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Bellringer

Explore the concepts of ideal gas behavior, gas properties, and the gas laws that govern their relationships. Understand topics such as pressure, volume, temperature, and the ideal gas law equation.

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Bellringer

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  1. Bellringer • An average human heart beats 60. times per minute. If the average person lives to the age of 78.33 years, how many times does the average heart beat?

  2. I. Physical Properties Ch. 12 - Gases

  3. A. Ideal Gas vs. Real Gas • Kinetic Molecular Theory (KMT) • The theory that explains the behavior of gases at the molecular level. • This theory makes some assumptions about a theoretical gas called an ideal gas.

  4. A. Ideal Gas vs. Real Gas

  5. A. Ideal Gas vs. Real Gas Gas behavior is most ideal… • at low pressures • at high temperatures • in nonpolar atoms/molecules

  6. B. Characteristics of Gases 1. Gases expand to fill any container. 2. Gases are fluids (like liquids). 3. Gases can be compressed. 4. Gases exert pressure. 5. Gases undergo diffusion & effusion.

  7. B. Characteristics of Gases • Diffusion • Spreading of gas molecules throughout a container until evenly distributed. • Effusion • Passing of gas molecules through a tiny opening in a container.

  8. ºC -273 0 100 K 0 273 373 C. Temperature • Always use absolute temperature (Kelvin) when working with gases. Gases have an avg. KE directly related to Kelvin temperature. If temp. goes up, KE goes up.

  9. D. Pressure SI unit for pressure is the Pascal: 1 Pa =Newton(N)/m2 Which shoes create the most pressure?

  10. D. Pressure Vacuum 760 mm Hg • Barometer: Measures atmospheric pressure • The pressure of the atmosphere at sea level will hold a column of mercury 760 mm Hg. • Meridan, ID: pressure = 745 mm Hg 1 atm Pressure

  11. D. Pressure • KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm (atmosphere) 760 mm Hg 760 torr 14.7 psi 29.94 in Hg

  12. D. Pressure • The column of mercury in a barometer is 745 mm high. What is the atmospheric pressure in kPa? 745 mm Hg 101.325 kPa 760 mm Hg = 99.3 kPa

  13. Standard Temperature & Pressure 0°C273 K 1 atm Or 101.325 kPaOr … Or E. STP STP

  14. V T P Describe the behavior of gases in regards to Pressure,Volume, &Temperature The Gas Laws:

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

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

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

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

  19. P2V2 n2T2 P1V1 n1T1 = Combined Gas Law P1V1 =P2V2 What other equations?

  20. G. 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: V1 /T1 = V2 / T2 473 cm3)/(309 K)=V2/(367 K) V2 = 562 cm3

  21. G. Gas Law Problems • A gas occupies 100. mL at 150. kPa. Find its volume at 1.50 x103 mm Hg. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 1.50 x103 mmHg =200. kPa P V WORK: P1 V1 = P2 V2 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL

  22. G. Gas Law Problems COMBINED GAS LAW • A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. 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: P1V1 = P2V2 (71.8 kPa)(7.84 cm3)/(298 K) =(101.325 kPa)V2 /(273 K) V2 = 5.09 cm3 T2 T1

  23. E. Gas Law Problems • On a spring morning, 20°C, you fill your tires to a pressure of 2.25 atm. As you ride along, the tire heats up to 45°C from the friction on the road. What is the pressure in the tires now in units of psi? GAY-LUSSAC’S LAW GIVEN: P1 = 2.25 atm T1 = 20°C = 293K P2 = ? T2 = 45°C = 318K P T WORK: P1/T1 = P2/T2 (2.25 atm)/(293 K)= ( ?)/ (318 K) P2 = 2.44 atm = 35.9 psi

  24. Ptotal = P1 + P2 + ... F. Dalton’s Law • The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor.

  25. F. Dalton’s Law & Water Displacement Ptotal = Pgas + P water vapor Measured at several temps. (see chart)

  26. F. Dalton’s Law • A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 mm Hg. What is the partial pressure of the dry gas? The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor. GIVEN: Pgas = ? Ptotal = 742.0 mm Hg PH2O = 42.18 mm Hg WORK: Ptotal = Pgas + PH2O 742.0 mm = Pgas+42.18 mm Pgas = 699.8 mm Hg Look up water-vapor pressure on the chart for 35.0°C. Sig Figs: Round to least number of decimal places.

  27. II. Ideal Gas Law Ch. 12 - Gases

  28. A. Ideal Gas Law This is where we ended with the Combined Gas Law: P1V1 P2V2 T1n1 T2n2 = Play video!

  29. A. Ideal Gas Law PV nT = R UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 LkPa/molK R=62.4 L.mmHg/mol.K You don’t need to memorize these values, But put them in you handouts!

  30. A. Ideal Gas Law PV=nRT “Piv-nert”

  31. B. 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

  32. WORK: 85 g 1 mol = 2.7 mol 32.00 g B. Ideal Gas Law Problems • 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=104.5 kPa R=8.315 LkPa/molK = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molKK V = 64 L

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