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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.

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The Gas Laws

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  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

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