4Learning Check C1Solve the combined gas laws for T2.
5Solution C1 Solve the combined gas law for T2. (Hint: cross-multiply first.)P1V1 = P2V2T T2P1V1T2 = P2V2T1T2 = P2V2T1P1V1
6Combined Gas Law Problem A sample of helium gas has a volume of L, a pressure of 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?
7Data Table Set up Data Table P1 = 0.800 atm V1 = 0.180 L T1 = 302 K P2 = atm V2= 90.0 mL T2 = ????
8Solution Solve for T2 T2 = 302 K x atm x mL = K atm mL Enter dataT2 = 302 K x atm x mL = Katm mLT2 = K = °C
9Calculation Solve for T2 T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm mLT2 = 604 K = °C
10Learning Check C2A gas has a volume of 675 mL at 35°C and atm pressure. What is the temperature in °C when the gas has a volume of L and a pressure of 802 mm Hg?
11Solution G9 T1 = 308 K T2 = ? V1 = 675 mL V2 = 0.315 L = 315 mL P1 = atm P2 = 802 mm Hg= 646 mm HgT2 = 308 K x mm Hg x mL646 mm Hg mLP inc, T inc V dec, T dec= K = - 95°C
12Volume and MolesHow 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?
13Learning 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 T3.___ At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.
14Solution 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 T3. (1) At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.
15Avogadro’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 gasV = V2n n2initial final
16STPThe volumes of gases can be compared when they have the same temperature and pressure (STP).Standard temperature 0°C or 273 KStandard pressure 1 atm (760 mm Hg)
17Learning 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 = KP2 = V2 = ?? T2 = KV2 = 15 L x atm x K = 6.8 Latm K
18Solution C4 P1 = 1.0 atm V1 = 15 L T1 = 273 K P2 = 2.0 atm V2 = ?? T2 = 248 KV2 = 15 L x atm x K = 6.8 L2.0 atm K
19Molar 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
20Molar Volume Factor 1 mole of a gas at STP = 22.4 L 22.4 L and 1 mole 1 mole L
21Learning Check C5 A.What is the volume at STP of 4.00 g of CH4? 1) L 2) L 3) 44.8 LB. How many grams of He are present in 8.0 L of gas at STP?1) g 2) g 3) 1.43 g
22Solution 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) = L16.0 g CH mole CH4B. How many grams of He are present in 8.0 L of gas at STP?8.00 L x 1 mole He x g He = g He22.4 He mole He
23Daltons’ Law of Partial Pressures Pressure each gas in a mixture would exert if it were the only gas in the containerDalton's Law of Partial PressuresThe total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture.PT = P P P
24Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N mmHg20.95% O mmHg0.94% Ar mmHg0.03% CO mmHgPAIR = PN + PO + PAr + PCO = 760 mmHgTotal Pressure mm Hg
25Learning Check C6A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?1) ) ) 760B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?1) ) )
26Solution C6A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?2) 156B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?1) 557
27Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not onthe types of particles.P = atm P = atm1 mole H20.5 mole O2+ 0.3 mole He+ 0.2 mole Ar
28Health NoteWhen a scuba diver is several hundred feetunder 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.
29Learning Check C7A 5.00 L scuba tank contains 1.05 mole of O2 and mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?
30Solution C7 P = nRT PT = PO + PHe V 2 PT = mol x L-atm x 298 K5.00 L (K mol)= 7.19 atm