Download presentation

Presentation is loading. Please wait.

Published byKeegan Romaine Modified over 2 years ago

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?

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-LUSSAC Worksheet

4
Provide Boyles Law formula. P 1 V 1 = P 2 V 2

5
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

6
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

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) (25.0 L) = (X) (15.3L) 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)(3.00L) = (X) (7.5L) 7.5 atm = (X) 7.5 1.0 atm

9
Provide Charles Law formula. V 1 V 2 T 1 T 2 =

10
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

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 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

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 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

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 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

14
Provide Gay-Lussac’s Law formula. P 1 P 2 T 1 T 2 =

15
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

16
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

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? 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

18
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

19
BOYLES CHARLES & GAY-LUSSAC Self Check

20
Provide Boyles Law formula. P 1 V 1 = P 2 V 2

21
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

22
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

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) (5.0 L) = (X) (2.5 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)(10.0L) = (X) (5.0 L) 8.5 atm = (X) 5.0 1.7 atm

25
Provide Charles Law formula. V 1 V 2 T 1 T 2 =

26
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

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 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

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 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

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 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

30
Provide Gay-Lussac’s Law formula. P 1 P 2 T 1 T 2 =

31
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

32
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

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? 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

34
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

Similar presentations

Presentation is loading. Please wait....

OK

Entry Task: March 22 Friday

Entry Task: March 22 Friday

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on types of parallelograms worksheets Keynote opening ppt on iphone Ppt on csr and sustainable development Download ppt on abdul kalam Ppt on national democratic alliance Ppt on forward rate agreement explain Ppt on credit default swaps index Ppt on force and pressure for class 9 Ppt on non agricultural activities in antigua Antibiotic slides ppt on diabetic foot ulcer