Temperature and Kinetic Theory of Gases slides

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Temperature and theTemperature and the Kinetic Theory of Gases

Physics Rapid Learning Series

Rapid Learning Centerwww.RapidLearningCenter.com/© Rapid Learning Inc. All rights reserved.

Wayne Huang, Ph.D.Keith Duda, M.Ed.

Peddi Prasad, Ph.D.Gary Zhou, Ph.D.

Michelle Wedemeyer, Ph.D.Sarah Hedges, Ph.D.



Learning Objectives

How gases cause pressureThe relationship between

By viewing this tutorial you will learn…

The relationship between temperature and kinetic energyThe relationships between common temperature scalesKinetic Molecular TheoryHow gas properties relate to each other

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How to apply several gas lawsThe difference between ideal and real gasesDiffusion and effusion

Concept Map

Physics

Studies

Previous content

New content

Matter and EnergyMatter and Energy

Gases

One state is

Volume

Pressure

MolecularMolecular

Rates of Effusion Rates of Effusion and Diffusion

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TemperatureTemperature

# of Moles Molar Mass and DensityMolar Mass and Density

Speeds

Gas Laws

Have properties

Related to each other with



Average KineticAverage Kinetic Energy and Temperature

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Definition - Kinetic Energy

Kinetic Energy (KE) - the energy due to motion of an object.j

m = mass of the object v = speed of the object

2mv21KE =

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p j

Thus, the kinetic energy of an object is proportional to the square of its speed.



Definition - Temperature

Temperature – Proportional to the average kinetic energy of the molecules.

Energy due to motion(Related to how fast the molecules are moving)

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As temperature increases

Molecular motion

increases

Temperature and Kinetic Energy

Increased average kinetic energy

Temperature is Decreased

Temperature is Increased

Loss of heat

Gain of heatenergy

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Decreased average kinetic energy

Loss of heatenergy



Definition - Average Kinetic Energy

Average Kinetic Energy (KEAVE) - The energy due to motion of an object.

3

R = universal gas constant = 8.31 (Joules/Kelvins *mole)T = temperature in Kelvins

RT23KE =

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Thus the temperature of an object is proportional to the average kinetic energy of its molecules. As the gas heats up, its molecules oscillate at a faster rate.

Calculating Average Kinetic Energy

3

Temperature is defined as proportional to average kinetic energy…how do you calculate it?

Avg. KE = Average Kinetic Energy (in J, Joules)RT

23KEAvg. = R = Gas constant (use 8.31 J/K mol)

T = Temperature (in Kelvin)

Find the average kinetic energy of a sample of O2 at 28°CExample:

( )3

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Avg. KE = ? JR = 8.31 J/K molT = 28°C + 273 = 301 K Avg. KE = 3752 J/mole

( ) 301KmoleKJ8.31

23KEAvg. ××=



Average Translational Kinetic Energy

The average translational kinetic energy Kave is related to the temperature T by the relationship:

3 kT23Kave =

ANRk =

k, a constant is the ratio of the universal gas constant to Avogadro’s number, NA.

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AN

TNR

23K

Aave =

Thus, the average translational kinetic energy per molecule is given by:

Kelvins, Celsius, and FahrenheitThere are three commonly used units for temperature.

Kelvin (K) Scale. It’s also referred to as the absolute scale, 0 K is the temperature at which molecules of an object have a kinetic energy of 0.

Celsius (C) Scale. This scale is based on the boiling and freezing points of water It is

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boiling and freezing points of water. It is commonly used in science.

Fahrenheit (F) Scale. This is the scale used on weather channels and is frequently used in the United States.



Definition- Kelvin Scale

Kelvin (K) – Temperature scale with b l tan absolute zero

Temperatures cannot fall below an absolute zeroA temperature scale with absolute zero is needed in Gas Law calculations because you can’t have negative pressures or volumes.

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K273C =+

Absolute vs. Relative Scales

Kelvin (K)0K is the temperature at which molecules of an object have zero kinetic energy (zero motion)

When in doubt use Kelvins for thermodynamics calculations.

Celsius (C)Water freezes at 0 °C

Water boils at 100 °C

T (Celsius) = T (Kelvins) - 273.15

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Fahrenheit (F)Water freezes at 32 °F

Water boils at 212 °F

T (Fahrenheit) = (9/5) * T (Celsius) + 32



Temperature Scales

Molecules are completely still(kinetic energy = 0

Increasing molecular motion

(kinetic energy)

0 Kelvins Water freezes273.15 K

0 °C32 °F

Water Boils 373.15 K100 °C212 °F

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Decreasing temperature Increasing temperature

Decreasing kinetic energy

Loss of heat energy

Increasing kinetic energy

Gain of heat energy

32 F 212 F

Note - Absolute Temperature

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Question: Temperature and Kinetic Energy

If the temperature of an ice cube is -8 °C, what is the average kinetic energy (in Joules) of the water molecules?

0 Joules because the water molecules are crystallized as ice and therefore not moving.

0 Joules because the temperature is below 0 K

Pick the best answer:

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temperature is below 0 K.

Greater than 0 Joules because the temperature is above absolute zero or 0 K.

Answer: Temperature and Kinetic Energy

If the temperature of an ice cube is -8 °C, what is the average kinetic energy (in Joules) of the water molecules?

0 Joules because the water molecules are crystalized as ice and therefore not moving.

0 Joules because the temperature is below 0 K

Pick the best answer:

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temperature is below 0 K

Greater than 0 Joules because the temperature is above absolute zero or 0 K



Note - Temperature Scales

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Gas BehaviorGas Behavior and Kinetic Molecular Theory

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Definition- Theory and KMT

Theory – An attempt to explain why or h b h i ti thhow behavior or properties are as they are. It’s based on empirical evidence.

Kinetic Molecular Theory (KMT) – An attempt to e plain gas beha ior

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attempt to explain gas behavior based upon the motion of molecules.

Assumptions of the KMT

All gases are made of atoms or molecules.

Gas particles are in constant, rapid, random motion

1

2 motion.

The temperature of a gas is proportional to the average kinetic energy of the particles.

Gas particles are not attracted nor repelled from one another.

3

4

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All gas particle collisions are perfectly elastic (no kinetic energy is lost to other forms).

The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant.

5

6



KMT and Gas Behavior

The Kinetic MolecularThe Kinetic Molecular Theory and its assumptions can be used to explain gas behavior.

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Definition - Pressure

Pressure – Force of gas particles running into a surface.

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Pressure and Number of Molecules

A b f C lli i P

If pressure is molecular collisions with the container…

As number of molecules increases, there are more molecules to collide with the wall

Collisions between molecules and the wall increase

Pressure increases

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As # of molecules increases, pressure increases.

Pressure (P) and # of molecules (n) are directly proportional (∝)

nP∝

Pressure and Volume

A l C lli i P

If pressure is molecular collisions with the container…

As volume increases, molecules can travel farther before hitting the wall

Collisions between molecules and the wall decrease

Pressure decreases

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As volume increases, pressure decreases.

Pressure and volume are inversely proportional.

V1P∝



Pressure and TemperatureIf temperature is related to molecular motion…and pressure is molecular collisions with the container…

As temperature increases, molecular motion increases

Collisions between molecules and the wall increase

Pressure increases

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As temperature increases, pressure increases.

Pressure and temperature are directly proportional.

TP∝

Pressure Inside and Outside a Container

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Definition- Atmospheric Pressure

Atmospheric Pressure Pressure dueAtmospheric Pressure – Pressure due to the layers of air in the atmosphere.

Less layers of air

Lower atmospheric Climb in

altitude

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air pressurealtitude

As altitude increases, atmospheric pressure decreases.

Pressure In Versus OutA container will expand or contract until the pressure inside = atmospheric pressure outside

Expansion will lower the internal pressure

Example: A bag of chips is bagged at sea level. What happens if the bag is then brought up to the top of a mountain.

Contraction will raise the internal pressure(Volume and pressure are inversely related)

The internal pressure is from low altitude (high pressure)

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The internal pressure is higher than the external pressure.The bag will expand in order to reduce the internal pressure.

( g p )The external pressure is high altitude (low pressure).Higher

pressure

Lowerpressure

Lower pressure



When Expansion Isn’t Possible

Example: An aerosol can is left in a car trunk in the summer. What happens?

Rigid containers cannot expand.

CanExplodes!

happens?

The temperature inside the can begins to rise.As temperature increases, pressure increases.

Higher pressure

Lowerpressure

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The internal pressure is higher than the external pressure.The can is rigid—it cannot expand, it explodes!

Soft containers or “movable pistons” can expand and contract.Rigid containers cannot.

Gas Laws

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General Strategy for Gas Law Problems

Id tif titi b th i it1

The following steps are a general way to approach these problems.

Identify quantities by their units.

Make a list of known and unknown quantities in symbolic form.Look at the list and choose the gas law that relates all the quantities together.Pl titi i d l

1

2

3

4

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Plug quantities in and solve.4

Pressure UnitsSeveral units are used when describing pressure.

Unit Symbol

atmospheres atm

Pascals, kiloPascals

millimeters of mercury

pounds per square inch

Pa, kPa

mm Hg

psi

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1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi



Definition- STP Conditions

Standard Temperat re and Press reStandard Temperature and Pressure (STP) – 1 atm (or the equivalent in another unit) and 0°C (273 K).

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Problems often use “STP” to indicate quantities… Don’t forget this “hidden” information when making your list!

KMT and Gas Laws

The Gas Laws are experimental observations of gas behavior that the Kinetic Molecular Theory explains.

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explains.



“Before” and “After” in Gas Laws

This section has 4 gas laws which have “before” and “after” conditions.

For example:

2

2

1

1

nP

nP

=

Where P1 and n1 are pressure and # of moles “before”

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and P2 and n2 are pressure and # of moles “after”

Both sides of the equation are talking about the same sample of gas—with the “1” variables before a change, and the “2” variables after the change

Avogadro’s LawAvogadro’s Law relates # of particles (moles) and pressure.

Where Temperature and Pressure are held constant

V = Volumen = # of moles of gas

2

2

1

1

nV

nV

=

Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles?

The two volume units must match!

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moles?

n1 = 0.15 molesV1 = 2.5 Ln2 = 0.55 molesV2 = ? L

0.55moleV

0.15mole2.5L 2=

2V0.15mole

2.5L0.55mole=

× V2 = 9.2 L



Boyles’ LawBoyles’ Law relates pressure and volume.

Where temperature and # of molecules are held constantP = pressureV = volume2211 VPVP =The two pressure units must match and the two volume units must match!

Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 745 mm Hg?

P it d t t h t

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P1 = 1.05 atmV1 = 2.5 LP2 = 745 mm HgV2 = ? L

V2 = 2.7 L

Pressure units need to match—convert one:

=0.980 atm

2V0.980atm2.5L1.05atm ×=×

2V0.980atm

2.5L1.05atm=

×

745 mm Hg= ______ atm

mm Hg

atm1

7600.980

Charles’ LawCharles’ Law relates temperature and pressure.

Where pressure and # of molecules are held constant

21 VV V = VolumeT = Temperature

2

2

1

1

TV

TV

=

The two volume units must match and temperature must be in Kelvin!

Example: What is the final volume if a 10.5 L sample of gas is changed from 25°C to 50°C?

T t d t b i K l i !

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V1 = 10.5 LT1 = 25°CV2 = ? LT2 = 50°C

V2 = 11.4 L

Temperature needs to be in Kelvin!

= 298 K

= 323 K

25°C + 273 = 298 K

50°C + 273 = 323 K

323KV

298K10.5L 2=

2V298K

10.5L323K=

×



Combined Gas LawThe combined gas law assumes that nothing is held constant.

P = PressureVPVP Each “pair” of unitsV = Volumen = # of molesT = Temperature22

22

11

11

TnVP

TnVP

=Each pair of units must match and temperature must be in Kelvin!

Example: What is the final volume if a 0.125 mole sample of gas at 1.7 atm, 1.5 L and 298 K is changed to STP and particles are added to make 0.225 moles?

P1 = 1.7 atmV1 = 1.5 L

STP is standard temperature (273 K) and pressure (1 atm)

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n1 = 0.125 moleT1 = 298 KP2 = 1.0 atmV2 = ? Ln2 = 0.225 moleT2 = 273 K V2 = 4.2 L


273K0.225moleV1.0atm

298K0.125mole1.5L1.7atm 2

××

=×

×

2V298K0.125mole1.0atm

1.5L1.7atm273K0.225mole=

×××××

Why You Only Really Need 1 out of the 4 Laws!

2211 VPVP=

The combined gas law can be used for all “before” and “after” gas law problems!

2211 TnTn

22

12

11

11

TnVP

TnVP

=

For example, if volume is held constant, then

and the combined gas law becomes:

21 VV =

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When two variables on opposites sides are the same, they cancel out and the rest of the equation can be used.

22

2

11

1

TnP

TnP

=



“Transforming” the Combined Gas Law

Watch as variables are held constant and the combined gas law “becomes” the other 3 laws.

VPVP22

22

11

11

TnVP

TnVP

=Hold pressure and temperature constant Avogadro’s Law

22

22

11

11

TnVP

TnVP

=Hold moles and temperature constant Boyles’ Law

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22

22

11

11

TnVP

TnVP

=Hold pressure and moles constant Charles’ Law

How to Memorize What’s Held Constant

How do you know what to hold constant for each law?

Hold Pressure and Temperature constantAvogadro’s Law

Hold moles and Temperature constantBoyles’ Law

Avogadro was a Professor at Turin University (Italy)

The last letter of his first name Robert is T

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Hold Pressure and moles constantCharles’ Law

The last letter of his first name, Robert, is T

Charles was from Paris



Example of Using only the Combined Law

Example: What is the final pressure if a 15.5 L sample of gas at 755 mm Hg and 298 K is changed to STP?

P 755 H“moles” is not mentioned in the problem—therefore


P1 = 755 mm HgV1 = 15.5 LT1 = 298 KP2 = 1.0 atmV2 = ? LT2 = 273 K

pit is being held constant.It is not needed in the combined law formula.

Pressure units must match!1 atm = 760 mm Hg= 760 mm Hg

22

22

11

11

TnVP

TnVP

=

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273KVHg760mm

298K15.5LHg755mm 2×

=×

2V298KHg760mm

15.5LHg755mm273K=

×××

Ideal Gas Law

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Definition - Ideal Gas

Ideal Gas – All of the assumptions of the Kinetic Molecular Theory (KMT)

lidare valid.

Ideal Gas Law – Describes properties of a gas under a set of conditions.

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nRTPV =

This law does not have “before” and “after”—there is no change in conditions taking place.

Definition- Gas Constant (R)

nRTPV =

Gas Constant (R) – Constant equal to the ratio of P×V to n×T for a gas.

Values for R

kPaL×

Use this one when the P unit is “kPa”

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8.31

0.0821

KmolekPaL×

×

KmoleatmL×

×

is kPa

Use this one when the P unit is “atm”



Memorizing the Ideal Gas Law

nRTPV =

Phony Vampires are not Real Things

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Ideal Gas Law ExampleAn example of the Ideal Gas Law:

P = PressureV = Volume

Choose your “R” based upon your “P” units.nRTPV = n = # of moles

R = Gas constantT = Temperature

upon your P units.

T must be in Kelvin!

nRTPV =

What is the pressure (in atm) of a gas if it is 2.75 L, has 0.25 moles and is 325 K?

Example:

P = ? Choose the “0.0821” for “R” since the problem asks for “ ”

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( ) 325KKmoleatmL0.08210.25moles2.75LP ××

××=×

P ? V = 2.75 Ln = 0.25 molesT = 325 K

Phydrogen = 2.43 atm

“atm”

R = 0.0821 (L×atm) / (mol×K)

( )2.75L

325KKmoleatmL0.08210.25moles

P××

××=



Definition - Molar Mass

Molar mass (MM) – Mass (m) per moles (n) of a substancemoles (n) of a substance.

nmMM =

Therefore: m

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Therefore:MM

n =

Ideal Gas Law and Molar MassThe Ideal Gas Law is often used to determine molar mass.

nRTPVmn =and RTmPVnRTPV = MM

n =and RTMM

PV =

A gas is collected. The mass is 2.889 g, the volume is 0.936 L, the temperature is 304 K and the pressure is 98.0 kPa. Find the molar mass

Example:

P = 98.0 kPaChoose the “8.31” for “R” since the problem uses “kPa”

( )kPaL2 889g

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P 98.0 kPa V = 0.936 Lm = 2.889 gT = 304 KMM = ? g/mole

MM = 79.6 g/moleR = 8.31 (L×kPa) / (mol×K)

( ) 304KKmolekPaL8.31

MM2.889g0.936L98.0kPa ××

×=×

( ) 304KKmolekPaL8.31

0.936L98.0kPa2.889gMM ××

××

=



Definition - Density

D i R i f lDensity – Ratio of mass to volume for a sample.

VmD =

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Ideal Gas Law and DensityUsing the density equation with the Ideal Gas Law:

VmD =andRT

MMmPV =

MMRT

VmP =

MMRTDP =

A gas is collected. The density is 3.09 g/L, the volume is 0.936 L, the temperature is 304 K and the pressure is 98.0 kPa. Find the molar mass

Example:

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P = 98.0 kPa V = 0.936 LD = 3.09 g/LT = 304 KMM = ? g/mole

MM = 79.6 g/mole

Choose the “8.31” for “R” since the problem uses “kPa”

R = 8.31 (L×kPa) / (mol×K)

( )MM

304KKmolekPaL8.31

Lg3.0998.0kPa

×××

=

( )98.0kPa

304KKmolekPaL8.31

Lg3.09MM

×××

=



Real Gases

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Real Gases

Real Gas – Two of the assumptions of the Kinetic Molecular Theory are not valid.

Wrong Assumption 1: Gas particles are not attracted nor repelled from one another

Wrong Assumption 2: The volume of gas particles

Reality: Gas particles do have attractions and repulsions towards one another.

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Wrong Assumption 2: The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant

Reality: Gas particles do take up space—thereby reducing the space available for other particles to be.



Real Gas LawThe Real Gas Law takes into account the deviations from the Kinetic Molecular Theory.

nRTPV =Ideal Gas Law nRTPV =

( ) nRTnbVV

anP 2

2

=−⎟⎟⎠

⎞⎜⎜⎝

⎛+

Ideal Gas Law

Real Gas LawAlso called “van der Waals equation”

Take into account the

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Take into account the change in pressure due

to particle attractions and repulsions

Takes into account the space the particles take up

“a” and “b” are constants that you look up for each gas!

Real Gas Law Example

Example: At what temperature would a 0.75 mole sample of CO2 be 2.75 L at 3.45 atm?

( ) nRTnbVV

anP 2

2

=−⎟⎟⎠

⎞⎜⎜⎝

⎛+

2.75 L at 3.45 atm? (van der Waals constants for CO2: a = 3.59 L2atm/mol2

b = 0.0427 L/molP = 3.45 atmV = 2.75 Ln = 0.75 moleT = ? Ka = 3.59 L2atm/mol2

b = 0.0427 L/mol

Choose the “0.0821” for “R” since the problem uses “atm”

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T = 164 K

( ) ( ) ( ) TKmolatmL0.08210.75molmol

L0.04270.75mol2.75L(2.75L)

molatmL3.59(0.75mol)

3.45atm 2

222

××××=×−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ ×+

R = 0.0821 (L×atm) / (mol×K)



Distribution of MolecularDistribution of Molecular Speeds and Mean Free Path

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Definition - Mean Free Path

Mean Free Path - (λ). The mean free path is the average distance traveled by an atom or molecule before it collidesan atom or molecule before it collides with another atom or molecule.

VNπd2

1λ2

= d = molecular diameterN/V = number of molecules

per unit volume

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Thus, the number of collisions a molecule undergoes increases as the number of molecules in a container increases and as the diameter of the molecules increases. In a reaction mixture, as the number of collisions increases the reaction rate increases.



Definition- Vrms

Root-Mean-Square Speed (vrms) –One measure of the average speed of

l l imolecules in a gas.

M3RTvrms =

R = 8.31 J/mol*k; T = temperature [K]; M = molar mass [kg/mol]

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[kg/mol]

Thus Vrms increases with temperature and decreases as the molecules increase in size.

61/4361/43

Molecular Speed

The pressure P exerted by n moles of a gas may be determined from the root-mean-square speed vrmsof its molecules.

P = pressure; n = moles of gas; V = volume; M = molar mass

3VnMvP

2rms=

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Note that the pressure increases with increasing average speed and molar mass.

Note that pressure decreases with increasing volume.



Maxwell Speed DistributionVave is just the average speed. Some molecules move much faster and some molecules move much slower.

The probability that a molecule in a gas at temperature T is moving at a given speed is shown by a probability function known as the Maxwell distribution.

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Diagram- Maxwell’s DistributionMaxwell's Speed Distribution Law

2

2.5

^-3

m/s

)

0

0.5

1

1.5

0

100

200

300

400

500

600

700

800

900

1000

1100

1200

Molecular Speed (m/s)

Pro

babi

lity

(10^ 0 deg C

200 deg C

N t th t i th f 0 C th d ith th

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Note that in the curve for 0 °C, the speed with the maximum probability is around 350 m/s.

For 200 °C, the speed with the maximum probability is higher, around 500 m/s.



Diffusion and Effusion

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Definition- Diffusion and Effusion

Diffusion – A gas spreads throughout a space.

Effusion A gas escapes through a

Perfume is sprayed in one corner of the room and a person on the other side smells it after a moment.

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Effusion – A gas escapes through a tiny hole.

Air leaks out of a balloon overnight and is flat the next day.



Diffusion, Effusion and Mass

Temperature is A heavier object Heavy molecules

The mass of a particle affects the rate of diffusion and effusion.

Temperature is proportional to average kinetic energy

A heavier object with the same kinetic energy as a lighter object moves slower than the lighter object

Heavy molecules move slower than smaller molecules

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Diffusion: If molecules move slower, it will take them longer to reach the other side of the room.

Effusion: If molecules move slower, it will take them longer to find the hole to escape through.

Both rates of diffusion and effusion are inversely proportional to molecular mass.

Effusion and Graham’s LawEffusion rates are related by Graham’s Law.

r1 = Rate of Effusion for molecule 1r2 = Rate of Effusion for molecule 2MM M l l f l l 1

21

MMMMr

= MM1 = Molecular mass for molecule 1MM2 = Molecular mass for molecule 212 MMr

Example: A gas molecule effuses 0.355 times as fast as O2. What is the molecular mass of the molecule?

Molecule 1 = O2

Molecule 2 = unknown moleculeIf the unknown molecule is 0.355 times as fast as O2,then make the rate of O2 = 1 and the

e32.00g/molMM

0.3551 2=

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r1 = 1r2 = 0.355MM1 = 32.00 g/moleMM2 = ? g/mole

then make the rate of O2 1 and the rate of unknown = 0.355

Molecular mass of O2:O 2 × 16.00 = 32.00 g/mole

e32.00g/molMM

0.3551 2

2

=⎟⎠⎞

⎜⎝⎛

( ) 2

2

MM0.355

1e32.00g/mol =⎟⎠⎞

⎜⎝⎛×

MM = 254 g/mole



Diffusion and Graham’s LawDistances traveled during diffusion:

d1 = Distance traveled for molecule 1d2 = Distance traveled for molecule 2MM M l l f l l 1

21

MMMM

dd

=MM1 = Molecular mass for molecule 1MM2 = Molecular mass for molecule 2

12 MMd

Example: A gas molecule is 4 times as heavy as O2. How far does it travel in the time that oxygen travels 0.25 m?

Molecule 1 = unknown moleculeMolecule 2 = O2 le128.00g/mo

e32.00g/mol0.25m

d1 =

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d1 = ?d2 = 0.25 mMM1 = 4×32.00 g/mole = 128 g/moleMM2 = 32.00 g/moleMolecular mass of O2:O 2 × 16.00 = 32.00 g/mole

41

0.25md1 =

0.25m21d1 ×=

d1 = 0.125 m

Real gases do not use 2 of the Real gases do

not use 2 of the Temperature is proportional to Temperature is proportional to

Rates of Effusion and Diffusion are

inversely

Rates of Effusion and Diffusion are

inversely

Learning Summary

assumptions of the KMT

assumptions of the KMT

average kinetic energy.

average kinetic energy.

inversely proportional to molecular mass

inversely proportional to molecular mass

Several Gas LawsSeveral Gas Laws

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Ideal gases follow the assumption of the Kinetic Molecular

Theory (KMT)

Ideal gases follow the assumption of the Kinetic Molecular

Theory (KMT)

Several Gas Laws are used to determine

properties under a set of conditions

Several Gas Laws are used to determine

properties under a set of conditions



Congratulations

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Temperature and Kinetic Theory of Gases slides

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Transcript of Temperature and Kinetic Theory of Gases slides