Pressure and Volume (Boyle’s Law) Chapter 6 Gases 6.3 Pressure and Volume (Boyle’s Law)

Boyle’s Law Boyle’s Law states that the pressure of a gas is inversely related to its volume when T and n are constant. if volume decreases, the pressure increases.

PV Constant in Boyle’s Law In Boyle’s Law, the product P x V is constant as long as T and n do not change. P1V1 = 8.0 atm x 2.0 L = 16 atm L P2V2 = 4.0 atm x 4.0 L = 16 atm L P3V3 = 2.0 atm x 8.0 L = 16 atm L Boyle’s Law can be stated as P1V1 = P2V2 (T, n constant)

Solving for a Gas Law Factor The equation for Boyle’s Law can be rearranged to solve for any factor. P1V1 = P2V2 Boyle’s Law To solve for V2 , divide both sides by P2. P1V1 = P2V2 P2 P2 V1 x P1 = V2 P2

Boyles’ Law and Breathing During an inhalation, the lungs expand. the pressure in the lungs decreases. air flows towards the lower pressure in the lungs. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

Boyles’ Law and Breathing During an exhalation, lung volume decreases. pressure within the lungs increases. air flows from the higher pressure in the lungs to the outside. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

Calculations with Boyle’s Law

Calculation with Boyle’s Law Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mm Hg after its pressure is changed to 2200 mm Hg at constant T? 1. Set up a data table: Conditions 1 Conditions 2 P1 = 550 mm Hg P2 = 2200 mm Hg V1 = 8.0 L V2 = ?

Calculation with Boyle’s Law (Continued) 2. When pressure increases, volume decreases. Solve Boyle’s Law for V2: P1V1 = P2V2 V2 = V1 x P1 P2 V2 = 8.0 L x 550 mm Hg = 2.0 L 2200 mm Hg pressure ratio decreases volume

Learning Check 1) pressure decreases 2) pressure increases For a cylinder containing helium gas indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant). 1) pressure decreases 2) pressure increases

Solution 1) Pressure decreases B 2) Pressure increases A For a cylinder containing helium gas indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant): 1) Pressure decreases B 2) Pressure increases A

Learning Check If a sample of helium gas has a volume of 120 mL and a pressure of 850 mm Hg, what is the new volume if the pressure is changed to 425 mm Hg ? 1) 60 mL 2) 120 mL 3) 240 mL

Solution P1 = 850 mm Hg P2 = 425 mm Hg V1 = 120 mL V2 = ?? V2 = V1 x P1 = 120 mL x 850 mm Hg = 240 mL P2 425 mm Hg Pressure ratio increases volume

Learning Check A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At 1.40 atm (T constant), is the new volume represented by A, B, or C?

Solution A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At a higher pressure (T constant), the new volume is represented by the smaller balloon A.

Learning Check A) 3.2 L B) 6.4 L C) 12.8 L If the sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm, what is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 L B) 6.4 L C) 12.8 L

Solution If the sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm, what is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 L V2 = V1 x P1 P2 V2 = 6.4 L x 0.70 atm = 3.2 L 1.40 atm Volume decreases when there is an increase in the pressure (temperature is constant.)

Learning Check 1) 200. mm Hg 2) 400. mm Hg 3) 1200 mm Hg A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant). 1) 200. mm Hg 2) 400. mm Hg 3) 1200 mm Hg

Solution 1) 200. mm Hg Data table Conditions 1 Conditions 2 P1 = 600. mm Hg P2 = ??? V1 = 12.0 L V2 = 36.0 L P2 = P1 x V1 V2 600. mm Hg x 12.0 L = 200. mm Hg 36.0 L