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2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0 °C?

2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0 °C?

Chemistry
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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Can you help me with the number 2 question? Can you explain step by step including the formula too? I need to plug in the fraction.
V₁=550.0mL T₁ = 77°c=(77+273)k = 350k Gas Laws Worksheet -86°C = 86+273) K = 359K Charles's Law (temperature, volume) V₁ = √₂ 1) A 550.0 mL sample of nitrogen gas is warmed from 77 °C to 86 °C. Find its new volume if the pressure = Constant remains constant. T Ta Va= Vita :(550.0 mL) x (359k) 350k V₁=1.00L 2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0 °C? +₁=0,00°C = 10.00+273) K=273k Ta=333.0°C (333.0 +273) K = 606K Va v V₂= V₁Ta - (1.00 L) x (6061) _ 273K = 564 ml 2.22 L Ta V₂ = 564m² Boyle's Law (pressure, volume) Pv=nr + >constant 3) Convert 338 L at 63.0 atm to its new volume at 1.00 atm. P.V₁ = Pay₂ V n P₁ = 63.0 atm √2=2013X10³/2 V₁ = 3382 21300 L or 2.13 x 10 L 1/₂ = ₁V₁ = (63.0 atm) x (338L) = (2.13×10¹ L Pa = 1.00A+m Pa 7,00 atm Tatm-760mm Hg 5730 mmHg 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. V₁ = √₂ Initial volume nitregen (V₁) = 14.0L Initial pressure nitregen (Pi) = 760.0mmits Final volume a) = na 26.6 L ^₁ 760.0mm Hg x 14.0 L 400.0mm Hg -_Pressure (pa) = 400.0 torr = 26.6 V₂= P₁V₁ Pa 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? P₂ = P₁V₁ Va P₁V₁=lava 7.540+m V₁ = 196.0L 1.00atm x 196.0L 26.0L P₁ = 19+m -564 ML = 2.22L T₁=37% 25+273,15 K = 298,15K 318 K 12 = XT = 16.0 atm X 298,15 K P. 1510 atm 7.54 atms 7.54 atm (160mmitty) Gay-Lussac's Law (temperature, pressure (5730mm Hg) 6) A gas has a pressure if 0.0370 atm at 50.0 °C. What is the pressure at 0.00 °C? P₂ = P₁T2=0,03700mx 273 k temy (+₁) = 50.00 ( Pressure (P=0.0370 atm .0313 atm 318.0266667 Ta-318K r=1mm 17g 4000 torr = 400,0mmity P₁ V₁ = Pav₂ 323 K Pa = 0.031272 446 0.0313atm 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=2610L P2=5 130 mm Hg (S0+273) k T2=0.00°C = (+373) K = 273 k = 323K Initial temp = 25°C Initial pressure (P)=15.0 atm Final pressure (Pa): 16.0 atm Final temperature (Ta):
Transcribed Image Text:V₁=550.0mL T₁ = 77°c=(77+273)k = 350k Gas Laws Worksheet -86°C = 86+273) K = 359K Charles's Law (temperature, volume) V₁ = √₂ 1) A 550.0 mL sample of nitrogen gas is warmed from 77 °C to 86 °C. Find its new volume if the pressure = Constant remains constant. T Ta Va= Vita :(550.0 mL) x (359k) 350k V₁=1.00L 2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0 °C? +₁=0,00°C = 10.00+273) K=273k Ta=333.0°C (333.0 +273) K = 606K Va v V₂= V₁Ta - (1.00 L) x (6061) _ 273K = 564 ml 2.22 L Ta V₂ = 564m² Boyle's Law (pressure, volume) Pv=nr + >constant 3) Convert 338 L at 63.0 atm to its new volume at 1.00 atm. P.V₁ = Pay₂ V n P₁ = 63.0 atm √2=2013X10³/2 V₁ = 3382 21300 L or 2.13 x 10 L 1/₂ = ₁V₁ = (63.0 atm) x (338L) = (2.13×10¹ L Pa = 1.00A+m Pa 7,00 atm Tatm-760mm Hg 5730 mmHg 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. V₁ = √₂ Initial volume nitregen (V₁) = 14.0L Initial pressure nitregen (Pi) = 760.0mmits Final volume a) = na 26.6 L ^₁ 760.0mm Hg x 14.0 L 400.0mm Hg -_Pressure (pa) = 400.0 torr = 26.6 V₂= P₁V₁ Pa 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? P₂ = P₁V₁ Va P₁V₁=lava 7.540+m V₁ = 196.0L 1.00atm x 196.0L 26.0L P₁ = 19+m -564 ML = 2.22L T₁=37% 25+273,15 K = 298,15K 318 K 12 = XT = 16.0 atm X 298,15 K P. 1510 atm 7.54 atms 7.54 atm (160mmitty) Gay-Lussac's Law (temperature, pressure (5730mm Hg) 6) A gas has a pressure if 0.0370 atm at 50.0 °C. What is the pressure at 0.00 °C? P₂ = P₁T2=0,03700mx 273 k temy (+₁) = 50.00 ( Pressure (P=0.0370 atm .0313 atm 318.0266667 Ta-318K r=1mm 17g 4000 torr = 400,0mmity P₁ V₁ = Pav₂ 323 K Pa = 0.031272 446 0.0313atm 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=2610L P2=5 130 mm Hg (S0+273) k T2=0.00°C = (+373) K = 273 k = 323K Initial temp = 25°C Initial pressure (P)=15.0 atm Final pressure (Pa): 16.0 atm Final temperature (Ta):
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