Nick Gross Jacob Shilling Garrett Buschmann

Kinetic Molecular Theory

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 The Kinetic Molecular Theory of Gases is five theories that describe the behavior of molecules in a gas.

A gas consists of a collection of small particles traveling in straight line motion and obeying Newton's Laws.

The molecules in a gas occupy no volume (that is, they are points).

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Collisions between molecules are perfectly elastic (that is, no energy is gained or lost during the collision).

There are no attractive or repulsive forces between the molecules.

The average kinetic energy of a molecule is 3

kT

/2. (

T

is the absolute temperature and

k

is the Boltzmann constant.) http://www.chm.davidson.edu/vce/kineticmoleculartheory/basicconcepts.html

States of Matter

Solid

 Particles are tightly packed in a pattern.

 Has the least amount of energy.

 Particles vibrate slightly but do not move chenistrychemistry.blogspot.com

blogs.msdn.com

-

Liquid

 Particles in a liquid are close together, but aren't rigid like solids.

 Molecules move around and slide around each other http://www.wisegeek.org/how-do-i-choose-the-best-electrolyte-drink.htm#slideshow

http://www.chem.purdue.edu/gchelp/liquids/character.html

Gas

 Particles in a gas have no definite shape.

 They take the shape and volume of the container that they are in.

 The particles vibrate and move freely at high speeds.

http:// www.123rf.com/photo_16248020_red-gas-container-isolated-on-white-background-3d-render.html

Gas Laws

 Boyles Gas Law  Charles’ Gas Law  Gay Lussac’s Gas Law  Combined Gas Law  Ideal Gas Law  Dalton’s Law of Partial Pressure  Avodadro’s Law

Gas Pressure Conversions

 Conversions must be made prior to starting a calculation.

 Calculations should be done in K for temperature, atm for pressure, and Liters for volume  1atm=101.3kPa=760torr=760mmhg  Atm means atmosphere, or the pressure of the earths atmosphere

Boyle’s Gas Law

 States that pressure and volume have an inverse relationship  P 1 V 1 =P 2 V 2 http://upload.wikimedia.org/wikipedia/commons/thumb/1/15/Boyles_Law_animated.gif/300px Boyles_Law_animated.gif

Boyle’s practice 1

 A sample of oxygen gas occupies a volume of 300mL at 750 torr pressure. What volume will it occupy at 800 torr? =  Formula: P 1 *V 1 = P 2 *V 2

Boyle’s practice 1

 A sample of oxygen gas occupies a volume of 300mL at 750 torr pressure. What volume will it occupy at 800 torr?

 300mL*750torr=

X

mL*800torr  22500mL*torr==

X

mL*800torr  22500mLtorr/800=

X

mL  281.25mL

 .281L

Boyle’s practice 2

 A sample of carbon dioxide occupies a volume of 4.0 Liters at 200 kPa pressure. If the gas was compressed to 1.5 liters, what would the new pressure be?

 Formula: P 1 *V 1 = P 2 *V 2

Boyle’s practice 2

 A sample of carbon dioxide occupies a volume of 4.0 Liters at 200 kPa pressure. If the gas was compressed to 1.5 liters, what would the new pressure be?

 200kPa*4.0L=

X

kPa*1.5L

 800Kpa*L/1.5L=

X

kPa 

X

=533.3kPa

Boyle’s practice 3

 Ammonia gas occupies a volume of 500mL at 720 mmHg. What volume will it occupy at 760mmHg?

Boyle’s practice 3

 Ammonia gas occupies a volume of 500mL at 720 mmHg. What volume will it occupied at 760mmHg?

 .5L*720mmHg=P 2 *760mmHg  360L*mmHg/760mmHg=P 2  Answer: 0.473L

Charles's Gas Law

 States that volume and temperature have a direct relationship  V 1 T 2 =V 2 T 1 http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Gases/Gas_Law s

Charles’ Practice 1

 A sample of nitrogen occupies a volume of 300mL at 30 °C. What volume will it occupy at 100 °C?

 .3L*373=V 2 *303  111.9K*L/303K=V 2  Volume= 0.37L

Charles’ Practice 2

 A sample of chlorine gas occupies a volume 2.0L at 350 Kelvin. At 1.0L what would its temperature be?

Charles’ Practice 2

 A sample of chlorine gas occupies a volume 2.0L at 350 Kelvin. At 1.0L what would its temperature be?  2.0L*T 2 = 1.0L*350K  T 2 = 350K*L/2.0L

 T 2 = 175K

Charles’ Practice 3

 A sample of neon gas occupies a volume 5.0L at 550 Kelvin. At 300 Kelvin what would its volume be?

Charles’ Practice 3

 A sample of neon gas occupies a volume 5.0L at 550 Kelvin. At 300 Kelvin what would its volume be?

 5.0L*300K=V 2 *550K  1500L*K/550K=V 2  Volume=2.72L

Gay-Lussac’s Gas Law

 Temperature and pressure have a direct relationship  P 1 T 2 =T 1 P 2 http://www.grc.nasa.gov/WWW/k-12/airplane/Animation/gaslab/Images/chtmmp.gif

Charles’ Practice 1

 A sample of nitrogen occupies a volume of 300mL at 30 °C. What volume will it occupy at 100 °C?

 Formulas : V 1 *T 2 = V 2 *T 1   Temperature must be in Kelvin K= °C+273

Gay-Lussac’s practice 1

 A tank of gas has a pressure of 3.0 atm at 25 °C. What will be the new pressure in the tank if its temperature is increased to 125 °C?  Formula: P 1 T 2 = P 2 T 1  Temperature must be in Kelvin  K= °C+273

Gay-Lussac’s practice 1

 A tank of gas has a pressure of 3.0 atm at 25 °C. What will be the new pressure in the tank if its temperature is increased to 125 °C?  Formula: P 1 T 2 = P 2 T 1  3.0atm*398K= P 2 *298K  P 2 = 1194Katm/298K  P 2 = 4.01atm

Gay-Lussac’s practice 2

 A tank of gas has a pressure of 1.0 atm at 25 °C. What will be the new temperature in the tank if its pressure is increased to 4.0atm?

 Formula: P 1 T 2 = P 2 T 1

Gay-Lussac’s practice 2

 A tank of gas has a pressure of 1.0 atm at 25 °C. What will be the new temperature in the tank if its pressure is increased to 4.0atm?

 Formula: P 1 T 2 = P 2 T 1  1.0atm* T 2 = 4.0atm*298K  T 2 =1192atmK/1.0atm

 T 2 =1192K

Gay-Lussac’s practice 3

 A tank of gas has a pressure of 2.0 atm at 30 °C. What will be the new pressure in the tank if its temperature is increased to 60 °C?

Gay-Lussac’s practice 3

 A tank of gas has a pressure of 2.0 atm at 30 °C. What will be the new pressure in the tank if its temperature is increased to 60 °C?

 2.0atm*333K= P 2 *303K  666atm*K/303K = P 2  Answer: 2.2atm

Combined Gas Law

 Gas law which combines Charles's law, Boyle's law, and Gay-Lussac's law.

 Temperature is always in Kelvin.

 Used when none of the values is constant

Combined Gas Law example problem #1 • • • • • • • • A container has a gas of volume 40L, at a pressure of 6atm and a temperature of 400 K. Find the temperature of the gas which has a volume 60 L at a pressure of 11atm. V1= 40L, P1 = 6atm, T1 = 400K, V2 = 60L, P2= 11atm, T2= ?

Substitute the values in below V1xP1/T1=V2xP2/T2 Final Temperature(T2) = (P2)(V2)(T1) / (P1)(V1) = (11atm x 60L x 400K) / (6atm x 40L) = 264000 / 240 Final Temperature(T2) = 1100 K

Combined Gas Law practice problem #1 • • A container has a gas of volume 22L, at a pressure of 3atm and a temperature of 400 K. Find the temperature of the gas which has a volume 50L at a pressure of 7atm.

Answer

• • • • • • • V1= 22L, P1 = 3atm, T1 = 400K, V2 = 50, P2= 7atm, T2= ?

Substitute the values in below V1xP1/T1=V2xP2/T2 Final Temperature(T2) = (P2)(V2)(T1) / (P1)(V1) = (7atm x 50L x 400K) / (3atm x 22L) = 140000 / 66 Final Temperature(T2) = 2121.21K

Combined Gas Law practice problem #2

• A container has a gas of volume 2L, at a pressure of 3atm and a temperature of 300 K. Find the temperature of the gas which has a volume 1.5L at a pressure of 1.2atm.

Answer

• .6K

Summation

 Increasing the pressure will increase the temperature and decrease volume of gas particles  Increasing the temperature will increase the pressure and volume of gas particles  Increasing the volume will decrease the pressure and increase the temperature of gas particles

Ideal Gas Law

 PV=nRT  P= Pressure in atm  V= Volume in L  n= moles  T= Temperature in K  R= Ideal gas constant https://encrypted tbn3.gstatic.com/images?q=tbn:ANd9GcQkcRI0PopuIU_Gu1MpACbg ILv2Rnkc3prMBpQD2r91s_vXBEWsSw

Ideal Gas Law example problem #1 • • • • • • • What is the volume from the 0.250 moles gas at 2 atm and 300K temperature?

P = 2atm, n = 0.250 moles, T = 300K, R = 8.314 L(atm)/mol(K) PV=nRT (V) = nRT / P = (0.250moles x 8.314 x 300K) / 2atm = 623.55 / 2 (V) = 311L

Ideal Gas Law practice problem #1 • Find the temperature from the .250L cylinder contaning 1.50 moles gas at 2.3 ATM?

Answer

• V = 0.250 L, n = 1.50 mol, P = 2.3atm, R =8.314 L(atm)/mol(K) • PV=nRT • T = PV / nR • = (2.3atm x 0.250L) / (1.50mol x 8.314) • = .0575 / 12.471

• (T) = .004 K

Ideal Gas Law Practice problem #1

• What is the volume from the .45 moles gas at 1.23 atm and 225K temperature?

Answer

• V=684.38L

Dalton’s Law of Partial Pressure

 States that in a mixture of two or more gases, the total pressure is the sum of the partial pressures  P total =P 1 +P 2 +P 3 +P 4 …..

http://img.docstoccdn.com/thumb/orig/104619699.png

Dalton’s Law of Partial Pressure Example 1

 In a mixture of Hydrogen, Oxygen, and Nitrogen the total pressure is 100 atm. The partial pressure of the Hydrogen is 10 atm, and the partial pressure of the Oxygen is 30 atm. What is the partial pressure of the Nitrogen?

P total =P 1 +P 2 +P 3 +P 4 …..

150atm= 10 atm H + 30 atm O + x atm N x = 60 atm N

Dalton’s Law of Partial Pressure Problem 1

 In a mixture of Hydrogen, Neon, and Radon the total pressure is 120 atm. The partial pressure of the Hydrogen is 20 atm, and the partial pressure of the Neon is 70 atm. What is the partial pressure of the Radon?

Dalton’s Law of Partial Pressure Problem 1 Answer

 Answer P total =P 1 +P 2 +P 3 +P 4 …..

120atm= 20 atm H + 70 atm Ne + x atm Rn x = 30 atm Rn

Dalton’s Law of Partial Pressure Practice Problem 2

 In a mixture of Oxygen, Helium, and Fluorine the total pressure is 40 atm. The partial pressure of the Oxygen is 10 atm, and the partial pressure of the Helium is 15 atm. What is the partial pressure of the Fluorine?

Dalton’s Law of Partial Pressure Problem 2 Answer

 Answer P total =P 1 +P 2 +P 3 +P 4 …..

40atm= 10 atm O + 15 atm He + x atm F x = 15 atm F

Calculating Collection Gas over Water

 Calculating gas over water is done using Daltons law of partial pressure, and the Ideal Gas Law http://crescentok.com/staff/jaskew/isr/tigerchem/gas_laws/dalton2.gif

https://www.boundless.com/chemistry/gase s/partial-pressure/collecting-gases-over water--35/

Graham’s Law of Effusion

 Effusion: Is the process where molecules of a gas in a container randomly pass through a tiny opening in the container http://img.sparknotes.com/content/testprep/bookimgs/sat2/chemistry/0003/sat117002_0510.gif

Graham’s Law of Diffusion

 Diffusion: Gradual mixing of two gases due to spontaneous and random motion http://www.google.com/url?sa=i&source=images&cd=&cad=rja&docid=HNdYZAWycZQxlM&tbnid =3GUDGeQSpT fUM:&ved=0CAgQjRwwAA&url=http%3A%2F%2Fwww.1728.org%2Fgraham.htm&ei=PdOoUbH oKYPTiwLrxIG4Cg&psig=AFQjCNE0xWYb-XnByJJNfInn3fnKJtRLcA&ust=1370105021720842

Graham’s Law Example 1

 Under the same conditions of temperature and pressure, how many times faster will hydrogen effuse compared to carbon dioxide?

rateH 2 / rateCO 2 = MCO 2 / MH 2 44.19 amu CO 2 / 2.016 amuH 2 rateH 2 = 4.67 m/s

Graham’s Law Example 2

What is the rate of diffusion of NH 3 compared to He? Does NH 3 effuse faster or slower than He?

MNH 3 / MHe 2 17.031 amu CO 2 / 4.003 amuH 2 Rate He = 4.67 m/s