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Gay-Lussac's Law (temperature, pressure) A gas has a pressure if 0.0370 atm at 50.0 °C. What is the pressure at 0.00 °C? P2 = 5

Gay-Lussac's Law (temperature, pressure) A gas has a pressure if 0.0370 atm at 50.0 °C. What is the pressure at 0.00 °C? P2 = 5

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 number 6? Was this the correct answer that shows my work includes the significant figures? Also, can you help me with Gay-Lussac'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 Ta=0.00 L=0.00+ 273.15k = 273.15 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 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 Ta=0.00 L=0.00+ 273.15k = 273.15 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 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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