Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 13 Gases and the Gas Laws Mythbuster – water heater Intro to gas laws.

Similar presentations


Presentation on theme: "Chapter 13 Gases and the Gas Laws Mythbuster – water heater Intro to gas laws."— Presentation transcript:

1 Chapter 13 Gases and the Gas Laws Mythbuster – water heater Intro to gas laws

2 Gas Laws in Action: workers steam cleaned this tanker car and then sealed up the container, they came back the following morning to this disaster. (video)video

3 Kinetic Theory Assumptions for an Ideal Gas Gas particles are in constant, random motion Gas particles themselves have virtually no volume Gas particles do not attract nor repel each other No kinetic energy is lost when particles collide If gases are at the same temp. they have the same KE

4 NOTE: Real gases (actual gases) do NOT obey all the assumptions made by the kinetic theory (only ideal gases behave this way- we will get exceptions later)

5 Factors (variables) that Affect Gases 1. Number of gas particles present 2. Temperature 3. Pressure 4. Volume of the sample

6 In a tied off balloon the pressure from the outside = pressure from the inside (in this chapter we will look at how changing the factors, changes these values) =

7 STP Standard temperature and standard pressure Standard temperature = 0° C (273 K) Standard pressure = 101.3 kPa (1 atm or 760 mm Hg)

8 Boyles Law States that the volume of a gas is inversely proportional to its pressure if the temperature remains constant As pressure goes up, volume goes down and vice versa if temperature is constant PV = k

9

10 PVP x V P 1 = 1 atm V 1 = 800 ml 800 P 2 = 2 atm V 2 = 400 ml 800 P 3 = 3 atm V 3 = 267 ml 800 P 1 V 1 = 800 P 2 V 2 = 800 So: P 1 V 1 = P 2 V 2 Boyles Law

11 If.600 L (V 1 )of a gas at 100.0 kPa (P 1 ) changes to 62.4 kPa.(P 2 ) What is the new volume if temperature remains constant? P 1 V 1 = P 2 V 2 (100.0 kPa) (.600L) = (62.4 kPa) (V 2 ).96153 L = V 2.962 L = V 2

12 Note: you do not need to convert units as long as they match on both sides of the equation A 185 ml sample has a pressure of 4.2 atm. What is it’s pressure when the volume is 250 ml if temperature remains constant? P 1 V 1 = P 2 V 2 (4.2 atm)(185 ml) = P 2 (250 ml) 3.1 atm = P 2

13 Charles Law Jacque Charles investigated the property of changing temperature on the volume of a gas (saw w/ each ° C change the volume changed by 1/273 rd ) Charles Law – volume of a fixed mass of gas is directly proportional to its kelvin temperature if the pressure is constant Ex. Helium balloon deflates when walking outside on a cold day

14

15 Charles Law: V 1 = V 2 T 1 T 2 *** “T” must be in Kelvin ( K = C +273) or V 1 T 2 = V 2 T 1

16 A balloon inflated in an air conditioned room at 28° C (T 1 ) has a volume of 4.0 L (V 1 ). If it is heated to a temperature of 58 °C ( T 2 ), what is the new volume (V 2 ) of the balloon if pressure remains constant? V 1 = V 2 T 1 T 2 T 1 = 28 + 273 = 301K T 2 = 58 + 273 = 331 K

17 V 1 T 2 = V 2 T 1 (4.0 L ) (331 K) = V 2 (301 K) 4.4 L = V 2

18 Adjust the volume of 609 ml of a gas at 83°C to standard temperature.

19 Gay-Lussac’s Law States that the pressure of a gas is directly proportional to the Kelvin temperature if volume is kept constant Ex. Spray paint can (rigid container) in a bonfire P 1 = P 2 T 1 T 2 or P 1 T 2 = P 2 T 1 “T” must be in Kelvin

20 The pressure of a gas in a tank is 3.20 atm (P 1 ) at 22.0 °C (T 1 ). If the temperature is raised to 60.0 °C (T 2 ), what is the new pressure (P 2 ) if volume is held constant? T1 = 22.0 + 273 = 295 K T2 = 60.0 + 273 = 333 K

21 P 1 = P 2 T 1 T 2 P 1 T 2 = P 2 T 1 (3.20 atm) (333K) = P 2 295 K 3.61 atm = P 2

22 The Combined Gas Law Many times it is hard to keep a variable constant (and only deal with changing 2 variables at a time), so we have to use all the laws together Combined Gas Law: all the variables of pressure, temperature and volume change (only thing that is constant is the number of particles)

23 P 1 V 1 = P 2 V 2 T 1 T 2 or P 1 V 1 T 2 = P 2 V 2 T 1

24 Find the volume of a gas at STP if it measures 806 ml at 26.0° C and 103.0 kPa P 1 V 1 = P 2 V 2 T 1 T 2 P 1 = 103.0 kPa V 1 = 806 ml T 1 = 26.0 + 273 = 299 K P 2 = 101.3 kPa (standard pressure) V 2 = ? T 2 = 273 K (standard temperature)

25 (103.0 kPa) (806 ml) = (101.3 kPa) (V 2 ) 299 K 273 K (103.0 kPa) (806ml) (273 K) = V 2 (299 K) (101.3 kPa) 748 ml = V 2

26 Gases and the MOLE Avogadro’s Principle: at equal temperatures and equal pressures, equal volumes of gases contain the same number of molecules Molar Volume: volume occupied by 1 mole of any gas under STP (0 °C, 101.3 kPa) is 22.4 L (conversion factor= 22.4 L/1 mole)

27 O2O2 He 1 mole O 2 at STP 6.02 x 10 23 molecules of O 2 32.0 g 22.4 L 1 mole He at STP 6.02 x 10 23 atoms He 4.00 g 22.4 L

28 What is the volume of 8.6 g of Cl 2 at STP? 1. Convert g  moles (molar mass) 2. Convert moles  volume (22.4 L/ mole 1) 8.6 g Cl 2 71.0 g Cl 2 1 mole Cl 2.12 moles 2).12 moles 1 mole 22.4 L 2.7 L

29 Ideal Gas Combines Avogadro’s principle, Boyles, Charles and Gay-Lussac’s Law into a statement w/ P, V, T and # moles Changing one variable will affect the other 3 variables Ideal Gas Equation: PV = nRT

30 n = # of mole R = Ideal Gas Constant ( experimentally determined constant based on Avogadro’s # and STP  dependent on unit used for pressure)

31 Pressure in: atm use: R=.0821 L· atm/ mole ·K kPa use : R = 8.314 L ·kPa/ mole· K mm Hg use: R = 62.4 L· mm Hg/ mole ·K

32 Calculate the number of moles of gas contained in a 3.0 L vessel at 30Ō K and a pressure of 1.50 atm. PV = nRT PV = n R=.0821 RT (1.50 atm) (3.0 L) = n (.0821 · 30ŌK) n =.18 moles

33 Applying the Ideal Gas Law Calculate molar mass: n (# of moles) = mass of gas = m Molar mass M So: PV = nRT PV = mRT or M M = mRT PV

34 Calculate density: D= m/V M = mRT PV (substitute D for m/v in this equation M = DRT P Or D = MP RT

35

36 Deviations from Ideal Behavior Ideal Gases: have no attractive forces and do not take up space (volume) Real Gases: Occupy definite volume (take up space)- but volume is small Under normal conditions real gases behave like ideal gases (follow all gas laws)

37 Under high pressures: particles are forced close together and can’t compress any further, also attractive forces take over So: real gases will liquefy instead of disappearing like Boyle predicted Same is true under really low temperatures

38 Gas Laws Test Formulas, R values and periodic table will be given to you > 40 questions 12 multiple choice 13 fill in the blank (need to know who did what law/PTV card/variables) 7 calculations (1 for each formula, 1 using 22.4 L= 1 mole, 1 PV= nRt)

39 Know: Who did each law What each law stands for Scenerios with each law Absolute zero STP Molar Volume /avogadros principle Variables on a gas Real gas vs ideal gas

40 When a sample of gas was placed in a sealed container with a volume of 3.35 L and heated to 105 C, the gas vaporized and the resulting pressure inside the container was 170.0 kPa. How many moles of the gas were present?

41 If you have 3.50 L of water vapor at STP, how many grams of water do you have?

42 Convert 3.5 L of gas at a temp of 25C and pressure of 96.6 kPa to STP.


Download ppt "Chapter 13 Gases and the Gas Laws Mythbuster – water heater Intro to gas laws."

Similar presentations


Ads by Google