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Entry Task: April 13 th Friday

Entry Task: April 13 th Friday. QUESTION: P1= 30 atm V1= 50 L T1= 293K P2= 25 atm V2= X L T2= 300K Solve for V2. REALLY QUICK- Discuss B,C, G-L #2 ws Discuss Ch. 14.2-3 reading notes HW: Combined and Ideal ws. Agenda.

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Entry Task: April 13 th Friday

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  1. Entry Task: April 13th Friday QUESTION: P1= 30 atm V1= 50 L T1= 293K P2= 25 atmV2= X L T2= 300K Solve for V2

  2. REALLY QUICK- Discuss B,C, G-L #2 ws • Discuss Ch. 14.2-3 reading notes • HW: Combined and Ideal ws Agenda

  3. Use the combined gas law equation with gas problems • P1V1/T1 = P2V2/T2 • Relate the amount of gas to its pressure, temperature and volume by using the ideal gas law • PV = nRT I can…

  4. Combined gas law states the relationship among pressure, volume, and temperature of a fixed amount of gas. Pressure-Volume, Volume-temperature, and Pressure-Temperature are brought together in ONE equation 14.2 The Combined Gas Law and Avogadro’s principle What does the combined gas law state?

  5. 14.2 The Combined Gas Law and Avogadro’s principle • Provide the equation for the combined gas law. P1 V1 = P2 V2 T1 T2

  6. 1. A helium- filled balloon at sea level has a volume of 2.1 L at 0.998 atm and 36°C. If it is released and rises to an elevation at which the pressure is 0.900 atm and the temperature is 28°C, what will be the new volume of the balloon. **Remember to change Celsius to kelvin. P1 = V1= T1= 36 + 273 = 309K 2.1 L 0.998 atm Application of Combined Gas LawFrom the Practice Problems provide the answers. Pg. 429 28 + 273 = 301K X L P2 = V2 = T2 = 0.900 atm

  7. Application of Combined Gas LawFrom the Practice Problems provide the answers. Pg. 429 P1 = V1= T1= 36 + 273 = 309K 2.1 L 0.998 atm 28 + 273 = 301K X L P2 = V2 = T2 = 0.900 atm (0.998 atm) (2.1L) = (0.900 atm) (X L) 301 K 309 K

  8. (0.998 atm) (2.1L) = (0.900 atm) (X L) 301 K 309 K (0.998 atm)(2.1L)(301K) GET X by its self!! = X L (309 K)(0.900 atm)

  9. (0.998)(2.1L)(301) = X L (309 )(0.900) 630.84 L DO the MATH = 2.3 L 278.1

  10. 2. At 0.00°C and 1.00 atm pressure, a sample of gas occupies 30.0 mL. If the temperature is increased to 30.0°C and the entire gas sample of transferred to a 20.0 mL container, what will be the gas pressure inside the container? P1 = V1= T1= 0.00 + 273 = 273K 30.0 mL 1.00 atm Application of Combined Gas LawFrom the Practice Problems provide the answers. Pg. 429 30 + 273 = 303K 20.0 mL P2 = V2 = T2 = X atm

  11. Application of Combined Gas LawFrom the Practice Problems provide the answers. Pg. 429 P1 = V1= T1= 0.00 + 273 = 273K 30.0 mL 1.00 atm 30 + 273 = 303K 20.0 mL P2 = V2 = T2 = X atm (1.00 atm) (30.0L) = (X atm) (20.0 L) 303 K 273 K

  12. (1.00 atm) (30.0L) = (X atm) (20.0 L) 303 K 273 K (1.00 atm)(30.0L)(303K) GET X by its self!! = X atm (273 K)(20.0 L )

  13. (1.00 atm)(30.0)(303) = X atm (273)(20.0) 9090 atm DO the MATH = 1.66 atm 5460

  14. 3. A sample of air in a syringe exerts a pressure of 1.02 atm at a temperature of 22.0°C. The syringe is placed in a boiling water bath at 100°C. The pressure of the air is increased to 1.23 atm by pushing the plunger in, which changes the volume to 0.224 mL. What was the original volume of the air? P1 = V1= T1= 22.0 + 273 = 295K X mL 1.02 atm Application of Combined Gas LawFrom the Practice Problems provide the answers. Pg. 429 0.224 mL P2 = V2 = T2 = 1.23 atm 100 + 273 = 373K

  15. Application of Combined Gas LawFrom the Practice Problems provide the answers. Pg. 429 P1 = V1= T1= 22.0 + 273 = 295K X mL 1.02 atm 0.224 mL P2 = V2 = T2 = 1.23 atm 100 + 273 = 373K (1.02 atm) (XmL) = (1.23atm) (0.224 mL) 373 K 295 K

  16. (1.02 atm) (XmL) = (1.23atm) (0.224 mL) 373 K 295 K (295K)(1.23 atm)(0.224 mL) GET X by its self!! = X mL (1.02 atm)(373 K)

  17. (295)(1.23)(0.224 mL) = X mL (1.02)(373) 81.3 mL DO the MATH = 0.214 mL 380.5

  18. It states that equal volumes of gases at the same temperature and pressure contains equal number of particles 14.2 The Combined Gas Law and Avogadro’s principle State the Avogadro’s principle.

  19. One mole of gas contains 6.02 x1023 molecules at STP 14.2 The Combined Gas Law and Avogadro’s principle Define molar volume.

  20. Standard temperature 0.00°C and pressure 1.00 atm. 14.2 The Combined Gas Law and Avogadro’s principle What is STP?

  21. 22.4 L /1 mole 14.2 The Combined Gas Law and Avogadro’s principle What is the conversion factor between moles and volumes of gas?

  22. 14.2 The Combined Gas Law and Avogadro’s principle From the Practice Problems provide the answers. Pg. 431GO THROUGH THE PRACTICE PROBLEM FIRST!! 2.4 mol 22.4 L = 54 L 1 mol 1. Determine the volume of a container that holds 2.4 mol of gas at STP?

  23. 14.2 The Combined Gas Law and Avogadro’s principle From the Practice Problems provide the answers. Pg. 431GO THROUGH THE PRACTICE PROBLEM FIRST!! 0.0459 mol 22.4 L = 1.028 L 1 mol 2. What size container do you need to hold 0.0459 mol of N2 at STP?

  24. The number of moles is the fourth variable 14.3 The Ideal Gas Law What is the fourth variable?

  25. By changing the number of moles of gas, changes the other variables. 14.3 The Ideal Gas Law How will changing the number of gas molecules in a system change the other variables?

  26. PV=nRT 14.3 The Ideal Gas Law P is pressure V is volume n is the number of moles R is a gas constant T is temperature in Kelvin Provide the equation for the ideal gas law. Explain what each term means.

  27. 0.0821 Latm/mol-K 8.314 LkPa/mol-K 62.4 L mmHg/mol-K 14.3 The Ideal Gas Law What is the “R” value for: atm___________________________ kPa____________________________ mmHg_________________________

  28. An ideal gas is one whose particles take up no space and have no intermolecular forces An ideal gas follows all the gas laws under all pressures and temperatures 14.3 The Ideal Gas Law What does the term “ideal gas” mean?

  29. At high temperatures and pressures, real gases do not follow the gas laws. 14.3 The Ideal Gas Law When is the ideal gas law NOT likely to work for a real gas?

  30. 14.3 The Ideal Gas LawFrom the Practice Problems provide the answers. Pg. 437 25 + 273=298 0.0821 X 0.044L 3.81 atm (X)(0.0821)(298) (3.81 atm)(0.044L) = 4. If the pressure exerted by a gas at 25°C in a volume of 0.044 L is 3.81 atm. How many moles of gas are present? P= V= T= R= n=

  31. (X)(0.0821)(298) (3.81 atm)(0.044L) = X (3.81 atm)(0.044L) = Get X by itself! (0.0821)(298) 0.168 = 0.0069 mols of gas 24.47

  32. 14.3 The Ideal Gas LawFrom the Practice Problems provide the answers. Pg. 437 2.49 mol X 1.00 L 8.314 143 kPa (2.49 mol)(8.314)(X) (143 kPa)(1.00L) = 5. Determine the Celsius temperature of 2.49 moles of gas contained in 1.00-L vessel at a pressure of 143 kPa P= V= T= R= n=

  33. (2.49 mol)(8.314)(X) (143 kPa)(1.00L) = X (143)(1.00) = Get X by itself! (2.49)(8.314) 143 = 6.91K subtract 273 = -266°C 20.70

  34. X L 0.900 atm 0.323 mol 0.0821 265 K 14.3 The Ideal Gas LawFrom the Practice Problems provide the answers. Pg. 437 (0.323 mol)(0.0821)(265) (0.900 atm)(XL) = 6. Calculate the volume that a 0.323-mol sample of a gas will occupy at 265K and a pressure of 0.900 atm. P= V= T= R= n=

  35. (0.323 mol)(0.0821)(265) (0.900 atm)(XL) = (0.323)(0.0821)(265) = X Get X by itself! (0.900) 7.03 = 7.81 L 0.900

  36. 20+273=293K 14.3 The Ideal Gas LawFrom the Practice Problems provide the answers. Pg. 437 X atm 0.505 L 0.0821 0.108 mol (0.108 mol)(0.0821)(293) (X atm)(0.505L) = 7. What is the pressure in atmospheres of a 0.108 molsample of helium gas at a temperature of 20°C if its volume is 0.505 L? P= V= T= R= n=

  37. (0.108 mol)(0.0821)(293) (X atm)(0.505L) = (0.108)(0.0821)(293) = X Get X by itself! (0.505) 2.596 = 5.14 atm 0.505

  38. 14.3 The Ideal Gas LawFrom the Practice Problems provide the answers. Pg. 437 0.988 atm X K 1.20 L 0.0821 0.0470 mol (0.0470 mol)(0.0821)(X) (0.988 atm)(1.20L) = 8. Determine the Kelvin temperature required for 0.0470 mol of gas to fill a balloon to 1.20 L under 0.988 atm pressure. P= V= T= R= n=

  39. (0.0470 mol)(0.0821)(X) (0.988 atm)(1.20L) = (0.988)(1.20) = X Get X by itself! (0.0470)(0.0821) 1.186 = 307K 0.00386

  40. Homework: Combined and Ideal gas law ws

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