)• Defined as force divided by area. • The force comes from atoms’ or molecules’
collisions with the wall of the container. • The greater the number of collisions or the
more energy with each collision, the greater the pressure.
Pressure (P)• Defined as force divided by area. • The force comes from atoms’ or molecules’
collisions with the wall of the container. • The greater the number of collisions or the
more energy with each collision, the greater the pressure.
Pressure Units
UnitUnit
Symbol 1 atm =Atmospheres atm --
Pascals Pa 101,325 Pa
Kilopascals kPa 101.3 kPa
Pounds per square inch
lbs. or psi 14.7 psi
Millimeters mercury
mm Hg 760 mm Hgin.2
Pressure0 atmOuter Space (a vacuum)
1,072 atmAt the Bottom of Mariana Trench
1 atmRegular Atmosphere (at sea level)
0.33 atmTop of Mt. Everest
Gas Properties
Property SymbolUsualUnit
Unit Symbol
Pressure P kilopascal kPa
Volume V liters L
Temperature T Kelvin K
Moles n moles mol
STP = Standard Temperature and Pressure
Temperature is 0°C = 273.15 Kand
Pressure is 1 atm = 101.3 kPa
Gas Laws• Most of the gas laws deal with taking a
quantity of gas and changing one property (pressure, temperature, or volume) and predicting how the other properties will change in response.
Boyle’s LawWhen given a certain amount of gas, if you
increase the pressure, the volume decreases. If you decrease the pressure, the volume
increases.Mathematically: P1V1 = P2V2
P1V1 = P2 V2
Boyle’s LawWhen given a certain amount of gas, if you
increase the pressure, the volume decreases. If you decrease the pressure, the volume
increases.Mathematically: P1V1 = P2V2
P1V1 = P2 V2
This assumes a constant temperature (T)
Boyle’s Law ExampleYour nephew is playing with a balloon in the car as
your family drives over a mountain pass. The balloon initially had a volume of 1 L when the car was at the bottom of the mountain (and the air pressure was 100 kPa).
Now that your family is at the top the air pressure is 70 kPa. What is the new volume of the balloon?
P1V1 = P2 V2
Boyle’s Law ExampleP1 = 100 kPa P2 = 70 kPa
V1 = 1 L V2 = ?
(100 kPa)(1 L) = (70 kPa)•V2
100 kPa•L = 70 kPa•V2
1.43 L = V2
P1V1 = P2V2
70 kPa 70 kPa
Charles’ LawWhen given a certain amount of gas, if you increase
the temperature, the volume increases. If you decrease the temperature, the volume
decreases.Mathematically: V1 V2
T1 T2
=
V1
T1
V2
T2
=
You must use Kelvin temperatures!
Charles’ LawWhen given a certain amount of gas, if you increase
the temperature, the volume increases. If you decrease the temperature, the volume
decreases.Mathematically: V1 V2
T1 T2
=
V1
T1
V2
T2
=
This assumes a constant pressure (P)
You must use Kelvin temperatures!
Charles’ Law ExampleIf 1.0 L of gas is contained within a piston at 27 ˚C (300 K), what will new volume be if the gas is cooled to -23 ˚C (250 K)? Assume that the pressure is constant.
V1
T1
V2
T2
=
Charles’ Law Example
V1
T1
V2
T2
=
V1 = 1.0 L V2 = ?
T1 = 300 K T2 = 250 K
1.0 L300 K
V2
250 K=
1.0300
V2
250= (250)(250) V2 = 0.83 L
Which Law?
7) Oxygen gas at 47 ˚C occupies a volume of 0.5 L. To what temperature should the oxygen gas be lowered to bring the volume to 0.2 L? Assume constant pressure.
V1 V2
T1 T2
=
Charles’ Law
V1
T1
V2
T2
Which Law?
9) A nitrous oxide sample (N2O) occupies a volume of 360 mL when under 70 kPa of pressure. How much volume will it occupy at 420 kPa? Assume constant temperature.
Boyle’s Law
P1V1
P2
V2
P1V1 = P2V2
Gay–Lussac’s LawWhen given a certain amount of gas, if you increase
the temperature, the pressure increases. If you decrease the temperature, the pressure
decreases.Mathematically: P1 P2
T1 T2
=
P1
T1
P2
T2
=
You must use Kelvin temperatures!
Gay–Lussac’s LawWhen given a certain amount of gas, if you increase
the temperature, the pressure increases. If you decrease the temperature, the pressure
decreases.Mathematically: P1 P2
T1 T2
=
P1
T1
P2
T2
=
This assumes a constant volume (V)
You must use Kelvin temperatures!
Gay–Lussac’s Law ExampleA tire that has already been inflated to its maximum volume, begins a drive at a temperature of 300K with an air pressure of 36 psi. During the hot day the air in the tire’s pressure increases to 42 psi. What will the new temperature be?
P2
T1
P1
T2
Gay–Lussac’s Law ExampleA tire that has already been inflated to its maximum volume, begins a drive at a temperature of 300K with an air pressure of 36 psi. During the hot day the air in the tire’s pressure increases to 42 psi. What will the new temperature be?
P2
T1
P1
T2
P1
T1
P2
T2
=
Gay–Lussac’s Law Example
P1 T1
P2 T2
=
P1 = 36 psi P2 = 42 psiT1 = 300 K T2 = ?
36 psi 300 K
42 psi T2
=
0.12 42 T2
= (T2)(T2)
T2 = 350 K0.12 (T2) = 42
0.12 0.12
Avogadro’s LawThe volume of a gas at Standard Temperature
and Pressure (STP) is directly proportional to the moles of the gas.
V = k n
At STP there are 22.4 L per mole of gas.
# of moles
Avogadro’s Law ExampleHow many liters will 3 moles of gas occupy
at STP?
V = k n
V = 67.2 LV = (22.4 )(3 mol) L
mol
Combined Gas LawThis combines Boyle’s, Charles’, and Gay-Lussac’s
Gas Laws. Mathematically:
P1 V1 P2 V2
T1 T2
=
Cancel out the properties that remain constant.
Combined Gas LawThis combines Boyle’s, Charles’, and Gay-Lussac’s
Gas Laws. Mathematically:
P1 V1 P2 V2
T1 T2
=
Cancel out the properties that remain constant.
Combined Gas Law Example
P1V1 T1
P2V2 T2
=
P1 = 50 kPa P2 = 200 kPaV1 = ? V2 = 0.12 LT1 = 300 K T2 = 400 K
(50)V1
300(200)(0.12) (400)=
(V1)0.167 = 0.06 V1 = 0.36 L 0.167 0.167
Ideal Gas LawIf we know 3 of the 4 gas properties (P, V, T, and n)
we can solve for the missing one by using the formula:
PV = nRT
R is called the gas constant. R = 8.314 kPa•L
mol•K
Ideal Gas LawA cylinder is filled with 0.2 moles of gas. The sealed
cylinder has a volume 3.0 L and is heated with 3,000 J to a temperature of 300K. What is the pressure inside the cylinder?
PV = nRT kPa•Lmol•K
P•(3.0 L) = (0.2 mol)(8.314 )(300 K)
P•(3.0 L) = 499 kPa•L
P = 166 kPa
3.0 L 3.0 L
Gas Laws SummaryGas Law FormulaBoyle’s Law P1V1 = P2V2
Charles’ Law
Gay-Lussac’s Law
Avogadro’s Law V = k nCombined Gas Law
Ideal Gas Law PV = nRT
V1 V2
T1 T2
=
P1 P2
T1 T2
=
P1 V1 P2 V2
T1 T2
=
What is the change in volume?
CH4(g) + H2O (g) CO (g) + 3 H2 (g)Methane water carbon hydrogen
monoxide
+ +
Vapor PressureFor a liquid as the temperature increases, its vapor pressure increases.
When the vapor pressure is equal to the external pressure, the liquid has reached its boiling point.
Effects of Vapor Pressure: • Warm water evaporates faster than cold water. • At higher altitudes, water boils at a lower
temperature.
Dynamic EquilibriumDynamic equilibrium occurs when the rate of condensation equals the rate of evaporation.
LiquidVapor
(or gas)
evaporation
condensation
Note: the amounts of liquid and vapor can be completely different.
Dalton’s Law of Partial PressuresThe total pressure exerted by a mixture of gases is equal to the sum of all partial pressures exerted by each individual gas.
Ptot = P1 + P2 + P3 + …
For example: our classroom
Ptot = PN2 + PO2 + PAr + PCO2 + PH2O + …1 atm = 0.79 atm + 0.19 atm + 0.01 atm + 0.003 atm + …
Graham’s Law of Diffusion• From the simulations we saw, that lighter gas
molecules move faster than heavier gas molecules.
• If we want to directly compare the speeds of gas molecules we can use:
These are the molar masses
vA MB
vB MA
=These are the average molecular speeds