Gas Laws Pressure and Volume (Boyle’s Law) Temperature and Volume (Charles’ Law)

P and V Changes The pressure of a gas is inversely proportional to its volume P1 P2 V1 V2

Pressure and Volume Experiment Pressure Volume P x V (atm) (L) (atm x L) 1 ↑ 1.0 ↓ 16 16 2 2.0 8.0 16 Volume is inversely proportional 3 4.0 4.0 _____ to Pressure 4 8.0 2.0 _____ P1V1 = 1.0 atm x 16 L = 16 atm L (K) P2V2 = 2.0 atm x 8.0 L = 16 atm L (K) Boyle's Law: When Temperature and amount of gas remains constant P x V = k (constant) Pressure x Volume = Constant P1V1 = k = P2V2 P1V1 = P2V2

Use this equation Initial Final P1V1 = P2V2 To find How the volume of a gas changes when its pressure changes. Final V Initial V x Initial P V2 = V1 x P1 P2 or How the pressure of a gas changes when its volume changes. Final P Initial P x Initial V P2 = P1 x V1 V2

Freon gas is used in refrigeration systems. Boyle’s Law Problem #1 The pressure of a gas is inversely proportional to its volume P1V1 = P2V2 Freon gas is used in refrigeration systems. What is the final volume (L) of a 1.6 L sample of Freon gas if its initial pressure of 50 mm Hg is changed to 200 mm of Hg at constant T? Remember must convert units to atmospheres and liters Prepare a data table DATA TABLE Initial conditions Final conditions P1 P2 V1 V2

Boyle’s Law Calculation #1 The pressure of a gas is inversely proportional to its volume P1V1 = P2V2 Freon gas is used in refrigeration systems. What is the final volume (L) of a 1.6 L sample of Freon gas if its initial pressure of 50. mm Hg is changed to 200. mm of Hg at constant T? Remember you must convert units to atmospheres and liters Convert mm Hg to atm 50 mm Hg (1 atm/760 mm Hg) = .066 atm 200 mm Hg (1 atm/760 mm Hg) = .26 atm Prepare a data table DATA TABLE Initial conditions Final conditions P1 = .066 atm P2 = .26 atm V1 = 1.6 L V2 = ?

1 Freon-12, CCl2F2, is used in refrigeration systems. Find New Volume (V2) 1 Freon-12, CCl2F2, is used in refrigeration systems. What is the final volume (L) of a 1.6 L sample of Freon gas if its initial pressure of 50. mm Hg is changed to 200. mm of Hg at constant T? Convert mm Hg to atm 50 mm Hg (1 atm/760 mm Hg) = .066 atm 200 mm Hg (1 atm/760 mm Hg) = .26 atm Prepare a data table DATA TABLE Initial conditions Final conditions P1 = .066 atm P2 = .26 atm V1 = 1.6 L V2 = ? Solve for V2: P1V1 = P2V2 V2 = V1 x P1 P2 V2 = 1.6 L x .066atm = 0.4 L .26 atm

Learning Check #1 Remember you must convert units to atmospheres and liters. A sample of nitrogen gas has a volume of 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? P1V1 = P2V2 V2 = V1 x P1 P2 Prepare a data table. There are no units to convert in this problem so you are ready to go. DATA TABLE Initial conditions Final conditions P1 = .70 atm P2 = 1.40 atm V1 = 6.4 L V2 = ? L

A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. Solution #1 A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? Prepare a data table: DATA TABLE Initial conditions Final conditions P1 = .70 atm P2 = 1.4 atm V1 = 6.4 L V2 = ? P1V1 = P2V2 V2 = V1 x P1 P2 V2 = 6.4 L x 0.70 atm = 3.2 L 1.40 atm

Learning Check #2 A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? P1V1 = P2V2 P2 = P1 x V1 V2 Remember you must convert units to atmospheres and liters. Convert mm Hg to atm 600. mm Hg (1 atm/760 mm Hg) = .789 atm

Solution #2 Pressure decrease when volume increases. A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? Prepare a data table: DATA TABLE Initial conditions Final conditions P1 = .789 atm P2 = ? In atm V1 = 12.0 L V2 = 36.0L Remember you must convert units to atmospheres and liters. .789 atm x 12.0 L = .263 or 2.63 x 10-1 atm 36.0 L Pressure decrease when volume increases.

Gas Laws Pressure and Volume (Boyle’s Law) Temperature and Volume (Charles’ Law)

Charles’ Law V = .250 L V = .175L T = 293 K T = 273K Observe the V and T of the balloons. How does volume change with temperature?

Volume and Temperature The volume of a gas is directly proportional to its absolute (K) temperature Experiment Volume Temp. V/T (L) (K) (L/K) 1 0.7 73 0.01 2 1.7 173 0.01 3 2.7 273 _____ 4 3.7 373 _____ V1/ T1 = 0.7 L / 73K = .01 (b) V2/ T2 = 1.7 L / 173K = .01 (b) Charles’ Law : V = bT V = b V1 = b = V2 or V1 = V2 T T1 T2 T1 T2

You must always convert units to the following: Charles’ Law: V and T At constant pressure, the volume of a gas is directly proportional to its absolute (K) temperature You must always convert units to the following: (V) is volume of gas in Liters (T) is the Temperature in Kelvin V1 = V2 T1 T2   Initial Final

Learning Check #1 Use Charles’ Law to complete the statements below: If final Temp. is higher than initial Temp, then the final Vol. is (greater, or less) than the initial Vol. If final Volume is less than initial Volume, then the final Temp. is (higher, or lower) than the initial Temp. The volume of a gas is directly proportional to its absolute (K) temperature

Solution #1 V1 = V2 T1 T2 1. If final T is higher than initial T, final V is (greater) than the initial V. 2. If final V is less than initial V, final T is (lower) than the initial T. The volume of a gas is directly proportional to its absolute (K) temperature

You must always convert units to the following: Charles’ Law Problem #1 A balloon has a volume of 785 mL on a fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon? V1 = V2 T1 T2 You must always convert units to the following: (V) is volume of gas in Liters (T) is the Temperature in Kelvin

Charles’ Law Calculation #1 Remember you must convert units to Kelvin and liters. Prepare a data table: DATA TABLE Initial conditions Final conditions V1 = 785 ml = .785 L V2 = ? L T1 = 21 °C + 273= 294 K T2 = 0 °C = 273 K V1 = V2 V2 = V1 x T2 T1 T2 T1 V2 = .785 L 273 K .729L or 7.29 x 10-1 L 294 K Check your answer. If temperature decreases, Volume should decrease

Learning Check #2 1) 716K 2) 443K 3) 191K V1 = V2 T2 = T1 x V2 A sample of O2 gas has a volume of 420 mL at a temperature of 18°C. What temperature (K) is needed to change the volume to 640 mL? 1) 716K 2) 443K 3) 191K V1 = V2 T2 = T1 x V2 T1 T2 V1

Solution #2 A sample of O2 gas has a volume of 420 mL at a temperature of 18°C. What temperature (in K) is needed to change the volume to 640 mL? Remember you must convert units to Kelvin and liters. Prepare a data table: DATA TABLE Initial conditions Final conditions V1 = 420 ml = .420 L V2 = 640 ml = .640 L T1 = 18 °C + 273 = 291 K T2 = ? K T2 = 291 K .640 L 443K .420 L V1 = V2 T2 = T1 x V2 T1 T2 V1