1 / 35

The Gas Laws

The Gas Laws. Characteristics of Gases. highly compressible. occupy the full volume of their containers. exert a uniform pressure on all inner surfaces of a container diffuse (mix) easily and quickly have very low densities. Kinetic Molecular Theory.

Download Presentation

The Gas Laws

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. The Gas Laws

  2. Characteristics of Gases • highly compressible. • occupy the full volume of their containers. • exert a uniform pressure on all inner surfaces of a container • diffuse (mix) easily and quickly • have very low densities.

  3. Kinetic Molecular Theory • Gases consist of a large number of molecules in constant random motion. • Volume of individual molecules negligible compared to volume of container. • Intermolecular forces (forces between gas molecules) negligible. • Collision of gas particles are elastic so no kinetic energy is lost • As temperature increases the gas particles move faster, hence increased kinetic energy.

  4. Four Physical Quantities for Gases

  5. Temperature K = ºC + 273 ºF -459 32 212 ºC -273 0 100 K 0 273 373 Always use absolute temperature (Kelvin) when working with gases.

  6. Kelvin Practice Absolute zero is –273C or 0 K What is the approximate temperature for absolute zero in degrees Celsius and kelvin? Calculate the missing temperatures 0C = _______ K 100C = _______ K 100 K = _______ C –30C = _______ K 300 K = _______ C 403 K = _______ C 25C = _______ K 0 K = _______ C 273 373 –173 243 27 130 298 –273

  7. Pressure Pressure (P ) is defined as the force exerted per unit area. The atmospheric pressure is measured using a barometer. Which shoes create the most pressure?

  8. Pressure • KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi • 1 atm = 760 mmHg = 760 torr = 101325 Pa.

  9. Pressure Aneroid Barometer Mercury Barometer • Barometer • measures atmospheric pressure

  10. STP STP Standard Temperature & Pressure 273 K 101.325 kPa

  11. STP SLC Standard Laboratory Conditions 25°C or 298 K 101.325 kPa

  12. V T P The Gas Laws -BOYLES-CHARLE-GAY-LUSSAC

  13. Boyle’s Law P V The pressure and volume of a gas are inversely related at constant mass & temp. PV = k

  14. A. Boyle’s Law

  15. Practice A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

  16. 726 mmHg x 946 mL P1 x V1 = 154 mL V2 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P2 = = 4460 mmHg

  17. Charles’ Law V T The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

  18. Charles’ Law

  19. Practice • A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? • If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?

  20. V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K 3.5 L / 300 K = V2 / 200 K V2 = (3.5 L/300 K) x (200 K) = 2.3 L • A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? • If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be? V1 = 1 L, T1 = 22°C = 295 K V2 = ?, T2 = 100 °C = 373 K V1/T1 = V2/T2, 1 L / 295 K = V2 / 373 K V2 = (1 L/295 K) x (373 K) = 1.26 L For more lessons, visit www.chalkbored.com

  21. 1.54 L x 398.15 K V2 x T1 = 3.20 L V1 A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? T2 = = 192 K

  22. Gay-Lussac’s Law P T The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

  23. Gay-Lussac’s Law

  24. Combined Gas Law P1V1 T1 P2V2 T2 = P1V1T2 =P2V2T1

  25. E. Gas Law Problems • A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK: P1V1T2 = P2V2T1 (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3

  26. E. Gas Law Problems • A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa P V WORK: P1V1T2 = P2V2T1 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL

  27. Practice • A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.

  28. E. Gas Law Problems • A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW P T V GIVEN: V1=7.84 cm3 P1=71.8 kPa T1=25°C = 298 K V2=? P2=101.325 kPa T2=273 K WORK: P1V1T2 = P2V2T1 (71.8 kPa)(7.84 cm3)(273 K) =(101.325 kPa)V2 (298 K) V2 = 5.09 cm3

  29. E. Gas Law Problems • A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? P T WORK: P1V1T2 = P2V2T1 (765 torr)T2 = (560. torr)(309K) T2 = 226 K = -47°C

  30. Avogadro’s Principle V n • Equal volumes of all gases contain equal numbers of moles at constant temp & pressure.

  31. The Ideal Gas Equation The gas laws can be combined into a general equation that describes the physical behavior of all gases. Charles’s law Avogadro’s law Boyle’s law 11.5 rearrangement PV = nRT R is the proportionality constant, called the gas constant.

  32. B. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R = 8.3145 J/mol·KR=0.0821 Latm/molK

  33. R = 0.0821 liter·atm/mol·K R = 8.3145 J/mol·KR = 8.2057 m3·atm/mol·KR = 62.3637 L·Torr/mol·K or L·mmHg/mol·K

  34. B. Ideal Gas Law • Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm

  35. B. Ideal Gas Law WORK: 85 g 1 mol = 2.7 mol 32.00 g • Find the volume of 85 g of O2 at 25°C and 104.5 kPa. IDEAL GAS LAW GIVEN: V=? n=85 g T=25°C = 298 K P=104.5 kPa R=8.315dm3kPa/molK = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molKK V = 64 dm3

More Related