1 / 34

Entry Task: Oct 26 th Friday

Entry Task: Oct 26 th Friday. Question: You can just write the variables. If the pressure in the balloon is 1 atm at 23C and it was placed in the oven with a temperature of 85C. What is the final pressure?. Agenda:. Discuss B, C, and G-L worksheet In-class worksheet on B, C, and G-L

sezja
Download Presentation

Entry Task: Oct 26 th Friday

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Entry Task: Oct 26th Friday Question: You can just write the variables. If the pressure in the balloon is 1 atm at 23C and it was placed in the oven with a temperature of 85C. What is the final pressure?

  2. Agenda: • Discuss B, C, and G-L worksheet • In-class worksheet on B, C, and G-L • Homework: Ch. 10 sec. 4-6 reading notes

  3. BOYLES CHARLES & GAY-LUSSACWorksheet

  4. Provide Boyles Law formula. P1V1 = P2 V2

  5. 1.  If some neon gas at 75.0 kPa were allowed to shrink from 6.0 dm3 to 3.0 dm3 without changing the temperature,  what pressure would the neon gas exert under these new conditions? (75.0 kPa) (X) (3.0 dm3) (6.0 dm3) = 450 kPa = (X) 3.0 150 kPa

  6. 2.  A quantity of gas under a pressure of 2.25 atm has a volume of 345 cm3.   The pressure is increased to 3.10 atm, while the pressure remains constant.   What is the new volume? (2.25 atm) (3.10 atm) (X) (345 cm3) = 776.25 cm3 = (X) 3.10 250 cm3

  7. 3.  A 25.0 L sample of gas exerts a pressure of 135 kPa. What pressure will the gas exert if its volume is reduced to 15.3 L?(constant temperature) (135 kPa) (X) (15.3L) (25.0 L) = 3375 kPa = (X) 15.3 221 kPa

  8. 4. A container that has 3.00 L of a gas is at 2.50 atm. What pressure is obtained when the volume is 7.5 L? (2.50 atm) (X) (7.5L) (3.00L) = 7.5 atm = (X) 7.5 1.0 atm

  9. Provide Charles Law formula. V1 V2T1 T2 =

  10. 1.  The temperature inside my refrigerator is about 60 Celsius. If I place a balloon in my fridge that initially has a temperature of 300 C and a volume of 1.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator? X L T1 =30 + 273 = 303K V1 = 1.5 L T2 = 6 + 273 = 279 K V2 = X L 1.5 L = 303K 279K 418.5 = (X) (303) 418.5= X 303 1.38 L OR 1 L

  11. 2. A man heats a balloon in the oven. If the balloon initially has a volume of 0.25 liters and a temperature of 30 0C, what will the volume of the balloon be after he heats it to a temperature of 325 0C? T1 =30 + 273 = 303K V1 = 0.25 L T2 = 325 + 273 = 598K V2 = X L 303K 598K = 0.25 L X L 149.5 = (X) (303) 149.5= X 303 0.49 L

  12. 3. On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250 mL bag at a temperature of 23 0C, and I leave it in my car which has a temperature of 400 C, what will the new volume of the bag be? T1 =23 + 273 = 296K V1 = 250 mL T2 = 40 + 273 = 313 K V2 = X L 296K 313K = 250 mL X L 78250 = (X) (296) 78250 = X 296 264 mL OR 260 ml

  13. 4. A soda bottle is flexible enough that the volume of the bottle can change even without opening it. If you have an empty soda bottle (volume of 2 L) at of a temperature 17 0C, what will the new volume be if you put it in your freezer (-4 0C)? T1 =17 + 273 = 290K V1 = 2 L T2 = -4 + 273 = 269 K V2 = X L 290K 269K = 2 L X L 538 = (X) (290) 538= X 290 1.85 L OR 2L

  14. Provide Gay-Lussac’s Law formula. P1 P2T1 T2 =

  15. 1. The pressure inside a container is 625 mmHg at a temperature of 47oC. What would the pressure be at 70oC? P1 = 625 mmHg T1 = 47 + 273 = 320K P2 = X T2 = 70 + 273 = 343K 625 mmHg X mmHg = 320 K 343 K 214375 = (X) (320) 214375 = X 320 669.9 or 670 mmHg

  16. 2. A rigid container is at a temperature of 12oC. When heated to 125oC, the pressure was 360 kPa. What was the initial pressure? P1 = X kPa T1 = 12 + 273 = 285K P2 = 360 kPa T2 = 125 + 273 = 398K X kPa 360 kPa = 285 K 398 K 102600 = (X) (497) 102600= X 398 258 kPa OR 260 kPa

  17. 3. If a gas is cooled from 225 K to 125 K and the volume is kept constant what final pressure would result if the original pressure was 630.0 mm Hg? P1 = 630 mmHg T1 = 225 K P2 = X mmHg T2 = 125 K 630 mmHg X mmHg = 225 K 125 K 78750 = (X) (225) 78750= X 225 350 mmHg

  18. 4. A gas has a pressure of 0.135 atm at 45.0 °C. What is the pressure at -12˚C ? P1 = 0.135 atm T1 = 45 + 273= 318 K P2 = X atm T2 = -12 + 273= 261 K 0.135 atm X atm = 318 K 261 K 35.24 = (X) (318) 35.24= X 318 0.11 atm

  19. BOYLES CHARLES & GAY-LUSSACSelf Check

  20. Provide Boyles Law formula. P1V1 = P2 V2

  21. 1.  If some neon gas at 30.0 kPa were allowed to expand from 7.7 dm3 to 12.0 dm3 without changing the temperature,  what pressure would the neon gas exert under these new conditions? (30.0 kPa) (X) (12.0 dm3) (7.7 dm3) = 231 kPa = (X) 12.0 19.3 kPa

  22. 2.  A quantity of gas under a pressure of 3.2 atm has a volume of 650 cm3.   The pressure is increased to 4.3 atm, while the pressure remains constant.   What is the new volume? (3.20 atm) (4.30 atm) (X) (650 cm3) = 2080 cm3 = (X) 4.30 480 cm3

  23. 3.  A 5.0 L sample of gas exerts a pressure of 175 kPa. What pressure will the gas exert if its volume is reduced to 2.5 L?(constant temperature) (175 kPa) (X) (2.5 L) (5.0 L) = 875 kPa = (X) 2.5 350 kPa

  24. 4.  10.0 L of a gas is at 0.85 atm. What pressure is obtained when the volume is 5.0 L? (0.85 atm) (X) (5.0 L) (10.0L) = 8.5 atm = (X) 5.0 1.7 atm

  25. Provide Charles Law formula. V1 V2T1 T2 =

  26. 1.  The temperature inside my refrigerator is about 90 Celsius. If I place a balloon in my fridge that initially has a temperature of 220 C and a volume of 0.3 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator? X L T1 =22 + 273 = 295K V1 = 0.3 L T2 = 9 + 273 = 282 K V2 = X L 0.3 L = 295K 282K 84.6 = (X) (295) 84.6= X 295 0.286 L or 0.3 L

  27. 2. A man heats a balloon in the oven. If the balloon initially has a volume of 0.70 liters and a temperature of 27 0C, what will the volume of the balloon be after he heats it to a temperature of 35 0C? T1 =27 + 273 = 300K V1 = 0.7 L T2 = 35 + 273 = 308K V2 = X L 300K 308K = 0.7 L X L 215.6 = (X) (300) 215.6= X 300 0.76 L

  28. 3. On hot days, you may have noticed that potato chip bags seem to “inflate”, even though they have not been opened. If I have a 250.0 mL bag at a temperature of 22.0 0C, and I leave it in my car which has a temperature of 50.00 C, what will the new volume of the bag be? T1 =22 + 273 = 295K V1 = 250 mL T2 = 50 + 273 = 323 K V2 = X L 295K 323K = 250 mL X L 80750 = (X) (295) 80750= X 295 274 mL

  29. 4. A soda bottle is flexible enough that the volume of the bottle can change even without opening it. If you have an empty soda bottle (volume of 2 L) at room temperature (25 0C), what will the new volume be if you put it in your freezer (-4 0C)? T1 =25 + 273 = 298K V1 = 2 L T2 = -4 + 273 = 269 K V2 = X L 298K 269K = 2 L X L 538 = (X) (298) 538= X 298 1.81 L OR 2 L

  30. Provide Gay-Lussac’s Law formula. P1 P2T1 T2 =

  31. 1. The pressure inside a container is 650 mmHg at a temperature of 75oC. What would the pressure be at 55oC? P1 = 650 mmHg T1 = 75 + 273 = 348K P2 = X T2 = 55 + 273 = 328K 650 mmHg X mmHg = 348 K 328 K 213200 = (X) (348) 267960= X 348 613 OR 610 mmHg

  32. 2. A rigid container is at a temperature of 3.0oC. When heated to 75oC, the pressure was 2.5 kPa. What was the initial pressure? P1 = X kPa T1 = 3.0 + 273 = 276K P2 = 2.5 kPa T2 = 75 + 273 = 348K X kPa 2.5 kPa = 276 K 348 K 690 = (X) (348) 690 = X 348 1.98 or 2.0 kPa

  33. 3. If a gas is cooled from 303.0 K to 200.5 K and the volume is kept constant what final pressure would result if the original pressure was 550.0 mm Hg? P1 = 550 mmHg T1 = 303.0 K P2 = X mmHg T2 = 200.5 K 550 mmHg X mmHg = 303.0 K 200.5 K 110275 = (X) (303) 110275= X 303 363.9 mmHg

  34. 4. A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at 22˚C ? P1 = 0.370 atm T1 = 50 + 273= 323 K P2 = X atm T2 = 22 + 273= 295 K 0.370 atm X atm = 323 K 295 K 109.15 = (X) (323) 109.15 = X 323 0.338 atm OR 0.34 atm

More Related