Unit 6 Gases and Gas Laws. Gases in the Atmosphere The atmosphere of Earth is a layer of gases...
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Transcript of Unit 6 Gases and Gas Laws. Gases in the Atmosphere The atmosphere of Earth is a layer of gases...
![Page 1: Unit 6 Gases and Gas Laws. Gases in the Atmosphere The atmosphere of Earth is a layer of gases surrounding the planet that is retained by Earth's gravity.](https://reader030.fdocuments.net/reader030/viewer/2022012901/56649e4e5503460f94b45005/html5/thumbnails/1.jpg)
Unit 6Gases and Gas Laws
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Gases in the Atmosphere
• The atmosphere of Earth is a layer of gases surrounding the planet that is retained by Earth's gravity.
• By volume, dry airis 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% CO2, and small amounts of other gases.
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Air Pollution
• Human activity has polluted the air with other gases:– Sulfur Oxides (SO2 & SO3) – produced from
coal burning. Contribute to acid rain.– Nitrogen Oxides (NO & NO2) – produced by burning
fossil fuels. Contribute to acid rain.– Carbon Monoxide (CO) –
emitted by motor vehicles.– Ground-level Ozone (O3) –
produced when products offossil fuel combustion reactin the presence of sunlight.
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The Ozone Layer• O3 in the troposphere (ground-level ozone) is a
pollutant, but O3 in the stratosphere is a necessary part of our atmosphere.
• Stratospheric O3 protectsus by absorbing UV light.
• CFCs destroy stratospheric O3, and have been banned in the US.
• Ozone: Good up high,bad nearby.
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Atmospheric Pressure
• Atmospheric pressure is the force per unit area exerted on a surface by the weight of the gases that make up the atmosphere above it.
Pressure =Force
Area
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Measuring Pressure
• A common unit of pressure is millimeters of mercury (mm Hg).
• 1 mm Hg is also called 1 torr in honor of Evangelista Torricelli whoinvented the barometer (used tomeasure atmospheric pressure).
• The average atmospheric pressure at sea level at 0°C is 760 mm Hg, so one atmosphere (atm) of pressure is 760 mm Hg.
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Measuring Pressure (continued)
• The pressure of a gas sample in the laboratory is often measured with a manometer.
• the difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere.
For this sample, the gas has a larger pressure than the atmosphere.
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Measuring Pressure (continued)
•Pressure can also be measured in pascals (Pa): 1 Pa = 1 N/m2.
•One pascal is verysmall, so usually kilopascals (kPa) are used instead.
•One atm is equal to 101.3 kPa.
1 atm = 760 mm Hg (Torr) = 101.3 kPa
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Units of Pressure
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Converting PressureSample Problem
The average atmospheric pressure in Denver, CO is 0.830 atm. Express this pressure in:
a. millimeters of mercury (mm Hg)
b. kilopascals (kPa)
0.830 atmatm
mm Hg1
760x =
0.830 atmatm
kPa1
101.3x =
631 mm Hg
84.1 kPa
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Dalton’s Law of Partial Pressures
• Dalton’s law of partial pressures - the total pressure of a gas mixture is the sum of the partial pressures of the component gases.
PT = P1 + P2 + P3 …
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Dalton’s Law of Partial PressuresSample Problem
A container holds a mixture of gases A, B & C. Gas A has a pressure of 0.5 atm, Gas B has a pressure of 0.7 atm, and Gas C has a pressure of 1.2 atm.
a.What is the total pressure of this system?
b. What is the total pressure in mm Hg?
PT = P1 + P2 + P3 …PT = 0.5 atm+ 0.7 atm + 1.2 atm = 2.4 atm
2.4 atmatm
mm Hg1
760x = 1800 mm Hg
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The Kinetic-Molecular Theory: A Model for Gases
• Matter is composed of particles which are constantly moving.
• The average kinetic energyof a particle is proportionalto its Kelvin temperature.
• The size of a particle is negligibly small.
• Collisions are completelyelastic – energy may beexchanged, but not lost(like billiard balls.)
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Ideal Gases
• The kinetic-molecular theory assumes:1) no attractions between gas molecules2) gas molecules do not take up space
• An Ideal Gas is a hypothetical gas that perfectly fits the assumptions of the kinetic-molecular theory.
• Many gases behave nearlyideally if pressure is not veryhigh and temperature is not very low.
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Properties of Gases: Fluidity
• Gas particles glide easily past one another.
Because liquids and gases flow, they are both referred to as fluids.
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Properties of Gases: Expansion
• Since there’s no significantattraction between gasmolecules, they keep moving around and spreading out until they fill their container.
• As a result, gases take the shape and the volume of the container they are in.
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Properties of Gases: Low Density
• Gas particles are very far apart.There is a lot of unoccupied space in the structure of a gas.
• Since gases do not have a lot of mass in a given volume, they have a very low density
• The density of a gas is about 1/1000 the density of the same substance as a liquid or solid.
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Properties of Gases: Compressibility
• Because there is a lot of unoccupied space in the structure of a gas, the gas molecules can easily be squeezed closer together.
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Diffusion and Effusion
• Diffusion is the gradual mixing of two or more gases due to their spontaneous, random motion.
• Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.
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Rate of Diffusion
• Light molecules move faster than heavy ones. • The greater the molar mass of a gas,
the slower it will diffuse and/or effuse.
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Graham’s Law of Effusion
• Graham’s Law of Effusion states that the rate of effusion is inversely proportional to the square root of the molar mass of the gas.
• The ratio of effusion rates of two different gases is given by the following equation:
Agas
B gas
B gas
Agas
MassMolar
MassMolar
rate
rate
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Graham’s Law of EffusionSample Problem
Calculate the molar mass of a gas that effuses at a rate 0.462 times N2 .
Solution:
(0.462)2 = 28.0 g/molMMunknown
MMunknown = = 131 g/mol
Agas
B gas
B gas
Agas
MassMolar
MassMolar
rate
rate
28.0 g/mol(0.462)2
= 28.0 g/mol(0.213)
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Gases and Pressure• Gas pressure is caused by collisions of the gas
molecules with each other and with the walls of their container.
• The greater the number of collisions, the higher the pressure will be.
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Pressure – Volume Relationship
• When the volume of a gas is decreased, more collisions will occur.
• Pressure is caused by collisions.
• Therefore, pressure will increase.
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Boyle’s Law
• Boyle’s Law – The volume of a fixed mass of gas varies inversely with the pressure at a constant temperature.
• P1 and V1 representinitial conditions, andP2 and V2 representanother set of conditions.
P1V1 = P2V2
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Boyle’s LawSample Problem
A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant?Solution:P1V1 = P2V2(0.947 atm)(150.0 mL) = (0.987 atm)V2
V2 =(0.947 atm)(150.0 mL)
(0.987 atm)= 144 mL
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Volume – Temperature Relationship
• the pressure of gas inside and outside the balloon are the same.
• at low temperatures, the gas molecules don’t move as much – therefore the volume is small.
• at high temperatures, the gas molecules move more – causing the volume to become larger.
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• Charles’s Law – The volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature.
Charles’s Law
V1 V2=T1 T2
• V1 and T1 represent initial conditions, and V2 and T2 represent another set of conditions.
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The Kelvin Temperature Scale
• Absolute zero – The theoretical lowest possible temperature where all molecular motion stops.
• The Kelvin temperature scale starts at absolute zero (-273oC.)
• This gives the followingrelationship between the two temperature scales:
K = oC + 273
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Charles’s LawSample Problem
A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?Solution:V1 V2=T1 T2
752 mL V2=298 K 323 K
K = oC + 273T1 = 25 + 273 = 298T2 = 50 + 273 = 323
752 mLV2 =298 K
323 Kx = 815 mL
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Pressure – Temperature Relationship
• Increasing temperature means increasing kinetic energy of the particles.
• The energy and frequency of collisions depend on the average kinetic energy of the molecules.
• Therefore, if volume is kept constant, the pressure of a gas increases with increasing temperature.
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Gay-Lussac’s Law
• Gay-Lussac’s Law – The pressure of a fixed mass of gas varies directly with the Kelvin temperature.
• P1 and T1 representinitial conditions.P2 and T2 representanother set of conditions.
P1 P2=T1 T2
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Gay-Lussac’s LawSample Problem
The gas in a container is at a pressure of 3.00 atm at 25°C. What would the gas pressure in the container be at 52°C?Solution:P1 P2=T1 T2
3.00 atm P2=298 K 325 K
K = oC + 273T1 = 25 + 273 = 298T2 = 52 + 273 = 325
3.00 atmP2 =298 K
325 Kx = 3.27 atm
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The Combined Gas Law
• The combined gas law is written as follows:
• Each of the other simple gas laws can be obtained from the combined gas law when the proper variable is kept constant.
P1 P2=T1 T2
V1 V2
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The Combined Gas LawSample Problem
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?Solution:
(1.08 atm) (0.855 atm)=
298 K 283 K
K = oC + 273T1 = 25 + 273 = 298T2 = 10 + 273 = 283
(1.08 atm)V2 =(298 K)
(283 K)= 60.0 L
P1 P2=T1 T2
V1 V2
(50.0 L) V2
(50.0 L)(0.855 atm)
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Avogadro’s Law
• In 1811, Amedeo Avogadro discovered that the volume of a gas is proportional to the number of molecules (or number of moles.)
• Avogadro’s Law - equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, or:
V1 V2=n1 n2
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The Ideal Gas Law
• All of the gas laws you have learned so far can be combined into a single equation, the ideal gas law:
• R represents the ideal gas constant which has a value of 0.0821 (L•atm)/(mol•K).
PV = nRT
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The Ideal Gas LawSample Problem
What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K?Solution:
PV = nRTP (10.0 L) = (0.500 mol)(0.0821 L•atm/mol•K)
P =(0.500 mol) (298 K)
(10.0 L)= 1.22
atm
(298 K)
(0.0821 L•atm/mol•K)
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Molar Mass of a Gas
• The ideal gas law can beused in combination withmass measurements to calculate the molar mass of an unknown gas.
• Molar mass is calculated by dividing the mass (in grams) by the amount of gas (in moles.)
Molar Massg
=mol
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Molar Mass of a GasSample Problem
Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg.Solution:
PV = nRT
PVRT=
886 mmHg
T =
P =
(1.17 atm)
(328 K)
(0.225 L) =
31.8 g/mol
55 + 273 = 328 K
(0.0821 L•atm/mol•K)n =
760. mmHg1 atm
1.17 atm=
0.00978 mol
gramsmole=MM =
0.311 g0.00978 mol =
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Standard Molar Volume
• Standard Temperature and Pressure (STP) is 0oC (273 K) and 1 atm.
• The Standard Molar Volume of a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L.
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Molar Volume Conversion Factor
• Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa.
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Molar Volume ConversionSample Problem
a. What quantity of gas, in moles, is contained in 5.00 L at STP?
b. What volume does 0.768 moles of a gas occupy at STP?
5.00 LL
mol22.41x = 0.223 mol
0.768 molmol
L1
22.4x = 17.2 L
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The Mole Map Revisited
• As you recall, you can convert between number of particles, mass (g), and volume (L) by going through moles.
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Stoichiometry Revisited• Remember that you can use mole ratios and
volume ratios (gases only) as conversion factors:
2CO(g) + O2(g) → 2CO2(g)2 molecules 1 molecule 2 molecules2 mole 1 mole 2 mol2 volumes 1 volume 2 volumes
• Example: What volume of O2 is needed to react completely with 0.626 L of CO at the same temperature & pressure conditions to form CO2?
0.626 L COL COL O2
21x = 0.313 L O2
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Gas StoichiometrySample Problem
What volume (in L) of H2 at 355 K and 738 mmHg is required to synthesize 35.7 g of methanol, given:
CO(g) + 2H2(g) → CH3OH(g)Solution:First, use stoichiometry to solve for moles of H2:
Then, use the ideal gas law to find the volume of H2:
35.7 g CH3OHg CH3OH
mol CH3OH32.01 = 2.23 mol H2
66.9 L H2
mol H22mol CH3OH1
nRTP=
(0.971 atm)(355 K)(2.23mol)
=(0.0821 L•atm/mol•K)
V =