Unit 10 Gas Laws

I. Kinetic Theory Particles in an ideal gas… 1.gases are hard, small, spherical particles 2.don’t attract or repel each other. 3.are in constant, random, straight-line motion. 4.indefinite shape and volume. 5.have “perfectly” elastic collisions.

A. Graham’s Law DiffusionDiffusion –The tendency of molecules to move toward areas of lower concentration. Ex: air leaving tire when valve is opened EffusionEffusion –Passing of gas molecules through a tiny opening in a container

A. Graham’s Law Which one is Diffusion and which one is Effusion? Diffusion Effusion Tiny opening

II. Factors Affecting Gas Pressure A. Amount of Gas Add gas - ↑ pressure Remove gas - ↓ pressure Ex: pumping up a tire adding air to a balloon aerosol cans

B. Volume Reduce volume - ↑ pressure Increase volume - ↓ pressure Ex: piston in a car II. Factors Affecting Gas Pressure

C. Temperature Increase Temp. - ↑ pressure Decrease Temp. - ↓ pressure Ex: Helium balloon on cold/hot day, bag of chips II. Factors Affecting Gas Pressure

Gas Pressure - collision of gas molecules with the walls of the container

Atmospheric Pressure- collision of air molecules with objects

Atmospheric pressure is measured with a barometer. Increase altitude – decrease pressure Ex. Mt. Everest – atmospheric pressure is 253 mm Hg Vacuum- empty space with no particles and no pressure Ex: space

Gas Pressure (Cont.) -- 3 ways to measure pressure: »atm (atmosphere) »mm Hg »kPa (kilopascals) U-tube Manometer

III. Variables that describe a gas VariablesUnits Pressure (P) –kPa, mm Hg, atm Volume (V) – L, mL, cm 3 Temp (T) –°C, K (convert to Kelvin) K = °C Mole (n) - mol

Draw on the Left Side of Your Spiral PressureVolume Temperature kPa Mole

How pressure units are related: 1 atm = 760 mm Hg = kPa How can we make these into conversion factors? 1 atm101.3 kPa 760 mm Hg 1 atm

Guided Problem: 1. Convert 385 mm Hg to kPa 385 mm Hg 2. Convert 33.7 kPa to atm 33.7 kPa x kPa = 51.3 kPa 760 mm Hg x =.33 atm kPa 1 atm

Standard Temperature and Pressure Standard pressure – 1 atm, 760 mmHg, or101.3 kPa Standard temp. – 0° C or 273K STP

Gases (cont.) Kelvin Temperature scale is directly proportional to the average kinetic energy

A. Boyle’s Law IV. Gas Laws P V The pressure and volume of a gas are inversely related -at constant mass & temp P 1 × V 1 = P 2 × V 2

10. The pressure on 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temp remains constant? P 1 = 105 kPa P 2 = 40.5 kPa V 1 = 2.5 L V 2 = ? P 1 × V 1 = P 2 × V 2 (105) (2.5) = (40.5)(V 2 ) = 40.5 (V 2 ) 6.48 L = V 2 Example Problems pg 335 # 10 &11

11. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure in the container if the temp remains constant? P 1 = 205 kPa P 2 = ? V 1 = 4.0 LV 2 = 12.0 L P 1 × V 1 = P 2 × V 2 (205) (4.0) = (P 2 )(12) 820 = (P 2 ) L = P 2 Example Problems pg 335 # 10 &11

B. Charles’ Law The volume and temperature (in Kelvin) of a gas are directly related –at constant mass & pressure V T V 1 = V 2 ***Temp must be in Kelvin K = °C T1T1 T2T2

Example Problems pg. 337 # 12 & If a sample of gas occupies 6.80 L at 325°C, what will be its volume at 25°C if the pressure does not change? V 1 = 6.8LV 2 = ? T 1 = 325°C = 598 K T 2 = 25°C = 298 K 6.8 = V × V 2 = V 2 = 3.39 L

13. Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant? V 1 = 5.0L V 2 = ? T 1 = -50°C = 223 K T 2 = 100°C = 373 K 5 = V (223) V 2 = V 2 = 8.36 L Example Problems pg. 337 # 12 & 13

P T C. Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related –at constant mass & volume P 1 = P 2 T 1 T 2 ***Temp must be in Kelvin K = °C + 273

Example Problems 1. The gas left in a used aerosol can is at a pressure of 103 kPa at 25°C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928°C? P 1 = 103 kPaP 2 = ? T 1 = 25°C = 298 K T 2 = 928°C = 1201 K 103 = P × P 2 = 123,703 P 2 = 415 kPa

Example Problem pg. 338 # A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change? P 1 = 6.58 kPaP 2 = ? T 1 = 539 K T 2 = 211 K 6.58 = P × P 2 = P 2 = 2.58 kPa

D. Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 Combines the 3 gas laws as follows: The other laws can be obtained from this law by holding one quantity (P,V or T) constant. Use this law also when none of the variables are constant.

How to remember each Law! P V T Gay-Lussac Boyles Charles Cartesian Divers Balloon and flask Demo Fizz Keepers

E. Ideal Gas Law The 4 th variable that considers the amount of gas in the system P 1 V 1 T 1 n = P 2 V 2 T 2 n Equal volumes of gases contain equal numbers of moles (varies directly).

E. Ideal Gas Law You don’t need to memorize this value! You can calculate the # of n of gas at standard values for P, V, and T PV Tn = R (1 atm)(22.4L) (273K)(1 mol) = R UNIVERSAL GAS CONSTANT R= atm∙L/mol∙K

E. Ideal Gas Law P= pressure in atm V = volume in liters n = number of moles R= atm∙L/mol∙K T = temperature in Kelvin PV=nRT

E. Example Problems 1. At what temperature will 5.00g of Cl 2 exert a pressure of 900 mm Hg at a volume of 750 mL? 2. Find the number of grams of CO 2 that exert a pressure of 785 mm Hg at a volume of 32.5 L and a temperature of 32 degrees Celsius. 3. What volume will 454 g of H 2 occupy at 1.05 atm and 25°C.

F. Dalton’s Partial Pressure Law The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. P total = P 1 + P 2 + P

F. Dalton’s Law Example problem: 1. Air contains oxygen, nitrogen, carbon dioxide, and trace amounts of other gases. What is the partial pressure of oxygen (P O 2 ) if the total pressure is kPa. And the partial pressures of nitrogen, carbon dioxide, and other gases are kPa, kPa, and 0.94 kPa. P O 2 = P total – (P N 2 + P CO 2 + P others ) = kPa – (79.10 kPa kPa kPa) = kPa

F. Dalton’s Law 2. A container holds three gases : oxygen, carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm respectively. What is the total pressure of the container? 3. A gas mixture contains oxygen, nitrogen and carbon dioxide. The total pressure is 50.0 kPa. If the carbon dioxide has a partial pressure of 21 kPa and the nitrogen has a partial pressure of 15 kPa, what is the partial pressure of the oxygen? 4. A container contains two gases – helium and argon, at a total pressure of 4.00 atm. Calculate the partial pressure of helium if the partial pressure of the argon is 1.5 atm.