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Chapter 5 Chapter 5 The Gas Laws The Gas Laws

1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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Page 1: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

1

Chapter 5Chapter 5

The Gas LawsThe Gas Laws

Page 2: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

2

PressurePressure Force per unit area.Force per unit area. Gas molecules fill container.Gas molecules fill container. Molecules move around and hit the Molecules move around and hit the

insides of the container. These insides of the container. These collisions cause the pressure.collisions cause the pressure.

Measured with a barometer or a Measured with a barometer or a manometer.manometer.

Page 3: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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BarometerBarometer The pressure of the The pressure of the

atmosphere at sea atmosphere at sea level will hold a level will hold a column of mercury column of mercury 760 mm Hg.760 mm Hg.

1 atm = 760 mm Hg1 atm = 760 mm Hg

1 atm Pressure

760 mm Hg

Vacuum

Page 4: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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ManometerManometer

Gas

h

Column of Column of mercury to mercury to measure measure pressure.pressure.

‘‘h’ is how h’ is how much lower much lower the pressure is the pressure is than outside. than outside.

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ManometerManometer h is how much h is how much

higher the gas higher the gas pressure is than pressure is than the atmosphere.the atmosphere.

h

Gas

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Pressure ReadingsPressure Readings

The difference in the height of mercury on both sides is the difference in pressure between the two gases.

The gas with the higher pressure will be able to move the mercury more than the gas with the lower pressure.

Remember our “Law of Phyics” The stronger force always wins !!

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Units of pressureUnits of pressure 1 atmosphere = 760 mm Hg = 760 torr1 atmosphere = 760 mm Hg = 760 torr 1 atm = 101,325 Pascals = 101.325 kPa1 atm = 101,325 Pascals = 101.325 kPa

The above values are Standard The above values are Standard Pressure and need to be known. Pressure and need to be known.

Page 8: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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The Gas LawsThe Gas Laws Boyle’s LawBoyle’s Law Pressure and volume are inversely Pressure and volume are inversely

related at constant temperature. OR related at constant temperature. OR

As one goes up, the other goes down.As one goes up, the other goes down. PV= k was Boyle’s data and it turned PV= k was Boyle’s data and it turned

into Pinto P11VV11 = P = P22 V V22

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Boyle’s Law Boyle’s Law The following two graphs show the The following two graphs show the

data when Pressure v. Volume are data when Pressure v. Volume are plotted.plotted.

In the first graph, it shows Pressure In the first graph, it shows Pressure graphed v. Volume.graphed v. Volume.

The second graph shows the straight The second graph shows the straight line plot of Volume v. 1 / Pressureline plot of Volume v. 1 / Pressure

Page 10: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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V

P (at constant T)

Page 11: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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V

1/P (at constant T)

Slope = k

Page 12: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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Molar VolumeMolar Volume Standard Pressure = 1 atmosphereStandard Pressure = 1 atmosphere Standard Temperature = 273 KStandard Temperature = 273 K

Together they from STP conditionsTogether they from STP conditions

1 mole of any gas at STP occupies 1 mole of any gas at STP occupies 22.4 Liters of Volume22.4 Liters of Volume

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ExamplesExamples 20.5 L of nitrogen at 742 torr are 20.5 L of nitrogen at 742 torr are

compressed to 9.8 atm. What is the compressed to 9.8 atm. What is the new volume?new volume?

Pressure is changing and a new Pressure is changing and a new volume is needed = Boyle’s Lawvolume is needed = Boyle’s Law

Make sure the units match up as torr Make sure the units match up as torr does not equal atm.does not equal atm.

(20.5 L) (742 torr) = (7448 torr) V(20.5 L) (742 torr) = (7448 torr) V22

VV22 = 2.042 L = 2.042 L

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More examplesMore examples 30.6 mL of carbon dioxide at 740 torr 30.6 mL of carbon dioxide at 740 torr

is to 750 mL What is the final is to 750 mL What is the final pressure in kPa ? pressure in kPa ?

(30.6 mL) (740 torr) = (750 mL) P(30.6 mL) (740 torr) = (750 mL) P22

PP22 = 30.19 torr (needs to be in kPa) = 30.19 torr (needs to be in kPa)

30.19 torr = 4.025 kPa30.19 torr = 4.025 kPa

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Charles’ LawCharles’ Law Volume of a gas varies directly with Volume of a gas varies directly with

the absolute temperature at constant the absolute temperature at constant pressure.pressure.

V = kT (if T is in Kelvin)V = kT (if T is in Kelvin)

VV1 1 = V = V22

T T11 = T = T22

Graphically it looks like this.Graphically it looks like this.

Page 16: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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V (

L)

T (ºC)

HeH2O

CH4

H2

-273.15ºC

Charles’ Law Graph

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ExamplesExamples What would the final volume be if 247 What would the final volume be if 247

mL of gas at 22ºC is heated to 98ºC ?mL of gas at 22ºC is heated to 98ºC ?

247 / 295 = V247 / 295 = V2 2 / 371 V/ 371 V22 = 310.6 mL = 310.6 mL

Remember to convert °C to KRemember to convert °C to K

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Avogadro's LawAvogadro's Law At constant temperature and At constant temperature and

pressure, the volume of gas is pressure, the volume of gas is directly related to the number of directly related to the number of moles. When air is added to a ball to moles. When air is added to a ball to pump it up, it is putting more moles pump it up, it is putting more moles of air in it !!of air in it !!

V = k n (n is the number of moles)V = k n (n is the number of moles)

VV1 1 = V = V22

n n11 = n = n22

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Gay- Lussac LawGay- Lussac Law At constant volume, pressure and At constant volume, pressure and

absolute temperature are directly absolute temperature are directly related.related.

P = k TP = k T

PP1 1 = P = P22

T T11 = T = T22

Page 20: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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Combined Gas LawCombined Gas Law If the moles of gas remains constant, If the moles of gas remains constant,

use this formula and cancel out the use this formula and cancel out the other things that don’t change.other things that don’t change.

PP1 1 VV11 = P = P22 V V22

.. T T11 T T22

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ExamplesExamples A deodorant can has a volume of 175 A deodorant can has a volume of 175

mL and a pressure of 3.8 atm at 22ºC. mL and a pressure of 3.8 atm at 22ºC. What would the pressure be if the What would the pressure be if the can was heated to 100.ºC?can was heated to 100.ºC?

What volume of gas could the can What volume of gas could the can release at 22ºC and 743 torr?release at 22ºC and 743 torr?

Page 22: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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Ideal Gas LawIdeal Gas Law PV = nRTPV = nRT R is the ideal gas law constant.R is the ideal gas law constant. R = ? It depends on the pressure unit. R = ? It depends on the pressure unit.

0.08205 L atm / mol K OR0.08205 L atm / mol K OR

62.36 L torr / mol K62.36 L torr / mol K OR OR

8.314 J / mol K8.314 J / mol K The other laws tell you about a gas The other laws tell you about a gas

when it changes. when it changes.

Page 23: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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ExamplesExamples A 47.3 L container containing 1.62 mol of A 47.3 L container containing 1.62 mol of

He is heated until the pressure reaches He is heated until the pressure reaches 1.85 atm. What is the temperature?1.85 atm. What is the temperature?

Kr gas in a 18.5 L cylinder exerts a Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8ºC. What is pressure of 8.61 atm at 24.8ºC. What is the mass of Kr?the mass of Kr?

A sample of gas has a volume of 4.18 L A sample of gas has a volume of 4.18 L at 29ºC and 732 torr. What would its at 29ºC and 732 torr. What would its volume be at 24.8ºC and 756 torr?volume be at 24.8ºC and 756 torr?

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Solved AnswersSolved Answers1. PV = nRT1. PV = nRT

(1.85)(47.3) = 1.62 R T(1.85)(47.3) = 1.62 R T

T = 658.32 K or 385.32 °CT = 658.32 K or 385.32 °C

2. PV = nRT2. PV = nRT

(8.61) (18.5) = n (.08205)(297.8)(8.61) (18.5) = n (.08205)(297.8)

n = 6.52 moles = 546.28 gramsn = 6.52 moles = 546.28 grams

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Gas Density and Molar MassGas Density and Molar Mass Sometimes we have to combine our Sometimes we have to combine our

gas laws with formulas that we gas laws with formulas that we learned earlier, such as : learned earlier, such as :

Density = m / V OR moles = g / MDensity = m / V OR moles = g / M PV = nRT substitute for n.PV = nRT substitute for n. MM = g RT / PV = g RT / PV

Another algebra manipulation isAnother algebra manipulation is P x P x MM = gRT/ V = P x M = DRT = gRT/ V = P x M = DRT

Page 26: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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Examples Examples What is the density of ammonia at What is the density of ammonia at

23ºC and 735 torr?23ºC and 735 torr?

Page 27: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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A compound has the empirical A compound has the empirical formula CHCl. A 256 mL flask at formula CHCl. A 256 mL flask at 100.ºC and 750 torr contains .80 g of 100.ºC and 750 torr contains .80 g of the gaseous compound. What is the the gaseous compound. What is the empirical formula?empirical formula?

Page 28: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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Gases and StoichiometryGases and Stoichiometry Reactions happen in molesReactions happen in moles At Standard Temperature and At Standard Temperature and

Pressure (STP, 0ºC and 1 atm) 1 Pressure (STP, 0ºC and 1 atm) 1 mole of gas occuppies 22.42 L.mole of gas occuppies 22.42 L.

If not at STP, use the ideal gas law to If not at STP, use the ideal gas law to calculate moles of reactant or calculate moles of reactant or volume of product.volume of product.

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ExamplesExamples Mercury can be achieved by the Mercury can be achieved by the

following reaction:following reaction:

What volume of oxygen gas can be What volume of oxygen gas can be produced from 4.10 g of mercury (II) produced from 4.10 g of mercury (II) oxide at STP?oxide at STP?

At 400.ºC and 740 torr?At 400.ºC and 740 torr?

O + Hg 2 HgO 2 2heat

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ExamplesExamples Using the following reactionUsing the following reaction

Calculate the mass of sodium hydrogen Calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L carbonate necessary to produce 2.87 L of carbon dioxide at 25ºC and 2.00 atm.of carbon dioxide at 25ºC and 2.00 atm.

If 27 L of gas are produced at 26ºC and If 27 L of gas are produced at 26ºC and 745 torr when 2.6 L of HCl are added 745 torr when 2.6 L of HCl are added what is the concentration of HCl?what is the concentration of HCl?

NaCl(aq) + CO (g) +H O(l)2 2

NaHCO (s) + HCl 3

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ExamplesExamples Consider the following reactionConsider the following reaction

What volume of NO at 1.0 atm and What volume of NO at 1.0 atm and 1000º C can be produced from 10.0 L 1000º C can be produced from 10.0 L of NHof NH33 and excess O and excess O22 at the same at the same

temperature and pressure?temperature and pressure? What volume of OWhat volume of O22 measured at STP measured at STP

will be consumed when 10.0 kg NHwill be consumed when 10.0 kg NH33 is is

reacted?reacted?

4NH (g) + 5 O 4 NO(g) + 6H O(g)3 22 ( )g

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The Same reactionThe Same reaction

What mass of HWhat mass of H22O will be produced O will be produced

from 65.0 L of Ofrom 65.0 L of O22 and 75.0 L of NH and 75.0 L of NH33

both measured at STP? both measured at STP? What volume of NO would be What volume of NO would be

produced?produced? What mass of NO is produced from What mass of NO is produced from

500. L of NH3 at 250.0ºC and 3.00 500. L of NH3 at 250.0ºC and 3.00 atm?atm?

4NH (g) + 5 O 4 NO(g) + 6H O(g)3 22 ( )g

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Dalton’s LawDalton’s Law The total pressure in a container is The total pressure in a container is

the sum of the pressure each gas the sum of the pressure each gas would exert if it were alone in the would exert if it were alone in the container.container.

The total pressure is the sum of the The total pressure is the sum of the partial pressures.partial pressures.

PPTotalTotal = P = P11 + P + P22 + P + P33 + P + P44 + P + P55 ... ...

Make sure that the units match up.Make sure that the units match up.

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The mole fractionThe mole fraction Ratio of moles of the substance to Ratio of moles of the substance to

the total moles.the total moles.

Symbol is Greek letter chi Symbol is Greek letter chi It is the same as % except you don’t It is the same as % except you don’t

multiply by 100 %multiply by 100 %

= n= n11 = P= P1 1

n nTotal Total PPTotalTotal

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ExamplesExamples The partial pressure of nitrogen in air The partial pressure of nitrogen in air

is 592 torr. Air pressure is 752 torr, is 592 torr. Air pressure is 752 torr, what is the mole fraction of what is the mole fraction of nitrogen?nitrogen?

What is the partial pressure of What is the partial pressure of nitrogen if the container holding the nitrogen if the container holding the air is compressed to 5.25 atm?air is compressed to 5.25 atm?

Page 36: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit the insides of the container

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ExamplesExamples

3.50 L

O2

1.50 L

N2

2.70 atm When these valves are opened, what is When these valves are opened, what is

each partial pressure and the total each partial pressure and the total pressure?pressure?

4.00 L

CH4

4.58 atm 0.752 atm

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Vapor PressureVapor Pressure Water evaporates!Water evaporates! When that water evaporates, the When that water evaporates, the

vapor has a pressure.vapor has a pressure. Gases are often collected over water Gases are often collected over water

so the vapor pressure of water must so the vapor pressure of water must be subtracted from the total be subtracted from the total pressure.pressure.

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ExampleExample NN22O can be produced by the O can be produced by the

following reactionfollowing reaction

What volume of NWhat volume of N22O collected over O collected over

water at a total pressure of 94 kPa water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g and 22ºC can be produced from 2.6 g of NHof NH44NONO33? ( the vapor pressure of ? ( the vapor pressure of

water at 22ºC is 21 torr)water at 22ºC is 21 torr)

NH NO NO (g) + 2H O4 heat

23 2( ) ( )s l

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Kinetic Molecular TheoryKinetic Molecular Theory Explains why ideal gases behave the Explains why ideal gases behave the

way they do.way they do. Assumptions that simplify the Assumptions that simplify the

theory, but don’t work in real gases.theory, but don’t work in real gases. The particles are so small we can The particles are so small we can

ignore their volume.ignore their volume. The particles are in constant motion The particles are in constant motion

and their collisions cause pressure. and their collisions cause pressure.

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Kinetic Molecular TheoryKinetic Molecular Theory The particles do not affect each other, The particles do not affect each other,

neither attracting or repelling.neither attracting or repelling. The average kinetic energy is The average kinetic energy is

proportional to the Kelvin proportional to the Kelvin temperature.temperature.

Appendix 2 shows the derivation of Appendix 2 shows the derivation of the ideal gas law and the definition of the ideal gas law and the definition of temperature.temperature.

We need the formula KE = 1/2 mvWe need the formula KE = 1/2 mv22

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What it tells usWhat it tells us (KE)(KE)avgavg = 3/2 RT R is in Joules ! = 3/2 RT R is in Joules ! This the meaning of temperature.This the meaning of temperature. u is the particle velocity.u is the particle velocity. u is the average particle velocity.u is the average particle velocity. u u 22 is the average particle velocity is the average particle velocity

squared.squared. the root mean square velocity is the root mean square velocity is

u u 2 = 2 = uurmsrms

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Combine these two equationsCombine these two equations (KE)(KE)avgavg = N = NAA(1/2 mu (1/2 mu 22 ) )

(KE)(KE)avgavg = 3/2 RT = 3/2 RT

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Combine these two equationsCombine these two equations (KE)(KE)avgavg = N = NAA(1/2 mu (1/2 mu 22 ) )

(KE)(KE)avgavg = 3/2 RT = 3/2 RT

Where M is the molar mass in Where M is the molar mass in kg/mole, and R has the units 8.3145 kg/mole, and R has the units 8.3145 J/K mol.J/K mol.

The velocity (u) will be in m/sThe velocity (u) will be in m/s

u = 3RT

Mrms

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Example Example Calculate the root mean square Calculate the root mean square

velocity of carbon dioxide at 25ºC.velocity of carbon dioxide at 25ºC. Calculate the root mean square Calculate the root mean square

velocity of hydrogen at 25ºC.velocity of hydrogen at 25ºC. Calculate the root mean square Calculate the root mean square

velocity of chlorine; Clvelocity of chlorine; Cl22 at 25ºC. at 25ºC.

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Range of velocitiesRange of velocities The average distance a molecule The average distance a molecule

travels before colliding with another travels before colliding with another is called the mean free path and is is called the mean free path and is small (near 10small (near 10-7-7))

Temperature is an average. There are Temperature is an average. There are molecules of many speeds in the molecules of many speeds in the average.average.

Shown on a graph called a velocity Shown on a graph called a velocity distributiondistribution

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num

ber

of p

arti

cles

Molecular Velocity

273 K

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47

num

ber

of p

arti

cles

Molecular Velocity

273 K

1273 K

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48

num

ber

of p

arti

cles

Molecular Velocity

273 K

873 K

1273 K

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VelocityVelocity Average increases as temperature Average increases as temperature

increases.increases. Spread increases as temperature Spread increases as temperature

increases.increases.

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EffusionEffusion Passage of gas through a small hole, Passage of gas through a small hole,

into a vacuum.into a vacuum. The effusion rate measures how fast The effusion rate measures how fast

this happens.this happens. Graham’s Law the rate of effusion is Graham’s Law the rate of effusion is

inversely proportional to the square inversely proportional to the square root of the mass of its particles.root of the mass of its particles.

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EffusionEffusion Passage of gas through a small hole, Passage of gas through a small hole,

into a vacuum.into a vacuum. The effusion rate measures how fast The effusion rate measures how fast

this happens.this happens. Graham’s Law the rate of effusion is Graham’s Law the rate of effusion is

inversely proportional to the square inversely proportional to the square root of the mass of its particles.root of the mass of its particles.

Rate of effusion for gas 1

Rate of effusion for gas 2

M

M

2

1

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DerivingDeriving The rate of effusion should be The rate of effusion should be

proportional to uproportional to urmsrms

Effusion Rate 1 = uEffusion Rate 1 = urms rms 11

Effusion Rate 2 = u Effusion Rate 2 = urms rms 22

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DerivingDeriving The rate of effusion should be The rate of effusion should be

proportional to uproportional to urmsrms

Effusion Rate 1 = uEffusion Rate 1 = urms rms 11

Effusion Rate 2 = u Effusion Rate 2 = urms rms 22

effusion rate 1

effusion rate 2

u 1

u 2

3RT

M

3RT

M2

M

Mrms

rms

1 2

1

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DiffusionDiffusion The spreading of a gas through a The spreading of a gas through a

room.room. Slow considering molecules move at Slow considering molecules move at

100’s of meters per second.100’s of meters per second. Collisions with other molecules slow Collisions with other molecules slow

down diffusions.down diffusions. Best estimate is Graham’s Law.Best estimate is Graham’s Law.

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ExamplesExamples A compound effuses through a porous A compound effuses through a porous

cylinder 3.20 time faster than helium. What is cylinder 3.20 time faster than helium. What is it’s molar mass?it’s molar mass?

If 0.00251 mol of NHIf 0.00251 mol of NH33 effuse through a hole effuse through a hole

in 2.47 min, how much HCl would effuse in in 2.47 min, how much HCl would effuse in the same time?the same time?

A sample of NA sample of N22 effuses through a hole in 38 effuses through a hole in 38

seconds. what must be the molecular weight seconds. what must be the molecular weight of gas that effuses in 55 seconds under of gas that effuses in 55 seconds under identical conditions? identical conditions?

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DiffusionDiffusion The spreading of a gas through a The spreading of a gas through a

room.room. Slow considering molecules move at Slow considering molecules move at

100’s of meters per second.100’s of meters per second. Collisions with other molecules slow Collisions with other molecules slow

down diffusions.down diffusions. Best estimate is Graham’s Law.Best estimate is Graham’s Law.

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Real GasesReal Gases Real molecules do take up space and Real molecules do take up space and

they do interact with each other they do interact with each other (especially polar molecules).(especially polar molecules).

Need to add correction factors to the Need to add correction factors to the ideal gas law to account for these.ideal gas law to account for these.

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Volume CorrectionVolume Correction The actual volume free to move in is less The actual volume free to move in is less

because of particle size.because of particle size. More molecules will have more effect.More molecules will have more effect. Corrected volume V’ = V - nbCorrected volume V’ = V - nb b is a constant that differs for each gas.b is a constant that differs for each gas.

P’ = nRTP’ = nRT

(V-nb) (V-nb)

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Pressure correctionPressure correction Because the molecules are attracted Because the molecules are attracted

to each other, the pressure on the to each other, the pressure on the container will be less than idealcontainer will be less than ideal

depends on the number of molecules depends on the number of molecules per liter.per liter.

since two molecules interact, the since two molecules interact, the effect must be squared.effect must be squared.

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Pressure correctionPressure correction Because the molecules are attracted Because the molecules are attracted

to each other, the pressure on the to each other, the pressure on the container will be less than idealcontainer will be less than ideal

depends on the number of molecules depends on the number of molecules per liter.per liter.

since two molecules interact, the since two molecules interact, the effect must be squared.effect must be squared.

Pobserved = P’ - a

2

( )Vn

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AltogetherAltogether PPobsobs= nRT - a n = nRT - a n 22

V-nb VV-nb V

Called the Van der Waal’s equation if Called the Van der Waal’s equation if

rearrangedrearranged

Corrected Corrected Corrected Corrected Pressure Pressure Volume Volume

( )

P + an

V x V - nb nRTobs

2

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Where does it come fromWhere does it come from a and b are determined by a and b are determined by

experiment.experiment. Different for each gas.Different for each gas. Bigger molecules have larger b.Bigger molecules have larger b. a depends on both size and polarity.a depends on both size and polarity. once given, plug and chug.once given, plug and chug.

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ExampleExample Calculate the pressure exerted by Calculate the pressure exerted by

0.5000 mol Cl0.5000 mol Cl22 in a 1.000 L container in a 1.000 L container

at 25.0ºCat 25.0ºC Using the ideal gas law.Using the ideal gas law. Van der Waal’s equationVan der Waal’s equation

– a = 6.49 atm La = 6.49 atm L22 /mol /mol22

– b = 0.0562 L/molb = 0.0562 L/mol