Slide 1

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? Slide 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. Slide 3 BOYLES CHARLES & GAY-LUSSAC Worksheet Slide 4 Provide Boyles Law formula. P 1 V 1 = P 2 V 2 Slide 5 1. If some neon gas at 121 kPa were allowed to expand from 3.7 dm 3 to 6.0 dm 3 without changing the temperature, what pressure would the neon gas exert under these new conditions? (121 kPa) (3.7 dm 3 ) = (X) (6.0 dm 3 ) 4477 kPa = (X) 6.0 75 kPa Slide 6 2. A quantity of gas under a pressure of 1.78 atm has a volume of 550 cm 3. The pressure is increased to 2.50 atm, while the pressure remains constant. What is the new volume? (1.78 atm) (550 cm 3 ) = (2.50 atm) (X) 979 cm 3 = (X) 2.50 392 cm 3 Slide 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) (14.2 L) = (X) (8. 53 L) 1775 kPa = (X) 8.53 208 kPa Slide 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)(5.00L) = (X) (10.0 L) 5.4 atm = (X) 10.0 0.540 atm Slide 9 Provide Charles Law formula. V 1 V 2 T 1 T 2 = Slide 10 1. The temperature inside my refrigerator is about 4 0 Celsius. If I place a balloon in my fridge that initially has a temperature of 22 0 C and a volume of 0.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator? T 1 =22 + 273 = 295K V 1 = 0.5 L T 2 = 4 + 273 = 277 K V 2 = X L 405.44 = (X) (295) 295K 0.47 L or 0.5 L 277K = 405.44= X 295 0.5 L X L Slide 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 0 C, what will the volume of the balloon be after he heats it to a temperature of 250 0 C? T 1 =20 + 273 = 293K V 1 = 0.4 L T 2 = 250 + 273 = 523K V 2 = X L 209.2 = (X) (293) 293K 0.4 L 0.7 L 523K X L = 209.2= X 293 Slide 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 0 C, and I leave it in my car which has a temperature of 60 0 C, what will the new volume of the bag be? T 1 =19 + 273 = 292K V 1 = 250 mL T 2 = 60 + 273 = 333 K V 2 = X L 83250 = (X) (292) 292K 250 mL 285 mL 333K X L = 83250= X 292 Slide 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 0 C), what will the new volume be if you put it in your freezer (-4 0 C)? T 1 =25 + 273 = 298K V 1 = 2 L T 2 = -4 + 273 = 269 K V 2 = X L 538 = (X) (298) 298K 2 L 1.81 L OR 2 L 269K X L = 538= X 298 Slide 14 Provide Gay-Lussacs Law formula. P 1 P 2 T 1 T 2 = Slide 15 1. The pressure inside a container is 770 mmHg at a temperature of 57 o C. What would the pressure be at 75 o C? P 1 = 770 mmHg T 1 = 57 + 273 = 330K P 2 = X T 2 = 75 + 273 = 348K 267960 = (X) (330) 770 mmHg 330 K 812 OR 810 mmHg X mmHg 348 K = 267960= X 330 Slide 16 2. A rigid container is at a temperature of 112 o C. When heated to 224 o C, the pressure was 288 kPa. What was the initial pressure? P 1 = X kPa T 1 = 112 + 273 = 385K P 2 = 288 kPa T 2 = 224 + 273 = 497K 110880 = (X) (497) X kPa 385 K 223 kPa 288 kPa 497 K = 110880 = X 497 Slide 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? P 1 = 750 mmHg T 1 = 323 K P 2 = X mmHg T 2 = 273.15 K 204862.5 = (X) (323) 750 mmHg 323 K 634.3 mmHg X mmHg 273.15 K = 204862.5 = X 323 Slide 18 4. A gas has a pressure of 0.370 atm at 50.0 C. What is the pressure at 22C ? P 1 = 0.370 atm T 1 = 50 + 273= 323 K P 2 = X atm T 2 = 22 + 273= 295 K 109.15 = (X) (323) 0.370 atm 323 K 0.338 atm OR 0.34 atm X atm 295 K = 109.15 = X 323 Slide 19 BOYLES CHARLES & GAY-LUSSAC Worksheet #2 Slide 20 Provide Boyles Law formula. P 1 V 1 = P 2 V 2 Slide 21 1. If some neon gas at 75.0 kPa were allowed to shrink from 6.0 dm 3 to 3.0 dm 3 without changing the temperature, what pressure would the neon gas exert under these new conditions? (75.0 kPa) (6.0 dm 3 ) = (X) (3.0 dm 3 ) 450 kPa = (X) 3.0 150 kPa Slide 22 2. A quantity of gas under a pressure of 2.25 atm has a volume of 345 cm 3. The pressure is increased to 3.10 atm, while the pressure remains constant. What is the new volume? (2.25 atm) (345 cm 3 ) = (3.10 atm) (X) 776.25 cm 3 = (X) 3.10 250 cm 3 Slide 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) (25.0 L) = (X) (15.3L) 3375 kPa = (X) 15.3 221 kPa Slide 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)(3.00L) = (X) (7.5L) 7.5 atm = (X) 7.5 1.0 atm Slide 25 Provide Charles Law formula. V 1 V 2 T 1 T 2 = Slide 26 1. The temperature inside my refrigerator is about 6 0 Celsius. If I place a balloon in my fridge that initially has a temperature of 30 0 C and a volume of 1.5 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator? T 1 =30 + 273 = 303K V 1 = 1.5 L T 2 = 6 + 273 = 279 K V 2 = X L 418.5 = (X) (303) 303K 1.38 L OR 1 L 279K = 418.5= X 303 1.5 L X L Slide 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 0 C, what will the volume of the balloon be after he heats it to a temperature of 325 0 C? T 1 =30 + 273 = 303K V 1 = 0.25 L T 2 = 325 + 273 = 598K V 2 = X L 149.5 = (X) (303) 303K 0.25 L 0.49 L 598K X L = 149.5= X 303 Slide 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 0 C, and I leave it in my car which has a temperature of 40 0 C, what will the new volume of the bag be? T 1 =23 + 273 = 296K V 1 = 250 mL T 2 = 40 + 273 = 313 K V 2 = X L 78250 = (X) (296) 296K 250 mL 264 mL OR 260 ml 313K X L = 78250 = X 296 Slide 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 0 C, what will the new volume be if you put it in your freezer (-4 0 C)? T 1 =17 + 273 = 290K V 1 = 2 L T 2 = -4 + 273 = 269 K V 2 = X L 538 = (X) (290) 290K 2 L 1.85 L OR 2L 269K X L = 538= X 290 Slide 30 Provide Gay-Lussacs Law formula. P 1 P 2 T 1 T 2 = Slide 31 1. The pressure inside a container is 625 mmHg at a temperature of 47 o C. What would the pressure be at 70 o C? P 1 = 625 mmHg T 1 = 47 + 273 = 320K P 2 = X T 2 = 70 + 273 = 343K 214375 = (X) (320) 625 mmHg 320 K 669.9 or 670 mmHg X mmHg 343 K = 214375 = X 320 Slide 32 2. A rigid container is at a temperature of 12 o C. When heated to 125 o C, the pressure was 360 kPa. What was the initial pressure? P 1 = X kPa T 1 = 12 + 273 = 285K P 2 = 360 kPa T 2 = 125 + 273 = 398K 102600 = (X) (497) X kPa 285 K 258 kPa OR 260 kPa 360 kPa 398 K = 102600 = X 398 Slide 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? P 1 = 630 mmHg T 1 = 225 K P 2 = X mmHg T 2 = 125 K 78750 = (X) (225) 630 mmHg 225 K 350 mmHg X mmHg 125 K = 78750 = X 225 Slide 34 4. A gas has a pressure of 0.135 atm at 45.0 C. What is the pressure at -12C ? P 1 = 0.135 atm T 1 = 45 + 273= 318 K P 2 = X atm T 2 = -12 + 273= 261 K 35.24 = (X) (318) 0.135 atm 318 K 0.11 atm X atm 261 K = 35.24 = X 318