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Ch Gases I. Physical Properties
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A. 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 avg. KE directly related to Kelvin temperature.
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B. Real Gases Particles in a REAL gas… have their own volume
attract each other Gas behavior is most ideal… at low pressures at high temperatures in nonpolar atoms/molecules
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C. Characteristics of Gases
Gases expand to fill any container. random motion, no attraction Gases are fluids (like liquids). no attraction Gases have very low densities. no volume = lots of empty space
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C. Characteristics of Gases
Gases can be compressed. no volume = lots of empty space Gases undergo diffusion & effusion. random motion
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D. Temperature Always use absolute temperature (Kelvin) when working with gases. ºF ºC K -459 32 212 -273 100 273 373 K = ºC + 273
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E. Pressure Which shoes create the most pressure?
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Standard Temperature & Pressure
F. STP STP Standard Temperature & Pressure 0°C K 1 atm kPa -OR-
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II. The Gas Laws BOYLES CHARLES GAY-LUSSAC
Ch Gases II. The Gas Laws BOYLES CHARLES GAY-LUSSAC P V T
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A. Boyle’s Law P V PV = k
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A. Boyle’s Law The pressure and volume of a gas are inversely related at constant mass & temp P V PV = k
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A. Boyle’s Law Boyle's Law Video
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B. Charles’ Law V T
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B. Charles’ Law The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure V T
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B. Charles’ Law Charles’ Law
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C. Gay-Lussac’s Law P T
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C. Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume P T
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C. Gay-Lussac’s Law Gay-Lussac’s Law
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Equations for the three Gas Laws
Charles’s Law Boyle’s Law Gay-Lussac’s Law V1/T1 = V2/T2 P1V1 = P2V2 P1/T1 = P2/T2
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D. Combined Gas Law P1V1 T1 = P2V2 T2
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V1/T1 = V2/T2 E. Gas Law Problems
A gas occupies 473 L at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 L T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK: V1/T1 = V2/T2 (473 L)(367 K)=V2(309 K) V2 = 562 L
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P1V1 = P2V2 E. Gas Law Problems
A gas occupies 100. mL at 1.5 atm. Find its volume at 2 atm. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 1.5 atm V2 = ? P2 = 2 atm P V WORK: P1V1 = P2V2 (1.5 atm)(100.mL)=(2 atm)V2 V2 = 75.0 mL
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E. Gas Law Problems COMBINED GAS LAW P T V V1 = 7.84 mL
A gas occupies 7.84 mL at 0.71 atm & 25°C. Find its volume at STP. COMBINED GAS LAW GIVEN: V1 = 7.84 mL P1 = 0.71 atm T1 = 25°C = 298 K V2 = ? P2 = 1 atm T2 = 273 K P T V WORK: P1V1T2 = P2V2T1 (0.71 atm)(7.84 mL)(273 K) =(1 atm) V2 (298 K) V2 = 5.09 mL
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P1/T1 = P2/T2 E. Gas Law Problems
A gas’ pressure is 1 atm at 23°C. At what temperature will the pressure be atm? GAY-LUSSAC’S LAW GIVEN: P1 = 1 atm T1 = 23°C = 296K P2 = atm T2 = ? P T WORK: P1/T1 = P2/T2 (1atm)T2 = (0.737 atm)(296K) T2 = 218 K = -55°C
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Ch Gases III. Ideal Gas Law
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A. Avogadro’s Principle
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A. Avogadro’s Principle
Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas V n
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UNIVERSAL GAS CONSTANT
B. Ideal Gas Law PV nT V n PV T = k = R UNIVERSAL GAS CONSTANT R= Latm/molK R=8.315 dm3kPa/molK
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UNIVERSAL GAS CONSTANT
B. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R= Latm/molK R=8.315 dm3kPa/molK
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B. Ideal Gas Law IDEAL GAS LAW P = ? atm n = 0.412 mol
Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW GIVEN: P = ? atm n = mol T = 16°C = 289 K V = 3.25 L R = Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm
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B. Ideal Gas Law IDEAL GAS LAW V = ? n = T = 25°C = 298 K P = 1 atm
Find the volume of 2.7 mol of O2 at 25°C and 1 atm. IDEAL GAS LAW GIVEN: V = ? n = T = 25°C = 298 K P = 1 atm R = Latm/molK WORK: 85 g 1 mol = 2.7 mol 32.00 g 2.7 mol PV = nRT (1)V=(2.7mol) (8.315) (298) V = 64 L
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