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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?

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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

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BOYLES CHARLES & GAY-LUSSAC Worksheet

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Provide Boyles Law formula. P 1 V 1 = P 2 V 2

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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

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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

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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

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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

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Provide Charles Law formula. V 1 V 2 T 1 T 2 =

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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

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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

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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

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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

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Provide Gay-Lussac’s Law formula. P 1 P 2 T 1 T 2 =

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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

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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

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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

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4. A gas has a pressure of 0.135 atm at 45.0 °C. What is the pressure at -12˚C ? 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

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BOYLES CHARLES & GAY-LUSSAC Self Check

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Provide Boyles Law formula. P 1 V 1 = P 2 V 2

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1. If some neon gas at 30.0 kPa were allowed to expand from 7.7 dm 3 to 12.0 dm 3 without changing the temperature, what pressure would the neon gas exert under these new conditions? (30.0 kPa) (7.7 dm 3 ) = (X) (12.0 dm 3 ) 231 kPa = (X) 12.0 19.3 kPa

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2. A quantity of gas under a pressure of 3.2 atm has a volume of 650 cm 3. The pressure is increased to 4.3 atm, while the pressure remains constant. What is the new volume? (3.20 atm) (650 cm 3 ) = (4.30 atm) (X) 2080 cm 3 = (X) 4.30 480 cm 3

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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) (5.0 L) = (X) (2.5 L) 875 kPa = (X) 2.5 350 kPa

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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)(10.0L) = (X) (5.0 L) 8.5 atm = (X) 5.0 1.7 atm

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Provide Charles Law formula. V 1 V 2 T 1 T 2 =

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1. The temperature inside my refrigerator is about 9 0 Celsius. If I place a balloon in my fridge that initially has a temperature of 22 0 C and a volume of 0.3 liters, what will be the volume of the balloon when it is fully cooled by my refrigerator? T 1 =22 + 273 = 295K V 1 = 0.3 L T 2 = 9 + 273 = 282 K V 2 = X L 84.6 = (X) (295) 295K 0.286 L or 0.3 L 282K = 84.6= X 295 0.3 L X L

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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 0 C, what will the volume of the balloon be after he heats it to a temperature of 35 0 C? T 1 =27 + 273 = 300K V 1 = 0.7 L T 2 = 35 + 273 = 308K V 2 = X L 215.6 = (X) (300) 300K 0.7 L 0.76 L 308K X L = 215.6= X 300

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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 0 C, and I leave it in my car which has a temperature of 50.0 0 C, what will the new volume of the bag be? T 1 =22 + 273 = 295K V 1 = 250 mL T 2 = 50 + 273 = 323 K V 2 = X L 80750 = (X) (295) 295K 250 mL 274 mL 323K X L = 80750= X 295

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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

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Provide Gay-Lussac’s Law formula. P 1 P 2 T 1 T 2 =

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1. The pressure inside a container is 650 mmHg at a temperature of 75 o C. What would the pressure be at 55 o C? P 1 = 650 mmHg T 1 = 75 + 273 = 348K P 2 = X T 2 = 55 + 273 = 328K 213200 = (X) (348) 650 mmHg 348 K 613 OR 610 mmHg X mmHg 328 K = 267960= X 348

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2. A rigid container is at a temperature of 3.0 o C. When heated to 75 o C, the pressure was 2.5 kPa. What was the initial pressure? P 1 = X kPa T 1 = 3.0 + 273 = 276K P 2 = 2.5 kPa T 2 = 75 + 273 = 348K 690 = (X) (348) X kPa 276 K 1.98 or 2.0 kPa 2.5 kPa 348 K = 690 = X 348

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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? P 1 = 550 mmHg T 1 = 303.0 K P 2 = X mmHg T 2 = 200.5 K 110275 = (X) (303) 550 mmHg 303.0 K 363.9 mmHg X mmHg 200.5 K = 110275 = X 303

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4. A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at 22˚C ? 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

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Ch. 12 & 13 - Gases II. The Gas Laws P V T.

Ch. 12 & 13 - Gases II. The Gas Laws P V T.

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