chapter 13 gas laws - bohs cp chemistry -...
TRANSCRIPT
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Chapter 13 Gas Laws
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Pressure and Force
• Pressure is the force per unit area on a surface.
Pressure = Force
Area
Chapter 13 – Gases and Pressure
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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 air is 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% CO2, and small amounts of other gases.
Chapter 13 – Gases and Pressure
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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.
Chapter 13 – Gases and Pressure
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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 who invented the barometer (used to measure 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.
Chapter 13 – Gases and Pressure
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Measuring Pressure (continued)
•Pressure can also be measured in pascals (Pa): 1 Pa = 1 N/m2.
•One pascal is very small, so usually kilopascals (kPa) are used instead.
•One atm is equal to 101.3 kPa.
1 atm = 760 mm Hg (Torr) = 101.3 kPa
Chapter 13 – Gases and Pressure
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Units of Pressure
Chapter 13 – Gases and Pressure
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Converting Pressure Sample 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 atm atm
mm Hg
1
760 x =
0.830 atm atm
kPa
1
101.3 x =
631 mm Hg
84.1 kPa
Chapter 13 – Gases and Pressure
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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 …
Chapter 13 – Gases and Pressure
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Dalton’s Law of Partial Pressures Sample 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 atm atm
mm Hg
1
760 x = 1800 mm Hg
Chapter 13 – Gases and Pressure
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#3 Gases and Pressure WS Explain how to calculate the partial pressure of a dry gas that is collected over water when the total pressure is atmospheric pressure.
Chapter 13 – Gases and Pressure
Dalton’s Law of Partial Pressures Sample Problem
• Because water molecules at a
liquid surface evaporates…gas
collected is not pure!
• The gas collected is a mixture of
dry gas and water vapor
• Use Dalton’s Law to solve for
partial pressure:
pdry gas= Patm – pwater
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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.
Chapter 13 –The Gas Laws
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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.
• This relationship between pressure and volume is inversely proportional.
Chapter 13 –The Gas Laws
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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 represent
initial conditions, and P2 and V2 represent another set of conditions.
P1V1 = P2V2
Chapter 13 –The Gas Laws
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Boyle’s Law Sample 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
Chapter 13 –The Gas Laws
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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.
Chapter 13 –The Gas Laws
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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.
Chapter 13 –The Gas Laws
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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 following relationship between the two temperature scales:
K = oC + 273
Chapter 13 –The Gas Laws
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Charles’s Law Sample 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 + 273 T1 = 25 + 273 = 298 T2 = 50 + 273 = 323
752 mL V2 =
298 K 323 K x = 815 mL
Chapter 13 –The Gas Laws
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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.
Chapter 13 –The Gas Laws
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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 represent
initial conditions. P2 and T2 represent another set of conditions.
P1 P2 =
T1 T2
Chapter 13 –The Gas Laws
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The Combined Gas Law
• The combined gas law is written as follows:
• Each of the other gas laws can be obtained from the combined gas law when the proper variable is kept constant.
P1 P2 =
T1 T2
V1 V2
Chapter 13 –The Gas Laws
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The Combined Gas Law Sample 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 + 273 T1 = 25 + 273 = 298 T2 = 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)
Chapter 13 –The Gas Laws
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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
Chapter 13 –The Gas Laws
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Standard Molar Volume
• Standard Temperature and Pressure (STP) is 0oC 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.
Chapter 13 –Gas Volumes and the Ideal Gas Law
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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.
Chapter 13 –Gas Volumes and the Ideal Gas Law
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Molar Volume Conversion Sample 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 L L
mol
22.4
1 x = 0.223 mol
0.768 mol mol
L
1
22.4 x = 17.2 L
Chapter 13 –Gas Volumes and the Ideal Gas Law
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Volume Ratios
• You can use the volume ratios as conversion factors just like you would use mole ratios.
2 CO(g) + O2(g) → 2 CO2(g) 2 molecules 1 molecule 2 molecules 2 mole 1 mole 2 mol 2 volumes 1 volume 2 volumes
• Example: What volume of O2 is needed to react completely with 0.626 L of CO to form CO2?
0.626 L CO
L CO
L O2
2
1 x = 0.313 L O2
Chapter 13 –Gas Volumes and the Ideal Gas Law
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The Mole Map
• You can now convert between number of particles, mass (g), and volume (L) by going through moles.
Chapter 13 –Gas Volumes and the Ideal Gas Law
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Gas Stoichiometry Sample Problem
Assume that 5.61 L H2 at STP reacts with excess CuO according to the following equation:
CuO(s) + H2(g) → Cu(s) + H2O(g)
a. How many moles of H2 react?
b. How many grams of Cu are produced?
5.61 L H2 L H2
mol H2
22.4
1 x = 0.250 mol H2
5.61 L H2 L H2
mol H2
22.4
1 x mol H2
mol Cu
1
1 x mol Cu
g Cu
1
63.5 x = 15.9 g Cu
Chapter 13 –Gas Volumes and the Ideal Gas Law
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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
Chapter 13 –Gas Volumes and the Ideal Gas Law
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The Ideal Gas Law Sample 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 = nRT
P (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)
Chapter 13 –Gas Volumes and the Ideal Gas Law
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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.
Chapter 13 –Diffusion and Effusion
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Graham’s Law of Effusion
• Light molecules move faster than heavy ones. • Graham’s
law of effusion says the greater the molar mass of a gas, the slower it will effuse.
Chapter 13 –Diffusion and Effusion