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Nick Gross Jacob Shilling Garrett Buschmann Kinetic Molecular Theory 1. 2. 3. 4. 5. 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 straightline motion and obeying Newton's Laws. The molecules in a gas occupy no volume (that is, they are points). 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 3kT/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 P1V1=P2V2 http://upload.wikimedia.org/wikipedia/commons/thumb/1/15/Boyles_Law_animated.gif/300pxBoyles_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: P1*V1 = P2*V2 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=XmL*800torr 22500mL*torr==XmL*800torr 22500mLtorr/800=XmL 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: P1*V1 = P2*V2 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=XkPa*1.5L 800Kpa*L/1.5L=XkPa 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=P2*760mmHg 360L*mmHg/760mmHg=P2 Answer: 0.473L Charles's Gas Law States that volume and temperature have a direct relationship V1T2=V2T1 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=V2*303 111.9K*L/303K=V2 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*T2 = 1.0L*350K T2 = 350K*L/2.0L T2 = 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=V2 *550K 1500L*K/550K=V2 Volume=2.72L Gay-Lussac’s Gas Law Temperature and pressure have a direct relationship P1T2=T1P2 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 : V1*T2 = V2*T1 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: P1 T2 = P2 T1 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: P1 T2 = P2 T1 3.0atm*398K= P2 *298K P2 = 1194Katm/298K P2 = 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: P1 T2 = P2 T1 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: P1 T2 = P2 T1 1.0atm* T2 = 4.0atm*298K T2 =1192atmK/1.0atm T2 =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= P2 *303K 666atm*K/303K = P2 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://encryptedtbn3.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 Ptotal=P1+P2+P3+P4….. 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? Ptotal=P1+P2+P3+P4….. 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 Ptotal=P1+P2+P3+P4….. 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 Ptotal=P1+P2+P3+P4….. 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- 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 =3GUDGeQSpTfUM:&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? rateH2 / rateCO2 = MCO / MH 2 44.19 amu CO2 / 2.016 amuH2 rateH2 = 4.67 m/s 2 Graham’s Law Example 2 What is the rate of diffusion of NH3 compared to He? Does NH3 effuse faster or slower than He? MNH3 / MHe2 17.031 amu CO2 / 4.003 amuH2 Rate He = 4.67 m/s