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1 GASES Chemistry I – Chapter 14 Chemistry I Honors – Chapter 13 SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, turn off the backgrounds (Tools>Options>Print>UNcheck "Background Printing")!

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2 GAS: One of Three STATES OF MATTER

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3 General Properties of Gases There is a lot of “free” space in a gas.There is a lot of “free” space in a gas. Gases can be expanded infinitely.Gases can be expanded infinitely. Gases fill containers uniformly and completely.Gases fill containers uniformly and completely. Gases diffuse and mix rapidly.Gases diffuse and mix rapidly.

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4 Importance of Gases Airbags fill with N 2 gas in an accident.Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3.Gas is generated by the decomposition of sodium azide, NaN 3. 2 NaN 3 ---> 2 Na + 3 N 22 NaN 3 ---> 2 Na + 3 N 2

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5 Our Atmosphere : Gravity holds the atmosphere close to the Earth. It is divided into various layers according to differences in temperature: The Troposphere:The Troposphere: –We live in the Troposphere –It contains more than 75% of the atmosphere’s mass The Stratosphere:The Stratosphere: –Most of the clouds and rain are in this layer –It extends up to 48km above the earth! The ozone layer is between the Stratosphere and MesosphereThe ozone layer is between the Stratosphere and Mesosphere The Mesosphere:The Mesosphere: –This layer extends to 80km above the Earth! The ThermosphereThe Thermosphere –Here the air is very thin –Over 99.99% of the atmosphere lies below this layer –It is comprised of:

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7 /C003124/en/atmos.htmhttp://library.thinkquest.org /C003124/en/atmos.htm

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8 The Ionosphere: Reflects radiowaves back to Earth so signals can be sent around the worldThe Ionosphere: Reflects radiowaves back to Earth so signals can be sent around the world The Exosphere: This layer begins at ~480km above the Earth and fades away into spaceThe Exosphere: This layer begins at ~480km above the Earth and fades away into space

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9 Did you know... About 99% of the atmosphere is made up of oxygen (21%) and nitrogen (78%).About 99% of the atmosphere is made up of oxygen (21%) and nitrogen (78%). The remainder is made up of Ar (argon), CO 2 and very small amounts of H 2, NH 3, O 3 (ozone), CH 4, CO, He, Ne, Kr and Xe. also gases such as SO 2, NO 2 are put into the air from vehicles & industryThe remainder is made up of Ar (argon), CO 2 and very small amounts of H 2, NH 3, O 3 (ozone), CH 4, CO, He, Ne, Kr and Xe. also gases such as SO 2, NO 2 are put into the air from vehicles & industry About 20 % of the Earth's population breathes severely contaminated air, largely CO 2 and SO 2 resulting from industrial processes. This increases the number of respiratory conditions, especially amongst children and elders. 13 % of the British children experience asthma caused by air contaminationAbout 20 % of the Earth's population breathes severely contaminated air, largely CO 2 and SO 2 resulting from industrial processes. This increases the number of respiratory conditions, especially amongst children and elders. 13 % of the British children experience asthma caused by air contamination

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10 Properties of Gases Gas properties can be modeled using math. Model depends on— V = volume of the gas (L)V = volume of the gas (L) T = temperature (K)T = temperature (K) –ALL temperatures in the entire chapter MUST be in Kelvin!!! No Exceptions! n = amount (moles)n = amount (moles) P = pressure (atmospheres)P = pressure (atmospheres)

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11 Temperature Average kinetic energy of the particles of a substanceAverage kinetic energy of the particles of a substance Without particles there is no temperatureWithout particles there is no temperature No temperature in a vacuumNo temperature in a vacuum Must be measured in Kelvin to avoid negative values while doing calculations C° = KMust be measured in Kelvin to avoid negative values while doing calculations C° = K

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12 PRESSURE : A FORCE per unit area Measured in pascals (Pa) = 1 N/m 2Measured in pascals (Pa) = 1 N/m 2 = 100 g / m 2 = 100 g / m 2 A normal day has a pressure of 100 kPaA normal day has a pressure of 100 kPa = 100 x 1000 Pa x 100 g / m 2 1 kPa 1 kPa = kg / m 2 Atmospheric pressure varies with altitude – the lower the altitude, the longer and heavier is the column of air above an area of the earth.

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14 Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643) Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). (Just like a straw in a soft drink) P of Hg pushing down related to Hg densityHg density column heightcolumn height

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15PressureColumn height measures Pressure of atmosphere * 1 standard atmosphere (atm) = 760 mm Hg (or torr) = inches Hg = 14.7 pounds/in2 (psi) = * kPa (SI unit is PASCAL) = about 34 feet of water! * Memorize these!

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16 Pressure Conversions A.The air pressure in a cave underwater is 2.30 atm, what is the pressure in kPa? kPa B. What is 475 mm Hg expressed in kPa? kPa 760 mm Hg = 233 kPa = kPa 475 mm Hg x 2.30 atm x 1 atm

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17 Pressure Conversions A.The air pressure in a cave underwater is 2.30 atm, what is the pressure in kPa? kPa B. What is 475 mm Hg expressed in kPa? kPa 760 mm Hg = 233 kPa = kPa475 mm Hg x 2.30 atm x 1 atm

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18 Pressure Conversions A. What is 3.71 atm expressed in kPa? B. The pressure of a tire is measured as 830 torr. What is this pressure in atm?

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19 Crushed Can Demo :Crushed Can Demo : Why did the can implode? What happened to the pressure inside once the temperature decreased? Why did the can implode? What happened to the pressure inside once the temperature decreased? Lower temperature = slower moving particles Lower temperature = slower moving particles = less collisions with the wall = less collisions with the wall T P T P Did the # of particles/moles change? Did the # of particles/moles change? Did the air pressure change(# particles colliding)? Did the air pressure change(# particles colliding)? Did the volume change? Did the volume change? Only because the pressure outside is much greater then inside – the ouside pressure crushed the can. Only because the pressure outside is much greater then inside – the ouside pressure crushed the can. T α P

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20 Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. P 1 P 2 = T 1 T 2 If temperature goes up, the pressure goes up!If temperature goes up, the pressure goes up! Joseph Louis Gay- Lussac ( ) P 1 = T 1 P 2 T 2 Charles’ Law h?v=tPH57yp0x1U h?v=tPH57yp0x1U

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21 Gas Pressure, Temperature, and Kinetic Molecular Theory P proportional to T As the temperature increases the particles move faster increasing # of collisions = ↑ pressure

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22 A gas has a pressure of 3.0 atm at 127º C. What is its pressure at 227º C? T 1 = 127°C = 400K P 1 = 3.0 atm T 2 = 327°C = 600K P 2 = ? What law applies to this situation? Gay-Lussac’s Law: T and P – Gay-Lussac’s Law: 3.0 atm = P 2 P 2 = 3.0 atm x 600 K P 2 = 4.50 atm 3.0 atm = P 2 P 2 = 3.0 atm x 600 K P 2 = 4.50 atm 400K 600K 400 K 400K 600K 400 K P 1 = P 2 T 1 T 2

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23 Charles’s Law If n and P are constant, then V α T V and T are directly proportional. V 1 V 2 = T 1 T 2 If temperature goes up, the volume goes up!If temperature goes up, the volume goes up! Jacques Charles ( ). Isolated boron and studied gases. Balloonist. V 1 = T 1 V 2 T 2

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24 Charles’s original balloon Modern long-distance balloon

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25 Charles’s Law Boyle’s Law – Vacuum Demo - 0 K -273 °C

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26 A gas has a volume of 3.0 L at 127°C. What is its volume at 227 °C? T 1 = 127°C = 400K V 1 = 3.0 L T 2 = 227°C = 500K V 2 = ? Which law applies to this situation? Charles’s Law: T and V - Charles’s Law: 3.0 L = V 2 V 2 = 3.0 L x 500 K V 2 = 3.75 L 3.0 L = V 2 V 2 = 3.0 L x 500 K V 2 = 3.75 L 400 K 500 K 400 K 400 K 500 K 400 K V 1 = V 2 T 1 T 2

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27 Boyle’s Law P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down. P 1 V 1 = P 2 V 2 Robert Boyle ( ). Son of Earl of Cork, Ireland.

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28 Volume decreases As Pressure Increases

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29 Boyle’s Law and Kinetic Molecular Theory P proportional to 1/V Smaller volume means the particles are forced closer together – increasing collisions with each other and with the walls of the container.

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30 Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

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31 Boyle’s Law: - Volume Pressure PV = k PV = k Temperature is constant Temperature is constant

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33 A gas has a volume of 3.0 L at 2 atm. What is its volume at 4 atm? P 1 = 2 atm V 1 = 3.0 L P 2 = 4 atm V 2 = ? Which law applies to this situation? P and V = Boyle’s Law P and V = Boyle’s Law 2 atm x 3.0 L = 4 atm x V 2 V 2 = 2 atm x 3.0 L V 2 = 2 atm x 3.0 L 4 atm 4 atm V 2 = 1.5 L V 2 = 1.5 L P 1 V 1 = P 2 V 2

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34 SUMMARY of Gas Laws LAW RELAT- IONSHIP LAWCONSTANT Charles’ V TV TV TV T V 1 /T 1 = V 2 /T 2 P, n Gay- Lussac’s P TP TP TP T P 1 /T 1 = P 2 /T 2 V, n Boyle’s P VP VP VP V P 1 V 1 = P 2 V 2 T, n

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35 Combined Gas Law The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! V 1 P 1 V 2 P 2 = T 1 T 2

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36 Combined Gas Law If you only need one of the gas laws, you can cover up the item that is constant and you will get that gas law! = P1P1 V1V1 T1T1 P2P2 V2V2 T2T2 Boyle’s Law Charles’ Law Gay-Lussac’s Law

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37 Combined Gas Law Problem A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? Given: P 1 = atm V 1 = 180 mL T 1 = 29°C = 302 K P 2 = 3.20 atm V 2 = 90 mL T 2 = ??

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38 Calculation P 1 = atm V 1 = 180 mL T 1 = 302 K P 2 = 3.20 atm V 2 = 90 mL T 2 = ?? P 1 V 1 P 2 V 2 = P 1 V 1 T 2 = P 2 V 2 T 1 T 1 T 2 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = 3.20 atm x 90.0 mL x 302 K atm x mL T 2 = 604 K = 331 °C T 2 = 604 K = 331 °C = 604 K

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39 Learning Check A gas has a volume of 675 mL at 35°C and atm pressure. What is the temperature in °C when the gas has a volume of L and a pressure of 802 mm Hg?

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40 One More Practice Problem A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

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41 And now, we pause for this commercial message from STP STP in chemistry stands for Standard Temperature and Pressure Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

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42 Try This One A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

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43 Ideal Gases, Avogadro’s Theory and the Ideal Gas Law What are three things that make the behaviour of a gas ideal? 1) How do they collide? 2) Does their size matter? 3) How do they move?

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44 What are three things that make the behaviour of a gas ideal? 1)How do they collide? ELASTICALLY no attraction, ELASTICALLY no attraction, no energy loss, perfect bounces no energy loss, perfect bounces 2) Does their size matter? NO too small, spaces between too big too small, spaces between too big 3) How do they move? straight lines and randomly straight lines and randomly

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45 FYI: What you will see in university

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46 Avogadro’s Theory Equal volumes of gases at the same T and P have the same number of molecules. Volume is proportional to molesVolume is proportional to moles V and n are directly relatedV and n are directly related V α nV α n twice the volume twice the volume twice as many molecules

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47 Ideal Gas Law: Generally speaking, an ideal gas obeys all gas laws perfectly under all conditions: 1) Charles’s Law V α T (V / T = k ) 2) Gay-Lussac’s Law P α T (P / T = k ) 3) Boyle’s Law V α 1 / p ( V x P = k ) 4) Avogadro’s Theory V α n 4) Avogadro’s Theory V α n In other words, V α n α T α 1/p In other words, V α n α T α 1/p All these variables can be combined to include an equation with moles, n using a constant… V = constant x n x T x 1/p The constant is the letter R The constant is the letter R

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48 The Constant R: Is called the Ideal GAS CONSTANTIs called the Ideal GAS CONSTANT Relates P, T, V, and molesRelates P, T, V, and moles Is found by substituting in known values for P, T, V and n (moles) Ex. You could sub in values at STP: T = 273 K, P = 1 atm, 1 mole gas = 22.4 LIs found by substituting in known values for P, T, V and n (moles) Ex. You could sub in values at STP: T = 273 K, P = 1 atm, 1 mole gas = 22.4 L R with different units : These #’s are always provided J / K mol or J K -1 mol J / K mol or J K -1 mol L atm/K mol mainly used L atm/K mol mainly used Pa m 3 K -1 mol Pa m 3 K -1 mol L Torr K -1 mol L Torr K -1 mol -1 Ideal Gas Law:

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49 Brings together gas properties. Can be derived from experiment and theory. BE SURE YOU KNOW THIS EQUATION! P V = n R T V= R x n x T x 1/p Multiply both sides by p, rearrange n… IDEAL GAS LAW

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50 Using PV = nRT P = Pressure V = Volume T = Temperature N = number of moles R is the constant called the Ideal Gas Constant R = L atm Mol K

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51 REMINDER: 1 standard atmosphere (atm) 1 standard atmosphere (atm) 1 atm = 760 mm Hg (or torr) 1 atm = 760 mm Hg (or torr) = kPa (SI unit is PASCAL) = kPa (SI unit is PASCAL) Ex kPa = kPa kPa / atm kPa / atm = atm Ex torr = 1040 torr 760 mm Hg / atm = 1.50 atm = 1.50 atm

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52 Using PV = nRT How much N 2 is required to fill a small room with a volume of 27,000 L to 745 mm Hg at 25 o C? Solution Solution 1. Given: convert to proper units V = L V = L T = 25 o C = 298 K T = 25 o C = 298 K P = 745 mm Hg (760 mm Hg/atm) = 0.98 atm P = 745 mm Hg (760 mm Hg/atm) = 0.98 atm And we always know R = Latm / molK

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53 Using PV = nRT How much N 2 is required to fill a small room with a volume of L to P = 745 mm Hg at 25 o C? Solution Solution 1. Divide both sides by RT to solve for n. 2. Now plug in those values and solve for the unknown. PV = nRT 1. Divide both sides by RT to solve for n. 2. Now plug in those values and solve for the unknown. PV = nRT n = PV / RT n = PV / RT n = 1.1 x 10 3 mol (or about 30 kg of gas) RT RT

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54 Learning Check Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. EX. Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure in atm in the tank in the dentist office? If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure in atm in the tank in the dentist office? Given: n=2.86 mol, V=20.0L T=23°C K P=? PV = nRT P = nRT / V P = 2.86 mol ( L atm/mol K)(296 K) 20.0L P = 3.48 atm P = 3.48 atm

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55 Using PV = nRT When Given Mass 0.78 grams of hydrogen gas is produced at 22°C and 125 kPa. What volume of hydrogen is expected? Given: T = 22°C = 295 K, Given: T = 22°C = 295 K, P = 125 kPa = 125 kPa/101.3 = 1.23 atm n = find n using molar mass first P = 125 kPa = 125 kPa/101.3 = 1.23 atm n = find n using molar mass first n = mass PV = nRT n = mass PV = nRT Mm V = nRT/P Mm V = nRT/P n = 0.78g = mol ( L atm/mol K )295K 2.02 g/mol 1.23 atm n = 0.78g = mol ( L atm/mol K )295K 2.02 g/mol 1.23 atm = mol = 0.76 L = mol = 0.76 L

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56 A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder? Step 1) find moles using PV = nRT Step 1) find moles using PV = nRT n = mol n = mol Step 2) find mass using m = n x Mm m = 40.3 g m = 40.3 g

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57 Deviations from Ideal Gas Law Real molecules have volume. The ideal gas consumes the entire amount of available volume. It does not account for the volume of the molecules themselves. There are intermolecular forces. An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions. –Otherwise a gas could not condense to become a liquid.

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58 Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N mm Hg 20.95% O mm Hg 0.94% Ar 7.1 mm Hg 0.03% CO mm Hg P AIR = P N + P O + P Ar + P CO = 760 mm Hg Total Pressure760 mm Hg

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59 Dalton’s Law of Partial Pressures What is the total pressure in the flask? P total in gas mixture = P A + P B +... Therefore, P total = P H 2 O + P O 2 = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures. 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) 0.32 atm 0.16 atm 0.32 atm 0.16 atm

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60 Dalton’s Law John Dalton

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61 Health Note When a scuba diver is several hundred feet under water, the high pressures cause N 2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O 2 in scuba tanks used for deep descents.

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62 Collecting a gas “over water” Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.

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63 Table of Vapor Pressures for Water

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64 Solve This! A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the H 2 gas? 768 torr – 17.5 torr = torr

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65 GAS DENSITY Highdensity Lowdensity 22.4 L of ANY gas AT STP = 1 mole

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66 Gases and Stoichiometry 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

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67 Gases and Stoichiometry 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Solution 1.1 g H 2 O 2 1 mol H 2 O 2 1 mol O L O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 = 0.36 L O 2 at STP

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68 Gas Stoichiometry: Practice! A. What is the volume at STP of 4.00 g of CH 4 ? B. How many grams of He are present in 8.0 L of gas at STP?

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69 What if it’s NOT at STP? 1. Do the problem like it was at STP. (V 1 ) 2. Convert from STP (V 1, P 1, T 1 ) to the stated conditions (P 2, T 2 )

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70 Try this one! How many L of O 2 are needed to react 28.0 g NH 3 at 24°C and atm? 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g)

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71 GAS DIFFUSION AND EFFUSION diffusion is the gradual mixing of molecules of different gases.diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container.effusion is the movement of molecules through a small hole into an empty container. HONORS only

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72 GAS DIFFUSION AND EFFUSION Graham’s law governs effusion and diffusion of gas molecules. Thomas Graham, Professor in Glasgow and London. Rate of effusion is inversely proportional to its molar mass. HONORS only

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73 GAS DIFFUSION AND EFFUSION Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to Tproportional to T inversely proportional to M.inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T. He HONORS only

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74 Gas Diffusion relation of mass to rate of diffusion HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube. HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube. HONORS only

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