1 / 8

Gay-Lussac’s Law

Gay-Lussac’s Law. Gay-Lussac’s Law. The relationship among pressure and temperature, at constant volume, can be mathematically represented by an equation known as Gay-Lussac’ law . P 1 = P 2 T 1 T 2 where:

gella
Download Presentation

Gay-Lussac’s Law

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. Gay-Lussac’s Law

  2. Gay-Lussac’s Law • The relationship among pressure and temperature, at constant volume, can be mathematically represented by an equation known as Gay-Lussac’ law. P1 = P2 T1 T2 where: P1 is the initial pressure and P2 is the new pressure. T1 is the initial temperature and T2 is the new temperature. V1 and V2 are the same (constant volume.)

  3. Gay-Lussac’s Law At constant volume, when temperature is increased, the pressure will increase. At constant volume, when temperature is decreased, the pressure will decrease. 22.4 L

  4. Gay-Lussac’s Law • As the Kelvin temperature of the gas increases, the pressure of the gas increases and vise versa. P = k T P P1 • What temperature is this? P2 • T (K) T2 T1 -273oC

  5. Gay-Lussac’s Law Ex. (1) If a rigid container of He at STP were cooled to 200. K, then what would be the new pressure in atmospheres? = P1V1 P1 P2V2 P2 T1 T2 = (1.00 atm) (X) (273 K) (200. K) = 0.733 atm X

  6. Gay-Lussac’s Law Ex. (2) If a rigid container of oxygen gas (O2) at 1.04 atm and 32.8oC were heated to 50.0oC, what would be new pressure? = P1V1 P1 P2V2 P2 T1 T2 = (1.04 atm) (X) (305.8 K) (323.0 K) = 1.10 atm X

  7. Gay-Lussac’s Law Ex. (3) What would be the new temperature of a rigid container of a gas at 101.3 kPa and 293 K if the gas pressure increased to 115 kPa? = P1V1 P1 P2V2 P2 T1 T2 = (101.3 kPa) (115 kPa) (293 K) (X) = 333 K X

  8. Gay-Lussac’s Law Ex. (4) What would have been the initial temperature of a rigid container of neon (Ne) at 1.00 atm if the final temperature was 293 K and the final pressure was 1.10 atm? = P1V1 P1 P2V2 P2 T1 T2 = (1.00 atm) (1.10 atm) (X) (293 K) = 266 K X

More Related