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. V₁= = √₂ T₁ T₂ 564 ml 2.22 L 350 k 2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0 °C? V₂ =V₁ Ta V₁=1.00L t₁ T₁ = 77°c =(77+273)k = 350k k Gas Laws Worksheet -86°C = (86+273)K= 359 K 26.6 L V₂= V₁T₂ : (550.0 mL) x (359k) T₁ Boyle's Law (pressure, volume) 3) Convert 338 L at 63.0 atm to its new volume at 1.00 atm. P.V₁ = Pal 5730 mmHg P₁=63.0 atm V₁=338 L 21300 L or 2.13 x 10 L V27 PV₁ = (63.0 atm) x (338L) = 2.13X10¹ L) Pa = 1.00A+M Pa 7.00 atm 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. .0313 atm 564 mL +₁=0,00°c=10.00+273) K=273 k Ta=333.0°C (333.0 +273 K = 606 K 2.222 V₂= V₁Ta -(1.00L) x (6061) - 273K 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? 7) If a gas in a closed container with an 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?

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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?
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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