Homework Help is Here – Start Your Trial Now!
Science
Chemistry
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. mm 179.

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. mm 179.

Introduction to General, Organic and Biochemistry
Introduction to General, Organic and Biochemistry
11th Edition
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Chapter5: Gases, Liquids, And Solids
Section: Chapter Questions
Problem 5.103P
See similar textbooks
icon
Related questions
Question
100%
Can you help me with the number 4 question? Can you explain step by step including the formula (Boyle’s law)? I need to plug in the fraction to give the correct answer.
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):
Expert Solution
steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Mole Concept
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Introduction to General, Organic and Biochemistry
Introduction to General, Organic and Biochemistry
Chemistry
ISBN:
9781285869759
Author:
Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:
Cengage Learning
Chemistry: The Molecular Science
Chemistry: The Molecular Science
Chemistry
ISBN:
9781285199047
Author:
John W. Moore, Conrad L. Stanitski
Publisher:
Cengage Learning
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Chemistry: Principles and Practice
Chemistry: Principles and Practice
Chemistry
ISBN:
9780534420123
Author:
Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:
Cengage Learning
General Chemistry - Standalone book (MindTap Cour…
General Chemistry - Standalone book (MindTap Cour…
Chemistry
ISBN:
9781305580343
Author:
Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:
Cengage Learning
Physical Chemistry
Physical Chemistry
Chemistry
ISBN:
9781133958437
Author:
Ball, David W. (david Warren), BAER, Tomas
Publisher:
Wadsworth Cengage Learning,
SEE MORE TEXTBOOKS