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Unit 9: The Gas Laws Chemistry. ~78% The Atmosphere  an “ocean” of gases mixed together Composition nitrogen (N 2 )………….. oxygen (O 2 )…………… argon (Ar)……………...

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Presentation on theme: "Unit 9: The Gas Laws Chemistry. ~78% The Atmosphere  an “ocean” of gases mixed together Composition nitrogen (N 2 )………….. oxygen (O 2 )…………… argon (Ar)……………..."— Presentation transcript:

1 Unit 9: The Gas Laws Chemistry

2 ~78% The Atmosphere  an “ocean” of gases mixed together Composition nitrogen (N 2 )………….. oxygen (O 2 )…………… argon (Ar)……………... carbon dioxide (CO 2 )… water vapor (H 2 O)……. Trace amounts of: ~21% ~1% ~0.04% ~0.1% He, Ne, Rn, SO 2, CH 4, N x O x, etc.

3 Depletion of the Ozone Layer O 3 depletion is caused by chlorofluorocarbons (CFCs). Uses for CFCs: refrigerants Ozone (O 3 ) in upper atmosphere blocks ultraviolet (UV) light from Sun. UV causes skin cancer and cataracts. O 3 is replenished with each strike of lightning. aerosol propellants -- banned in U.S. in 1996 CFCs

4 Ozone Hole Grows Larger Ozone hole increased 50% from

5 Monthly Change in Ozone TOMS Total Ozone Monthly Averages

6 Greenhouse Effect Light from the sun: UV, viz, IR Light is absorbed by Earth and re- radiated into the atmosphere as heat (IR) Some heat (IR) lost to outer space Kelter, Carr, Scott, Chemistry A Wolrd of Choices 1999, page 392 Greenhouse gases absorb heat (IR), and radiate it back into the atmosphere

7 Greenhouse Gases

8 Why more CO 2 in atmosphere now than 500 years ago? -- burning of fossil fuelsdeforestation urban sprawl wildlife areas rain forests coal petroleum natural gas wood * The burning of ethanol won’t slow greenhouse effect… C 2 H 5 OH + O 2  CO 2 + H 2 O

9 Carbon Dioxide Levels Zumdahl, Zumdahl, DeCoste, World of Chemistry  2002, page Year Atmospheric CO 2 (ppm)

10 What can we do? 1. Reduce consumption of fossil fuels. At home: On the road: 2. Support environmental organizations. 3. Rely on alternate energy sources. insulate home; run dishwasher full; avoid temp. extremes (A/C & furnace); wash clothes on “warm,” not “hot” bike instead of drive; carpool; energy-efficient vehicles solar, wind energy, hydroelectric power

11 The Kinetic Molecular Theory (KMT) deals w/“ideal” gas particles… 1. …are so small that they are assumed to have zero volume 2. …are in constant, straight-line motion 3. …experience elastic collisions in whichelastic collisions no energy is lost 4. …have no attractive or repulsive forces toward each other 5.…have an average kinetic energy (KE) that is proportional to the absolute temp. of gas (i.e., Kelvin temp.) explains why gases behave as they do as Temp., KE

12 Collisions of Gas Particles

13 Theory “works,” except at high pressures and low temps. (attractive forces become significant) N 2 can be pumped into tires to increase tire life KMT “works” liquid nitrogen (N 2 ); the gas condenses into a liquid at –196 o C KMT starts to break down –196 o C H 2 O freezes 37 o C100 o C0oC0oC H 2 O boils body temp. liquid N 2

14 Collisions 8 3 Elastic collision Inelastic collision

15 Model Gas Behavior All collisions must be elastic Take one step per beat of the metronome Container –Class stands outside tape box Higher temperature –Faster beats of metronome Decreased volume –Divide box in half More Moles –More students are inside box  Mark area of container with tape on ground.  Add only a few molecules of inert gas  Increase temperature  Decrease volume  Add more gas  Effect of diffusion  Effect of effusion (opening size)

16 ** Two gases w/same # of particles and at same temp. and pressure have the same kinetic energy. KE is related to mass and velocity: KE = ½ m v 2 More massive gas particles are _______ than less massive gas particles (on average). = ½ m1m1 m2m2 v1v1 v2v2 KE 1 KE same temp. To keep same KE, as m, v must OR as m, v must. slower same KE

17 (2 g/mol) Particle-Velocity Distribution (different gases, same T and P) # of particles Velocity of particles (m/s) (SLOW)(FAST) CO 2 (28 g/mol) N2N2 H2H2 (44 g/mol) CO 2 N2N2 H2H2

18 Particle-Velocity Distribution (same gas, same P, different T) # of particles Velocity of particles (m/s) (SLOW)(FAST) O 10 o C O 50 o C O 100 o C O 10 o C O 50 o C O 100 o C

19 Graham’s Law Consider two gases at same temp. Gas 1: KE 1 = ½ m 1 v 1 2 Gas 2: KE 2 = ½ m 2 v 2 2 Since temp. is same, then…KE 1 = KE 2 ½ m 1 v 1 2 = ½ m 2 v 2 2 m 1 v 1 2 = m 2 v 2 2 Divide both sides by m 1 v 2 2 … Take sq. rt. of both sides to get Graham’s Law: ** To use Graham’s Law, both gases must be at same temp.

20 diffusion:effusion: particle movement from high to low conc. (perfume) For gases, rates of diffusion & effusion obey Graham’s law: diffusion of gas particles through an opening (balloon) NET MOVEMENT more massive = slow; less massive = fast

21 irrelevant, so long as they are the same On avg., carbon dioxide travels at 410 m/s at 25 o C. Find avg. speed of chlorine at 25 o C. CO 2 Cl 2 = 320 m/s **Hint: Put whatever you’re looking for in the numerator. (the algebra is easier) use molar masses

22 At a certain temp., fluorine gas travels at 582 m/s and a noble gas travels at 394 m/s. What is the noble gas? He Ne Ar Kr Xe Rn 86 (222) F 2 mm = 38 g/mol = 82.9 g/mol best guess = Kr

23 He Ne Ar Kr Xe Rn 86 (222) mm = 16 g/mol = 39.9 g/mol Ar CH 4 moves 1.58 times faster than which noble gas? “Ar?” “Aahhrrrr! Buckets o’ blood! Swab de decks, ye scurvy dogs!” Ne 2 or Ar?

24 mm = 36.5 g/mol HCl and NH 3 are released at same time from opposite ends of 1.20 m horiz. tube. Where do gases meet? HCl NH 3 more massive travels slower mm = 17 g/mol less massive travels faster ABC A

25 Gas Pressure Pressure occurs when a force is dispersed over a given surface area. If F acts over a large area… But if F acts over a small area… F A = P A P = F (e.g., your weight)

26 = 1.44 x 10 6 in 2 At sea level, air pressure is standard pressure: 1 atm = kPa = 760 mm Hg = 14.7 lb/in 2 = 2 x 10 7 lb. Find force of air pressure acting on a baseball field tarp… 100 ft. 2 F = P A= 14.7 lb/in 2 (1.44 x 10 6 in 2 ) F = 2 x 10 7 lb.= 10,000 tons Key: Gases exert pressure in all directions. A = 10,000 ft 2 A = 10,000 ft. ft

27 Atmospheric pressure changes with altitude: As altitude, pressure. barometer: device to measure air pressure Vacuum (nothing) air pressure mercury (Hg)

28 Pressure and Temperature STP (Standard Temperature and Pressure) standard temperaturestandard pressure 0 o C 273 K Equations / Conversion Factors: K = o C atm = kPa = 760 mm Hg 1 atm kPa 760 mm Hg

29 101.3 kPa = 139 kPa Convert 25 o C to Kelvin. How many kPa is 1.37 atm? K = o C atm = kPa = 760 mm Hg K = o C + 273= = 298 K 1.37 atm How many mm Hg is kPa? = 1737 mm Hg 1 atm kPa kPa 760 mm Hg

30 Bernoulli’s Principle For a fluid traveling // to a surface: LIQUID OR GAS -- FAST-moving fluids exert ______ pressure -- SLOW-moving fluids exert ______ pressure FAST SLOW LOW HIGH HIGH P LOW P NET FORCE

31 roof in hurricane or tornado… SLOW FAST LOW P HIGH P

32 airplane wing / helicopter propeller AIR PARTICLES FAST SLOW Resulting Forces (BERNOULLI’S PRINCIPLE) (GRAVITY) frisbee HIGH P LOW P FAST, LOW P SLOW, HIGH P

33 CURTAIN creeping shower curtain HIGH P SLOW COLD LOW P FAST WARM

34 windows and high winds FAST SLOW windows burst outwards TALL BUILDING LOW P HIGH P

35 measures the pressure of a confined gas manometer: CONFINED GAS AIR PRESSURE Hg HEIGHT DIFFERENCE SMALL + HEIGHT = BIG differential manometer manometers can be filled with any of various liquids

36 96.5 kPa233 mm Hg Atmospheric pressure is 96.5 kPa; mercury height difference is 233 mm. Find confined gas pressure, in atm. X atm 96.5 kPa 233 mm Hg B S SMALL + HEIGHT = BIG atm atm += atm + X atm =

37 101.3 kPa Manometers HW # kPa X atm 0 mm Hg SMALL + HEIGHT = BIG If H = 0, then S = B 126 kPa = X atm = 1.24 atm 126 kPa 1 atm

38 = mm Hg 1 atm Manometers HW # atm kPa X mm Hg SMALL + HEIGHT = BIG 0.78 atm + X mm Hg = kPa Convert units into mm Hg… = mm Hg 0.78 atm 760 mm Hg B S kPa kPa 760 mm Hg mm Hg + X mm Hg = mm Hg X = mm Hg

39 = 22.4 L The Ideal Gas Law P V = n R T P = pres. (in kPa) V = vol. (in L or dm 3 ) T = temp. (in K) n = # of moles of gas (mol) R = universal gas constant = L ∙ kPa/mol ∙ K 32 g oxygen at 0 o C is under kPa of pressure. Find sample’s volume. P V = n R T T = 0 o C = 273 K PP

40 54.0 kPa 0.25 g carbon dioxide fills a 350 mL container at 127 o C. Find pressure in mm Hg. = 54.0 P V = n R T T = 127 o C = 400 K VV V = 350 mL/1000 = L kPa = 405 mm Hg

41 P, V, T Relationships At constant P, as gas T, its V ___. balloon placed in liquid nitrogen (T decreases from 20 o C to –200 o C)

42 P, V, T Relationships (cont.) At constant V, as gas T, its P ___. blown-out truck tire KABLOOEY!

43 P, V, T Relationships (cont.) At constant T, as P on gas, its V ___. Gases behave a bit like jacks-in- the-box (or little-brothers-in-the-box)

44 Derivation of the Combined Gas Law Consider a sample of gas under the conditions of P 1, V 1, and T 1. P 1 V 1 = n R T 1 Now, change the gas sample’s conditions to P 2, V 2, and T 2. P 2 V 2 = n R T 2 (A) (B) But n and R are the same for both (A) and (B). Solve each equation for the quantity n x R. n R = P 1 V 1 T1T1 = P 2 V 2 T2T2 This is the combined gas law.

45 P 1 V 1 = P 2 V 2 The Combined Gas Law P = pres. (any unit)1 = initial conditions V = vol. (any unit)2 = final conditions T = temp. (K) A gas has vol. 4.2 L at 110 kPa. If temp. is constant, find pres. of gas when vol. changes to 11.3 L. 110(4.2) = P 2 (11.3) P 2 = 40.9 kPa 11.3

46 Original temp. and vol. of gas are 150 o C and 300 dm 3. Final vol. is 100 dm 3. Find final temp. in o C, assuming constant pressure. 423 K 300(T 2 ) = 423(100) T 2 = 141 K 300 T 2 = –132 o C

47 A sample of methane occupies 126 cm 3 at –75 o C and 985 mm Hg. Find its vol. at STP. 198 K 985(126)(273) = 198(760)(V 2 ) 198(760) V 2 = 225cm 3 Researchers at U of AK, Fairbanks, say methane has surfaced in the Arctic due to global warming. This methane could account for up to 87% of the observed spike in atmospheric methane.

48 Derivation of the Density of Gases Equation Consider a sample of gas under two sets of conditions. P 1 V 1 = n R T 1 P 2 V 2 = n R T 2 P1P1 T1D1T1D1 Density of Gases Equation P 1 V 1 = R T 1 mass mm ( ) P 2 V 2 = R T 2 mass mm ( ) P 1 = T 1 mass V1V1 ( ) R mm ( ) P 2 = T 2 mass V2V2 R mm D1D1 = R D2D2 = P2P2 T2D2T2D2 Solve each of these for the ratio R/mm...

49 Density of Gases Density formula for any substance: For a sample of gas, mass is constant, but pres. and/or temp. changes cause gas’s vol. to change. Thus, its density will change, too. ORIG. VOL. NEW VOL. ORIG. VOL. NEW VOL. If V (due to P or T ), then… D If V (due to P or T ), then… D

50 Density of Gases Equation: ** As always, T’s must be in K. A sample of gas has density g/cm 3 at –18 o C and 812 mm Hg. Find density at 113 o C and 548 mm Hg. 812(386)(D 2 ) = 255(0.0021)(548) 255 K 386 K 812 (0.0021) = (D 2 ) 386 D 2 = 9.4 x 10 –4 g/cm 3 (386) 812 (386) 812

51 A gas has density 0.87 g/L at 30 o C and kPa. Find density at STP. 303 K 131.2(273)(D 2 ) = 303(0.87)(101.3) (0.87) = (D 2 ) 273 D 2 = 0.75 g/L (273) (273) Find density of argon at STP.

52 Find density of nitrogen dioxide at 75 o C and atm. 348 K NO 2 D of NO STP… 1(348)(D 2 ) = 273(2.05)(0.805) 1 (2.05) = (D 2 ) 348 D 2 = 1.29 g/L (348) 1 1 NO 2 participates in reactions that result in smog (mostly O 3 )

53 A gas has mass 154 g and density 1.25 g/L at 53 o C and 0.85 atm. What vol. does sample occupy at STP? 326 K Find STP… 0.85(273)(D 2 ) = 326(1.25)(1) 0.85 (1.25) = (D 2 ) 273 D 2 = g/L (273) 0.85 (273) 0.85 Find vol. when gas has that density. = 87.7 L

54 Dalton’s Law of Partial Pressure In a gaseous mixture, a gas’s partial pressure is the one the gas would exert if it were by itself in the container. The mole ratio in a mixture of gases determines each gas’s partial pressure. John Dalton (1766–1844) Since air is ~80% N 2, (i.e., 8 out of every 10 air-gas moles is a mole of N 2 ), then the partial pressure of N 2 accounts for ~80% of the total air pressure. At sea level, where P ~100 kPa, N 2 accounts for ~80 kPa.

55 Total pressure of mixture (3.0 mol He and 4.0 mol Ne) is 97.4 kPa. Find partial pressure of each gas. = 41.7 kPa = 55.7 kPa Dalton’s Law: the total pressure exerted by a mixture of gases is the sum of all the partial pressures P Tot = P A2 + P B2 + …

56 80.0 g each of He, Ne, and Ar are in a container. The total pressure is 780 mm Hg. Find each gas’s partial pressure. 80 g He= 20 mol He 80 g Ne= 4 mol Ne 80 g Ar= 2 mol Ar Total: 26 mol P He = 20 / 26 of total P Ne = 4 / 26 of total P Ar = 2 / 26 of total P He = 600 mm Hg P Ne = 120 mm Hg P Ar = 60 mm Hg Total pressure is 780 mm Hg

57 Two 1.0 L containers, A and B, contain gases under 2.0 and 4.0 atm, respectively. Both gases are forced into Container Z ( w /vol. 2.0 L). Find total pres. of mixture in Z. ABZ P1P1 V1V1 V2V2 P2P2 A B 2.0 atm 4.0 atm 1.0 L 1.0 L 2.0 L 1.0 atm 2.0 atm = PRESSURES IN ORIG. CONTAINERS VOLUMES OF ORIG. CONTAINERS VOLUME OF FINAL CONTAINER PARTIAL PRESSURES IN FINAL CONTAINER TOTAL PRESSURE (P tot ) IN LAST CONTAINER 3.0 atm 1.0 L 2.0 atm 1.0 L 4.0 atm 2.0 L

58 Find total pressure (P Tot )of mixture in Z. P1P1 V1V1 V2V2 P2P2 A B C AB ZC 1.3 L 2.6 L 3.8 L 2.3 L 3.2 atm1.4 atm 2.7 atm X atm 3.2 atm1.3 L 2.3 L 1.81 atm 1.4 atm2.6 L1.58 atm 2.7 atm3.8 L4.46 atm 7.85 atm =

59 Gas Stoichiometry Find vol. hydrogen gas made when 38.2 g zinc react w/excess hydrochloric acid. Pres.=107.3 kPa; temp.= 88 o C. Zn (s) + 2 HCl (aq)  ZnCl 2 (aq) + H 2 (g) 38.2 g Zn excess V = X L H g Zn = mol H 2 P = kPa T = 88 o C 361 K Not at STP! P V = n R T = 16.3 L ZnH2H2

60 What mass solid magnesium is req’d to react w /250 mL carbon dioxide at 1.5 atm and 77 o C to produce solid magnesium oxide and solid carbon? 2 Mg (s) + CO 2 (g)  2 MgO (s) + C (s) X g Mg V = 250 mL P = 1.5 atm T = 77 o C mol CO 2 P V = n R T = 0.63 g Mg = mol CO L kPa 350 K CO 2 Mg

61 Vapor Pressure -- a measure of the tendency for liquid particles to enter gas phase at a given temp. -- a measure of “stickiness” of liquid particles to each other more “sticky” less likely to vaporize In general: LOW v.p. not very “sticky” more likely to vaporize In general: HIGH v.p.

62 A liquid’s VP depends on the temp. AND what substance it is. TEMPERATURE ( o C) PRESSURE (kPa) CHLOROFORM ETHANOL WATER _______ substances evaporate easily (have high v.p.’s). BOILING  Volatile vapor pressure = confining pressure (usually from atmosphere)

63 At sea level and 20 o C… WATER AIR PRESSURE (~100 kPa) ETHANOL VAPOR PRES. (~5 kPa) V. P. (~10 kPa) NET PRESSURE (~95 kPa) NET PRESSURE (~90 kPa)


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