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

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.

Let’s Try a Practice Problem Your local weather report announces that the barometric pressure is 773.21 mmHg. Convert this pressure to (atm). 1.00 atm 773.21 mmHg X ---------------- = 1.02 atm 760 mmHg

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 = 101.3 kPa = 14.7 psi

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 V1 V2 Charles’s Law: ------ = ------ T1 T2 V1 V2 Avogadro’s Law: ------ = ------- n1 n2

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 - 1.00 atm = 1.13 atm 10 m 1.13 atm X ---------- = 11.3 m below the surface 1 atm

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)? V1 V2 88.2 mL V2 Charles Law: ------ = ------ = --------------- = ----------------- T1 T2 (35oC+273) (155oC +273) V2 = 123 mL ( or 1.23X102 mL)

Let’s Try Another! A chemical reaction occurring in a cylinder equipped with a movable piston produces 0.621 mol of gaseous product. If the cylinder contained 0.120 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.) V1 V2 2.18 L V2 Avogadro’s Law: ------ = ------- = -------------- = --------------- n1 n2 0.120 mol 0.621 mol V2 = 11.3 L

5. 1-5. 3 pgs. 238 – 239 #’s 26(a), 28(b), 32, 34 & 36 Read 5. 4-5 5.1-5.3 pgs. 238 – 239 #’s 26(a), 28(b), 32, 34 & 36 Read 5.4-5.6 pgs. 206-219