Gases – Kinetic Theory revisited (assumptions for “ Ideal” Gases)

1. Gases – Kinetic Theory revisited (assumptions for “Ideal” Gases) - Small molecules far apart (compressible) - Constant rapid straight-line motion (collisions create gas pressure) • No attraction between molecules - Kinetic energy (molecule speed -heat) proportional to Kelvin temperature

2. Kinetic Molecular “theory” What Are the Characteristics Of an ideal gas?

3. Variables that describe gases Volume (V) – in liters Pressure (P)– in Kilopascals or atmospheres Temperature (T) – in Kelvin’s Number of molecules (n) – in moles Gas pressure (due to collisions) depends on… Number of molecules - double the molecules - pressure doubles (2 x more collisions) Volume – double volume, pressure becomes half (Molecules more spread out, less collisions) Temperature – double the Kelvin Temp – pressure doubles (Molecules moving 2 x faster)

4. THE GAS LAWS Mathematical relationships between Pressure, volume and temperature Pressure vs. Volume (Boyle’s law) Volume of a gas is inverselyproportional to its pressure (if temp is kept constant) Ex: If pressure is doubled, volume becomes one-half Or If volume is doubled, pressure becomes one-half Why? (molecules confined to smaller space collide more, and vice versa)

5. Pressure x Volume = K (constant) PV = k Ex: PressureVolumek 1 50 6 300 2 100 3 300 3 150 2 etc. P1 x V1 = 300 = P2 x V2 Equation: P1V1 = P2V2 Ex: If pressure is reduced to 25, what is the new volume? P1V1 = P2V2 (50)(6) = (25) V2 (50)(6) = V2 (25) V2 = 12

6. Volume vs. Temperature (Charles law) Volume of a gas is directly proportional to Kelvin temperature (if pressure is constant) As Kelvin temperature increases, Volume increases Why? (Molecules move faster, push out volume) Volume = K constant V = k Temp (Kelvins) T 0.5 L = 1.0 L k = 0.0025 200 K 400 K V1 = V2 T1 T2 If temp is reduced to 50 K, what is the new volume? 0.5 L = V2 (0.5 L)(50 K) = (200 K )V2 200 K 50 K (0.5 L)(50 K) = V2 (200 K) V2 = 0.125 L

7. Charles law and the Kelvin scale How much volume could a gas occupy if it had no KE, so it produces no pressure?

8. Practice problems P1V1 = P2 V2 • A 150. mL sample of a gas at standard pressure (1 atm) is compressed to 125 mL. What is its new pressure? • A 150 mL sample of a gas at standard temp (00C) is heated to 250C. What is its new volume? P1V1 = P2V2 (1atm)(150 mL) = P2(125 mL) (1atm)(150 mL) = P2 (125 mL) P2 = 1.20 atm V1 T2 = V2 T1 V1 = V2 T1 T2 K = C + 273 K = 25 + 273 (150 mL) (298 K) = V2 (273 K) (150 mL) = V2 (273 K) (298 K) V2 = 164 mL

9. Pressure vs. Temperature (Gay-Lussac’s law) pressure is directly proportional to Kelvin Temp As Kelvin temp increases, pressure increases P/T=K P1 = P2 T1 T2 (KMT: faster molecules collide more, increase pressure) Ex: an “empty” 0.5 liter aerosol can at 250C is heated to 8000C. What is the final pressure inside the can? P1 = 1 atm V1 and V2 = 0.5 L T1 = 250C + 273 = 298 K P2 = ? V is constant T2 = 800 + 273 = 1073 K P1 = P2(1 atm) = P2 P2 = 3.6 atm T1 T2 (298 K) (1074 K)

10. Combined laws (Three equations in one!) P V and T P1V1 = P2V2 T1 T2 Ex: 100 mL of a gas at STP has its temp increased to 546 K while its pressure is increased to 2 atmospheres. What is the new volume? (1 atm)(100 mL) = (2 atm) V2 (273 K) (546 K) (1 atm)(100 mL) (546 K) = (273 K)(2 atm) V2 (1 atm)(100 mL) (546 K) = V2 100 mL = V2 (273 K)(2 atm) Why didn’t the volume change?

11. If variables are P1V1 = P2V2 P1V1 = P2V2 Held constant: T1 T2 Boyles law P1V1 = P2V2V1 = V2 Charles T1 T2 T1 T2 P1V1 = P2V2P1= P2 etc. T1 T2 T1 T2

12. Practice problems: 3. An 200 mL aerosol can contains gas under pressure at room temperature (270C). If its initial pressure is 500 kpa. What pressure will the gas exert when its temperature is increased to 627 0C? P1 T2 = P2 T1 P1 = P2 T1 T2 (500 kPa) = P2 (300 K) (900 K) (500 kPa)(900 k) = P2 (300 K) P2 = 1500 kPa

13. 4) A 5.0 Liter He balloon is released and rises into the air. The temperature decreases from 300 k to 250 k, while the pressure decreases from 100 kpa to 50 kpa. What is the final volume of the balloon? P1V1 = P2V2 T1 T2 P1V1 T2 = V2 P2T1 (100)(5) = (50)V2 (300) (250) (100)(5)(250) = V2 (300)(50) V2 = 8.3 L Is this answer reasonable? P T V

14. KINETIC MOLECULAR THEORY Applied to Gases (Ideal gases) Ideal (theoretical) Gases: KMT explains gases properties: TheoryProperty Molecules are far apart explains gases can be compressed (or expand: no definite volume or shape) Moving fast and collide with things explains gas pressure Also explains gas laws: Boyles (P vs. V – T constant) increased volume, molecules collide less, pressure lowers Charles (V vs. T – P constant) increased temp, molecules move faster, collide more, volume expands G-L (P vs. T – V constant) increased temp, molecules move faster, collide more, pressure increases

15. Other assumptions Molecules act like they have no volume (because they are far apart relative to their small size) Molecules have no attraction for each other (because they move so fast, the weak forces are not noticeable) Gases which have the above characteristics will follow the gas law: P1V1 = P2V2 T1 T2 Gases act most ideal when the molecules are Far apart (at low pressure) Moving fast (at high temperature) Also, Small molecules like H2 and He are more ideal (Must move faster to make up for their tiny size)

16. DEVIATIONS Gases will Deviate (don’t follow gas laws) when molecules are Moving slowly (low temperature) sticky forces attract molecules; volume dec Close together (high pressure) larger molecules interferes with dec in volume Large molecules like CO2 (mass = 44) move slowly, and have stronger sticky forces Examples: 1) Tire pump gets hot when pressure goes up, from pumping Why? As molecules are pushed together the attraction forces pull molecules together and they speed up (temp increases) 2) Air cools as it rises up (its cold on mountain tops) Why? As air rises, the pressure on it decreases. So it expands (As Pressure decreases, Volume increases) The attractive force slows molecules down as they move away from each other and temperature drops. 3) Gas liquefaction: Gases can be liquefied at very low temperatures and high pressures (Molecules close together, moving slow can stick together and liquefy)

17. PARTIAL PRESSURES (DALTON’S LAW) (Each molecule is free to move on its own and produces its own pressure!) In a mixture of gases, each gas produces its own part of the total pressure Ex: 2 moles of O2 gas are mixed with 1 mole of H2 gas. If the total pressure is 100 kpa, how much pressure does each gas exert? Since 2/3 rds of the molecules are O2 and 1/3 rd are H2 The oxygen will produce 2/3 rds of the pressure 2 x (100 kpa) = 67 kpa 3 and hydrogen 1/3 rd of the pressure: 1 x (100 kpa) = 33 kpa 3

18. H2 O2 AVOGADRO’S HYPOTHESIS Equal volumes of gases, at the same temperature and Pressure Must have equal numbers of molecules Or Two gas samples with equal number of molecules Under same conditions of Temp and Press Will occupy equal volumes 1 liter 250C 1 atmosphere Ex: Two balloons one with hydrogen, the other with oxygen both occupy 1 liter at the same temp and pressure. If the hydrogen balloon contains 0.25 moles of gas, how much oxygen gas is contained in the other balloon?

19. GAS DIFFUSION (Graham’s law) – diffusion: movement of gas as it expands in open space Graham’s law: gas molecules diffuse at a rate which is inversely proportional to the square root of its mass KE = ½ MV2 (v = velocity) kinetic energy of a molecule proportional it mass and speed Small molecules must move faster to make up for lower mass Large molecules can move slower, since they have greater mass

20. Gas Density 22.4 liters = 1 mole = molar mass (gram formula mass) D= M/V or Dimensional analysis What is the density of H2 gas (at STP) in g/L? Density = Mass Volume 0.0893 g/L Density = 2.00 grams = 22.4 liters

21. 63 grams Kr x 1 mL______ 0.00375 grams 0.0893 grams = 1 liter 1) What is the mass of 50 liters of H2 gas at STP? H2 gas density = 0.0893 g/L 0.0893 grams 50 liters H2 x ____________ = 4.5 grams H2 1 liter H2 2) What is the volume of 63 grams of Kr gas at STP? (Reference Table S: density of Kr = 0.00375 g/mL ) = 16800 mL