Homework Help is Here – Start Your Trial Now!
Science
Chemistry
2= 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 = Pay 21300 L P₁ V₁ Pa V₂ = 273K = 2.22L P₁ =63.0 atm V/2=2,13x10³2 V₁=3382 or 2.13 x 10 L (63.0 atm) x (338L) = (2.13X10¹L Pa = 1.06A+M

2= 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 = Pay 21300 L P₁ V₁ Pa V₂ = 273K = 2.22L P₁ =63.0 atm V/2=2,13x10³2 V₁=3382 or 2.13 x 10 L (63.0 atm) x (338L) = (2.13X10¹L Pa = 1.06A+M

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 5.118P: 5-118 Isooctane, which has a chemical formula C8H18 is the component of gasoline from which the term...
See similar textbooks
icon
Related questions
Question
100%
Can you help me with the number 3 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
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

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
  • SEE MORE 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 for Engineering Students
Chemistry for Engineering Students
Chemistry
ISBN:
9781337398909
Author:
Lawrence S. Brown, Tom Holme
Publisher:
Cengage Learning
Introductory Chemistry: An Active Learning Approa…
Introductory Chemistry: An Active Learning Approa…
Chemistry
ISBN:
9781305079250
Author:
Mark S. Cracolice, Ed Peters
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781133611097
Author:
Steven S. Zumdahl
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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
SEE MORE TEXTBOOKS