Combined Gas Law and Avogadro’s Hypothesis

Combined Gas Law Recall…. P x V = constant And V/T = constant P/T = constant It follows then that……

Combined Gas Law The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. P1 V1 P2 V2 = T1 T2

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

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 Problems 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? Set up Data Table P1 = 0.800 atm V1 = 0.180 L T1 = 302 K P2 = 3.20 atm V2= 0.090 L T2 = ??

Calculation = 604 K P1 = 0.800 atm V1 = 180 mL T1 = 302 K P1 V1 P2 V2 = P1 V1 T2 = P2 V2 T1 T1 T2 T2 = P2 V2 T1 P1 V1 T2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL T2 = 604 K - 273 = 331 °C = 604 K

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

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 V a number of moles (n) V = constant x n V1 = V2 n1 n2 When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V a number of moles (n) V = constant x n V1 = V2 n1 n2 initial final

Avogadro’s Law (continued) Equal volumes of gases at same temperature and pressure have equal numbers of molecules. Gas molecules may break up when they react.

4NH3 + 5O2 4NO + 6H2O 4 mole NH3 4 mole NO At constant T and P Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH3 + 5O2 4NO + 6H2O 4 mole NH3 4 mole NO At constant T and P 1 volume NH3 1 volume NO

New Combined Gas Law Boyle’s law V a 1/P Charles’s law V a T Avogadro’s law V a n Therefore it follows that…… V a nT P

The General Gas Equation P1V1 n1T1 = P2V2 n2T2 = P2 T2 P1 T1 If we hold the number of moles and volume constant:

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) or 101.3 kPa

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 = 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