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**Entry Task: March 22 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?

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**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.

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**BOYLES CHARLES & GAY-LUSSAC Worksheet**

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**Provide Boyles Law formula.**

P1V1 = P2 V2

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**75 kPa (121 kPa) 4477 kPa = (X) (3.7 dm3) = (X) (6.0 dm3) 6.0**

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? (3.7 dm3) = (X) (6.0 dm3) (121 kPa) 4477 kPa = (X) 6.0 75 kPa

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**392 cm3 979 cm3 = (X) (550 cm3) = (2.50 atm) (X) (1.78 atm) 2.50**

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? (550 cm3) = (2.50 atm) (X) (1.78 atm) 979 cm = (X) 2.50 392 cm3

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**208 kPa (125 kPa) 1775 kPa = (X) (14.2 L) = (X) (8. 53 L) 8.53**

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) (14.2 L) = (X) (8. 53 L) (125 kPa) 1775 kPa = (X) 8.53 208 kPa

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**0.540 atm 5.4 atm = (X) 10.0 (1.08 atm) (5.00L) = (X) (10.0 L)**

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

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**Provide Charles Law formula.**

V1 V2 T T2 =

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**1. The temperature inside my refrigerator is about 40 Celsius**

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? T1 = = 295K V1 = 0.5 L T2 = = 277 K V2 = X L X L 0.5 L = 295K 277K = (X) (295) 405.44= X 295 0.47 L or 0.5 L

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**2. A man heats a balloon in the oven**

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 = = 293K V1 = 0.4 L T2 = = 523K V2 = X L 293K 523K = 0.4 L X L = (X) (293) 209.2= X 293 0.7 L

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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 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 = = 292K V1 = 250 mL T2 = = 333 K V2 = X L 292K 333K = 250 mL X L = (X) (292) 83250= X 292 285 mL

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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 0C), what will the new volume be if you put it in your freezer (-4 0C)? T1 = = 298K V1 = 2 L T2 = = 269 K V2 = X L 298K 269K = 2 L X L 538 = (X) (298) 538= X 298 1.81 L OR 2 L

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**Provide Gay-Lussac’s Law formula.**

P1 P2 T T2 =

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**812 OR 810 mmHg = P1 = 770 mmHg T1 = 57 + 273 = 330K P2 = X**

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 = = 330K P2 = X T2 = = 348K 770 mmHg X mmHg = 330 K 348 K = (X) (330) 267960= X 330 812 OR 810 mmHg

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**223 kPa = P1 = X kPa T1 = 112 + 273 = 385K P2 = 288 kPa**

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 = = 385K P2 = 288 kPa T2 = = 497K X kPa 288 kPa = 385 K 497 K = (X) (497) = X 497 223 kPa

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**634.3 mmHg = P1 = 750 mmHg T1 = 323 K P2 = X mmHg T2 = 273.15 K 323 K**

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

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**0.338 atm OR 0.34 atm = P1 = 0.370 atm T1 = 50 + 273= 323 K P2 = X atm**

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

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**BOYLES CHARLES & GAY-LUSSAC Worksheet #2**

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**Provide Boyles Law formula.**

P1V1 = P2 V2

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**150 kPa 450 kPa = (X) (6.0 dm3) = (X) (3.0 dm3) (75.0 kPa) 3.0**

1. If some neon gas at 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? (6.0 dm3) = (X) (3.0 dm3) (75.0 kPa) 450 kPa = (X) 3.0 150 kPa

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**250 cm3 776.25 cm3 = (X) (345 cm3) = (3.10 atm) (X) (2.25 atm) 3.10**

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? (345 cm3) = (3.10 atm) (X) (2.25 atm) cm = (X) 3.10 250 cm3

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**221 kPa (135 kPa) 3375 kPa = (X) (25.0 L) = (X) (15.3L) 15.3**

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) (25.0 L) = (X) (15.3L) (135 kPa) 3375 kPa = (X) 15.3 221 kPa

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**1.0 atm 7.5 atm = (X) 7.5 (2.50 atm) (3.00L) = (X) (7.5L)**

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? (X) (7.5L) (2.50 atm) (3.00L) = 7.5 atm = (X) 7.5 1.0 atm

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**Provide Charles Law formula.**

V1 V2 T T2 =

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**1. The temperature inside my refrigerator is about 60 Celsius**

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? T1 = = 303K V1 = 1.5 L T2 = = 279 K V2 = X L X L 1.5 L = 303K 279K = (X) (303) 418.5= X 303 1.38 L OR 1 L

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**2. A man heats a balloon in the oven**

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 = = 303K V1 = 0.25 L T2 = = 598K V2 = X L 303K 598K = 0.25 L X L = (X) (303) 149.5= X 303 0.49 L

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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 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 = = 296K V1 = 250 mL T2 = = 313 K V2 = X L 296K 313K = 250 mL X L = (X) (296) 78250 = X 296 264 mL OR 260 ml

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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 0C, what will the new volume be if you put it in your freezer (-4 0C)? T1 = = 290K V1 = 2 L T2 = = 269 K V2 = X L 290K 269K = 2 L X L 538 = (X) (290) 538= X 290 1.85 L OR 2L

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**Provide Gay-Lussac’s Law formula.**

P1 P2 T T2 =

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**669.9 or 670 mmHg = P1 = 625 mmHg T1 = 47 + 273 = 320K P2 = X**

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 = = 320K P2 = X T2 = = 343K 625 mmHg X mmHg = 320 K 343 K = (X) (320) = X 320 669.9 or 670 mmHg

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**258 kPa OR 260 kPa = P1 = X kPa T1 = 12 + 273 = 285K P2 = 360 kPa**

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 = = 285K P2 = 360 kPa T2 = = 398K X kPa 360 kPa = 285 K 398 K = (X) (497) = X 398 258 kPa OR 260 kPa

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**350 mmHg = P1 = 630 mmHg T1 = 225 K P2 = X mmHg T2 = 125 K 225 K 125 K**

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 mm Hg? P1 = 630 mmHg T1 = 225 K P2 = X mmHg T2 = 125 K 630 mmHg X mmHg = 225 K 125 K = (X) (225) = X 225 350 mmHg

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**0.11 atm = P1 = 0.135 atm T1 = 45 + 273= 318 K P2 = X atm**

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

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Boyle’s Law - Review P1V1 = P2V2.

Boyle’s Law - Review P1V1 = P2V2.

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