G - L’s Law – Pressure vs. Temperature

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

= P1V1 P2V2 BOYLE’S LAW – Pressure vs. Volume Pressure (kPa) Volume (mL)

= V1 V2 T1 T2 CHARLES’S LAW – Temp vs. Volume Volume (mL) Temperature (K)

Joseph Gay-Lussac (1778-1850) Determined that temperature and pressure of a gas is a direct relationship (volume and amount of gas are held constant)

**As with Charles’ Law, temperature in Kelvin. = P1 P2 T1 T2

If a 12. 0 L sample of gas is found to have a pressure of 101 If a 12.0 L sample of gas is found to have a pressure of 101.3 kPa at 0.0°C, calculate the new pressure at 128°C if the volume is held constant. P1 = P2 T1 T2 0.0°C + 273 = 273 K 128°C +273 = 401 K 101.3 149 kPa P2 (401) = 273

combined gas law P1 V1 = P2 V2 T1 T2 Boyle’s Law: Pressure α ___1___ volume Charles’ Law: Volume α temperature Gay-Lussac’s Law: Pressure α temperature P1 V1 = P2 V2 T1 T2 combined gas law

If a gas occupies a volume of 25. 0 L at 25. 0°C and 1 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 0.750 atm. P1 V1 = P2 V2 T1 T2 25°C + 273 = 298 K 128°C +273 = 401 K (1.25) 25 = 56.1 L V2 (401) 298 (0.750)

A gas has a volume of 125 L at 325 kPa and 58 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 101.3 kPa P1 V1 = P2 V2 T1 T2 58°C + 273 = 331 K 18.5 K T2 (101.3) 22.4 (331) = 325 (125) 18.5 K - 273 = -254°C

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: P1 = 6.00 atm V1 = 145L T1 = 5.20°C P2 = 1.00 atm T2 = 16.0°C If the maximum volume of the lift bag is 750 L, will the bag burst at the surface? Find: V2

Ptotal = P1 + P2 + P3 + P4 … Dalton’s 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. Ptotal = P1 + P2 + P3 + P4 …

1.0 mol H2 2.0 mol He 3.0 mol gas

2.0 L of O2 at an original pressure of 202.6 kPa The following gases are placed in a 10.00 litre container and held at a constant temperature: 2.0 L of O2 at an original pressure of 202.6 kPa 3.00 L of Ne at an original pressure of 303.9 kPa What pressure is produced inside the container? HINT: Each gas is will expand when put into the 10.00 L container, so each will exert a partial pressure less than its original pressure.

P1V1 = P2V2 Oxygen: (202.6)(2.00) = P2 (10.00) P2 = 40.5 kPa Neon: (303.9)(3.00) = P2 (10.00) P2 = 91.2 kPa Total pressure = 40.5 + 91.2 = 131.7 kPa

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