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Pressure: The Result of Molecular Collisions and The Simple Gas Laws

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Presentation on theme: "Pressure: The Result of Molecular Collisions and The Simple Gas Laws"— Presentation transcript:

1 5.1-5.3 Pressure: The Result of Molecular Collisions and The Simple Gas Laws

2 Pressure A constant force on the surfaces exposed to gas. This force is the result of multiple molecular collisions. P = F / A (pressure = force / unit area) Pressure depends on many factors: 1.) Concentration of gas molecules 2.) Volume 3.) Temperature How does pressure change with altitude? Pressure decreases with increasing altitude.

3 Let’s Try a Practice Problem
Your local weather report announces that the barometric pressure is mmHg. Convert this pressure to (atm) atm mmHg X = 1.02 atm 760 mmHg

4 Pressure Unit Relation
There are several units used for pressure. These include: mmHg (millimeters of mercury) torr Pa (pascal) or kPa (kilopascal) atm (atmosphere) psi (pounds per square inch) 760 mmHg = 760 torr = 1 atm = kPa = 14.7 psi

5 Simple Gas Laws The following three gas laws, can be used when two out of the four variables (pressure (P), volume (V), Temperature (T), and moles of gas (n)) are held constant. Boyle’s Law : P1V1 = P2V2 V V2 Charles’s Law: = T T2 V V2 Avogadro’s Law: = n n2

6 Let’s Try a Few Sample Problems
A snorkeler takes a syringe filled with 16 mL of air from the surface, where the pressure is 1 atm, to an unknown depth. The volume of the air in the syringe at this depth is 7.5 mL. What is the pressure at this depth? If the pressure increases by 1 atm for every additional 10 m of depth, how deep is the snorkeler? Step 1: Choose The Gas Law Boyles Law = P1V1 = P2V2 Step 2: Solve for P2 (1 atm) (16 mL air) = (P2) (7.5 mL air) P2 = 2.13 atm ( or 2 atm correct # of sig. figs.) Step 3: Solve for depth 2.13 atm atm = 1.13 atm 10 m 1.13 atm X = 11.3 m below the surface 1 atm

7 Another Sample Problem
A gas cylinder with a movable piston has an initial volume of 88.2 mL. If we heat the gas from 35oC to 155oC. What is its final volume (in mL)? V V mL V2 Charles Law: = = = T T (35oC+273) (155oC +273) V2 = 123 mL ( or 1.23X102 mL)

8 Let’s Try Another! A chemical reaction occurring in a cylinder equipped with a movable piston produces mol of gaseous product. If the cylinder contained mol of gas before the reaction and had an initial volume of 2.18 L, what was the volume after the reaction? (Assume constant pressure and temperature and that the initial amount of gas completely reacts.) V V L V2 Avogadro’s Law: = = = n n mol mol V2 = 11.3 L

9 5. 1-5. 3 pgs. 238 – 239 #’s 26(a), 28(b), 32, 34 & 36 Read 5. 4-5
pgs. 238 – 239 #’s 26(a), 28(b), 32, 34 & 36 Read pgs


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