1 / 17

G - L’s Law – Pressure vs. Temperature

G - L’s Law – Pressure vs. Temperature. Experiment to develop the relationship between the pressure and temperature of a gas. Include : Gay-Lussac’s Law , Combined Gas Law, partial pressure Additional KEY Terms. BOYLE’S LAW – Pressure vs. Volume. P 1 V 1. P 2 V 2. =. Pressure ( kPa ).

gus
Télécharger la présentation

G - L’s Law – Pressure vs. Temperature

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. G - L’s Law – Pressure vs. Temperature

  2. Experimentto develop the relationship between the pressure and temperature of a gas. • Include: Gay-Lussac’s Law, Combined Gas Law, partial pressure • Additional KEY Terms

  3. BOYLE’S LAW – Pressure vs. Volume P1V1 • P2V2 = Pressure (kPa) Volume (mL)

  4. CHARLES’S LAW – Temp vs. Volume • T1 • T2 V1 • V2 = Volume (mL) Temperature (K)

  5. Joseph Gay-Lussac (1778-1850) Determined that temperature and pressure of a gas is a direct relationship (volume and amount of gas are held constant)

  6. **As with Charles’ Law, temperature in Kelvin. • T1 • T2 P1 • P2 =

  7. If a 12.0 L sample of gas is found to have a pressure of 101.3 kPa at 0.0°C, calculate the new pressure at 128°C if the volume is held constant. P1 • P2 = • T1 • T2 0.0°C + 273 = 273 K128°C +273 = 401 K • 149 kPa 101.3 • P2 = • (401) • 273

  8. Boyle’s Law: Pressure α ___1___ • volume • Charles’ Law: Volume α temperature • Gay-Lussac’s Law: Pressure α temperature P1 V1 • P2 • V2 = • T1 • T2 combined gas law

  9. If a gas occupies a volume of 25.0 L at 25.0°C and 1.25 atm, calculate the volume at 128°C and 0.750 atm. P1 V1 • P2 • V2 = • T1 • T2 25°C + 273 = 298 K128°C +273 = 401 K (1.25) 25 = • 56.1 L • V2 • (401) • 298 (0.750)

  10. A gas has a volume of 125 L at 325 kPa and 58.0°C, calculate the temperature in Celsius to produce a volume of 22.4 L at 101.3 kPa P1 V1 • P2 • V2 = • T1 • T2 58°C + 273 = 331 K (101.3) • (331) 22.4 • 18.5 K • T2 = • 325 (125) 18.5 K - 273 = -254°C

  11. A bag contains 145 L of air at the bottom of a lake, at a temperature of 5.20°C and a pressure of 6.00 atm. When the bag is released, it ascends to the surface where the pressure is 1.00 atm and 16.0°C. Given: P1 = 6.00 atmV1 = 145LT1 = 5.20°C P2 = 1.00 atmT2 = 16.0°C If the maximum volume of the lift bag is 750 L, will the bag burst at the surface? Find: V2

  12. Dalton’s Law of Partial Pressure: Each gas in a mixture exerts pressure independently. Total pressure = sum of the pressures of each gas (partial pressures) Partial pressure depends on the number of gas particles present, the temperature and volume of container. Ptotal = P1 + P2 + P3 + P4 …

  13. 1.0 mol H2 2.0 mol He 1.0 mol H2 2.0 mol He 3.0 mol gas

  14. The following gases are placed in a 10.00 litre container and held at a constant temperature: 2.0 L of O2 at an original pressure of 202.6 kPa 3.00 L of Ne at an original pressure of 303.9 kPa What pressure is produced inside the container? HINT: Each gas is will expand when put into the 10.00 L container, so each will exert a partial pressure less than its original pressure.

  15. P1V1= P2V2 Oxygen:(202.6)(2.00) = P2 (10.00)P2= 40.5 kPa Neon:(303.9)(3.00) = P2 (10.00)P2= 91.2 kPa Total pressure = 40.5 + 91.2 = 131.7 kPa

  16. CAN YOU / HAVE YOU? • Experimentto develop the relationship between the pressure and temperature of a gas. • Include: Gay-Lussac’s Law, Combined Gas Law, partial pressure • Additional KEY Terms

More Related