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GAS LAWS Kinetic Molecular Theory Particles in an ideal gas… –have no volume. –have elastic collisions. –are in constant, random, straight-line motion.

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Presentation on theme: "GAS LAWS Kinetic Molecular Theory Particles in an ideal gas… –have no volume. –have elastic collisions. –are in constant, random, straight-line motion."— Presentation transcript:

1

2 GAS LAWS

3 Kinetic Molecular Theory Particles in an ideal gas… –have no volume. –have elastic collisions. –are in constant, random, straight-line motion. –don’t attract or repel each other. –have an average KE directly related to Kelvin temperature.

4 Real Gases Particles in a REAL gas… –have their own volume –attract and repel each other Gas behavior is most ideal… –at low pressures –at high temperatures ***Most real gases act like ideal gases except under high pressure and low temperature.

5 Characteristics of Gases Gases expand to fill any container. –Take the shape and volume of their container. Gases are fluids (like liquids). –Little to no attraction between the particles Gases have very low densities. = lots of empty space between the particles

6 Characteristics of Gases Gases can be compressed. –lots of empty space between the particles –Indefinite density Gases undergo diffusion. –random motion –scatter in all directions

7 Pressure Which shoes create the most pressure?

8 Pressure- how much a gas is pushing on a container. Atmospheric pressure- atmospheric gases push on everything on Earth UNITS AT SEA LEVEL 1 atm =101.3 kPa (kilopascal)= 760 mmHg =760 torr

9 Pressure Barometer –measures atmospheric pressure Mercury Barometer Aneroid Barometer

10 Pressure Manometer –measures contained gas pressure C. Johannesson U-tube ManometerBourdon-tube gauge

11 C. Johannesson Temperature= how fast the molecules are moving ºF ºC K K = ºC Always use absolute temperature (Kelvin) when working with gases.

12 Standard Temperature & Pressure 0°C 273 K 1 atm101.3 kPa 760 mm Hg -OR- STP -OR-

13 Volume = how much space a gas occupies Units –L, mL, cm mL = 1 L 1 mL = 1 cm 3

14 BASIC GAS LAWS

15 Charles’ Law T  V (temperature is directly proportional to volume) T ↑ V↑ & T↓ V↓ V 1 = V 2 T 1 T 2 T is always in K –K = °C –P and n = constant V T

16 Timberlake, Chemistry 7 th Edition, page 259 (Pressure is held constant) T 1 T 2 V 1 V 2 = Charles’ Law

17 Timberlake, Chemistry 7 th Edition, page 254

18 Courtesy Christy Johannesson Charles’ Law The egg out of the bottle

19 Mrs. Rodriguez inflates a balloon for a party. She is in an air- conditioned room at 27.0 o C, and the balloon has a volume of 4.0 L. Because she is a curious and intrepid chemistry teacher, she heats the balloon to a temperature of 57.0 o C. What is the new volume of the balloon if the pressure remains constant? Charles’ Law Problem Given Unkown Equation  Substitute and Solve  T 1 = 27.0 o C +273= 300 K V 1 = 4.0 L T 2 = 57.0 o C +273= 330 K V 2 = ? L P 1 V 1 = P 2 V 2 T 1 V 1 T L = V 2 = 300 K 330K 4.4 L

20 A 25 L balloon is released into the air on a warm afternoon (42º C). The next morning the balloon is recovered on the ground. It is a very cold morning and the balloon has shrunk to 22 L. What is the temperature in º C? Charles’ Law Learning Check Given Unkown Equation  Substitute and Solve  V 1 = 25 L T 1 = 42 o C +273= 315 K V 2 = 22 L T 2 = ? ºC P 1 V 1 = P 2 V 2 T 1 V 1 T 2 25 L = 22 L = 315 K T K -273 = 4.2 ºC

21 Boyle’s Law P↓ V ↑ & P↑ V ↓ P  1/V (pressure is inversely proportional to volume) P 1 V 1 = P 2 V 2 –T and n = constant P V

22 Timberlake, Chemistry 7 th Edition, page 253 P 1 V 1 = P 2 V 2 (Temperature is held constant) Boyle’s Law

23 Timberlake, Chemistry 7 th Edition, page 254  Marshmallows in a vacuum Boyle’s Law

24 Mechanics of Breathing Timberlake, Chemistry 7 th Edition, page 254 Boyle’s Law

25 A balloon is filled with 30.L of helium gas at 1.00 atm. What is the volume when the balloon rises to an altitude where the pressure is only 0.25 atm? GivenUnkown Equation Substitute and Solve Boyle’s Law Problem  V atm = 30 L x 1.0 atm = 120 L V 1 = 30 L P 1 = 1 atm P 2 =.25atm V 2 = ? L P 1 V 1 = P 2 V 2 T 1 T 2

26 A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. Given Unkown Equation Substitute and Solve Boyle’s Law Learning Check  V 2 x 200. kPa = L x 150. kPa= 75.0 mL  L V 1 = 100. mL = L P 1 = 150. kPa P 2 = 200. kPa V 2 = ? L P 1 V 1 = P 2 V 2 T 1 T 2

27 AVOGADRO’S LAW V  n  V  n  V  n (direct) V 1 = V 2 n 1 n 2 –T & P Constant n V

28 A 3.0 liter sample of gas contains 7.0 moles. How much gas will there be, in order for the sample to be 2.3 liters? P & T do not change GivenUnkown Equation Substitute and Solve Avogadro’s Law Problem V 1 = 3.0 L n 1 = 7.0 mol V 2 = 2.3 L n 2 = ? mol P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T L = 2.3 L = 7.0 mol n 2 mol 5.4 mol

29 Gay-Lussac’s Law P 1 = P 2 T 1 T 2 –V & n constant Direct relationship P  T  P  T  P T

30 Gay-Lussac Law Collapsing Barrel

31 Gay-LussacLaw Gay-Lussac Law Tank car implosion

32 COMBINED IDEAL GAS LAW P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2 If P, V, n, or T are constant then they cancel out of the equation. n usually constant (unless you add or remove gas), so P 1 V 1 = P 2 V 2 T 1 T 2

33 Combined Gas Law Problem Ms. Evans travels to work in a hot air balloon from the Rocky Mountains. At her launch site, the temperature is 5.00 °C, the atmospheric pressure is atm, and the volume of the air in the balloon is L. When she lands in Plano, the temperature is 28.0 °C and the atmospheric pressure is kPa. What is the new volume of the air in the balloon? Given Unkown Equation  Substitute and Solve  V 2 x 1 atm = L x atm = 104 L  301K 278 K T 1 = 5.0 o C +273= 278 K P 1 = atm V 1 = L T 2 = 28.0 o C +273= 301 K P 2 = kPa = 1 atm V 2 = ? L V 1 x P 1 = V 2 x P 2 T 1 T 2

34 Combined Gas Law Learning Check Nitrogen gas is in a 7.51 L container at 5.  C and 0.58 atm. What is the new volume of the gas at STP? Given Unkown Equation  Substitute and Solve  V 2 x 1.0 atm = 7.51L x 0.58 atm = 4.3 L  273 K 278 K T 1 = 5.0 o C +273= 278 K P 1 = 0.58 atm V 1 = 7.51 L T 2 = 273 K P 2 = 1 atm V 2 = ? L V 1 x P 1 = V 2 x P 2 T 1 T 2

35 Ideal Gas Law (“Pivnert”) PV=nRT R = The Ideal Gas Constant R = (L*atm) R = 62.4 (L*mm Hg) (mol*K) (mol*K) R = 8.31 (L*kPa) (mol*K) V has to be in Liters, n in Moles, T in Kelvin, P can be in atm, kPa or mmHg * Choose which R to used based on the units of your pressure. P V = n R T (atm) (L) = (moles) (L*atm/mol*K) (K) (kPa) (L) = (moles) (L*kPa/mol*K) (K) mm Hg (L) = (moles) (L*mmHg/mol*K) (K)

36 Ideal Gas Law Problem A rigid steel cylinder with a volume of 20.0 L is filled with nitrogen gas to a final pressure of atm at 27.0 o C. How many moles of gas does the cylinder hold? Given Unkown Equation  Substitute and Solve n 0821 atm L/K Mole x 300 K = atm x 20.0L= 162 moles V = 20.0 L P = atm T =27.0 o C +273= 300 K moles of nitrogen? PV=nRT R=.0821 atm L/K Mole

37 Ideal Gas Law Learning Check A balloon contains 2.00 mol of nitrogen at a pressure of atm and a temperature of 37  C. What is the volume of the balloon? Given Unkown Equation  Substitute and Solve atm x V= 2.00 mol x atm L/K Mole x 310 K = 51.9 L n = 2.00 mol P = atm T =37.0 o C +273= 310 K V in L? PV=nRT R=.0821 atm L/K Mole

38 Dalton’s Law of Partial Pressure The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. P total = P gas 1 + P gas 2 + P­ gas 3 + … A metal container holds a mixture of 2.00 atm of nitrogen, 1.50 atm of oxygen and 3.00 atm of helium. What is the total pressure in the canister? 6.5 atm

39 Welcome to Mole Island 1 mol = 6.02 x particles

40 1 mol = molar mass Welcome to Mole Island

41 1 mole = 22.4 STP Welcome to Mole Island

42 Gas Stoichiometry Moles  Liters of a Gas: –2 –2C 4 H 10 (g) + 13O 2 (g) 8CO 2 (g) + 10H 2 O(g) Courtesy Christy Johannesson 2 mol + 13 mol 8 mol + 10 mol Avogadro’s principle states that one mole of any gas occupies 22.4 L at STP. Thus when gases are involved, the coefficients in a balanced chemical equation represent not only molar amounts but also relatives volumes 2 L + 13 L 8 L + 10 L Recall: The coefficients in a chemical reaction represent molar amounts of substances taking part in the reaction.

43 Gas Stoichiometry Problem In the following combustion reaction, what volume of methane (CH 4 ) is needed to produce 26 L of water vapor? –CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) Courtesy Christy Johannesson x L 26 L 1 L 2 L x L = 26 L 1L 2L x= 13 L 1 mol 2 mol

44 Gas Stoichiometry use ideal gas law –Looking for grams or moles of gas? Step 1: start with ideal gas law to find moles of gas Step 2: 1change to grams of gas Courtesy Christy Johannesson PV=nRT Grams/mol?1) Use Ideal Gas Law 2) Do stoichiometry calculations

45 Example 1 How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? PV=nRT 4 Al (s) + 3 O 2(g)  2 Al 2 O 3(s) Given liters: Start with Ideal Gas Law and calculate moles of O 2. Courtesy Christy Johannesson Given Unkown V O2 = 15.0 L O 2 grams of Al 2 O 3 ? R=.0821 atm L/K Mole P O2 = 97.3 kPa= atm T O2 = 21 o C +273= 294 K Step 1: Calculate moles of O 2  n = PV = atm x 15.0 L = mol O 2 RT atm L/K Mole 294 K Step 2: Calculate mass of Al 2 O mol O 2 = X mol Al 2 O 3 = mol Al 2 O 3 3moleO 2 2 mole Al 2 O 3 Use stoich to convert moles of O 2 to grams Al 2 O g Al 2 O mol Al 2 O 3 x g Al 2 O 3= 1 mol Al 2 O 3

46 Gas Stoichiometry use ideal gas law Courtesy Christy Johannesson PV=nRT Looking for volume of gas? Step 1: start with stoichiometry conversion to find moles of gas Step 2: use ideal gas law to find the volume Liters ? 1) Do stoichiometry calculations 2) Use Ideal Gas Law

47 What volume of CO 2 forms from 5.25 g of CaCO 3 at kPa & 25ºC? CaCO 3  CaO + CO 2 Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law to find volume. Given Unkown PV=nRT m = 5.25 g CaCO 3 volume of CO 2 ? R=.0821 atm L/K Mole P = kPa = 1 atm T = 25.0 o C +273= 298 K Step 1: Calculate moles of CO 2  5.25 g CaCO 3 x 1 mole CaCO 3 = mol CaCO 3  100 g CaCO 3  1 mole CO 2 = 1mole CaCO 3 ; mol CO 2 Step 2: Calculate volume of CO 2 V = nRT = mol CO 2 x.0821 atm L/K Mole x 298 K = 1.28 L P1 atm Example 2


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