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Chapter 5 Chapter 5 The Gas Laws The Gas Laws

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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 sides. n Collisions are the

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 sides. n Collisions are the

2

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

sides.sides. Collisions are the force.Collisions are the force. Container has the area.Container has the area. Measured with a barometer.Measured with a barometer.

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 sides. n Collisions are the

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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 sides. n Collisions are the

4

ManometerManometer

Gas

h

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

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

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

5

ManometerManometer h is how much h is how much

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

h

Gas

Page 6: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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Units of pressureUnits of pressure 1 atmosphere = 760 mm Hg1 atmosphere = 760 mm Hg 1 mm Hg = 1 torr1 mm Hg = 1 torr 1 atm = 101,325 Pascals = 101.325 kPa1 atm = 101,325 Pascals = 101.325 kPa Can make conversion factors from Can make conversion factors from

these.these. What is 724 mm Hg in kPa?What is 724 mm Hg in kPa? in torr?in torr? in atm?in atm?

Page 7: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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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.related at constant temperature. PV= kPV= k As one goes up, the other goes As one goes up, the other goes

down.down. PP11VV11 = P = P22 V V22

GraphicallyGraphically

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 sides. n Collisions are the

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V

P (at constant T)

Page 9: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

9

V

1/P (at constant T)

Slope = k

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 sides. n Collisions are the

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PV

P (at constant T)

CO2

O2

22.4

1 L

atm

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 sides. n Collisions are the

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ExamplesExamples 20.5 L of nitrogen at 25ºC and 742 20.5 L of nitrogen at 25ºC and 742

torr are compressed to 9.8 atm at torr are compressed to 9.8 atm at constant T. What is the new volume?constant T. What is the new volume?

30.6 mL of carbon dioxide at 740 torr 30.6 mL of carbon dioxide at 740 torr is expanded at constant temperature is expanded at constant temperature to 750 mL. What is the final pressure to 750 mL. What is the final pressure in kPa? in kPa?

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 sides. n Collisions are the

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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

GraphicallyGraphically

Page 13: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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V (

L)

T (ºC)

H2O

HeCH4

H2

-273.15ºC

Page 14: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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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 , if the pressure is held constant?if the pressure is held constant?

Page 15: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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ExamplesExamples At what temperature would 40.5 L of At what temperature would 40.5 L of

gas at 23.4ºC have a volume of 81.0 gas at 23.4ºC have a volume of 81.0 L at constant pressure? L at constant pressure?

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 sides. n Collisions are the

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Avogadro's LawAvogadro's Law Avagadro’sAvagadro’s At constant temperature and At constant temperature and

pressure, the volume of gas is directly pressure, the volume of gas is directly related to the number of moles.related to the number of moles.

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

Page 17: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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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 18: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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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

Page 19: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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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 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 sides. n Collisions are the

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Ideal Gas LawIdeal Gas Law PV = nRTPV = nRT V = 22.42 L at 1 atm, 0ºC, n = 1 mole, V = 22.42 L at 1 atm, 0ºC, n = 1 mole,

what is R?what is R? R is the ideal gas constant.R is the ideal gas constant. R = 0.08206 L atm/ mol KR = 0.08206 L atm/ mol K Tells you about a gas is NOW.Tells you about a gas is NOW. The other laws tell you about a gas The other laws tell you about a gas

when it changes. when it changes.

Page 21: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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Ideal Gas LawIdeal Gas Law An An equation of stateequation of state.. Independent of how you end up Independent of how you end up

where you are at. where you are at. Does not depend on the path.Does not depend on the path. Given 3 you can determine the Given 3 you can determine the

fourth.fourth. An Empirical Equation - based on An Empirical Equation - based on

experimental evidence.experimental evidence.

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Ideal Gas LawIdeal Gas Law A hypothetical substance - the ideal A hypothetical substance - the ideal

gasgas Think of it as a limit.Think of it as a limit. Gases only approach ideal behavior Gases only approach ideal behavior

at low pressure (< 1 atm) and high at low pressure (< 1 atm) and high temperature.temperature.

Use the laws anyway, unless told to Use the laws anyway, unless told to do otherwise.do otherwise.

They give good estimates.They give good estimates.

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 sides. n Collisions are the

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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?

Page 24: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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Gas Density and Molar MassGas Density and Molar Mass D = m/VD = m/V Let Let MM stand for molar mass stand for molar mass MM = m/n = m/n n= PV/RTn= PV/RT MM = m = m

PV/RT PV/RT MM = mRT = m RT = DRT = mRT = m RT = DRT

PV PV V P V P P P

Page 25: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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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? 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 molecular formula?molecular formula?

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 sides. n Collisions are the

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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 occupies 22.42 L.mole of gas occupies 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.

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 sides. n Collisions are the

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

following reactionfollowing reaction

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

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

HgO Hg(l) + O (g) heat

2

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 sides. n Collisions are the

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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

Page 29: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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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 1000ºC can be produced from 10.0 L 1000ºC 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

Page 30: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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ExamplesExamples

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 NH500. L of NH33 at 250.0ºC and 3.00 atm? at 250.0ºC and 3.00 atm?

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

Page 31: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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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 ... ...

For each P = nRT/VFor each P = nRT/V

Page 32: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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Dalton's LawDalton's Law PPTotalTotal = n = n11RT + nRT + n22RT + nRT + n33RT +...RT +...

V V V V V V In the same container R, T and V are the In the same container R, T and V are the

same.same.

PPTotalTotal = (n = (n11+ n+ n22 + n + n33+...)RT+...)RT

V V

PPTotalTotal = (n = (nTotalTotal)RT)RT

V V

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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

= 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 35: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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ExamplesExamples

3.50 L

O2

1.50 L

N2

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

partial pressure and the total pressure?partial pressure and the total 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 pressure be subtracted from the total pressure to find the pressure of the gas.to find the pressure of the gas.

It must be given.It must be given.

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

following reactionfollowing reaction

what volume of N what volume of N22O collected O collected

over water at a total pressure of 94 over water at a total pressure of 94 kPa and 22ºC can be produced from kPa and 22ºC can be produced from 2.6 g of NH2.6 g of NH44NONO33? ( the vapor ? ( the vapor

pressure of water at 22ºC is 21 torr)pressure of 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 Theory tells why the things happen.Theory tells why the things happen. 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 = 3/2 RT 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 of the squared is the average of the squared particle velocity.particle velocity.

the root mean square velocity is the root mean square velocity is

u u 2 = 2 = uurmsrms

Page 41: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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Combine these two equationsCombine these two equations

For a mole of gas For a mole of gas NNA A is Avogadro's number is Avogadro's number

2A

1 3N ( mu ) = RT

2 2

avg

3(KE) = RT

2

2avg A

1(KE) =N ( mu )

2

2

A

3RTu =

N m

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Combine these two equationsCombine these two equations

m is kg for one particle, so Nm is kg for one particle, so Naam is kg m is kg

for a mole of particles. We will call it for a mole of particles. We will call it MM

Where Where MM is the molar mass in is the molar mass in kgkg/mole, /mole, and R has the units 8.3145 J/Kmol.and R has the units 8.3145 J/Kmol.

The velocity will be in m/sThe velocity will be in m/s

2

A

3RTu = u

N rmsm

3RT u = rms M

Page 43: 1 Chapter 5 The Gas Laws. 2 Pressure n Force per unit area. n Gas molecules fill container. n Molecules move around and hit sides. n Collisions are the

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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 at 250ºC.velocity of chlorine at 250ºC.

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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 K2RT

Mv2

2

3 2

evRT

Μ4 f(v)

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46

num

ber

of p

arti

cles

Molecular Velocity

273 K

1273 K

2RT

Mv2

2

3 2

evRT

Μ4 f(v)

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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.

2RT

Mv2

2

3 2

evRT

Μ4 f(v)

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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.

1

2

2 gasfor effusion of Rate

1 gasfor effusion of Rate

M

M

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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 Helium effuses through a porous cylinder Helium effuses through a porous cylinder

3.20 time faster than a compound. What is 3.20 time faster than a compound. 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 The actual volume free to move in is

less because of particle size.less because of particle size. More molecules will have more effect.More molecules will have more effect. Bigger molecules have more effectBigger molecules 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.

PP’’ = nRT= 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 ideal gascontainer will be less than ideal gas

Depends on the type of moleculeDepends on the type of molecule 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. 2

observed V

nP'-a P

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AltogetherAltogether

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

2

observed V

na-

nb-V

nRT P

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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. Look them upLook them up 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