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Properties and Laws of Gases

This text provides an overview of the physical properties and kinetic molecular theory of gases, including their behavior, temperature, pressure, and volume. It also explores the characteristics and laws governing gas behavior, such as Boyle's Law, Charles' Law, and the Ideal Gas Law. Gas stoichiometry problems are also included to demonstrate the application of these principles.

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Properties and Laws of Gases

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  1. Ch. 11 - Gases I. Physical Properties

  2. Kinetic Molecular Theory • Particles in an ideal gas… • have no volume. • have elastic collisions. • are in constant, random, straight-line motion. • don’t attract or repel each other. • have an avg. KE directly related to Kelvin temperature.

  3. Real Gases • Particles in a REAL gas… • have their own volume • attract each other • Gas behavior is most ideal… • at low pressures • at high temperatures • in nonpolar atoms/molecules

  4. Characteristics of Gases • Gases expand to fill any container. • random motion, no attraction • Gases are fluids (like liquids). • no attraction • Gases have very low densities. • no volume = lots of empty space

  5. Characteristics of Gases • Gases can be compressed. • no volume = lots of empty space • Gases undergo diffusion & effusion. • random motion

  6. Temperature K = ºC + 273 ºF -459 32 212 ºC -273 0 100 K 0 273 373 • Always use absolute temperature (Kelvin) when working with gases.

  7. Pressure Which shoes create the most pressure?

  8. Pressure Aneroid Barometer Mercury Barometer • Barometer • measures atmospheric pressure

  9. Pressure U-tube Manometer Bourdon-tube gauge • Manometer • measures contained gas pressure

  10. Pressure • KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi

  11. STP Standard Temperature & Pressure 0°C273 K 1 atm101.325 kPa -OR- STP

  12. Boyle’s Law P V PV = k

  13. Boyle’s Law P V • The pressure and volume of a gas are inversely related • at constant mass & temp PV = k

  14. Charles’ Law V T

  15. Charles’ Law V T • The volume and absolute temperature (K) of a gas are directly related • at constant mass & pressure

  16. Gay-Lussac’s Law P T

  17. Gay-Lussac’s Law P T • The pressure and absolute temperature (K) of a gas are directly related • at constant mass & volume

  18. Combined Gas Law P1V1 T1 P2V2 T2 = P T V T PV T PV = k P1V1T2 =P2V2T1

  19. Gas Law Problems • A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK: P1V1T2 = P2V2T1 (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3 C. Johannesson

  20. Gas Law Problems • A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa P V WORK: P1V1T2 = P2V2T1 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL C. Johannesson

  21. Gas Law Problems • A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW P T V GIVEN: V1=7.84 cm3 P1=71.8 kPa T1=25°C = 298 K V2=? P2=101.325 kPa T2=273 K WORK: P1V1T2 = P2V2T1 (71.8 kPa)(7.84 cm3)(273 K) =(101.325 kPa)V2 (298 K) V2 = 5.09 cm3 C. Johannesson

  22. Gas Law Problems • A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? P T WORK: P1V1T2 = P2V2T1 (765 torr)T2 = (560. torr)(296K) T2 = 217K = -56°C C. Johannesson

  23. Avogadro’s Principle V n • Equal volumes of gases contain equal numbers of moles • at constant temp & pressure • true for any gas

  24. Ideal Gas Law = k V n Merge the Combined Gas Law with Avogadro’s Principle: PV T PV nT = R UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK C. Johannesson

  25. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 dm3kPa/molK C. Johannesson You don’t need to memorize these values!

  26. Ideal Gas Law Problems • Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm C. Johannesson

  27. Ideal Gas Law Problems WORK: 85 g 1 mol = 2.7 mol 32.00 g • Find the volume of 85 g of O2 at 25°C and 104.5 kPa. GIVEN: V=? n=85 g T=25°C = 298 K P=104.5kPa=1.031 atm R=0.0821 L*atm/molK = 2.7 mol PV = nRT (1.031)V=(2.7) (0.0821) (298) atmmolLatm/molKK V = 64.071L = 64 L (s/fs) C. Johannesson

  28. Gas Stoichiometry • Moles  Liters of a Gas: • STP - use 22.4 L/mol • Non-STP - use ideal gas law • Non-STP • Given liters of gas? • start with ideal gas law • Looking for liters of gas? • start with stoichiometry conv. C. Johannesson

  29. Gas Stoichiometry Problem • What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? CaCO3 CaO + CO2 5.25 g ? Lnon-STP Looking for liters: Start with stoich and calculate moles of CO2. 5.25 g CaCO3 1 mol CaCO3 100.09g CaCO3 1 mol CO2 1 mol CaCO3 = 1.26 mol CO2 Plug this into the Ideal Gas Law to find liters.

  30. Gas Stoichiometry Problem • What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? GIVEN: P=103 kPa V = ? n=1.26 mol T=25°C = 298 K R=8.315dm3kPa/molK WORK: PV = nRT (103 kPa)V=(1mol)(8.315dm3kPa/molK)(298K) V = 1.26 dm3 CO2 C. Johannesson

  31. Gas Stoichiometry Problem • How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al + 3 O2 2 Al2O3 15.0 L non-STP ? g GIVEN: P=97.3 kPa V = 15.0 L n=? T=21°C = 294 K R=8.315dm3kPa/molK WORK: PV = nRT (97.3 kPa) (15.0 L)= n (8.315dm3kPa/molK) (294K) n = 0.597 mol O2 Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT  C. Johannesson

  32. Gas Stoichiometry Problem • How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al + 3 O2 2 Al2O3 15.0L non-STP ? g Use stoich to convert moles of O2 to grams Al2O3. 0.597 mol O2 2 mol Al2O3 3 mol O2 101.96 g Al2O3 1 mol Al2O3 = 40.6 g Al2O3

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