Warm-Up 1)Convert the following pressures to pressures in standard atmospheres: A. 151.98 kpa B. 456 torr Conversions 1 atm=101.3 kpa= 760 mm Hg= 760 torr Standard temp. & pressure = 1 atm & 0 C (STP) 2) Sketch a phase diagram 151.98 kpa x 1 atm = 1.5 atm 101.3kPa 456 torr x 1 atm = 0.6 atm 760 torr
Gases Chapter 11
The Nature of Gases Gases expand to fill their containers Gases are fluid they flow Gases have low density 1/1000 the density of the equivalent liquid or solid Gases are compressible Gases effuse and diffuse
ki net ic Kinetic Molecular Theory (not on sheet) 1. pertaining to motion. 2. caused by motion. 3. characterized by movement: Running and dancing are kinetic activities. Origin: 1850 55; < Gk kīnētikós moving, equiv. to kīnē- (verbid s. of kīneîn to move) + -tikos Source: Websters Dictionary
Kinetic Molecular Theory (gases) Gas particles are ALWAYS in motion. Volume of individual gas particles is zero. Collisions of gas particles with container walls cause the pressure exerted by gas. Particles exert no forces on each other. Average kinetic energy is proportional temperature (move faster if hotter)
Kinetic Energy of Gas Particles At the same conditions of temperature, all gases have the same average kinetic energy. At the same 1 temperature, 2 m = small mass molecules KE move mv FASTER than large molecules v = velocity 2
GAS DIFFUSION AND EFFUSION Diffusion is the gradual mixing of molecules of different gases. (smaller molecules mix faster, increase temp speeds up process) Effusion is the movement of molecules through a small hole into an empty container. (smaller molecules move faster)
GAS DIFFUSION AND EFFUSION Graham s law calculates the effusion & diffusion RATE of gas molecules. Rate of effusion is inversely proportional to its molar mass- lighter moves faster! Thomas Graham, 1805-1869. Professor in Glasgow and London.
Graham s Law of Diffusion/effusion Rate gas 1 Rate gas 2 M M 2 1 M 1 = Molar Mass of gas 1 M 2 = Molar Mass of gas 2
Example: Will a balloon filled with He or N 2 deflate faster if P & T are constant? He Molar mass= 4g N 2 molar mass = 28g. He effuses more rapidly than N 2 at same T because it is lighter. He
Graham s Law Rate gas 1 Rate gas 2 M M 2 1 Compare the rates of effusion of Helium (gas 1) & Nitrogen (gas 2) Rate He = 28.0g N 2 = 5.3g = 2.7 Rate N 2 4.0g He 2.0g 1 Helium gas effuses nearly three times faster than nitrogen gas.
Gas Laws
Variables that Describe a Gas Pressure (atm, mmhg, kpa) Volume (L) Temperature ( C or K) Number of moles (mol)
What happens if you decrease the volume of the container? Pressure may increase Temperature may increase increase the temperature? Pressure may increase Volume may increase Increase the pressure? Temperature may increase Volume may decrease
Boyle s Law Boyle s Law states that the volume of a gas varies inversely with its pressure if temperature is held constant. What does that mean? If volume goes up the pressure goes down! (or vice versa!) P 1 V 1 = P 2 V 2
Lets Practice Boyle s Law A sample of oxygen occupies a volume of 250.0mL at 740.0 torr pressure. What volume will it occupy at 800.0 torr pressure? Identify what you know: V 1 =250.0 ml P 1 =740.0 torr P 2 = 800.0 torr V 2 =? Solve for what you don t know P 1 V 1= V 2 P 2 (740 torr) (250 ml) = V 2 (800.0 torr) V 2 = 231 ml
Charles Law The volume of a gas varies directly with the Kelvin temperature (when pressure is constant) V 1 = V 2 T 1 T2 Temperature must be converted into Kelvin K= o C + 273
Lets Practice Charles Law A sample of nitrogen occupies a volume of 250.0mL at 25 o C. What volume will it occupy at 95 o C? 1) Identify what you know: V 1 =250.0 ml T 1 = 25 o C (298 K) T 2 = 95 o C (368 K) V 2 =? 2) Convert Temp into kelvin: K= 25 o C + 273= 298 K & K= 95 o C + 273= 368 K 3) Solve: V 1 = V 2 250mL = V 2 V 2 = 309 ml T 1 T 2 298K 368K
Gay-Lussac s Law Temperature and Pressure are directly proportional if mass and volume are kept constant. When one increases, the other increases and vice versa. Why? Think collisions remember volume is constant. P 1 = P 2 T 1 T 2 Temperature must be in Kelvin. Example: Tire will not stretch with energy increase. Aerosol can will not expand with energy increase.
Let s Practice A gas has a pressure of 6.58 kpa at 540 K. What will the pressure be at 210 K if the volume remains constant? Identify what you know: P 1 = 6.58 kpa; P 2 =? T 1 = 540 K T 2 = 210 K Solve for what you don t know P 1 = P 2 6.58 kpa = P 2 T 1 T 2 540 K 210K P 2 = (6.58 kpa) (210K) = 2.56 KPa 540 K
The Combined Gas Law If you write Boyle s, Charles and Gay- Lussac s gas laws as one formula, then all three variables can change; only mass remains constant. P 1 V 1 = P 2 V 2 T 1 T 2 Why are we interested? It allow us to predict what will happen to a gas if the conditions change.
So, what s the relationship? between volume, temperature, and pressure P 1 V 1 = P 2 V 2 T 1 T 2 AKA: The Combined Gas Law Allows you to predict what will happen to a gas if some of the conditions change!
Lets Practice! A gas with a volume of 4.0L at 90.0kPa expands until the pressure drops to 20.0kPa. What is the new volume if the temperature remains constant? Identify what you know V 1 =4.0L P 1 =90.0kPa P 2 =20.0kPa T 1 =T 2 Solve for what you don t know V 2 = P 1 V 1 x T 2 T 1 P 2 V 2 = 18.0L
Now you try! A gas with a volume of 3.00x10 2 ml at 150.0 C is heated until its volume is 6.00x10 2 ml.what is the new temperature of the gas if the pressure remains constant at 1.0 atm during the heating. T 2 = 300.0 C
Ideal Gases vs. Real Gases Particles have no volume Particles are not attracted or repelled to each other Ideal gases can never be liquefied or solidified Particles have volume Particles experience intermolecular forces Real gases can be liquefied and solidified
Real Gases Act Like Ideal Gases When temperatures are very high and pressures are very low Can you explain why? High Temperatures means lots of kinetic energy so particles are moving rapidly and randomly Low Pressures means particles do not hit the container (and each other) often, so they cannot feel attraction and repulsion
The Ideal Gas Law Allows us to solve for a property of an ideal gas when properties are constant! PV=nRT P=pressure V=volume T=temperature n=number of moles R= 8.31 L x kpa or 0.08206 L x atm K x mol K x mol
Lets Practice Determine the volume occupied by 0.582 mol of a gas at 15 C if the pressure is 81.1 kpa. Identify what you know V=? n=0.582 mol T=15 C + 273 = 288 K P=81.1 kpa. R= 8.31 L x kpa / K x mol
PV=nRT 81.1 Kpa x V=(0.582 mol) (8.31 L x kpa / K x mol) (288K) 81.1 KPa x V=1392.89 L x kpa V=17.2 L
Two rules! Temp= K, Volume =L, and pressure= atm or kpa If you are given grams convert into moles. 36 grams of H 2 O x 1 mole H 2 O =2 mol H 2 O 18 g H 2 O
Dalton s Law of Partial Pressures the total pressure of a mixture of gases is equal to the sum of the partial pressures of all the gases present. (at constant T & V) Formula: P total = P 1 + P 2 + P 3 + P 4
Dalton s Law Example #1: What is the partial pressure of oxygen at 101.3kPa of total pressure if the partial pressure of nitrogen is 79.10 kpa and carbon dioxide is 0.040kPa?
Solution: P total = P o + P N 2 + P CO 2 P o = P total (P N 2 + P CO 2 ) Po = 101.30kPa (79.10kPa + 0.040 kpa) =22.16 kpa
Dalton Example #2 What is the partial pressure of each gas if you have a mixture of 2.0 moles of O 2 and 2.0 moles of CO 2 that exert at total pressure of 700 torr? O 2 = 2.0 mol O 2 x 700 torr = 350 torr 4.0 total mol CO 2 = 2.0 mol CO 2 x 700 torr = 350 torr 4.0 total mol
Now you try! Show all your work on your worksheet. Ideal Gas Law: PV=nRT Make sure: Lets Party 1) Temp is in Kelvin (C+273= K) 2) Volume is in L (1000mL=1L) 350 ml x 1 L. 3) Pressure is in atm or kpa 1000 ml 4) 101.3Kpa=1 atm=760 mmhg=760 torr 5) grams-> mol (use molar mass) 36 grams of H 2 O x 1 mole H 2 O =2 mol H 2 O 18 g H 2 O