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Entry Task: Nov 1 st Block 2

Entry Task: Nov 1 st Block 2. Question: Calculate the number of moles, if pressure is 2 atm , volume is 500.0 ml and temperature is 300K. You have 5 minutes!!. Agenda:. D iscuss Ch. 10 sec 7-10 In Class practice on Grahams Law HW: Grahams Law ws. BREAK OUT AP EQUATION SHEET.

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Entry Task: Nov 1 st Block 2

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  1. Entry Task: Nov 1st Block 2 Question: Calculate the number of moles, if pressure is 2 atm, volume is 500.0 ml and temperature is 300K. You have 5 minutes!!

  2. Agenda: • Discuss Ch. 10 sec 7-10 • In Class practice on Grahams Law • HW: Grahams Law ws

  3. BREAK OUT AP EQUATION SHEET Ideal gas law Van der Waals equation Daltons Partial pressure Moles= molar mass/molarity Kelvin/Celsius Combination gas law These formulas are rarely or not at all on the AP Exam

  4. BREAK OUT AP EQUATION SHEET Density of gas Root mean Speed of gas Kinetic energy of gas molecules and moles of gas Grahams Law Osmotic pressure and Beers Law These formulas are rarely or not at all on the AP Exam

  5. Chapter 10Gases

  6. I can… • Calculate the rate of diffusion of a gas with Grahams Law • Apply the Kinetic-Molecular theory to describe how real gases behave.

  7. Kinetic Molecular Theory • This theory presents physical properties of gases in terms of the motion of individual molecules. • Gas molecules are in continuous rapid random motion • Particle volume is negligible compared to gas volume • Gas molecules experience no attraction or repulsion • Gas collisions are perfectly elastic • Average Kinetic Energy  Kelvin Temperature

  8. Kinetic Molecular Theory • Maxwell speed distribution curves.

  9. Kinetic Molecular Theory

  10. Speed of a Gas Particle • Converting • KE average = ½ mu2 • Particle speed is calculated by getting the root mean speed (urms) is the square root of 3 times R time T divided by its molar mass

  11. 10.13 • Sample: What is rms speed (u) of an N2 molecule at 25°C? Convert g to kg 3*8.314 kg-m2/s2-mol-K * 298K 0.028 kg/mol = 7432.72 0.028 = 515 or 5.15 x 102 m/s

  12. 10.13 problem • Sample: What is rms speed (u) of an He molecule at 25°C? Convert g to kg 3*8.314 kg-m2/s2-mol-K * 298K 0.004 kg/mol = 7432.72 0.004 = 1363 or 1.36 x 103 m/s

  13. Another one • What is the rms speed of O2 at 35°C? Convert g to kg 3*8.314 kg-m2/s2-mol-K * 308K 0.032 kg/mol = 7682.2 0.032 = 489 or 4.9 x 102 m/s

  14. One more • What is the rms speed of CO2 at 20°C? Convert g to kg 3*8.314 kg-m2/s2-mol-K * 293K 0.0439 kg/mol = 7308.0 0.04329 = 408 or 4.1 x 102 m/s

  15. Effusion The escape of gas molecules through a tiny hole into an evacuated space.

  16. Diffusion The spread of one substance throughout a space or throughout a second substance.

  17. Gas Diffusion and Effusion Graham's law governs effusion and diffusion of gas molecules. Rate of effusion is inversely proportional to its molar mass. Thomas Graham (1805 - 1869)

  18. Graham’s Law • The rate of diffusion/effusion is proportional to the mass of the molecules • The rate is inversely proportional to the square root of the molar mass of the gas 80 g F I N I S H 250 g S T A R T Large molecules move slower than small molecules

  19. 10.14 problem Calculate the ratio of the effusion rate of N2 and O2, rN2/rO2 Step 1) Write given information N2 GAS 1 = Nitrogen GAS 2 = Oxygen O2 M1 = 28 g M2 = 32 g r1= x r2= x Step 2) Equation Step 3) Substitute into equation and solve r1 32.0 g 1.07 = r2 28 g 1 N2diffuses 1.07 times faster than O2

  20. 2 17 He Cl 4.0026 35.453 Find the relative rate of diffusion of helium and chlorine gas Step 1) Write given information GAS 1 = helium GAS 2 = chlorine He Cl2 M1 = 4.0 g M2 = 71.0 g r1= x r2= x Step 2) Equation Step 3) Substitute into equation and solve r1 71.0 g 4.21 = r2 4.0 g 1 He diffuses 4.21 times faster than Cl2

  21. 9 10 F Ne 18.9984 20.1797 If fluorine gas diffuses at a rate of 363 m/s at a certain temperature, what is the rate of diffusion of neon gas at the same temperature? Step 1) Write given information GAS 1 = Neon GAS 2 = Fluorine Ne F2 M1 = 20.18 g M2 = 38.0 g v1= x v2 = 363 m/s Step 2) Equation Step 3) Substitute into equation and solve v1 38.0 g 498 m/s = 363 m/s 20.18 g Rate of diffusion of Ne = 498 m/s

  22. 1 8 H O 1.00794 15.9994 Graham’s Law A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? Put the gas with the unknown speed as “Gas 1”.

  23. 8 O 15.9994 1 Graham’s Law H H2 = 2 g/mol 1.0 An unknown gas diffuses 4.0 times faster than O2. Find its molar mass. The first gas is “Gas 1” and the second gas is “Gas 2”. The ratio “v1/v2” is 4.0. Square both sides to get rid of the square root sign.

  24. Behavior of Real Gases • Deviations result from assumptions about ideal gases. • Molecules in gaseous state do not exert any force, either attractive or repulsive, on one another. • Volume of the molecules is negligibly small compared with that of the container.

  25. Real Gases In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

  26. Behavior of Real Gases • At higher pressures, particles are much closer together and attractive forces become more important than at lower pressures.

  27. Behavior of Real Gases • The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value.

  28. Deviations from Ideal Behavior The assumptions made in the kinetic-molecular model break down at high pressure and/or low temperature.

  29. Corrections for Nonideal Behavior • The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account. • The corrected ideal-gas equation is known as the van der Waals equation.

  30. (P + ) (V−nb) = nRT n2a V2 The van der Waals Equation

  31. Real Gases: Deviations from Ideal Behavior • The van der Waals Equation • General form of the van der Waals equation: Corrects for molecular volume Corrects for molecular attraction

  32. Using Van der Waal’s equation If 1.000 mol of an ideal gas were confined to 22.41 L at 0.0°C, it would exert a pressure of 1.000 atm. Use the van der Waals equation and the constants in Table 10.3 to estimate the pressure exerted by 1.000 mol of Cl2 (g) in 22.41 L at 0.0°C? n= 1.000 mol, R = 0.0821 L-atm/mol-K, T=273 K , V= 22.41 L and a=6.49L2atm/mol2, and b = 0.0562 L/mol (1.000 mol)(0.0821L-atm/mol-K)(273) = 22.41 22.35 = 1.003 (correction for atm) 22.41 L - (1.000 mol)(0.0562 L/mol) - 1.003 - 0.0013 = 0.990 atm (1.000)2(6.49L2atm/mol) (22.41L)2 = 0.0013 Correction for molecular attractions

  33. What is the rms speed of CO2 at 35°C? Convert g to kg 3*8.314 kg-m2/s2-mol-K * 308K 0.044 kg/mol = 7682.136 0.044 = 418 or 4.18 x 102 m/s

  34. What is the rms speed of Cl2at 40°C? Convert g to kg 3*8.314 kg-m2/s2-mol-K * 313K 0.0709 kg/mol = 7806.846 0.0709 = 333 or 3.33 x 102 m/s

  35. What is the rms speed of CH4 at 25°C? Convert g to kg 3*8.314 kg-m2/s2-mol-K * 298K 0.016 kg/mol = 7432.716 0.016 = 682 or 6.82 x 102 m/s

  36. Find the relative rate of diffusion of Kr and Ar? Step 1) Write given information GAS 1 = Ar GAS 2 = Kr M1 = 39.9 g M2 = 83.8 g r1= x r2= x Step 2) Equation Step 3) Substitute into equation and solve r1 83.8 g 1.45 = r2 39.9 g 1 Ardiffuses 1.45 times faster than Kr

  37. Graham’s Law If fluorine gas diffuses at a rate of 363 m/s at a certain temperature, what is the rate of diffusion of chlorine gas at the same temperature? Put the gas with the unknown speed as “Gas 1”.

  38. Graham’s Law • A molecule of oxygen gas has an average speed of 242 m/s at a given temp and pressure. What is the average speed of neon molecules at the same conditions? Put the gas with the unknown speed as “Gas 1”.

  39. Graham’s Law An unknown gas diffuses 2.0 times faster than F2. Find its molar mass. The first gas is “Gas 1” and the second gas is “Gas 2”. The ratio “v1/v2” is 2.0. Square both sides to get rid of the square root sign.

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