Entry Task: April 11 th -12 th Block #2

1. Entry Task: April 11th -12th Block #2 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. 14 sec. 2 &3 reading/math.

3. BOYLES CHARLES & GAY-LUSSACWorksheet

4. Provide Boyles Law formula. P1V1 = P2 V2

5. 1.  If some neon gas at 121 kPa were allowed to expand from 3.7 dm3 to 6.0 dm3 without changing the temperature,  what pressure would the neon gas exert under these new conditions? (121 kPa) (X) (6.0 dm3) (3.7 dm3) = 4477 kPa = (X) 6.0 75 kPa

6. 2.  A quantity of gas under a pressure of 1.78 atm has a volume of 550 cm3.   The pressure is increased to 2.50 atm, while the pressure remains constant.   What is the new volume? (1.78 atm) (2.50 atm) (X) (550 cm3) = 979 cm3 = (X) 2.50 392 cm3

7. 3.  A 14.2 L sample of gas exerts a pressure of 125 kPa. What pressure will the gas exert if its volume is reduced to 8.53 L?(constant temperature) (125 kPa) (X) (8. 53 L) (14.2 L) = 1775 kPa = (X) 8.53 208 kPa

8. 4.  5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is 10.0 L? (1.08 atm) (X) (10.0 L) (5.00L) = 5.4 atm = (X) 10.0 0.540 atm

9. Provide Charles Law formula. V1 V2T1 T2 =

10. 1.  The temperature inside my refrigerator is about 40 Celsius. If I place a balloon in my fridge that initially has a temperature of 220 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator? X L T1 =22 + 273 = 295K V1 = 0.5 L T2 = 4 + 273 = 277 K V2 = X L 0.5 L = 295K 277K 405.44 = (X) (295) 405.44= X 295 0.47 L or 0.5 L

11. 2. A man heats a balloon in the oven. If the balloon initially has a volume of 0.4 liters and a temperature of 20 0C, what will the volume of the balloon be after he heats it to a temperature of 250 0C? T1 =20 + 273 = 293K V1 = 0.4 L T2 = 250 + 273 = 523K V2 = X L 293K 523K = 0.4 L X L 209.2 = (X) (293) 209.2= X 293 0.7 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 19 0C, and I leave it in my car which has a temperature of 600 C, what will the new volume of the bag be? T1 =19 + 273 = 292K V1 = 250 mL T2 = 60 + 273 = 333 K V2 = X L 292K 333K = 250 mL X L 83250 = (X) (292) 83250= X 292 285 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 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

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

15. 1. The pressure inside a container is 770 mmHg at a temperature of 57oC. What would the pressure be at 75oC? P1 = 770 mmHg T1 = 57 + 273 = 330K P2 = X T2 = 75 + 273 = 348K 770 mmHg X mmHg = 330 K 348 K 267960 = (X) (330) 267960= X 330 812 OR 810 mmHg

16. 2. A rigid container is at a temperature of 112oC. When heated to 224oC, the pressure was 288 kPa. What was the initial pressure? P1 = X kPa T1 = 112 + 273 = 385K P2 = 288 kPa T2 = 224 + 273 = 497K X kPa 288 kPa = 385 K 497 K 110880 = (X) (497) 110880 = X 497 223 kPa

17. 3. If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what final pressure would result if the original pressure was 750.0 mm Hg? P1 = 750 mmHg T1 = 323 K P2 = X mmHg T2 = 273.15 K 750 mmHg X mmHg = 323 K 273.15 K 204862.5 = (X) (323) 204862.5 = X 323 634.3 mmHg

18. 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

19. BOYLES CHARLES & GAY-LUSSACWorksheet #2

20. Provide Boyles Law formula. P1V1 = P2 V2

21. 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

22. 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

23. 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

24. 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

25. Provide Charles Law formula. V1 V2T1 T2 =

26. 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

27. 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

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 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

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 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

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

31. 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

32. 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 kPaOR 260 kPa

33. 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

34. 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