Pages Chp 11 Gas Laws

Boyle’s Law P V PV = k

Boyle’s Law The pressure and volume of a gas are inversely related (at constant mass & temp) P V PV = k

Boyle’s Law

No sound … In all movies, the air is being pumped out from the jar, which DECREASES PRESSURE. You can easily see that the VOLUME of a gas INCREASES as you DECREASE PRESSURE!

Boyle’s Law P 1 V 1 = P 2 V 2 Initial Pressure and Volume Changed Pressure and Volume

Boyle’s Law Practice Problem #1 A gas occupies a volume of 458 mL at a pressure of 1.01 kPa and temperature of 295 K. When the pressure is changed, the volume becomes 477 mL, If there has been no change in temperature, what is the new pressure? Practice Problem #1 A gas occupies a volume of 458 mL at a pressure of 1.01 kPa and temperature of 295 K. When the pressure is changed, the volume becomes 477 mL, If there has been no change in temperature, what is the new pressure? P1P1P1P1 V1V1V1V1 V2V2V2V2 P 1 V 1 = P 2 V 2

Boyle’s Law P 1 V 1 = P 2 V 2 P 2 = (1.01 kPa) (458 mL) (477 mL) (1.01 kPa)(458 mL) = P 2 (477 mL) P 2 = kPa

V T Charle’s Law

V T The volume of a fixed amount of gas is directly proportional to its absolute temperature. (at constant pressure)

Charle’s Law No sound … it is easy to see that as the gas TEMPERATURE INCREASES, the VOLUME INCREASES!

Charle’s Law = Initial Volume and Temperature Changed Volume and Temperature V1V1V1V1 T2T2T2T2 V2V2V2V2 T1T1T1T1

Charle’s Law Practice Problem #6 What will be the volume of a gas sample at 309 K if its volume at 215 K is 3.42 L Practice Problem #6 What will be the volume of a gas sample at 309 K if its volume at 215 K is 3.42 L T1T1T1T1 V1V1V1V1 T2T2T2T2 = V1V1V1V1 T2T2T2T2 V2V2V2V2 T1T1T1T1

Charle’s Law = V1V1V1V1 T2T2T2T2 V2V2V2V2 T1T1T1T L 215 K 309 K = V2V2V2V2 V2V2V2V2 (3.42 L) (309 K) = 215 K = 4.92 L

Dalton’s Law P total = P 1 + P 2 + P 3... The sum of the partial pressures of all the components in a gas mixture is equal to the total pressure of the gas mixture.

Dalton’s Law Practice Problem #13 What is the pressure of a mixture of helium, nitrogen, and oxygen if their partial pressures are 600. mm Hg, 150. mm Hg, and 102 mm Hg? Practice Problem #13 What is the pressure of a mixture of helium, nitrogen, and oxygen if their partial pressures are 600. mm Hg, 150. mm Hg, and 102 mm Hg? P1P1P1P1 P2P2P2P2 P3P3P3P3 P total = P 1 + P 2 + P 3

Dalton’s Law Practice Problem #13 What is the pressure of a mixture of helium, nitrogen, and oxygen if their partial pressures are 600. mm Hg, 150. mm Hg, and 102 mm Hg? Practice Problem #13 What is the pressure of a mixture of helium, nitrogen, and oxygen if their partial pressures are 600. mm Hg, 150. mm Hg, and 102 mm Hg? P total = P 1 + P 2 + P 3 P total = P total = 852 mm Hg

*Dalton’s Law Practice Problem #16 The barometer shows the atmospheric pressure to be 762 mm Hg. What is the partial pressure of nitrogen if nitrogen makes up 78% of the air. Practice Problem #16 The barometer shows the atmospheric pressure to be 762 mm Hg. What is the partial pressure of nitrogen if nitrogen makes up 78% of the air. 762 mm Hg (0.78) = 594 mm Hg

Gay-Lussac’s Law The pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature. = P1P1 T1T1 T2T2 P2P2

= kPV PTPT VTVT T Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

Avogadro’s Law V = k 3 n Equal volumes of gases at the same temperature and pressure contain an equal number of particles. volume Constant (molar volume of a gas at STP) Number of moles

If you keep pressure and temperature constant, what is the only way to increase the volume? Avogadro’s Law V = k 3 n BY INCREASING THE NUMBER OF PARTICLES!