Gases The Combined Gas Law Volume and Moles (Avogadro’s Law) Partial Pressures

Combined Gas Law T1 T2 Rearrange the combined gas law to solve for V2 P1V1 = P2V2 T1 T2 Rearrange the combined gas law to solve for V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1

Combined Gas Law T1 T2 Isolate V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1 P1V1 = P2V2 T1 T2 Isolate V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1

Learning Check C1 Solve the combined gas laws for T2.

Solution C1 Solve the combined gas law for T2. (Hint: cross-multiply first.) P1V1 = P2V2 T1 T2 P1V1T2 = P2V2T1 T2 = P2V2T1 P1V1

Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Data Table Set up Data Table P1 = 0.800 atm V1 = 0.180 L T1 = 302 K P2 = 3.20 atm V2= 90.0 mL T2 = ?? ??

Solution Solve for T2 T2 = 302 K x atm x mL = K atm mL Enter data T2 = 302 K x atm x mL = K atm mL T2 = K - 273 = °C

Calculation Solve for T2 T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T2 = 604 K - 273 = 331 °C

Learning Check C2 A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

Solution G9 T1 = 308 K T2 = ? V1 = 675 mL V2 = 0.315 L = 315 mL P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg T2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec = 178 K - 273 = - 95°C

Volume and Moles How does adding more molecules of a gas change the volume of the air in a tire? If a tire has a leak, how does the loss of air (gas) molecules change the volume?

Learning Check C3 True (1) or False(2) 1.___The P exerted by a gas at constant V is not affected by the T of the gas. 2.___ At constant P, the V of a gas is directly proportional to the absolute T 3.___ At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

Solution C3 True (1) or False(2) 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

Avogadro’s Law V1 = V2 n1 n2 initial final When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V1 = V2 n1 n2 initial final

STP The volumes of gases can be compared when they have the same temperature and pressure (STP). Standard temperature 0°C or 273 K Standard pressure 1 atm (760 mm Hg)

Learning Check C4 P1 = V1 = T1 = K P2 = V2 = ?? T2 = K A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C? P1 = V1 = T1 = K P2 = V2 = ?? T2 = K V2 = 15 L x atm x K = 6.8 L atm K

Solution C4 P1 = 1.0 atm V1 = 15 L T1 = 273 K P2 = 2.0 atm V2 = ?? T2 = 248 K V2 = 15 L x 1.0 atm x 248 K = 6.8 L 2.0 atm 273 K

Molar Volume At STP 4.0 g He 16.0 g CH4 44.0 g CO2 1 mole 1 mole 1mole (STP) (STP) (STP) V = 22.4 L V = 22.4 L V = 22.4 L

Molar Volume Factor 1 mole of a gas at STP = 22.4 L 22.4 L and 1 mole 1 mole 22.4 L

Learning Check C5 A.What is the volume at STP of 4.00 g of CH4? 1) 5.60 L 2) 11.2 L 3) 44.8 L B. How many grams of He are present in 8.0 L of gas at STP? 1) 25.6 g 2) 0.357 g 3) 1.43 g

Solution C5 A.What is the volume at STP of 4.00 g of CH4? 4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L 16.0 g CH4 1 mole CH4 B. How many grams of He are present in 8.0 L of gas at STP? 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 He 1 mole He

Daltons’ Law of Partial Pressures Pressure each gas in a mixture would exert if it were the only gas in the container Dalton's Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. PT = P1 + P2 + P3 + .....

Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N2 593.4 mmHg 20.95% O2 159.2 mmHg 0.94% Ar 7.1 mmHg 0.03% CO2 0.2 mmHg PAIR = PN + PO + PAr + PCO = 760 mmHg 2 2 2 Total Pressure 760 mm Hg

Learning Check C6 A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air? 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air? 1) 557 2) 9.14 3) 0.109

Solution C6 A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air? 1) 557

Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. P = 1.00 atm P = 1.00 atm 1 mole H2 0.5 mole O2 + 0.3 mole He + 0.2 mole Ar

Health Note When a scuba diver is several hundred feet under water, the high pressures cause N2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in scuba tanks used for deep descents.

Learning Check C7 A 5.00 L scuba tank contains 1.05 mole of O2 and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?

Solution C7 P = nRT PT = PO + PHe V 2 PT = 1.47 mol x 0.0821 L-atm x 298 K 5.00 L (K mol) = 7.19 atm