T₁ = 77°c=(77+273)K = 350k Gas Laws Worksheet -86°C = (86+273)K=359K Charles's Law (temperature, volume) 1) A 550.0 mL sample of nitrogen gas is warmed from 77 °C to 86 °C. Find its new volume if the pressure V₁ = √₂ remains constant. 564 ml T₁ V2= Vita (550.0 mL) x (359k) - √564 T₁ Ta m/
T₁ = 77°c=(77+273)K = 350k Gas Laws Worksheet -86°C = (86+273)K=359K Charles's Law (temperature, volume) 1) A 550.0 mL sample of nitrogen gas is warmed from 77 °C to 86 °C. Find its new volume if the pressure V₁ = √₂ remains constant. 564 ml T₁ V2= Vita (550.0 mL) x (359k) - √564 T₁ Ta m/
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Would you help me with the number 1 question? Also, can you show me the step-by-step and formula?
![co
564 ml
Charles's Law (temperature, volume)
1) A 550.0 mL sample of nitrogen gas is warmed from 77 °C to 86 °C. Find its new volume if the pressure
remains constant.
2.22 L
350 k
2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0 °C?
V₁=1.00L
4₁
V₂ = V₁ = 1
V₂ = V₁Ta - (1.00 L) x (6061) - 2.222
273K
21300 L
26.6 L
318 K
V₁=550.0 mL
T₁ = 77°c = (77+273)k
= 350k
Gas Laws Worksheet -86°C -86+273)K=359K
V₁
V₂
Boyle's Law (pressure, volume)
3) Convert 338 L at 63.0 atm to its new volume at 1.00 atm.
P.V₁
Pala
V2= Vita : (550.0 ML) x (359 k)
T₁
5730 mmHg
Та
.0313 atm
T₁ T₂
-564mL
+₁=0,00°C = 10.00+273) K=273k
Ta333.0°C (333.0 +273 K = 606 K
V₂=
Pa
4) A tank of nitrogen has a volume of 14.0 L and a pressure of 760.0 mm Hg. Find the volume of nitrogen
when its pressure is changed to 400.0 torr while the temperature is held constant.
P₁=63.0 atm
V₁ = 3382
or 2.13 x 10 L
= (63.0 atm) x (338L) = (2+13X10² L) Po=1.06A+m
7.00 atm
5) What pressure (mm Hg) is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder
whose volume is 26.0 liters?
Gay-Lussac's Law (temperature, pressure)
6) A gas has a pressure if 0.0370 atm at 50.0 °C. What is the pressure at 0.00 °C?
k
7) If a gas in a closed container, with an original temperature of 25.0 °C, is pressurized from 15.0
atmospheres to 16.0 atmospheres, what would the final temperature of the gas be?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe077a947-dd8a-4a6d-874b-21eab4c7694d%2Fd3491f91-22c8-4099-a10b-959ce0d8b679%2Fxkouhsg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:co
564 ml
Charles's Law (temperature, volume)
1) A 550.0 mL sample of nitrogen gas is warmed from 77 °C to 86 °C. Find its new volume if the pressure
remains constant.
2.22 L
350 k
2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0 °C?
V₁=1.00L
4₁
V₂ = V₁ = 1
V₂ = V₁Ta - (1.00 L) x (6061) - 2.222
273K
21300 L
26.6 L
318 K
V₁=550.0 mL
T₁ = 77°c = (77+273)k
= 350k
Gas Laws Worksheet -86°C -86+273)K=359K
V₁
V₂
Boyle's Law (pressure, volume)
3) Convert 338 L at 63.0 atm to its new volume at 1.00 atm.
P.V₁
Pala
V2= Vita : (550.0 ML) x (359 k)
T₁
5730 mmHg
Та
.0313 atm
T₁ T₂
-564mL
+₁=0,00°C = 10.00+273) K=273k
Ta333.0°C (333.0 +273 K = 606 K
V₂=
Pa
4) A tank of nitrogen has a volume of 14.0 L and a pressure of 760.0 mm Hg. Find the volume of nitrogen
when its pressure is changed to 400.0 torr while the temperature is held constant.
P₁=63.0 atm
V₁ = 3382
or 2.13 x 10 L
= (63.0 atm) x (338L) = (2+13X10² L) Po=1.06A+m
7.00 atm
5) What pressure (mm Hg) is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder
whose volume is 26.0 liters?
Gay-Lussac's Law (temperature, pressure)
6) A gas has a pressure if 0.0370 atm at 50.0 °C. What is the pressure at 0.00 °C?
k
7) If a gas in a closed container, with an original temperature of 25.0 °C, is pressurized from 15.0
atmospheres to 16.0 atmospheres, what would the final temperature of the gas be?
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