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G - Ls Law – Pressure vs. Temperature

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Experiment to develop the relationship between the pressure and temperature of a gas. Include: Gay-Lussacs Law, Combined Gas Law, partial pressure Additional KEY Terms

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Pressure (kPa) Volume (mL) BOYLES LAW – Pressure vs. Volume = P1V1P1V1 P2V2P2V2

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Temperature (K) Volume (mL) CHARLESS LAW – Temp vs. Volume = V1V1 V2V2 T2T2 T1T1

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Joseph Gay-Lussac ( ) Determined that temperature and pressure of a gas is a direct relationship (volume and amount of gas are held constant)

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= P1P1 P2P2 T2T2 T1T1 **As with Charles Law, temperature in Kelvin.

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If a 12.0 L sample of gas is found to have a pressure of kPa at 0.0°C, calculate the new pressure at 128°C if the volume is held constant. 0.0°C = 273 K 128°C +273 = 401 K = P1P1 P2P2 T2T2 T1T1 = 101.3P2P2 (401) kPa

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Boyles Law: Pressure α ___1___ volume Charles Law: Volume α temperature Gay-Lussacs Law: Pressure α temperature = P1P1 P2P2 T2T2 T1T1 V1V1 V2V2 combined gas law

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If a gas occupies a volume of 25.0 L at 25.0°C and 1.25 atm, calculate the volume at 128°C and atm. = P1P1 P2P2 T2T2 T1T1 V1V1 V2V2 = (1.25) (401) 298 (0.750) 25 V2V2 25°C = 298 K 128°C +273 = 401 K 56.1 L

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A gas has a volume of 125 L at 325 kPa and 58.0°C, calculate the temperature in Celsius to produce a volume of 22.4 L at kPa = P1P1 P2P2 T2T2 T1T1 V1V1 V2V2 = (101.3)(331) 325 (125) 22.4 T2T2 58°C = 331 K 18.5 K 18.5 K = -254°C

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A bag contains 145 L of air at the bottom of a lake, at a temperature of 5.20°C and a pressure of 6.00 atm. When the bag is released, it ascends to the surface where the pressure is 1.00 atm and 16.0°C. Given: P 1 = 6.00 atm V 1 = 145L T 1 = 5.20°C P 2 = 1.00 atm T 2 = 16.0°C Find: V 2 If the maximum volume of the lift bag is 750 L, will the bag burst at the surface?

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Daltons Law of Partial Pressure: Each gas in a mixture exerts pressure independently. Total pressure = sum of the pressures of each gas (partial pressures) Partial pressure depends on the number of gas particles present, the temperature and volume of container. P total = P 1 + P 2 + P 3 + P 4 …

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1.0 mol H mol He 1.0 mol H mol He 3.0 mol gas

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The following gases are placed in a litre container and held at a constant temperature: 2.0 L of O 2 at an original pressure of kPa 3.00 L of Ne at an original pressure of kPa What pressure is produced inside the container? HINT: Each gas is will expand when put into the L container, so each will exert a partial pressure less than its original pressure.

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P 1 V 1 = P 2 V 2 Oxygen: (202.6)(2.00) = P 2 (10.00) P 2 = 40.5 kPa Neon: (303.9)(3.00) = P 2 (10.00) P 2 = 91.2 kPa Total pressure = = kPa

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CAN YOU / HAVE YOU? Experiment to develop the relationship between the pressure and temperature of a gas. Include: Gay-Lussacs Law, Combined Gas Law, partial pressure Additional KEY Terms

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