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20.0 6mm Ha 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? P, V₁ = P₂ V₂ V₁196.04

20.0 6mm Ha 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? P, V₁ = P₂ V₂ V₁196.04

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...
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Can you help me with number 5? Was this the correct answer that shows my work includes the significant figures? Also, can you help me with Boyle's Law formula to plug in?

Charles's Law (temperature, volume) 1) A 550.0 mL sample of nitrogen gas is warmed from 77 °C to 86 °C. remains constant. C to 86 °C. V₂ = V₁ T₂ = (550.0 ML) X 350k T₁ 564 ml --та 2.22 L K=c't273.15 V₁ = 550.0mL T₁ = 77°C = 17+273) k T₁ = 350 k Gas Laws Worksheet T₂-86°C = 86 +273) K= 359k (564 m² 2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0 °C? 2 = 333.0°C (333.0+273.15k)- V₂ Vi 7₁ V₂ =V₁ Ta TI Ta t₁ = 0,00° C = (0.00+ 273.15 k) = 273.15k 2.219110379 V2= V₁T₂ 1 =(1.00L) x(606.15k) (273.15K) = 2.222 "Patta Pa Va V₂=564mL 26.6 L 760.0 mm Hgx 14.0L = 26.6L) 400.0 mm Hg Boyle's Law (pressure, volume) 3) Convert 338 L at 63.0 atm to its new volume at 1.00 atm. Pr=nrt constant P₁V₁ = P₂ V₂, P₁.V₁ 21300 L or 2.13 x 104 L Və 2.12:24x10 V₂ = P₁V₁ -(63.0 atm) x (3384) = 2.13 X 10 L 1.00 atm x (359x) 7.549+m こ T P₁ = 63.0 atm Pa = 1.00 atm, V2 = 2.13x10 "L V₁ = 338 L 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. P₁x V₁ = P₂xV₂ P₁V₁ = P₂ V₂ V₂₂ = P₁ V₁ Initial volume nitrogen (V₁) = 14.0L Pa Find its new volume if the pressure = constant V₁ = V₂ Ti Ta 1.00 atm x 196.0L 26.04 = P, V, Va Pressure (P2) = 400.0 torr. 1 torr=1mm Hg 400.0 Jorr = 400.0mm Hg 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? P₁V₁= P₂ V₂ V₁196.02 latm=760mm Hg P₂ = P₁V₁ 5730 mmHg Va 1₁-19 atm V2=26.04 12=5730 mm Hg Initial pressure nitrogen (P₁)=760.0mm Hg Final volume (V₂) 760mm Hg Tatm =(5730 mm Hg 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? P₁ - Pa Pa = P₁ T₂ 0.031275104 T₁ T2 T₁ 0.0370 atm x 273.15K 323.15 K .0313 atm Pa T₂ = P₂-T₂ P₁ Pressure (Pi) = 0.0370 atm Pa = 7.54 atm 273,15k L = 0.0313 atm Temp (Ti) = 50.0°C (S0 + 273.15) K = 323.15K P2= 0.0313 atm Tx 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? T₁=372 25+273,15K=298,15K Initial temp = 25c Ta=0.00 L=0.00+ 273.15k = 273.15 318 K Initial pressure (₁) = 15.0 atm atm K = C²7273,15 K 16.00 gtm x 2.98.15K - 318 K). Final pressure (P2) = 16.0 Final temperature (1₂) T₂= Parti 15.0 atm Ta= T₁ ра Pi 318.0266667
Transcribed Image Text:Charles's Law (temperature, volume) 1) A 550.0 mL sample of nitrogen gas is warmed from 77 °C to 86 °C. remains constant. C to 86 °C. V₂ = V₁ T₂ = (550.0 ML) X 350k T₁ 564 ml --та 2.22 L K=c't273.15 V₁ = 550.0mL T₁ = 77°C = 17+273) k T₁ = 350 k Gas Laws Worksheet T₂-86°C = 86 +273) K= 359k (564 m² 2) A gas occupies 1.00 L at 0.00°C. What is the volume at 333.0 °C? 2 = 333.0°C (333.0+273.15k)- V₂ Vi 7₁ V₂ =V₁ Ta TI Ta t₁ = 0,00° C = (0.00+ 273.15 k) = 273.15k 2.219110379 V2= V₁T₂ 1 =(1.00L) x(606.15k) (273.15K) = 2.222 "Patta Pa Va V₂=564mL 26.6 L 760.0 mm Hgx 14.0L = 26.6L) 400.0 mm Hg Boyle's Law (pressure, volume) 3) Convert 338 L at 63.0 atm to its new volume at 1.00 atm. Pr=nrt constant P₁V₁ = P₂ V₂, P₁.V₁ 21300 L or 2.13 x 104 L Və 2.12:24x10 V₂ = P₁V₁ -(63.0 atm) x (3384) = 2.13 X 10 L 1.00 atm x (359x) 7.549+m こ T P₁ = 63.0 atm Pa = 1.00 atm, V2 = 2.13x10 "L V₁ = 338 L 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. P₁x V₁ = P₂xV₂ P₁V₁ = P₂ V₂ V₂₂ = P₁ V₁ Initial volume nitrogen (V₁) = 14.0L Pa Find its new volume if the pressure = constant V₁ = V₂ Ti Ta 1.00 atm x 196.0L 26.04 = P, V, Va Pressure (P2) = 400.0 torr. 1 torr=1mm Hg 400.0 Jorr = 400.0mm Hg 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? P₁V₁= P₂ V₂ V₁196.02 latm=760mm Hg P₂ = P₁V₁ 5730 mmHg Va 1₁-19 atm V2=26.04 12=5730 mm Hg Initial pressure nitrogen (P₁)=760.0mm Hg Final volume (V₂) 760mm Hg Tatm =(5730 mm Hg 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? P₁ - Pa Pa = P₁ T₂ 0.031275104 T₁ T2 T₁ 0.0370 atm x 273.15K 323.15 K .0313 atm Pa T₂ = P₂-T₂ P₁ Pressure (Pi) = 0.0370 atm Pa = 7.54 atm 273,15k L = 0.0313 atm Temp (Ti) = 50.0°C (S0 + 273.15) K = 323.15K P2= 0.0313 atm Tx 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? T₁=372 25+273,15K=298,15K Initial temp = 25c Ta=0.00 L=0.00+ 273.15k = 273.15 318 K Initial pressure (₁) = 15.0 atm atm K = C²7273,15 K 16.00 gtm x 2.98.15K - 318 K). Final pressure (P2) = 16.0 Final temperature (1₂) T₂= Parti 15.0 atm Ta= T₁ ра Pi 318.0266667
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