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?
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
BOYLES CHARLES & GAY-LUSSAC
Worksheet
Provide Boyles Law formula.
P1V1 = P2 V2
1. If some neon gas at 75.0 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?
(75.0 kPa) (6.0 dm3) =
450 kPa
3.0
150 kPa
(X) (3.0 dm3)
= (X)
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?
(2.25 atm) (345 cm3) =
776.25 cm3
3.10
250
3
cm
(3.10 atm) (X)
= (X)
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) =
3375 kPa
15.3
221 kPa
(X) (15.3L)
= (X)
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) =
7.5 atm
7.5
1.0 atm
(X) (7.5L)
= (X)
Provide Charles Law formula.
V1 V2
=
T1 T2
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?
X L
1.5 L
=
303K
279K
T =30 + 273 = 303K
1
V1 = 1.5 L
T2 = 6 + 273 = 279 K
V2 = X L
418.5 = (X) (303)
418.5= X
303
1.38 L OR 1 L
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?
303K
T1 =30 + 273 = 303K
V1 = 0.25 L
T2 = 325 + 273 = 598K
V2 = X L
0.49 L
598K
=
0.25 L
X L
149.5 = (X) (303)
149.5= X
303
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?
296K
T1 =23 + 273 = 296K
V1 = 250 mL
T2 = 40 + 273 = 313 K
V2 = X L
313K
=
250 mL
X L
78250 = (X) (296)
78250 = X
296
264 mL OR 260 ml
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)?
290K
T1 =17 + 273 = 290K
V1 = 2 L
T2 = -4 + 273 = 269 K
V2 = X L
2L
269K
= XL
538 = (X) (290)
538= X
290
1.85 L OR 2L
Provide Gay-Lussac’s Law formula.
P1 P 2
=
T1 T2
1. The pressure inside a container is 625 mmHg at a
temperature of 47oC. What would the pressure be at 70oC?
625 mmHg
P1 = 625 mmHg
T1 = 47 + 273 = 320K
P2 = X
T2 = 70 + 273 = 343K
X mmHg
=
320 K
343 K
214375 = (X) (320)
214375 = X
320
669.9 or 670 mmHg
2. A rigid container is at a temperature of 12oC. When
heated to 125oC, the pressure was 360 kPa. What was the
initial pressure?
X kPa
P1 = X kPa
T1 = 12 + 273 = 285K
P2 = 360 kPa
T2 = 125 + 273 = 398K
360 kPa
=
285 K
398 K
102600 = (X) (497)
102600 = X
398
258 kPa OR 260 kPa
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?
630 mmHg
P1 = 630 mmHg
T1 = 225 K
P2 = X mmHg
T2 = 125 K
X mmHg
=
225 K
125 K
78750 = (X) (225)
78750 = X
225
350 mmHg
4. A gas has a pressure of 0.135 atm at 45.0 °C. What is the
pressure at -12˚C ?
0.135 atm
P1 = 0.135 atm
T1 = 45 + 273= 318 K
P2 = X atm
T2 = -12 + 273= 261 K
X atm
=
318 K
261 K
35.24 = (X) (318)
35.24 = X
318
0.11 atm
BOYLES CHARLES & GAY-LUSSAC
Self Check
Provide Boyles Law formula.
P1V1 = P2 V2
1. If some neon gas at 30.0 kPa were allowed to expand
from 7.7 dm3 to 12.0 dm3 without changing the
temperature, what pressure would the neon gas exert
under these new conditions?
(30.0 kPa) (7.7 dm3) =
231 kPa
12.0
19.3 kPa
(X) (12.0 dm3)
= (X)
2. A quantity of gas under a pressure of 3.2 atm has a
volume of 650 cm3. The pressure is increased to 4.3
atm, while the pressure remains constant. What is the
new volume?
(3.20 atm) (650 cm3) =
2080 cm3
4.30
480
3
cm
(4.30 atm) (X)
= (X)
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) =
875 kPa
2.5
350 kPa
(X) (2.5 L)
= (X)
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) =
8.5 atm
5.0
1.7 atm
(X) (5.0 L)
= (X)
Provide Charles Law formula.
V1 V2
=
T1 T2
1. The temperature inside my refrigerator is about 90 Celsius.
If I place a balloon in my fridge that initially has a temperature
of 220 C and a volume of 0.3 liters, what will be the volume of
the balloon when it is fully cooled by my refrigerator?
X L
0.3 L
=
295K
282K
T =22 + 273 = 295K
1
V1 = 0.3 L
T2 = 9 + 273 = 282 K
V2 = X L
84.6 = (X) (295)
84.6= X
295
0.286 L or 0.3 L
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 0C, what
will the volume of the balloon be after he heats it to a
temperature of 35 0C?
300K
T1 =27 + 273 = 300K
V1 = 0.7 L
T2 = 35 + 273 = 308K
V2 = X L
0.76 L
308K
=
0.7 L
X L
215.6 = (X) (300)
215.6= X
300
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 0C, and I leave it in my car which has a
temperature of 50.00 C, what will the new volume of the bag be?
295K
T1 =22 + 273 = 295K
V1 = 250 mL
T2 = 50 + 273 = 323 K
V2 = X L
323K
=
250 mL
X L
80750 = (X) (295)
80750= X
274 mL
295
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)?
298K
T1 =25 + 273 = 298K
V1 = 2 L
T2 = -4 + 273 = 269 K
V2 = X L
2L
269K
= XL
538 = (X) (298)
538= X
298
1.81 L OR 2 L
Provide Gay-Lussac’s Law formula.
P1 P 2
=
T1 T2
1. The pressure inside a container is 650 mmHg at a
temperature of 75oC. What would the pressure be at 55oC?
650 mmHg
P1 = 650 mmHg
T1 = 75 + 273 = 348K
P2 = X
T2 = 55 + 273 = 328K
X mmHg
=
348 K
328 K
213200 = (X) (348)
267960= X
348
613 OR 610 mmHg
2. A rigid container is at a temperature of 3.0oC. When
heated to 75oC, the pressure was 2.5 kPa. What was the
initial pressure?
X kPa
P1 = X kPa
T1 = 3.0 + 273 = 276K
P2 = 2.5 kPa
T2 = 75 + 273 = 348K
2.5 kPa
=
276 K
348 K
690 = (X) (348)
690 = X
348
1.98 or 2.0 kPa
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?
550 mmHg
P1 = 550 mmHg
T1 = 303.0 K
P2 = X mmHg
T2 = 200.5 K
X mmHg
=
303.0 K
200.5 K
110275 = (X) (303)
110275 = X
303
363.9 mmHg
4. A gas has a pressure of 0.370 atm at 50.0 °C. What is the
pressure at 22˚C ?
0.370 atm
P1 = 0.370 atm
T1 = 50 + 273= 323 K
P2 = X atm
T2 = 22 + 273= 295 K
X atm
=
323 K
295 K
109.15 = (X) (323)
109.15 = X
323
0.338 atm OR 0.34 atm