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II. The Gas Laws

V. T. P. II. The Gas Laws. A. Boyle’s Law. P. V. PV = k. A. Boyle’s Law. P. V. The pressure and volume of a gas are inversely related at constant mass & temp. PV = k. Example Problems pg 335 # 10 &11.

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II. The Gas Laws

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  1. V T P II. The Gas Laws

  2. A. Boyle’s Law P V PV = k

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

  4. Example Problems pg 335 # 10 &11 1. The pressure on 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temp remains constant? P1 = 105 kPa P2 = 40.5 kPa V1 = 2.5 L V2 = ? P1× V1 = P2× V2 (105) (2.5) = (40.5)(V2) 262.5 = 40.5 (V2) 6.48 L = V2

  5. Example Problems pg 335 # 10 &11 2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure in the container if the temp remains constant? P1 = 205 kPa P2 = ? V1 = 4.0 L V2 = 12.0 L P1 × V1 = P2 × V2 (205) (4.0) = (P2)(12) 820 = (P2) 12 68.3 kPa = P2

  6. B. Charles’ Law V T

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

  8. Example Problems pg. 337 # 12 & 13 3. If a sample of gas occupies 6.80 L at 325°C, what will be its volume at 25°C if the pressure does not change? V1= 6.8L V2 = ? T1 = 325°C = 598 K T2 = 25°C = 298 K 6.8 = V2 • 298 6.8 X 298 = V2 X 598 2026.4 = V2 X 598 3.39 L = V2

  9. Example Problems pg. 337 # 12 & 13 4. Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant? V1= 5.0L V2 = ? T1 = -50°C = 223 K T2 = 100°C = 373 K 5 = V2 223 373 5 X 373 = V2 X 223 1865 = V2 X 223 8.36L = V2

  10. C. Gay-Lussac’s Law P T

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

  12. Example Problems 5. The gas left in a used aerosol can is at a pressure of 103 kPa at 25°C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928°C? P1= 103 kPa P2 = ? T1 = 25°C = 298 K T2 = 928°C = 1201 K 103 = P2 298 1201 298 × P2 = 123,703 P2 = 415 kPa

  13. Example Problem pg. 338 # 14 6. A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change? • P1= 6.58 kPa P2 = ? • T1 = 539 K T2 = 211 K • 6.58 = P2 • 211 • 6.58 X 211 = P2 X 539 • 1388.38 = P2 X 539 • 2.58 kPa = P2

  14. D. Avogadro’s Law • Under the same condition of temperature and pressure, equal volumes of all gases contain the same number of molecules.

  15. E. Combined Gas Law P1V1 T1 P2V2 T2 = P T V T PV T PV = k P1V1T2 =P2V2T1

  16. F. 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

  17. F. 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

  18. F. 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

  19. F. 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

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