Gas Pressure Tire Whats going on inside Air Nitrogen 78 Oxygen 21 Argon 1 Carbon Dioxide lt1 Each of these particles are constantly flying around Like a lotto ball They slam against the container and keep the tire full The particles press against the walls ID: 1038376
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1. Gas Laws
2. Gas PressureJust means that gas is “pushing” on something.
3. Gas PressureTire What’s going on inside?Air:Nitrogen 78%Oxygen 21%Argon ~1%Carbon Dioxide <1%Each of these particles are constantly flying around. Like a lotto ball!They slam against the container and keep the tire “full”. The particles press against the walls.
4. Measuring Gas PressureAir:Nitrogen 78%Oxygen 21%Argon ~1%Carbon Dioxide <1%Think of a giant ball pit miles and miles up.At the bottom of the ball pit, is like us walking around. That’s the atmospheric pressure.
5. Measuring Gas PressureU-TubeCan’t use it to measure atmospheric pressure, because atmospheric pressure presses on everything equally.VacuumSo how do we measure it?VacuumIt pushes down on this side, and it moves up on the other side.
6. Measuring Gas PressureVacuumWe can measure that!Take a ruler and measure low to high in milimeters! The fluid that is contained in this U tube, is mercury. If we measure this at sea level, we get. 760mmHg between the bottom and the top.760 mmHg
7. Measuring Gas PressureWhat if we go up a mountain or down into a mine?Think about that ball pit again. If you’re at the bottom of the ball pit will it weigh more or less than at the top?Sea LevelMore Pressure760mmHgLess Pressure
8. Measuring Gas Pressure of Containers800 mmHg40 mmHgWhat if I snap off the vacuum bulb?Because atmospheric pressure is pushing down!760 mmHg
9. Measuring Gas PressureBarometerManometer
10. Gas Pressure ConversionsHow do we measure things? Lots of ways! Same goes with gas pressure.Gas Pressure UnitsmmHg atmosphere kilopascalTorr atm kPaConversions760 mmHg = 1 atm = 101.3kpa
11. Gas Pressure ConversionsThe pressure inside a car tire is 225 kPa. Express this value in both atm and mmHg.760 mmHg = 1 atm = 101.3 kPa225 kPa x 1 atm 101.3 kPa=2.22 atm225 kPa x 760 mmHg 101.3 kPa=1688 mmHg
12. Boyle’s LawIf we keep the temperature the same, we can predict what pressure and volume will do.
13. Boyle’s LawPressure and VolumeGas particles have a bunch of room.Gas particles are squeezed into smaller space.What about volume?V=HighV=LowAs pressure goes up, volume goes down. That means inverse relationship.P= LowP=High
14. Boyle’s Teeter Totter When volume is high, pressure is low When the volume is low, pressure is high An Inverse relationship. PressureVolume
15. Boyle’s LawBoyle’s law is explained by the equation P1V1=P2V2Let’s get right to it!At 1.70 atm, a sample of gas takes up 4.35 L. If the pressure on the gas is increased to 2.40 atm, what will the new volume be? P1V1 = P2V2(before) (after)What do you know?P1 (before pressure) = V1 (before volume)=P2 (after pressure) =V2 = ??(1.70 atm)(4.35L)=(2.40 atm)V27.40atm/L = (2.40atm)V2V2 =3.01L1.70 atm4.35 L2.4 atm
16. Boyle’s LawDoes that answer make sense?At 1.70 atm, a sample of gas takes up 4.35 L. If the pressure on the gas is increased to 2.40 atm, what will the new volume be?We increased the pressure, so we pushed down that piston. We squeezed the molecules into a smaller space. So the volume should go down!
17. Boyle’s LawIf I have 5.6 liters of gas in a piston at a pressure of 1.5 atm and compress the gas until its volume is 4.8 L, what will the new pressure inside the piston be? P1V1 = P2V2(before) (after)P1 (before pressure) =V1 (before volume)=P2 (after pressure) =V2 =(1.5atm)(5.6L) = (P2)(4.8L)8.4 atm/L = (4.8L)P21.8 atm = P21.5 atm5.6 L?4.8L
18. Charles’ LawCharles’ law relates volume and temperature, while keeping pressure the sameV1 = V2T1 T2
19. Charles’ LawHow could we test the theory that temperature and volume are related?Think about kinetic theory and molecules.
20. Charles’ LawHOTCOLDT= HighT = LowV= HighV = LowCharles’ law says that as the temp increases, so does volume. A direct relationship.What’s going on with the temp?
21. Charles’ LawSo now we can relate volume and temperature. V1 = V2T1 T2MUST ALWAYS USE KELVIN TEMPERATURE in gas lawsA balloon takes up 625 L at 0°C. If it is heated to 80°C, what will its new volume be?Must convert to Kelvin.0 °C + 273 = 273K80 °C + 273 = 353K625 L0 °C??V1 =T1 =T2 = V2 =80 °C
22. Charles’ LawV1 = V2T1 T2A balloon takes up 625 L at 0°C. If it is heated to 80°C, what will its new volume be?V1 = 625 LT1 = 273KT2 = 353KV2 = ??L625L = V2273K 353K2.29L/K= V2 353K808L = V2
23. Charles’ LawAt 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C?What’s the equation?V1 = V2T1 T2V1=T1=V2=T2=6.00 L27 °C??150.0 °CMust convert to Kelvin.27 °C + 273 = 300K150°C + 273 = 423K
24. Charles’ LawAt 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C?V1 = V2T1 T2V1=T1=V2=T2=6.00 L??300K423K6.00L = V2300K 423K0.02L/K = V2 423K8.46L = V2
25. Avogadro’s LawRelationship between:Amount of gas (n) and the Volume.What happens to one, when I change the other?I start with the first balloon, and then blow more air into it…will the volume increase?Yes, a direct relationship
26. Avogadro’s LawAs the amount (in moles) goes up, so does the volume.If we double the amount, it doubles the volume.
27. Avogadro’s LawWe only changed TWO things. The volume and the amount of particles. We didn’t mess with the pressure or the temperature, they were held constant. V1 = V2 n1 n2
28. Avogadro’s Law V1 = V2 n1 n2Let’s try!In a sample of gas, 50.0 g of oxygen gas (O2) take up 48L of volume. Keeping the pressure constant, the amount of gas is changed until the volume is 79 L. How many mols of gas are now in the container?n1= n2 = V1 = V2 = When doing Avogadro's law, “n” MUST be in moles!50g40L mol?79L
29. Avogadro’s Law V1 = V2 n1 n2Before Aftern1=50g n2 = g?V1 = 48L V2 = 79L When doing Avogadro's law, “n” MUST be in moles!50g O2 x 1 mol O2 32g O2= 1.6 mol O21.6mol1.6 mol O2 48L= n2 79L0.03 = n2 79L2.6 mol = n2
30. Gay-Lussac’s LawThe pressure and Kelvin temperature of a gas are directly proportional, when the volume remains constant.
31. Gay Lussac’s LawThis law only applies to gases held at a constant volume. Only the pressure and temperature will change.P1 = P2T1 T2Pi =initial pressurePf = final pressureTi = initial temperature (kelvin)Tf = final temperature (kelvin)The pressure in a sealed can of gas is 235 kPa when it sits at room temperature (20C). If the can is warmed to 48C, what will the new pressure inside the can be?
32. Gay Lussac’s LawThe pressure in a sealed can of gas is 235 kPa when it sits at room temperature (20°C). If the can is warmed to 48°C, what will the new pressure inside the can be?P1 = P2T1 T2Must convert to Kelvin20°C + 273 = 293K48°C + 273 = 321KP1 =P2 =T1 =T2 =235 kPa?20°C48°C
33. 235293= Pf 3210.80 = Pf 321257.5 kPa = PfP1 =P2 =T1 =T2 =235 kPa?The pressure in a sealed can of gas is 235 kPa when it sits at room temperature (20°C). If the can is warmed to 48°C, what will the new pressure inside the can be?P1 = P2T1 T2293K321K
34. How to use these formulasCharle’s LawV1 = V2T1 T2Avogadro’s LawV1 = V2n1 n2Gay Lussac’s LawP1 = P2T1 T2They are all pretty much the same equation, just different variables!
35. Combined Gas LawCharle’s LawV1 = V2T1 T2Boyle’s Law(P1)(V1) = (P2)(V2)Gay Lussac’s LawP1 = P2T1 T2What if I had a balloon. I wanted to increase the pressure and cool it down. What is the volume? Do we have an equation for that? P, T, V.We can combine the laws!Combined Gas Law(P1)(V1) = (P2)(V2) T1 T2
36. Combined Gas LawA 40.0L balloon is filled with air at sea level (1.00 atm, 25.0 °C). It's tied to a rock and thrown in a a cold body of water, and it sinks to the point where the temperature is 4.0 ° C and the pressure is 11.00 atm. What will its new volume be?(P1)(V1) = (P2)(V2) T1 T2Convert to Kelvin25°C + 273 = 298K4°C + 273 = 277KP1= 1 atmP2= 11 atmV1= 40 LV2= ??T1= 298KT2= 277K(1)(40) = (11)(V2) 298K 277K0.13 = (11)(V2) 277K36.01 = (11)(V2)3.27 L = V2P1 =P2 = V1 =V2 =T1 =T2 =1 atm11 atm40 L??25°C4°C
37. Ideal Gas LawHow can we describe what’s going on in this container? What variables can we think of?Temperature (T)313KPressure (P)3.18 atmVolume (V)95.2 LAmount of Gas (n)7.5 molDid you know that if we know 3 of the 4 variables, we can find the last one?
38. Ideal Gas LawIdeal gas law: PV = nRTTemperature (T)313KPressure (P)Volume (V)95.2 LAmount of Gas (n)7.5 molHow would we rearrange the problem to find P???P =nRT VWhat if we needed the amount of gas (n)?3.18 atm??PV = nRT
39. Ideal Gas LawPV = nRTSo what is R?R is a constant! For most cases, R = 0.0821 L▪atm/mol ▪KThose units look familiar.V = LP = atmT = Kn = molThe units on “R” MUST match the units in the problem!
40. Ideal Gas Law“R” will come in many forms.R = 62.4 L▪mmHg /K ▪molR = 8.31 L▪kPa /K ▪molNOT A BIG DEAL! The “R” constant will always be given, just use the right constant.
41. Ideal Gas Law2.3 moles of Helium gas are at a pressure of 1.70 atm, and the temperature is 41°C. What is volume of the gas?PV = nRTP = V = n =R = T = 0.0821 L▪atm/K ▪mol1.70 atm??2.3 mol41°CConvert to Kelvin41°C + 273 = 314K
42. Ideal Gas LawPV = nRTP = V = n =R = T = 0.0821 L▪atm/K ▪mol1.70 atm??2.3 mol314K2.3 moles of Helium gas are at a pressure of 1.70 atm, and the temperature is 41°C. What is volume of the gas?Rearrange the equation.V = nRT PV = (2.3 mol)(314K) x 0.0821 L ▪atm 1.70 atm K ▪ molV = 59.3 1.7V = 34.9 L
43. Ideal Gas LawAt a certain temperature, 3.24 moles of CO2 gas at 2.15 atm takes up a volume of 35.28 L. What is this temperature (in Celsius)?P = V =T = n =R = 2.15 atm35.28 L??3.24 mol0.0821 L▪atm/K ▪molDo the units given match the R?
44. Ideal Gas LawV =T = n =R = 2.15 atm35.28 L??3.24 mol0.0821 L▪atm/K ▪molAt a certain temperature, 3.24 moles of CO2 gas at 2.15 atm takes up a volume of 35.28 L. What is this temperature (in Celsius)?P = PV = nRTRearrange the equation.T = PV nRT = (2.15 atm)(35.28L) X K ▪ mol (3.24 mol) 0.0821 L ▪ atm
45. Ideal Gas LawCharle’s LawV1 = V2T1 T2Avogadro’s LawV1 = V2n1 n2Gay Lussac’s LawP1 = P2T1 T2Who wants to memorize all of these?!?!Ideal Gas LawPV = nRTCombined Law(P1)(V1) = (P2)(V2) T1 T2You don’t have to!
46. Gas LawJust memorize one!Ideal Gas LawPV = nRTCan use it for any of the gas law problems!Warning:If this blows your mind and you get totally confused, just memorize the equations.
47. Gas LawBefore AfterP1 = 3 atm P2 = 7atmT1 = ?? T2 = 150kRearrange the ideal equation so that the variables given are on the same “side”PV =nRTPV = nRTV VP = nRTT VTP= nRT VYou’ve found the equation you need to use. You don’t need “n, R, or V”. P1 = P2T1 T2
48. Gas LawP1 = 1,217 mmHgP2 = 732 mmHgV1 = ??V2 = 42LPV = nRTRearrange the equation so the variables you’re looking for are on the same side of the equation.Easy! PV is already on the same side. Now just double it.P1V1 = P2V2
49. Gas LawV1 = 7.5LV2 = 1.2Ln1= 32 moln2 = ? PV = nRTRearrange the equation so V and n are on the same side.PV = nRT P PV = nRT PV = nRTn PnV1 = V2n1 n2
50. Gas LawBefore AfterV1 = ? V2 = 54LP1 = 96 kPa P2 = 112 kPaT1 = 12K T2 = 42KPV = nRTRearrange so V, T, P are on same side.PV = nRT T T P1V1 = P2V2 T1 T2