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

Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2

Problem 1 Solve the combined gas laws for T 2.

Solution for Problem 1 Solve the combined gas law for T 2. (Hint: cross-multiply first.) P 1 V 1 = P 2 V 2 T 1 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 T 2 = P 2 V 2 T 1 P 1 V 1

Combined 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?

Data Table Set up Data Table P 1 = atm V 1 = L T 1 = 302 K P 2 = 3.20 atm V 2 = 90.0 mL T 2 = ?? ??

Solution Solve for T 2 Enter data T 2 = 302 K x atm x mL = K atm mL T 2 = K = °C

Calculation Solve for T 2 T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K atm mL T 2 = 604 K = 331 °C

Try this problem A 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?

Solution T 1 = 308 KT 2 = ? V 1 = 675 mLV 2 = L = 315 mL P 1 = atm P 2 = 802 mm Hg = 646 mm Hg T 2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec

Solution T 1 = 308 KT 2 = ? V 1 = 675 mLV 2 = L = 315 mL P 1 = atm P 2 = 802 mm Hg = 646 mm Hg T 2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec = 178 K = - 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?

Answer these questions 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 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 When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V 1 = V 2 n 1 n 2 initial final

STP The volumes of gases can be compared when they have the same temperature and pressure. If the temperature and pressure are: Standard temperature 0°C or 273 K Standard pressure 1 atm (760 mm Hg) then, it is represented by STP.

Try this problem 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? P 1 = V 1 = T 1 = K P 2 = V 2 = ?? T 2 = K V 2 = 15 L x atm x K = 6.8 L atm K

Solution P 1 = 1.0 atm V 1 = 15 L T 1 = 273 K P 2 = 2.0 atm V 2 = ?? T 2 = 248 K V 2 = 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 CH g CO 2 1 mole 1 mole1mole (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

Try these problems A.What is the volume at STP of 4.00 g of CH 4 ? 1) 5.60 L2) 11.2 L3) 44.8 L B. How many grams of He are present in 8.0 L of gas at STP? 1) 25.6 g2) g3) 1.43 g

Solution A.What is the volume at STP of 4.00 g of CH 4 ? 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 16.0 g CH 4 1 mole CH 4 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 Partial Pressure 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. P T = P 1 + P 2 + P

Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N mmHg 20.95% O mmHg 0.94% Ar 7.1 mmHg 0.03% CO mmHg P AIR = P N + P O + P Ar + P CO = 760 mmHg Total Pressure760 mm Hg

Dalton’s Law Check A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 1) ) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557 2) 9.143) 0.109

Solution A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 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 0.5 mole O mole He mole Ar 1 mole H 2