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Gas Laws. Main Points. Behavior of Gases Charles’ Law Amontons’ Law The General Gas Law. Objectives. State Charles’ Law, express it as a working formula and discuss it’s application to scuba. State Amontons’ Law, express it as a working formula and discuss it’s application to scuba.
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Main Points • Behavior of Gases • Charles’ Law • Amontons’ Law • The General Gas Law
Objectives • State Charles’ Law, express it as a working formula and discuss it’s application to scuba. • State Amontons’ Law, express it as a working formula and discuss it’s application to scuba. • State The General Gas law, express it as a working formula and discuss it’s application to scuba.
Volume and Temperature Charles’ Law • The volume of a gas at constant pressure is directly proportional to the absolute temperature V1 = V2 T1 T2 • Can be used to approximate changes in the volume of gas available from a cylinder as the temperature changes.
Absolute Temperatures • Absolute temperature scales must be used in gas law calculations • When working in Farenheit add 460 • Rankine scale • When working in Centigrade add 273 • Kelvin Scale • Gay Lussac found that over the temperature range of 0-100oC, for each degree rise in temperature, gases expanded approximately 1/270th of the volume at 0oC
Charles’ Law Problem (English) If a scuba cylinder delivers 40 ft3 of air at 78o F, how much air is available at 55o F? 40 FT3 = V2 78F+460 55F+460 40FT3 = V2 538 515 20600 =538V2 38.29FT3 = V2
Amonton’s Law • The pressure of a gas at constant volume is directly proportional to the absolute temperature P1 = P2 T1 T2 • Can be used to calculate the changes in pressure in a cylinder as temperature changes.
Amontons’ Problem (Imperial) A scuba cylinder contains 3,014.7 psia at 78o F. What will be the pressure in a car trunk at 115o F? 3014.7PSIA = P2 78F+460 115F+460 3014.7PSI = P2 538 575 1733452.5 = 538P2 1733452.5/538 = P2 3222 PSI =P2
General Gas Law • Combines Boyle’s law, Charles’ law and Amontons’ law. • Predicts the behavior of any quantity of gas in terms of pressure, volume and temperature. P1V1 = P2V2 T1 T2 • Absolute pressure and absolute must be used.
Example At the surface, a cylinder contains 95 ft3 of air at a temperature of 84F. What volume of gas is available to the diver at a depth of 112ffw and a temperature of 40F?
Step 1Determine Absolute Values P1 = 1ata V1 = 95 ft3 T1 = 84F + 460 = 544R P2 = (112ffw / 34ffw) + 1atm = 4.29 ata V2 = Unknown T2 40F + 460 = 500R
Apply the General Gas Law P1V1 = P2V2 T1 T2 1ata x 95 ft3 = 4.29 ata x V2 544R 500R V2 = 1ATA x 95FT3 x 500R 4.29ata x 544R V2= 47500 / 2333.76 = 20.35 ft3
Have We Covered: • Behavior of Gases • Charles’ Law • Amontons’ Law • The General Gas Law
Can you • State Charles’ Law, express it as a working formula and discuss it’s application to scuba? • State Amontons’ Law, express it as a working formula and discuss it’s application to scuba? • State The General Gas law, express it as a working formula and discuss it’s application to scuba?
Last Thoughts • Understanding the gas laws allows you to anticipate phenomena such as variation in cylinder temperature with a change in temperature.