Gas Pressure, and Gas Laws Chapter 17 Gases (p.300-301) Properties of Gases, Gas Pressure, and Gas Laws

Properties of Gases Expand to completely fill their container. Take the shape of their container. Low density. Much less than solid or liquid state. Compressible. Mixtures of gases are always homogeneous. 2

Air Pressure Pressure: a measure of the force applied by gas molecules on objects This can be felt by the walls of a container or on objects in air http://www.youtube.com/watch?v=t-Iz414g-ro

Pressure What causes gas pressure in a closed container? Pressure is the result of a force distributed over an area. Collisions between particles of a gas and the walls of the container cause the pressure in a closed container of gas.

Helium Filled Balloon The helium atoms in a balloon are constantly moving. When many particles collide with the walls of a container at the same time, they produce a measurable pressure. The more frequent the collisions, the greater the pressure is. The speed of the particles and their mass also affect the pressure.

Atmospheric Pressure Atmospheric pressure is the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth.

Factors That Affect Gas Pressure Factors that affect the pressure of an enclosed gas are its temperature, its volume, and the number of its particles.

Temperature P= 60 kPa P= 72 kPa Raising the temperature of a gas will increase its pressure if the volume of the gas and the number of particles are constant. P= 60 kPa P= 72 kPa

Effect of Temperature on Pressure in a Fixed Volume As the temperature rises, the average kinetic energy of the particles in the air increases. With increased kinetic energy, the particles move faster and collide more often with the inner walls of the container. Faster-moving particles hit the walls with greater force. More collisions and increased force cause the pressure of the gas in the container to rise. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/gasesv6.swf

Pressure and Volume Boyle’s Law http://videos.howstuffworks.com/hsw/17060-physical-science-gases-video.htm Reducing the volume of a gas increases its pressure if the temperature of the gas and the number of particles are constant. 10

Boyle’s Law Boyle’s law states that the pressure of a gas is inversely related to its volume when T and the amount of molecules are constant. if volume decreases, the pressure increases. 11

In the next lab, we will test… When you double the pressure on a gas, the volume is cut in half (as long as the temperature and amount of gas do not change). Tro's Introductory Chemistry, Chapter 11 12

Boyle’s Law Robert Boyle described the relationship between the pressure and volume of a gas. The graph shows an inverse relationship between the volume of a gas and the pressure of the gas.

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, number of molecules (n) constant) 14

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 15

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 Copyright © 2009 by Pearson Education, Inc. 16

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 Copyright © 2009 by Pearson Education, Inc. 17

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 is constant), is the new volume represented by A, B, or C? Copyright © 2009 by Pearson Education, Inc. 18

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. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/gasesv6.swf 19

Calculations with Boyle’s Law 20

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 mmHg after its pressure is changed to 2200 mmHg at constant T and n? 1. Set up a data table: Conditions 1 Conditions 2 P1 = 550 mmHg P2 = 2200 mmHg V1 = 8.0 L V2 = ? 21

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 mmHg = 2.0 L 2200 mmHg pressure ratio decreases volume 22

Example 1 If a sample of helium gas has a volume of 120 mL and a pressure of 850 mmHg, what is the new volume if the pressure is changed to 425 mmHg? 1) 60 mL 2) 120 mL 3) 240 mL 23

Example 1 Answer P1 = 850 mmHg P2 = 425 mmHg V1 = 120 mL V2 = ?? Choice 3) 240 mL P1 = 850 mmHg P2 = 425 mmHg V1 = 120 mL V2 = ?? V2 = V1 x P1 = 120 mL x 850 mmHg = 240 mL P2 425 mmHg Pressure ratio increases volume 24

Example 2 A Cylinder with a Movable Piston Has a Volume of 6.0 L at 4.0 atm. What Is the Volume at 1.0 atm? 25

Example 2: A Cylinder with a Movable Piston Has a Volume of 6.0 L at 4.0 atm. What Is the Volume at 1.0 atm? P1 = 4.0 atm P2 = 1.0 atm V1 = 6.0 L V2 = ?? V2 = V1 x P1 = 6.0 L x 4.0 atm = 24 L P2 1.0 atm Pressure ratio increases volume Check: Since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does. 26

Example 3 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 27

Example 3: Solution Choice 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.) P1 = 0.7 atm P2 = 1.40 atm V1 = 6.4 L V2 = ?? 28

Example 4 1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg A sample of oxygen gas has a volume of 12.0 L at 600. mmHg. What is the new pressure when the volume changes to 36.0 L? (T and n constant). 1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg Copyright © 2009 by Pearson Education, Inc. 29

Example 4: Solution Choice 1) 200. mmHg Data Table Conditions 1 Conditions 2 P1 = 600. mmHg P2 = ??? V1 = 12.0 L V2 = 36.0 L P2 = P1 x V1 V2 600. mmHg x 12.0 L = 200. mmHg 36.0 L 30

Temperature and Volume Charles’s Law http://videos.howstuffworks.com/hsw/17060-physical-science-gases-video.htm 31

Tro's Introductory Chemistry, Chapter 11 Standard Conditions When doing gas problems, always have temperatures in kelvin. K = °C + 273 Common reference points for comparing. Standard pressure = 1.00 atm. Standard temperature = 0 °C= 273 K. STP. Tro's Introductory Chemistry, Chapter 11 32

Charles’s Law In Charles’s Law, the Kelvin temperature of a gas is directly related to the volume. P and amount of molecules (n) are constant. when the temperature of a gas increases, its volume increases. Copyright © 2009 by Pearson Education, Inc. 33

Volume and Temperature As a gas is heated, it expands. This causes the density of the gas to decrease. Because the hot air in the balloon is less dense than the surrounding air, it rises. 34

Charles’s Law: V and T For two conditions, Charles’s law is written V1 = V2 (P and n constant) T1 T2 Rearranging Charles’s law to solve for V2: T2 x V1 = V2 x T1 T1 T1 V2 = V1 x T2 T1 35

Learning Check Solve Charles’s law expression for T2. V1 = V2 T1 T2 36

Solution V1 = V2 T1 T2 Cross-multiply to give: V1T2 = V2T1 Isolate T2 by dividing through by V1: V1 V1 T2 = T1 x V2 V1 37

E1: Using Charles’s Law A balloon has a volume of 785 mL at 21 °C. If the temperature drops to 0 °C, what is the new volume of the balloon (P constant)? 1. Set up data table: Conditions 1 Conditions 2 V1 = 785 mL V2 = ? T1 = 21 °C = 294 K T2 = 0 °C = 273 K Be sure to use the Kelvin (K) temperature in gas calculations. 38

E1: Solution 2. Solve Charles’s law for V2: V1 = V2 T1 T2 V2 = V1 x T2 V2 = 785 mL x 273 K = 729 mL 294 K 39

E2 A sample of oxygen gas has a volume of 420 mL at a temperature of 18 °C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? 1) 443 °C 2) 170 °C 3) - 82 °C 40

E2: Solution 1. Set up data table: Conditions 1 Conditions 2 V1 = 420 mL V2 = 640 mL T1 = 18 °C +273= 291 K T2 = ? 2. 170 °C T2 = T1 x V2 V1 T2 = 291 K x 640 mL = 443 K 420 mL = 443 K – 273 = 170 °C 41

Learning Check Use the gas laws to complete each sentence with 1) increases or 2) decreases. A. Pressure _______ when V decreases. B. When T decreases, V _______. C. Pressure _______ when V changes from 12 L to 24 L. D. Volume _______when T changes from 15 °C to 45 °C. 42

Solution Use the gas laws to complete each sentence with 1) increases or 2) decreases. A. Pressure 1) increases when V decreases. B. When T decreases, V 2) decreases. C. Pressure 2) decreases when V changes from 12 L to 24 L. D. Volume 1) increases when T changes from 15 °C to 45 °C. 43