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Transcript of S TATES OF M ATTER. State of Matter VolumeShapeDensity Compressibility Motion of Molecules Gas...
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STATES OF MATTER
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State of Matter
Volume Shape Density Compressibility Motion of Molecules
Gas
Liquid
Solid
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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
GASESChapter 5
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Elements that exist as gases at 250C and 1 atmosphere
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PHYSICAL CHARACTERISTICS OF GASES
Assume the shape of their container Most compressible state of matter Mix evenly and completely (Diffusion) Very low density compared with solids
and liquids.Density usually measured in g/L of gas
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Pressure = ForceArea
PRESSUREPressure (P): the force per unit area on a
surface.
What causes pressure?
collisions of the gas molecules with each other and with surfaces with which they come into contact.
depends on volume, temperature, and the number of molecules present.
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CALCULATING FORCE
The SI Unit for force is the Newton (N)
Consider a person with a mass of 51 kg. At Earth’s surface, gravity has an acceleration of 9.8 m/s2. What is the value of force?
Force = mass x acceleration
Force = 51 kg × 9.8 m/s2 = 500 kg • m/s2 = 500 N
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P = 1.5 N/cm2 P = 38.5 N/cm2 P = 77 N/cm2
The greater area the less pressure on the floor.
RELATIONSHIP BETWEEN PRESSURE, FORCE, AND AREA
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BAROMETER
Barometer
device used to measure atmospheric pressure
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Sea level 1 atm
4 miles 0.5 atm
10 miles 0.2 atm
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UNITS OF PRESSURE
1 atm = 101.3 kPa = 760 mmHg = 760 torr
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GAS VS. VAPOR
Gas:Substance found normally in the gaseous
state at normal temperature and pressure
Vapor:Gaseous form of any substance that is a
liquid or a solid at normal temperature and pressure
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GAS LAWS
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BOYLE’S LAW
As volume increases, pressure _________________
As volume decreases, pressure _________________
As pressure increases, volume ________________
As pressure decreases, volume _________________
PRESSURE AND VOLUME ARE: __________________ RELATED
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As P (h) increases V decreases
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Boyle’s Law
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BOYLE’S LAW EQUATION
P 1/V
P x V = constant
P1 x V1 = P2 x V2
Constant temperatureConstant amount of gas
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A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
P1 x V1 = P2 x V2
P1 = 726 mmHg
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 = P1 x V1
V2
726 mmHg x 946 mL154 mL
= = 4460 mmHg
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CHARLES’ LAW
As temperature increases, volume ________________
As temperature decreases, volume _________________
As volume increases, temperature _________________
As volume decreases, temperature _________________
TEMPERATURE AND VOLUME ARE: __________________ RELATED
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As T increases V increases
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Variation of gas volume with temperatureat constant pressure.
Charles’ Law
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CHARLES’ LAW EQUATION
V T
V / T = constant
T (K) = t (0C) + 273
Temperature must be in Kelvin
Constant pressureConstant amount of gas
V1 = V2
T1 T2
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A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
V1 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 = V2 x T1
V1
1.54 L x 398.15 K3.20 L
= = 192 K
V1/T1 = V2/T2
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AVOGADRO’S LAW
As amount increases, volume ________________
As amount decreases, volume _________________
As volume increases, amount _________________
As volume decreases, amount _________________
AMOUNT AND VOLUME ARE: __________________ RELATED
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AVOGADRO’S LAW EQUATION
V number of moles (n)
V /n = constant
V1 = V2
n1 n2
Constant temperatureConstant pressure
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If 22.3 mol of N2 gas has a volume of 15 L, how many moles of N2 gas will have a volume of 12 L?
V1 = 15 L
n1 = 22.3 mol
V2 = 12 L
n2 = ?
n2 = V2 x n1
V1
12 L x 22.3 mol15 L
= = 18 mol
V1/n1 = V2/n2
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COMBINED GAS LAW EQUATION
Boyle’s law and Charles’s law can be combined into a single equation that can be used for situations in which temperature, pressure, and volume, all vary at the same time.
1 1 2 2
1 2
PV P V
T T
The temperature MUST be in Kelvin
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A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C?
1 1 22
2 1
PV TV
P T 1 1 2 2
1 2
PV P V
T T
(1.08 atm)(50.0 L He)(283 K)
(0.855 atm)(298 K)60.0 L He1 1 2
22 1
PV TV
P T
V1 = 50.0 L
P1 = 1.08 atm
V2 = ?
P2 = .855 atm
T1 = 25oC T2 = 10oC
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IDEAL GAS EQUATION
Charles’ law: V T(at constant n and P)
Avogadro’s law: V n(at constant P and T)
Boyle’s law: V (at constant n and T)1P
V nT
P
V = constant x = RnT
P
nT
PR is the gas constant
PV = nRT
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0 0C and 1 atm
PV = nRT
R = PVnT
=(1 atm)(22.414L)
(1 mol)(273.15 K)
R = 0.0821 L • atm / (mol • K)
Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.
STP = STANDARD TEMPERATURE AND PRESSURE
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NUMERICAL VALUES OF THE GAS CONSTANT “R”
ALWAYS MATCH UP YOUR UNITS!!!!
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What is the volume (in liters) occupied by 49.8 g of HCl at STP?
PV = nRT
V = nRTP
T = 0 0C = 273.15 K
P = 1 atm
n = 49.8 g x 1 mol HCl36.45 g HCl
= 1.37 mol
V =1 atm
1.37 mol x 0.0821 x 273.15 KL•atmmol•K
V = 30.6 L
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Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?
PV = nRT n, V and R are constant
nRV
= PT
= constant
P1
T1
P2
T2
=
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
P2 = P1 x T2
T1
= 1.20 atm x 358 K291 K
= 1.48 atm
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Density (d) Calculations
d = mV =
PMRT
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRTP
M = d is the density of the gas in g/L
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V and T are
constant
P1 P2 Ptotal = P1 + P2
DALTON’S LAW OF PARTIAL PRESSURES
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Daltons Law of Partial Pressure
•Gases mix homogeneously (form a solution) in any proportions
•Each gas in a mixture behaves as if it were the only gas present (assuming no chemical reactions).
PT = PgasA + PgasB + PgasC + etc.
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Examples:
1. An organic chemist was considering the pressures exerted by three gases (M, N, L) in a flask. The total pressure inside the flask was 456 mmHg. If gas M contributes 200 mmHg, and gas L contributes 10 mmHg, what is the pressure exerted by gas N.
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Examples:
2. An organic chemist was considering the pressures exerted by three gases (M, N, L) in a flask. The total pressure inside the flask was 644 mmHg. If gas M contributes to 21% of the pressure, and gas N contributes 54% what are the pressures exerted by all three gases in mmHg.
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MOLE FRACTIONS
Mole fraction of gas A = Moles of gas A_____ Total number of moles of gas
GAS AMOUNT IN MOLES
A 0.235B 1.025C 2.35D 0.78
Examples:1. Four gases are found in an atmospheric sample of gas. The data below indicates their respective amount. Determine the mole fraction of each.
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GAS AMOUNT IN GRAMS
He 23.5
CO2 45.7
CH4 32.3
Steam
24.7
2. Four gases are found in an atmospheric sample of gas. The data below indicates their respective amount. Determine the mole fraction of each.
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Pi = Xi PT
HOW DO PARTIAL PRESSURE AND MOLE FRACTION RELATE?
Where: Pi is the pressure i
Xi is the mole fraction of i
PT is the total pressureExamples: 1. Determine the partial pressure of oxygen, nitrogen, and argon using the following data. Total pressure of the system is 760 mmHg.
GAS AMOUNT IN MOLES
O2 20
N2 79
Ar 1
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Example:
2. Determine the partial pressure of oxygen, nitrogen, and argon using the following data. Total pressure of the system is 760 mmHg.
GAS
AMOUNT IN GRAMS
O
2
45.6
N
2
32.2
Ar
100.76
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WHAT IS VAPOR PRESSURE?
The pressure that exists above the surface of a liquid from particles escaping the surface of the liquid
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VAPOR PRESSURE IS EFFECTED BY HOW VOLATILE THE LIQUID IS
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BeforeEvaporation
At Equilibrium
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COLLECTING GASES OVER WATER
•Dalton’s Law can be used to calculate the pressure of gas collected over water•Set-up for such a system is shown below
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•The collection flask initially contains all water•When a reaction takes place in a reaction chamber•Gas travels through tubing attached to the reaction chamber•The gas displaces the water in the collection flask •When all the water is displaced the flask is stoppered and the gas and some water vapor is collected
To find the pressure attributed by the gas collected you need to subtract the pressure due to water vapor at a specific temperature from the total pressure.
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Examples:1. A gas is collected by water displacement at 50oC and barometric pressure of 95 kPa. What is the pressure exerted by the dry gas? 2. Oxygen gas is collected by water displacement from the reaction of Na2O2 and water. The oxygen displaces 318 mL of water at 23oC and 1.000atm. What is the pressure of dry O2.
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Gas Stoichiometry
What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
g C6H12O6 mol C6H12O6 mol CO2 V CO2
5.60 g C6H12O6
1 mol C6H12O6
180 g C6H12O6
x6 mol CO2
1 mol C6H12O6
x = 0.187 mol CO2
V = nRT
P
0.187 mol x 0.0821 x 310.15 KL•atmmol•K
1.00 atm= = 4.76 L
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Kinetic Molecular Theory of Gases
1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume.
2. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic.
3. Gas molecules exert neither attractive nor repulsive forces on one another.
4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy
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**Gases behave ideally at very high temperatures and low pressures**
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Effect of intermolecular forces on the pressure exerted by a gas.
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DIFFUSION AND EFFUSION
DIFFUSION: the gradual mixing of two or more gases due to their spontaneous, random motion (kinetic properties)
EFFUSION: process when the molecules of a gas confined in a container randomly pass through a tiny opening in the container
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EFFUSION RATE VIDEO
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GRAHAM’S LAW OF EFFUSION
Graham’s law of effusion: the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.
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GRAHAM’S LAW- VISUAL PROBLEM
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GRAHAM’S LAW PROBLEM
Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.
Given: identities of two gases, H2 and O2
Unknown: relative rates of effusion
Hydrogen = Compound AOxygen = Compound B
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GRAHAM’S LAW PROBLEM
HINT: Always put the substance with the larger molar mass on top as compound B.
1. Calculate:
2. Rearrange the equation:rate of effusion of A = 3.98 rate of
effusion of B3. Write a sentence:
Hydrogen diffuses 3.98 times faster than Oxygen