# kinetic molecular theory & gases an honors/ap chemistry presentation

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Kinetic Molecular Theory

& Gases

Kinetic Molecular Theory

& GasesAn Honors/AP Chemistry

PresentationAn Honors/AP Chemistry

Presentation

Kinetic Molecular Theory

Kinetic Molecular Theory

Kinetic means motionSo the K.M.T. studies the motions of molecules.Solids - vibrate a littleLiquids - vibrate, rotate, and translate (a little)

Gases - vibrate, rotate, and translate (a lot)!

Kinetic means motionSo the K.M.T. studies the motions of molecules.Solids - vibrate a littleLiquids - vibrate, rotate, and translate (a little)

Gases - vibrate, rotate, and translate (a lot)!

Basic Assumptions of KMT

Basic Assumptions of KMT

Gases consist of large numbers of molecules in continuous random motion.

The volume of the molecules is negligible compared to the total volume.

Gases consist of large numbers of molecules in continuous random motion.

The volume of the molecules is negligible compared to the total volume.

Intermolecular interactions are negligible.

When collisions occur, there is a transfer of kinetic energy, but no loss of kinetic energy.

The average kinetic energy is proportional to the absolute temperature.

Intermolecular interactions are negligible.

When collisions occur, there is a transfer of kinetic energy, but no loss of kinetic energy.

The average kinetic energy is proportional to the absolute temperature.

Gas PropertiesGas Properties

Volume - amount of space (L or mL)

Temperature - relative amount of molecular motion (K)

Pressure - the amount of force molecules exert over a given area (atm, Torr, Pa, psi, mm Hg)

Moles - the number of molecules (mol)

Volume - amount of space (L or mL)

Temperature - relative amount of molecular motion (K)

Pressure - the amount of force molecules exert over a given area (atm, Torr, Pa, psi, mm Hg)

Moles - the number of molecules (mol)

Temperature ConversionsTemperature Conversions

C = 5/9(F-32)F = 9/5C + 32K = C + 273So what is the absolute temperature (K) of an object at -40 oF?

C = 5/9(F-32)F = 9/5C + 32K = C + 273So what is the absolute temperature (K) of an object at -40 oF?

Answer to Temperature Conversion

Answer to Temperature Conversion

-40 oF = -40 oC

-40 oC = 233 K or 230 K

-40 oF = -40 oC

-40 oC = 233 K or 230 K

Pressure ConversionsPressure Conversions

1 atm = 760 mm Hg = 760 Torr = 101,325 Pa = 14.7 psi

How many atmospheres is 12.0 psi?

How many Torr is 1.25 atm?

How many Pascals is 720 mm Hg?

1 atm = 760 mm Hg = 760 Torr = 101,325 Pa = 14.7 psi

How many atmospheres is 12.0 psi?

How many Torr is 1.25 atm?

How many Pascals is 720 mm Hg?

QuickTime™ and a decompressor

are needed to see this picture.

Answers to Pressure Conversions

Answers to Pressure Conversions

12.0 psi = .816 atm1.25 atm =950. Torr720 mm Hg = 96000 Pa

12.0 psi = .816 atm1.25 atm =950. Torr720 mm Hg = 96000 Pa

A BarometerA Barometer

A mercury barometer measures air pressure by allowing atmospheric pressure to press on a bath of mercury, forcing mercury up a long tube. The more pressure, the higher the column of mercury.

A mercury barometer measures air pressure by allowing atmospheric pressure to press on a bath of mercury, forcing mercury up a long tube. The more pressure, the higher the column of mercury.

More on the barometer

More on the barometer Although American meteorologists will sometimes measure the height in inches, typically this pressure is measured in mm Hg.

1 mm Hg = 1 Torr

Although American meteorologists will sometimes measure the height in inches, typically this pressure is measured in mm Hg.

1 mm Hg = 1 Torr

S.T.P.S.T.P.

When making comparisons we often use benchmarks or standards to compare against.

In chemistry Standard Temperature is 0 oC (273K) and Standard Pressure is 1 atm.

When making comparisons we often use benchmarks or standards to compare against.

In chemistry Standard Temperature is 0 oC (273K) and Standard Pressure is 1 atm.

Boyle’s LawBoyle’s Law

If the amount and temperature of the gas are held constant, then the volume of a gas is inversely proportional to the pressure it exerts.

Mathematically this means that the pressure times the volume is a constant.

P*V = kP1V1=P2V2

If the amount and temperature of the gas are held constant, then the volume of a gas is inversely proportional to the pressure it exerts.

Mathematically this means that the pressure times the volume is a constant.

P*V = kP1V1=P2V2

Boyle’s Law in Action

Boyle’s Law in Action

Sample QuestionsSample Questions

The volume of a balloon is 852 cm3 when the air pressure is 1.00 atm. What is the volume if the pressure drops to .750 atm?

A gas is trapped in a 2.20 liter space beneath a piston exerting 25.0 psi. If the volume expands to 2.75 L, what is the new pressure?

The volume of a balloon is 852 cm3 when the air pressure is 1.00 atm. What is the volume if the pressure drops to .750 atm?

A gas is trapped in a 2.20 liter space beneath a piston exerting 25.0 psi. If the volume expands to 2.75 L, what is the new pressure?

The Answers are…The Answers are…

P1V1=P2V2

(1atm)(852cm3)= (.750atm)*V2 V2 =1140cm3

(25.0psi)(2.20L)=P2(2.75L) P2 = 20.0 psi

P1V1=P2V2

(1atm)(852cm3)= (.750atm)*V2 V2 =1140cm3

(25.0psi)(2.20L)=P2(2.75L) P2 = 20.0 psi

Charles’ LawCharles’ Law

If the amount and the pressure of a gas are held constant, then the volume of a gas is directly proportional to its absolute temperature.

Mathematically, this means that the volume divided by the temperature is a constant.

V/T = kV1/T1=V2/T2

If the amount and the pressure of a gas are held constant, then the volume of a gas is directly proportional to its absolute temperature.

Mathematically, this means that the volume divided by the temperature is a constant.

V/T = kV1/T1=V2/T2

Charles Law in Action

Charles Law in Action

Sample QuestionsSample Questions

The volume of a balloon is 5.00 L when the temperature is 20.0 oC. If the air is heated to 40.0 oC, what is the new volume?

3.00 L of air are held under a piston at 0.00 oC. If the air is allowed to expand at constant pressure to 4.00 L, what is the new Celsius temperature of the gas?

The volume of a balloon is 5.00 L when the temperature is 20.0 oC. If the air is heated to 40.0 oC, what is the new volume?

3.00 L of air are held under a piston at 0.00 oC. If the air is allowed to expand at constant pressure to 4.00 L, what is the new Celsius temperature of the gas?

The Answers Are…The Answers Are…

V1/T1=V2/T2

5.00L/293K = V2/313K V2=5.34L

273K/3.00L = T2/4.00L T2=364K=91oC

V1/T1=V2/T2

5.00L/293K = V2/313K V2=5.34L

273K/3.00L = T2/4.00L T2=364K=91oC

The Gay-Lussac LawThe Gay-Lussac Law

If the amount and volume of the gas are held constant, then the pressure exterted by the gas is directly proportional to its absolute temperature.

Mathematically this means that the pressure divided by the temperature is a constant.

P/T = k P1/T1=P2/T2

If the amount and volume of the gas are held constant, then the pressure exterted by the gas is directly proportional to its absolute temperature.

Mathematically this means that the pressure divided by the temperature is a constant.

P/T = k P1/T1=P2/T2

The Gay-Lussac Law in Action

The Gay-Lussac Law in Action

Sample QuestionsSample Questions

A tank of oxygen is stored at 3.00 atm and -20 oC. If the tank is accidentally heated to 80 oC, what is the new pressure in the tank?

A piston is trapped in place at a temperature of 25 oC and a

pressure of 112 kPa. At what celcius temperature is the pressure 102 kPa?

A tank of oxygen is stored at 3.00 atm and -20 oC. If the tank is accidentally heated to 80 oC, what is the new pressure in the tank?

A piston is trapped in place at a temperature of 25 oC and a

pressure of 112 kPa. At what celcius temperature is the pressure 102 kPa?

The Answers are…The Answers are…

P1/T1=P2/T2

(3atm)/(253 K)= P2/ (353 K) P2 =4.19 atm

(298 K)/(112 kPa)=T2/(102kPa) T2 = 271 K = -2 oC

P1/T1=P2/T2

(3atm)/(253 K)= P2/ (353 K) P2 =4.19 atm

(298 K)/(112 kPa)=T2/(102kPa) T2 = 271 K = -2 oC

Avogadro’s LawAvogadro’s Law

If the temperature and the pressure of a gas are held constant, then the volume of a gas is directly proportional to the amount of gas.

Mathematically, this means that the volume divided by the # of moles is a constant.

V/n = k or V/m = kV1/n1=V2/n2 or V1/m1=V2/m2

If the temperature and the pressure of a gas are held constant, then the volume of a gas is directly proportional to the amount of gas.

Mathematically, this means that the volume divided by the # of moles is a constant.

V/n = k or V/m = kV1/n1=V2/n2 or V1/m1=V2/m2

Avogadro’s Law in Action

Avogadro’s Law in Action

Sample QuestionsSample Questions

The volume of a balloon is 5.00 L when there is .250 mol of air. If 1.25 mol of air is added to the balloon, what is the new volume?

3.00 L of air has a mass of about 4.00 grams. If more air is added so that the volume is now 24.0 L, what is the mass of the air now?

The volume of a balloon is 5.00 L when there is .250 mol of air. If 1.25 mol of air is added to the balloon, what is the new volume?

3.00 L of air has a mass of about 4.00 grams. If more air is added so that the volume is now 24.0 L, what is the mass of the air now?

The Answers Are…The Answers Are…

V1/n1=V2/n2 or V1/m1=V2/m2

5.00L/.250 mol = V2/1.50 mol V2=30.0 L

4.00g/3.00L = m2/24.0L m2=32.0 g

V1/n1=V2/n2 or V1/m1=V2/m2

5.00L/.250 mol = V2/1.50 mol V2=30.0 L

4.00g/3.00L = m2/24.0L m2=32.0 g

The Combined Gas LawThe Combined Gas Law

This law combines the inverse proportion of Boyle’s Law with the direct proportions of Charles’, Gay-Lussac’s, and Avogadro’s Laws.

P1V1/(n1T1) = P2V2/(n2T2)orP1V1/T1 = P2V2/T2

This law combines the inverse proportion of Boyle’s Law with the direct proportions of Charles’, Gay-Lussac’s, and Avogadro’s Laws.

P1V1/(n1T1) = P2V2/(n2T2)orP1V1/T1 = P2V2/T2

Four Gas Laws in OneFour Gas Laws in One

The combined gas law could be used in place of any of the previous 4 gas laws.

For example, in Boyle’s Law, we assume that the amount and temperature are constant. So if we cross them off of the combined gas law:

P1V1/(n1T1) = P2V2/(n2T2)P1V1 = P2V2

The combined gas law could be used in place of any of the previous 4 gas laws.

For example, in Boyle’s Law, we assume that the amount and temperature are constant. So if we cross them off of the combined gas law:

P1V1/(n1T1) = P2V2/(n2T2)P1V1 = P2V2

Another ExampleAnother Example

A sample of hydrogen has a volume of 12.8 liters at 104 oF and 2.40 atm. What is the volume at STP?

A sample of hydrogen has a volume of 12.8 liters at 104 oF and 2.40 atm. What is the volume at STP?

The answer is:The answer is:

P1V1/(n1T1) = P2V2/(n2T2)

P1=2.40atm,V1=12.8L, T1=104oF=40oC=313K, T2=273K, P2=1atm, n1=n2

(2.4atm)(12.8L)/(313K) = (1atm)V2/(273K)

V2 = 26.8 L

P1V1/(n1T1) = P2V2/(n2T2)

P1=2.40atm,V1=12.8L, T1=104oF=40oC=313K, T2=273K, P2=1atm, n1=n2

(2.4atm)(12.8L)/(313K) = (1atm)V2/(273K)

V2 = 26.8 L

The Ideal Gas LawThe Ideal Gas Law

If, P1V1/(n1T1) = P2V2/(n2T2)Then PV/(nT) = constantThat constant is R, the ideal gas law constant.

R = .0821 L*atm/(mol*K)R = 8.314 J/(mol*K)So, PV=nRT

If, P1V1/(n1T1) = P2V2/(n2T2)Then PV/(nT) = constantThat constant is R, the ideal gas law constant.

R = .0821 L*atm/(mol*K)R = 8.314 J/(mol*K)So, PV=nRT

But what about…But what about…

Since n = m/M, we can substitute into PV = nRT and get

PVM = mRTSince D = m/V, we can substitute in again and get

PM = DRT

Since n = m/M, we can substitute into PV = nRT and get

PVM = mRTSince D = m/V, we can substitute in again and get

PM = DRT

So which one is it?So which one is it?

Like a good carpenter, it is good to have many tools so that you can choose the right tool for the right job.

If I am solving a gas problem with density, I use PM = DRT.

If I am solving a gas problem with moles, I use PV = nRT.

If I am solving a gas problem with mass, I use PVM = mRT.

Like a good carpenter, it is good to have many tools so that you can choose the right tool for the right job.

If I am solving a gas problem with density, I use PM = DRT.

If I am solving a gas problem with moles, I use PV = nRT.

If I am solving a gas problem with mass, I use PVM = mRT.

Such as….Such as….

Under what pressure would oxygen have a density of 8.00 g/L at 300 K?

PM = DRTP(32 g/mol) = (8 g/L)(.0821 latm/molK)(300 K)

P = 6.16 atm

Under what pressure would oxygen have a density of 8.00 g/L at 300 K?

PM = DRTP(32 g/mol) = (8 g/L)(.0821 latm/molK)(300 K)

P = 6.16 atm

An Important NumberAn Important Number

What is the volume of 1 mole of a gas at STP?

PV = nRT V = nRT/PV = (1mol)*(.0821Latm/molK)(273K)/ (1atm)

V = 22.4 LThis is called the standard molar volume of an ideal gas.

What is the volume of 1 mole of a gas at STP?

PV = nRT V = nRT/PV = (1mol)*(.0821Latm/molK)(273K)/ (1atm)

V = 22.4 LThis is called the standard molar volume of an ideal gas.

Gas StoichiometryGas Stoichiometry

We had said that stoichiometry implied a ratio of molecules, or moles. Up until now we only used mole ratios.

However Avogadro said that the volume is directly proportional to the number of molecules.

This means that we can do stoichiometry with volume or moles.

We had said that stoichiometry implied a ratio of molecules, or moles. Up until now we only used mole ratios.

However Avogadro said that the volume is directly proportional to the number of molecules.

This means that we can do stoichiometry with volume or moles.

Example 1 of Gas Stoichiometry

Example 1 of Gas Stoichiometry

What volume of hydrogen is needed to synthesize 6.00 liters of ammonia?

N2 (g) + 3 H2 (g) --> 2 NH3 (g)

6.00 L H2 x (2 NH3/3 H2) = 4.00 L NH3

What volume of hydrogen is needed to synthesize 6.00 liters of ammonia?

N2 (g) + 3 H2 (g) --> 2 NH3 (g)

6.00 L H2 x (2 NH3/3 H2) = 4.00 L NH3

Example 2 of Gas Stoichiometry

Example 2 of Gas Stoichiometry

What mass of nitrogen is needed to synthesize 20.0 L of ammonia at 1.50 atm and 25 oC?

N2 (g) + 3 H2 (g) --> 2 NH3 (g) 20.0 L NH3 x (1 N2/2 NH3) = 10.0 L N2

PVM = mRT (1.5 atm)(10 L)(28 g/mol) = m(.0821Latm/molK)(298K)

m = 17.2 g N2

What mass of nitrogen is needed to synthesize 20.0 L of ammonia at 1.50 atm and 25 oC?

N2 (g) + 3 H2 (g) --> 2 NH3 (g) 20.0 L NH3 x (1 N2/2 NH3) = 10.0 L N2

PVM = mRT (1.5 atm)(10 L)(28 g/mol) = m(.0821Latm/molK)(298K)

m = 17.2 g N2

Dalton’s LawDalton’s Law

When we talk about air pressure, we need to understand that air is not oxygen.

Air is a solution of nitrogen (78.09%), oxygen (20.95%), argon (.93%), and CO2 (.03%).

So when we talk about air pressure, which gas are we talking about?

When we talk about air pressure, we need to understand that air is not oxygen.

Air is a solution of nitrogen (78.09%), oxygen (20.95%), argon (.93%), and CO2 (.03%).

So when we talk about air pressure, which gas are we talking about?

ALL OF THEM!ALL OF THEM!

Dalton’s Law of Partial Pressures states that the total pressure of a system is equal to the sum of the partial (or individual) pressures of each component.

Ptotal = P1 + P2 + … Px

So if air pressure is 1 atm, then we can assume that the N2 is .78 atm, the O2 is .21 atm, and the Ar is about .01 atm.

Dalton’s Law of Partial Pressures states that the total pressure of a system is equal to the sum of the partial (or individual) pressures of each component.

Ptotal = P1 + P2 + … Px

So if air pressure is 1 atm, then we can assume that the N2 is .78 atm, the O2 is .21 atm, and the Ar is about .01 atm.

A CorollaryA Corollary

If we extend Boyle’s Law and Avogadro’s Law, we could infer that, at constant temperature and volume, the pressure of a gas is directly proportional to its pressure.

P1/Ptotal = n1/ntotal

If we extend Boyle’s Law and Avogadro’s Law, we could infer that, at constant temperature and volume, the pressure of a gas is directly proportional to its pressure.

P1/Ptotal = n1/ntotal

An important exampleAn important example

A sample of CaCO3 is heated, releasing CO2, which is collected over water (a typical practice).

A sample of CaCO3 is heated, releasing CO2, which is collected over water (a typical practice).

QuickTime™ and a decompressor

are needed to see this picture.

The pressure in the collection bottle is the sum of the pressure of the CO2 plus the pressure of the water vapor (since some water always evaporates).Ptotal = PCO2 + PH2O

Sample Water Vapor Pressures

Sample Water Vapor Pressures

Water Vapor Pressure Table Temperature Pressure (°C) (mmHg)

Temperature Pressure (°C) (mmHg)

Temperature Pressure (°C) (mmHg)

0.0 4.6 5.0 6.5 10.0 9.2 12.5 10.9 15.0 12.8 15.5 13.2 16.0 13.6 16.5 14.1 17.0 14.5 17.5 15.0 18.0 15.5 18.5 16.0 19.9 16.5

19.5 17.0 20.0 17.5 20.5 18.1 21.0 18.6 21.5 19.2 22.0 19.8 22.5 20.4 23.0 21.1 23.5 21.7 24.0 22.4 24.5 23.1 25.0 23.8 26.0 25.2

27.0 26.7 28.0 28.3 29.0 30.0 30.0 31.8 35.0 42.2 40.0 55.3 50.0 92.5 60.0 149.4 70.0 233.7 80.0 355.1 90.0 525.8 95.0 633.9 100.0 760.0

So in our exampleSo in our example

If a total pressure of 365 Torr is collected at 25 oC in a 100 ml collection bottle:What is the partial pressure of CO2?

What mass of CaCO3 decomposed?

If a total pressure of 365 Torr is collected at 25 oC in a 100 ml collection bottle:What is the partial pressure of CO2?

What mass of CaCO3 decomposed?

Here’s how it worksHere’s how it works

Ptotal = PCO2 + PH2O 365 Torr = PCO2 + 23.8 TorrPCO2 = 341.2 Torr = .449 atm PVM = mRT(.449 atm)(.100 L)(44.0 g/mol) = m(.0821Latm/molK)(298K)

m = .0807 g CO2

Ptotal = PCO2 + PH2O 365 Torr = PCO2 + 23.8 TorrPCO2 = 341.2 Torr = .449 atm PVM = mRT(.449 atm)(.100 L)(44.0 g/mol) = m(.0821Latm/molK)(298K)

m = .0807 g CO2

Corollary ProblemCorollary Problem

A gas collection bottle contains .25 mol of He, .50 mol Ar, and .75 mol of Ne. If the partial pressure of Helium is 200 Torr:What is the total pressure in the system?

What are the partial pressures of Ne and Ar?

A gas collection bottle contains .25 mol of He, .50 mol Ar, and .75 mol of Ne. If the partial pressure of Helium is 200 Torr:What is the total pressure in the system?

What are the partial pressures of Ne and Ar?

The answers are…The answers are…

nHe = .25 mol, nAr = .50 mol, nNe = .75 mol, PHe = 200 Torr.

ntotal =1.50 mol Ptotal/Phe = ntotal/nHe

Ptotal/200Torr = 1.50 mol/.25 mol Ptotal = 1200 Torr PAr/Ptotal = nAr/ntotal

Par/1200 = .50 mol/1.50 mol PAr = 400 Torr PNe = 1200 Torr - 400 Torr - 200 Torr Pne = 600 Torr

nHe = .25 mol, nAr = .50 mol, nNe = .75 mol, PHe = 200 Torr.

ntotal =1.50 mol Ptotal/Phe = ntotal/nHe

Ptotal/200Torr = 1.50 mol/.25 mol Ptotal = 1200 Torr PAr/Ptotal = nAr/ntotal

Par/1200 = .50 mol/1.50 mol PAr = 400 Torr PNe = 1200 Torr - 400 Torr - 200 Torr Pne = 600 Torr

Temperature and Kinetic Energy

Temperature and Kinetic Energy

Earlier, I stated that temperature is a relative measure of molecular motion.

By definition, Kinetic energy is a measure of the energy of motion.

Pretty similar right?

Earlier, I stated that temperature is a relative measure of molecular motion.

By definition, Kinetic energy is a measure of the energy of motion.

Pretty similar right?

Yes they areYes they are

KEav = 3/2*R*TThe average kinetic energy depends only on the absolute temperature.

R, the Ideal Gas Law Constant, should be 8.314 J/molK, since we will want the energy in the proper SI unit of Joules.

KEav = 3/2*R*TThe average kinetic energy depends only on the absolute temperature.

R, the Ideal Gas Law Constant, should be 8.314 J/molK, since we will want the energy in the proper SI unit of Joules.

A Thought QuestionA Thought Question

Which of the following ideal gases would have the largest average kinetic energy at 25oC? He, N2, CO, or H2

Which of the following ideal gases would have the largest average kinetic energy at 25oC? He, N2, CO, or H2

They are all the same!

They are all the same!

Since Keav = 3/2*R*T, the mass does not make a difference (ideally).

KE = 3/2*(8.314J/molK)*(298K)

KE = 3716 J/mol

Since Keav = 3/2*R*T, the mass does not make a difference (ideally).

KE = 3/2*(8.314J/molK)*(298K)

KE = 3716 J/mol

Speed vs Kinetic Energy

Speed vs Kinetic Energy

In physics, you learned that KE = 1/2*m*v2. The velocity, v, describes the speed of an object in a specific direction. If the mass, m, is measured in kg and the velocity is measured in m/s, then the kinetic energy would be measured in Joules.

In physics, you learned that KE = 1/2*m*v2. The velocity, v, describes the speed of an object in a specific direction. If the mass, m, is measured in kg and the velocity is measured in m/s, then the kinetic energy would be measured in Joules.

Physics to ChemistryPhysics to Chemistry

Rewriting the physics version, we could say that v=√(2*KE/m).

In chemistry, the Kinetic energy is measured in J/mol, so the mass would have to be measured in Kg/mol which is essentially molar mass.

Rewriting the physics version, we could say that v=√(2*KE/m).

In chemistry, the Kinetic energy is measured in J/mol, so the mass would have to be measured in Kg/mol which is essentially molar mass.

Root Mean Square Speed

Root Mean Square Speed

In Chemistry, we are not worried about velocities in multiple directions. We want an average speed independent of direction.

We call this Vrms - the root mean square speed.

Vrms = √(3RT/M)

In Chemistry, we are not worried about velocities in multiple directions. We want an average speed independent of direction.

We call this Vrms - the root mean square speed.

Vrms = √(3RT/M)

A Thought Question Revisited

A Thought Question Revisited

Which of the following ideal gases would have the largest root mean square speed at 25oC? He, N2, CO, or H2

Which of the following ideal gases would have the largest root mean square speed at 25oC? He, N2, CO, or H2

This Time They Are Different

This Time They Are Different

VVrmsrms = √(3RT/ = √(3RT/MM))

For He, VFor He, Vrmsrms = √(3RT/ = √(3RT/MM) = ) = √(3*8.314J/molK*298K)/4g/mol = 1363 m/s√(3*8.314J/molK*298K)/4g/mol = 1363 m/s

For NFor N22, V, Vrmsrms = √(3RT/ = √(3RT/MM) = ) = √(3*8.314J/molK*298K)/28g/mol = 515 m/s√(3*8.314J/molK*298K)/28g/mol = 515 m/s

Since the molar mass is the same for NSince the molar mass is the same for N22 and CO, and CO, their Vtheir Vrmsrms would be the same, 515 m/s. would be the same, 515 m/s.

For HFor H22, V, Vrmsrms = √(3RT/ = √(3RT/MM) = ) = √(3*8.314J/molK*298K)/2g/mol = 1928 m/s√(3*8.314J/molK*298K)/2g/mol = 1928 m/s

Because HBecause H22 is the lightest, it moves the is the lightest, it moves the fastest.fastest.

VVrmsrms = √(3RT/ = √(3RT/MM))

For He, VFor He, Vrmsrms = √(3RT/ = √(3RT/MM) = ) = √(3*8.314J/molK*298K)/4g/mol = 1363 m/s√(3*8.314J/molK*298K)/4g/mol = 1363 m/s

For NFor N22, V, Vrmsrms = √(3RT/ = √(3RT/MM) = ) = √(3*8.314J/molK*298K)/28g/mol = 515 m/s√(3*8.314J/molK*298K)/28g/mol = 515 m/s

Since the molar mass is the same for NSince the molar mass is the same for N22 and CO, and CO, their Vtheir Vrmsrms would be the same, 515 m/s. would be the same, 515 m/s.

For HFor H22, V, Vrmsrms = √(3RT/ = √(3RT/MM) = ) = √(3*8.314J/molK*298K)/2g/mol = 1928 m/s√(3*8.314J/molK*298K)/2g/mol = 1928 m/s

Because HBecause H22 is the lightest, it moves the is the lightest, it moves the fastest.fastest.

And this leads us to…

And this leads us to…

Graham’s LawGraham’s LawThe rate of effusion The rate of effusion (or diffusion) is (or diffusion) is inversely proportional inversely proportional to the square root of to the square root of the molar mass.the molar mass.

Effusion is the process Effusion is the process of a gas escaping from of a gas escaping from one container through a one container through a small opening.small opening.

Diffusion is the Diffusion is the process of a gas process of a gas spreading out in a spreading out in a large container.large container.

Graham’s LawGraham’s LawThe rate of effusion The rate of effusion (or diffusion) is (or diffusion) is inversely proportional inversely proportional to the square root of to the square root of the molar mass.the molar mass.

Effusion is the process Effusion is the process of a gas escaping from of a gas escaping from one container through a one container through a small opening.small opening.

Diffusion is the Diffusion is the process of a gas process of a gas spreading out in a spreading out in a large container.large container.

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RateRate11*√*√MM1 1 = Rate= Rate22*√*√MM22

Rate vs. SpeedRate vs. Speed

When we say rate, we When we say rate, we are talking about an are talking about an amount of gas (moles, amount of gas (moles, grams, or even liters) grams, or even liters) per unit of time.per unit of time.

This is not the same as This is not the same as speed which is distance speed which is distance over time.over time.

However, the main idea However, the main idea is the same; lighter is the same; lighter gases move/effuse gases move/effuse faster.faster.

When we say rate, we When we say rate, we are talking about an are talking about an amount of gas (moles, amount of gas (moles, grams, or even liters) grams, or even liters) per unit of time.per unit of time.

This is not the same as This is not the same as speed which is distance speed which is distance over time.over time.

However, the main idea However, the main idea is the same; lighter is the same; lighter gases move/effuse gases move/effuse faster.faster.

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For exampleFor example

Under a given set of conditions, oxygen diffuses at 10 L/hr. A different gas diffuses at 20 L/hr under the same conditions. What is the molar mass of this gas?

Under a given set of conditions, oxygen diffuses at 10 L/hr. A different gas diffuses at 20 L/hr under the same conditions. What is the molar mass of this gas?

2 ways to solve this2 ways to solve this

By the equation: RateRate11*√*√MM1 1 = Rate= Rate22*√*√MM22

10 L/hr(√32g/mol) = 40 L/hr 10 L/hr(√32g/mol) = 40 L/hr *√*√MM22

MM22 = 2 g/mol = 2 g/mol

By Logic: If the rate of the unknown gas If the rate of the unknown gas

is 4 times faster, it must be is 4 times faster, it must be 4422, or 16, times lighter., or 16, times lighter.

32 g/mol divided by 16 is 2 32 g/mol divided by 16 is 2 g/mol.g/mol.

By the equation: RateRate11*√*√MM1 1 = Rate= Rate22*√*√MM22

10 L/hr(√32g/mol) = 40 L/hr 10 L/hr(√32g/mol) = 40 L/hr *√*√MM22

MM22 = 2 g/mol = 2 g/mol

By Logic: If the rate of the unknown gas If the rate of the unknown gas

is 4 times faster, it must be is 4 times faster, it must be 4422, or 16, times lighter., or 16, times lighter.

32 g/mol divided by 16 is 2 32 g/mol divided by 16 is 2 g/mol.g/mol.

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Real vs. IdealReal vs. Ideal

At the start of the presentation, we talked about the major assumptions of the Kinetic Molecular Theory.

If a gas obeys the KMT, it is ideal.

If it doesn’t obey the KMT, it is real.

At the start of the presentation, we talked about the major assumptions of the Kinetic Molecular Theory.

If a gas obeys the KMT, it is ideal.

If it doesn’t obey the KMT, it is real.

So What does that Mean?

So What does that Mean?

The molecules of an ideal The molecules of an ideal gas do not interact with gas do not interact with one another, except to one another, except to collide elastically.collide elastically.

The molecules of a real gas The molecules of a real gas will interact, to some will interact, to some degree. degree.

Since no gases are always Since no gases are always ideal, the trick is to make ideal, the trick is to make a real gas behave ideally.a real gas behave ideally.

The molecules of an ideal The molecules of an ideal gas do not interact with gas do not interact with one another, except to one another, except to collide elastically.collide elastically.

The molecules of a real gas The molecules of a real gas will interact, to some will interact, to some degree. degree.

Since no gases are always Since no gases are always ideal, the trick is to make ideal, the trick is to make a real gas behave ideally.a real gas behave ideally.

Real Gases Behaving Ideally

Real Gases Behaving Ideally

If we don’t want the molecules attracting or repelling one another, the first issue is to use a nonpolar gas.

If we use smaller amounts of the gas, there are less chances of them interacting.

If we don’t want the molecules attracting or repelling one another, the first issue is to use a nonpolar gas.

If we use smaller amounts of the gas, there are less chances of them interacting.

Real Gases Behaving Ideally

Real Gases Behaving Ideally

If we put the gas in larger volumes, the molecules will not interact as much.

Likewise, if we keep the gas under low pressure , the molecules will not interact as much.

This could also be stated by having molecules that have low densities.

If we put the gas in larger volumes, the molecules will not interact as much.

Likewise, if we keep the gas under low pressure , the molecules will not interact as much.

This could also be stated by having molecules that have low densities.

Real Gases Behaving Ideally

Real Gases Behaving Ideally

The smaller the molecules, the The smaller the molecules, the less likely they are to less likely they are to interact.interact.

Lastly, at higher temperatures Lastly, at higher temperatures the molecules are moving too the molecules are moving too fast to actually interact with fast to actually interact with one another - they are more one another - they are more likely to collide elastically.likely to collide elastically.

The smaller the molecules, the The smaller the molecules, the less likely they are to less likely they are to interact.interact.

Lastly, at higher temperatures Lastly, at higher temperatures the molecules are moving too the molecules are moving too fast to actually interact with fast to actually interact with one another - they are more one another - they are more likely to collide elastically.likely to collide elastically.

Phase DiagramsPhase Diagrams

A phase diagram shows how the different states of matter exist based on the pressure and temperature.

A phase diagram shows how the different states of matter exist based on the pressure and temperature.

A Typical Phase Diagram

A Typical Phase Diagram

Water is not Typical…

Water is not Typical…

…and Helium is weird!

…and Helium is weird!