Chapter 11
description
Transcript of Chapter 11
![Page 1: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/1.jpg)
CHAPTER 11Gases
![Page 2: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/2.jpg)
10. 1 Kinetic Molecular Theory State the kinetic-molecular theory of matter, and
describe how it explains certain properties of matter.
List the five assumptions of the kinetic-molecular theory of gases.
Define the terms ideal gas and real gas.
Describe each of the following characteristic properties of gases: expansion, density, fluidity, compressibility, diffusion, and effusion.
Describe the conditions under which a real gas deviates from “ideal” behavior.
![Page 3: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/3.jpg)
What is the Kinetic Molecular Theory?
Break it down: Kinetic: movement Molecular: particles Theory: tested ideas
Tested ideas about the movement of particles!
Describes microscopic properties of molecules and their interactions, leading to observerable macroscopic properties pressure, volume, temperature
![Page 4: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/4.jpg)
KMT of Gases Ideal gas: hypothetical gas that satisfies all 5 ideas
of KMT pressure is not too high temperature is not too low
1. Gases consist of large numbers of tiny particles that are far apart relative to their size.
Most of the volume is empty space
2. Collisions between gas particles and between particles and container walls are elastic collisions.
elastic collision when there is no net loss of total kinetic energy
![Page 5: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/5.jpg)
KMT cont.3. Gas particles are in continuous, rapid, random
motion. 4. There are no forces of attraction between gas
particles.5. The temperature of a gas depends on the
average kinetic energy of the particles of the gas.
The kinetic energy of any moving object is given by the following equation:
KE m 212
![Page 6: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/6.jpg)
Effusion Because gases have motion, they can
travel. Effusion: process by which gas particles
pass through a tiny opening
How does effusion & diffusion relate to KMT? Gases at the same temperature have the
same KE so… Heavier gases travel (effuse/diffuse) slower Lighter gases travel faster
![Page 7: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/7.jpg)
Gas Behavior KMT applies only to ideal gasses.
Which parts are not true for real gases?
![Page 8: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/8.jpg)
Phase Diagramsshows the relationship between the
physical state of a substance and the temperature and pressure of the substance.
![Page 9: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/9.jpg)
Phase Diagram for CO2
![Page 10: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/10.jpg)
Phase Diagram for H2O
![Page 11: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/11.jpg)
• A phase diagram has three lines:• One line shows the liquid-gas equilibrium.• Another line shows the liquid-solid equilibrium• A third line shows the solid-gas equilibrium.
• triple point, the temperature and pressure at which all three states of a substance coexist at equilibrium.
• Used to identify boiling point/freezing point at any pressure.
![Page 12: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/12.jpg)
critical point: temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure.
above the critical point the substance becomes s a supercritical fluid
![Page 13: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/13.jpg)
11.1 Gases and Pressures Define pressure, give units of pressure,
and describe how pressure is measured.
State the standard conditions of temperature and pressure and convert units of pressure.
Use Dalton’s law of partial pressures to calculate partial pressures and total pressures.
![Page 14: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/14.jpg)
Pressure (P) Volume (V)Temperature (T) Mols (n)
What causes pressure?• collisions of the gas molecules with each other and
with surfaces with which they come into contact.
• depends on volume (mL or L), temperature (oF, oC, K), and the number of molecules present (mol, mmol).
4 Variables of Gases
![Page 15: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/15.jpg)
Equation for PressurePressure (P): the force per unit area on a
surface.
Pressure = Force Area
What are two ways you could increase the pressure on an object? More force on a given area, the greater the pressure.
decrease the area the force acts on, the greater the pressure.
![Page 16: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/16.jpg)
Relationship Between Pressure, Force and Area
![Page 17: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/17.jpg)
Measuring Pressure barometer: device used to measure
atmospheric pressure Pressure of
atmosphere supports a column of Hg about 760 mm above surface of mercury in dish
Can change depending on weather & elevation
![Page 18: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/18.jpg)
Measuring Pressure
![Page 19: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/19.jpg)
Units for Measuring Pressure
mm Hg : millimeters of mercuryA pressure of 1 mm Hg is also called 1 torr in honor of
Torricelli for his invention of the barometer. atm : atmosphere of pressure kPa : kiloPascal
Others…psi : pounds per square inchBar torr
1 atm = 101.3 kPa = 760 mmHg = 760 torr
![Page 20: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/20.jpg)
STP STP : Standard Temperature & Pressure
1.0 atm (or any of units of equal value) 0 oC
Used by scientists to compare volumes of gases
![Page 21: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/21.jpg)
Pressure ConversionsThe average atmospheric pressure in Denver, Colorado is 0.830 atm. Express this pressure in:
a. millimeters of mercury (mm Hg) andb. kilopascals (kPa)
Given: atmospheric pressure = 0.830 atm
Unknown: a. pressure in mm Hg b. pressure in kPa
![Page 22: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/22.jpg)
Pressure Conversions Answers
A)
B)
760 mm Hgatm mm Hg; atm mm Hg
atm
101.325 kPaatm kPa; atm kPa
atm
760 mm Hg0.830 atm
a631
tmmm Hg
101.325 kPa0.830 atm
at84.1
m kPa
![Page 23: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/23.jpg)
Dalton’s Law of Partial Pressures
The pressure of each gas in a mixture is called the partial pressure.
John Dalton discovered that the pressure exerted by each gas in a mixture is independent of that exerted by other gases present.
Dalton’s law of partial pressures: the total pressure of a gas mixture is the sum of the partial pressures of each gas.
![Page 24: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/24.jpg)
Dalton’s Law of Partial Pressures
Dalton derived the following equation:
PT = P1 + P2 + P3 + …
Total Pressure = sum of pressures of each individual gas
![Page 25: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/25.jpg)
Dalton’s Law of Partial Pressures
![Page 26: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/26.jpg)
Gases Collected by Water Displacement
Water molecules at the liquid surface evaporate and mix with the gas molecules. Water vapor, like other gases, exerts a pressure known as vapor pressure.
Gases produced in the laboratory are often collected over water. The gas produced by the reaction displaces the water in the reaction bottle.
![Page 27: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/27.jpg)
Particle Model for a Gas Collected Over Water
![Page 28: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/28.jpg)
Gases Collected by Water Displacement (ctd)
Step 1: Raise bottle until water level inside matches the water level outside. (Ptot = Patm)
Step 2: Dalton’s Law of Partial Pressures states:
Patm = Pgas + PH2O
To get Patm, record atmospheric pressure.
Step 3: look up the value of PH2O at the temperature of the experiment in a table, you can then calculate Pgas.
![Page 29: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/29.jpg)
Dalton’s Law of Partial Pressures Sample Problem
KClO3 decomposes and the oxygen gas was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0°C. respectively. What was the partial pressure of the oxygen collected?
Given:PT = Patm = 731.0 torrPH2O = 17.5 torr (vapor pressure of water at 20.0°C, from table A-8 in your book)Patm = PO2 + PH2O
Unknown: PO2 in torr
![Page 30: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/30.jpg)
Dalton’s Law Sample Problem Solution
Solution:Patm = PO2 + PH2O
PO2 = Patm - PH2O
substitute the given values of Patm and into the equation: PO2 =731.0 torr – 17.5 torr = 713.5 torr
![Page 31: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/31.jpg)
Mole Fractions (X)
Mole fraction of a gas(XA) = Moles of gas A (nA)
Total number of moles of a gas(ntot)
mole fraction: ratio of the number of moles of one component of a mixture to the total number of moles
![Page 32: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/32.jpg)
Calculating Partial Pressure
PA = XA PT
Partial pressures can be determined from mole fractions using the following equation:
![Page 33: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/33.jpg)
11.2 The Gas LawsUse the kinetic-molecular theory to explain the relationships between gas volume, temperature and pressure.
Use Boyle’s law to calculate volume-pressure changes at constant temperature.
Use Charles’s law to calculate volume-temperature changes at constant pressure.
Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume.
Use the combined gas law to calculate volume-temperature-pressure changes.
![Page 34: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/34.jpg)
Boyle’s Law If you increase the pressure on a gas in a
flexible container, what happens to the volume? If you decrease the pressure, what happens the
volume?
Pressure and volume are ________ related.
P1V1 = P2V2
Variables: pressure & volumeConstant: temperature, amount of gas
![Page 35: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/35.jpg)
Boyle’s Law
![Page 36: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/36.jpg)
Boyle’s Law Video
![Page 37: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/37.jpg)
Boyle’s Law ProblemA 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?
P1 = 0.947 atm P2 = 0.987 atmV1 = 150.0 mL V2 = ?
![Page 38: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/38.jpg)
Boyle’s Law Problem Solution
1 12
2
PVVP 2
2(0.947 atm)(150.0 mL O )0.987 at
144 mL Om
![Page 39: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/39.jpg)
Charles’ Law
If you increase the temperature of gas, what will happen to the volume?
If you decrease the temperature of a gas, what will happen to the volume?
Volume and temperature are ______ related.
Variables: volume & temperature Constant: pressure & amount of gas
1 2
1 2
V VT T
![Page 40: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/40.jpg)
Charles’ Law
![Page 41: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/41.jpg)
Charles’ Law Video
![Page 42: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/42.jpg)
Temperature in Charles Law To Convert to Kelvin K = 273 + °C. absolute zero: when all motion stops
O K = -273 oC
![Page 43: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/43.jpg)
Charles’ Law ProblemA 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?
Temperature must be in KELVIN!!!V1 = 752 mL V2 = ?T1 = 25°C T2 = 50°C
![Page 44: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/44.jpg)
Charles’ Law Sample Problem Solution
1 22
1
VTVT
(752 mL Ne)(323 K) 298 K
815 mL Ne1 22
1
VTVT
![Page 45: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/45.jpg)
Gay-Lussac’s Law If you increase the temperature of a gas
what will happen to the pressure? If you decrease the temperature of gas
what will happen to the pressure? Pressure and temperature are _____ related.
Variables: pressure & temperature Constant: volume & amount of gas
1 2
1 2
P PT T
![Page 46: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/46.jpg)
Gay-Lussac’s Law
![Page 47: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/47.jpg)
GL Law Video
![Page 48: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/48.jpg)
Gay-Lussac’s Law ProblemThe gas in a container is at a pressure of
3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?Temperature must also be in KELVIN!!!
P1 = 3.00 atm P2 = ?T1 = 25°C T2 = 52°C
![Page 49: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/49.jpg)
Gay-Lussac’s Law Problem Solution
P2 = P1T2 = (3.00 atm) (325 K) = 3.27 atm T1 298 K
1 22
1
PTPT
![Page 50: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/50.jpg)
The Combined Gas LawConstant: amount of gas
combined gas law: used when pressure, temperature, and volume change within a system
1 1 2 2
1 2
PV PVT T
NOTE: P & V are directly related to T, while P is inversely related to V
![Page 51: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/51.jpg)
Combined Gas Law Problem
A helium-filled balloon has a volume of 50.0 L at 25.0°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C?
Temperature must be in KELVIN!!
P1 = 1.08 atm P2 = 0.855 atmV1 = 50.0 L V2 = ?T1 = 25.0°C T2 = 10.0°C
![Page 52: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/52.jpg)
Combined Gas Law Problem Solution
1 1 22
2 1
PVTVPT
(1.08 atm)(50.0 L He)(283 K) (0.855 atm)(298 K)
60.0 L He1 1 22
2 1
PVTVPT
![Page 53: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/53.jpg)
End of Material for Quiz #1
![Page 54: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/54.jpg)
11.3 Gas Volumes and the Ideal Gas Law
State Avogadro’s law and explain its significance.
Define standard molar volume of a gas and use it to calculate gas masses and volumes.
State the ideal gas law.
Using the ideal gas law, calculate pressure, volume, temperature, or amount of gas when the other three quantities are known.
![Page 55: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/55.jpg)
Avogadro’s Law If you increase the amount of moles, what
happens to the volume? If you decrease the amount of moles what
happens to the volume?
Amount of moles & volume are ____ related.
Variables: volume , moles Constants: pressure, temperature
V1 = V2 n1 n2
![Page 56: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/56.jpg)
Because of Avogadro’s law equal volumes of gases at constant temperature and pressure contain equal numbers of molecules.
Avogadro determined one mole of any gas (regardless of mass differences) will expand to the same volume every time
standard molar volume of a gas:22.41410 L (rounded to 22.4 L)
![Page 57: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/57.jpg)
Molar Volume Video
![Page 58: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/58.jpg)
Deriving the Ideal Gas LawReview: Write down the combined gas law; where
do you think “n” fits in?
If both sides must equal each other, we can set one side equal to a constant. We’ll call this constant “R.”
![Page 59: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/59.jpg)
The Ideal Gas Law Equation
PV = nRT ideal gas law: relates all variables –
pressure, volume, moles, temperature
![Page 60: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/60.jpg)
Deriving the Ideal Gas Law Constant
R: ideal gas constantIts value depends on the units chosen for
pressure, volume, and temperature in the rest of the equation.
What are the standard conditions for an ideal gas?
P = n = V = T =
Plug in values into the equation and calculate. What is the constant that you get?
Usually rounded to 0.0821 (Latm/molK)
![Page 61: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/61.jpg)
Numerical Values of The Gas Constant “R”
ALWAYS MATCH UP YOUR UNITS!!!!
![Page 62: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/62.jpg)
Gas Stoichiometry Avogadro’s law can be applied in calculating the
stoichiometry of reactions involving gases. The coefficients in chemical equations of gas
reactions reflect not only mole ratios, but also volume ratios (assuming conditions remain the same).Discovered by Dalton, while exploring why water was a ratio
of 2H to 1O example2H2(g) + O2(g) → 2H2O(g)
2 molecules 1 molecule 2 molecules2 mole 1 mole 2 mol2 volumes 1 volume 2 volumes
![Page 63: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/63.jpg)
Gas Stoichiometry ProblemNumber 1 on Practice Sheet What volume of nitrogen at STP would
be required to react with 0.100 mol of hydrogen to produce ammonia?
N2 + 3 H2 2 NH3
![Page 64: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/64.jpg)
Gas Stoichiometry Problem Solution
0.100 mol H2 x 1 mol N2 x 22.4 L N2
3 mol H2 1 mol N2
= 0.747 L N2
![Page 65: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/65.jpg)
Ideal Gas Law Sample Problem A sample of carbon dioxide with a mass
of 0.250 g was placed in a 350. mL container at 400 K. What is the pressure exerted by the gas?
P = ?V = 350. mL = 0.350 Ln = 0.250 g = ? molT = 400 K
![Page 66: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/66.jpg)
Ideal Gas Law Problem Solution
nRTPV
P = nRT = .00568 mol (.0821 Latm/molK) 400 K V .350 L
= 0.533 atm
![Page 67: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/67.jpg)
Gas Stoich and Ideal Gas Law
Number 2 on Practice Sheet What volume of nitrogen at 215OC and
715 mmHg would be required to react with 0.100 mol of hydrogen to produce ammonia?
N2 + 3 H2 2 NH3
Note: This system is NOT at STP!!
![Page 68: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/68.jpg)
Gas Stoichiometry Problem Solution
0.100 mol H2 x 1 mol N2 = 0.0333 mol N2
3 mol H2
P = 715 mmHgV = ?n = 0.0333 mol N2
R = 62.4 LmmHg/molKT = 25OC + 273 = 488 K
![Page 69: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/69.jpg)
11.4 Diffusion and Effusion Describe the process of diffusion.
State Graham’s law of effusion.
State the relationship between the average molecular velocities of two gases and their molar masses.
![Page 70: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/70.jpg)
Diffusion and EffusionREMEMBER:
EFFUSION: process when the molecules of a gas confined in a container randomly pass through a tiny opening in the container
DIFFUSION: the gradual mixing of two or more gases due to their spontaneous, random motion
![Page 71: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/71.jpg)
Graham’s Law of Effusion
![Page 72: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/72.jpg)
Graham’s Law Of EffusionGraham’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.
B
A
MAB M
rate of effusion of rate of effusion of
![Page 73: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/73.jpg)
Sample Problem What is the rate of effusion of hydrogen if oxygen has a velocity of 175 m/s at the same temperature and pressure.
![Page 74: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/74.jpg)
Substitute the given values into the equation:
Hydrogen rate of effusion is …
Graham’s Law of Effusion, continued
32.00 g/molrate of effusion of 32.00 g/mol 3.98rate of effusion of 2.02 g/mol2.02 g/mol
B
A
MAB M
![Page 75: Chapter 11](https://reader035.fdocuments.us/reader035/viewer/2022062315/568164d3550346895dd702a3/html5/thumbnails/75.jpg)
Graham’s Law- Visual Problem