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) 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) cm 3 = (X) 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) 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) 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 = = 303K V 1 = 1.5 L T 2 = = 279 K V 2 = X L = (X) (303) 303K 1.38 L OR 1 L 279K = 418.5= X 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 C? T 1 = = 303K V 1 = 0.25 L T 2 = = 598K V 2 = X L = (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 = = 296K V 1 = 250 mL T 2 = = 313 K V 2 = X L = (X) (296) 296K 250 mL 264 mL OR 260 ml 313K X L = = 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 = = 290K V 1 = 2 L T 2 = = 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 = = 320K P 2 = X T 2 = = 343K = (X) (320) 625 mmHg 320 K or 670 mmHg X mmHg 343 K = = 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 = = 285K P 2 = 360 kPa T 2 = = 398K = (X) (497) X kPa 285 K 258 kPa OR 260 kPa 360 kPa 398 K = = 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 mm Hg? P 1 = 630 mmHg T 1 = 225 K P 2 = X mmHg T 2 = 125 K = (X) (225) 630 mmHg 225 K 350 mmHg X mmHg 125 K = = X 225

18
4. A gas has a pressure of atm at 45.0 °C. What is the pressure at -12˚C ? P 1 = atm T 1 = = 318 K P 2 = X atm T 2 = = 261 K = (X) (318) atm 318 K 0.11 atm X atm 261 K = = 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) 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) 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) kPa

24
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) 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 = = 295K V 1 = 0.3 L T 2 = = 282 K V 2 = X L 84.6 = (X) (295) 295K L or 0.3 L 282K = 84.6= X 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 = = 300K V 1 = 0.7 L T 2 = = 308K V 2 = X L = (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 mL bag at a temperature of C, and I leave it in my car which has a temperature of C, what will the new volume of the bag be? T 1 = = 295K V 1 = 250 mL T 2 = = 323 K V 2 = X L = (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 = = 298K V 1 = 2 L T 2 = = 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 = = 348K P 2 = X T 2 = = 328K = (X) (348) 650 mmHg 348 K 613 OR 610 mmHg X mmHg 328 K = = 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 = = 276K P 2 = 2.5 kPa T 2 = = 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 K to K and the volume is kept constant what final pressure would result if the original pressure was mm Hg? P 1 = 550 mmHg T 1 = K P 2 = X mmHg T 2 = K = (X) (303) 550 mmHg K mmHg X mmHg K = = X 303

34
4. A gas has a pressure of atm at 50.0 °C. What is the pressure at 22˚C ? P 1 = atm T 1 = = 323 K P 2 = X atm T 2 = = 295 K = (X) (323) atm 323 K atm OR 0.34 atm X atm 295 K = = X 323

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google