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The Gas Laws

V. T. P. The Behavior of Gases. The Gas Laws. STP . Standard Temperature and Pressure: 273 K and 760 mm Hg Or 0 C and 1atm. Temperature Conversions. [° C] =  K  −  273 [ ° C] = 5/9( ° F-32) [K] =  °C  +  273 [ K] =  (°F  + 459.67) ×  5 ⁄ 9 [°F] =  K  ×  9 ⁄ 5  −  459.67

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The Gas Laws

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  1. V T P The Behavior of Gases The Gas Laws

  2. STP • Standard Temperature and Pressure: • 273 K and 760 mm Hg • Or 0 C and 1atm

  3. Temperature Conversions • [°C] = K − 273 • [°C] = 5/9(°F-32) • [K] = °C + 273 [K] = (°F + 459.67) × 5⁄9 • [°F] = K × 9⁄5 − 459.67 • [°F] = °C × 9⁄5 + 32

  4. A. Boyle’s Law P V PV = k

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

  6. A. Boyle’s Law

  7. Boyle’s Law Practice Problem • Given 1.53 L of a sample of SO2 at a pressure of 5.6 x 103 Pa. If the pressure is changed to 1.5 x 104Pa at a constant temp, what will be the new volume of the gas. GIVEN: V1=1.53 L P1=5.6 x 103Pa V2=? P2=1.5 x 104 Pa WORK: P1V1= P2V2 5.6 x 103Pa x 1.53 L 1.5 x 104 Pa = 0.57L

  8. 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: P1V1 = P2V2 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL

  9. B. Charles’ Law V T

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

  11. B. Charles’ Law

  12. Charles’ Law Practice • A sample of gas at 15 degrees C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38 degree C and 1 atm? (Hint** convert to K first) GIVEN: V1=2.58 L T1=15 C V2=? T2=38 C WORK: V1/T1= V2/T2 2.58 L = V2 288 K 311 K = 2.79 L

  13. C. Gay-Lussac’s Law P T

  14. C. Gay-Lussac’s Law P T • The pressure and absolute temperature (K) of a gas are directly related • at constant mass & volume P1/T1= P2/T2

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

  16. 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: P1T2 = P2T1 (765 torr)T2 = (560. torr)(296K) T2 = 217 K = -56°C

  17. 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

  18. 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: V1T2 = V2T1 (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3

  19. F. The First 4 Gas Laws • The Gas Laws Table

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