Guy-Lussac’s Law P1 / T1 = P2 / T2 EMPIRICAL GAS LAWS Boyle’s Law P1V1 = P2V2 Charles’ Law V1 / T1 = V2 / T2 Guy-Lussac’s Law P1 / T1 = P2 / T2 Combined Gas Law P1V1 / T1 = P2 V2 / T2 Ideal Gas Law PV = nRT Avogadro’s Law V1 / n1 = V2 / n2 Dalton’s Law Ptotal = Pgas A + Pgas B +… Dalton’s Law Pgas A= (n A/n total ) x P total Molar volume STP 22.4 L/mol Molar volume SATP 24.8 L/mol

Avogadro’s Hypothesis Avogadro pictured the moving molecule as occupying a small portion of the larger space apparently occupied by the gas. Thus the “volume” of the gas is related to the spacing between particles and not to the particle size itself. Imagine 3 balloons each filled with a different gas (He, Ar, & Xe). These gases are listed in increasing particle size, with Xe being the largest atom. According to Avogadro’s Hypothesis, the balloon filled with one mole of He will occupy that same volume as a balloon filled with one mole of Xe. So for a gas, the “volume” and the moles are directly related. V a n

Avogadro’s Hypothesis A sample of N2 gas at 3.0 atm and 20.0oC is known to occupy a volume of 1.43 L. What volume would a 0.179 mole sample of NH3 gas occupy at the same pressure and temperature? First calculate the number of moles of nitrogen gas: PV = nRT where P = 3.0 atm, V = 1.43 L, R = 0.082 L-atm/mol-K, and T = 20.0 oC + 273 = 293K n = PV / RT = (3.0 atm x 1.43L) / (0.082 L-atm/mol-K x 293K) = 0.179 moles of N2 So since the moles of N2 is 0.179 mol and the moles of ammonia is 0.179 mol according to Avogadro’s hypothesis the volume of NH3 at that pressure and that temperature is 1.43 L, the same!!!

At STP, gas molecules are so far apart that for 1 mole of gas, the overall volume does not change. STP : P = 1 atm or 101 kPa & T = 273 K (0oC)

At SATP, gas molecules are so far apart that for 1 mole of gas, the overall volume does not change. SATP : P = 100 kPA & T = 298 K

Implications of Avogadro’s Law: 1- When a chemical reaction is performed and more than one of the reactants/products are gases, they will all be at the SAME TEMPERATURE and PRESSURE (lab conditions) … simplifying from PV=nRT, it means that the volume of the gas is proportional to the number of moles V1/n1= V2/n2

Implications of Avogadro’s Law: 1- EXAMPLE:V1/n1= V2/n2 30.0 mL of nitrogen gas at STP reacts with hydrogen gas to make ammonia gas(N2 + 3H2 = 2NH3. How many mL of ammonia are created in those conditions? 2 ways to solve: A-Can use PV=nRT, solve for n, use mole ratios and then PV=nRT, solve for V B- Can use V1/n1= V2/n2

Implications of Avogadro’s Law: 2- If two different gases are placed in the same container, one after another, they will both expand and fill the container and be at the same temperature and pressure. Thus if volume of the container is known and the mass of the second gas is found, then the molar mass of the second gas (and hence the identity) can be deduced. In this case, V1 and V2 are the same (same container) and so n of one gas is the same as n for the other gas (V1/n1=V2/n2)

Implications of Avogadro’s Law: 2- Fluorine gas is placed in a 44.8L container at STP. The same container is emptied and refilled with a mysterious gas under the same atmospheric conditions. The mysterious gas in the same container weighs 40.0g. Determine the molar mass of the mysterious gas. Solve for n in PV=nRT for the fluorine gas. The key is that n is the same for the mysterious gas. Since we know m and n, we can use find mm since mm=m/n.

IDEAL GAS LAW . What is the pressure inside a gas balloon if it filled with 852 g of Xe gas at 25.0oC and occupies a volume of 7.00 L? P = ? 852 g Xe ( 1 mol / 131 g) = 6.50 mol V = 7.00 L T = 25oC + 273 = 298 K P = nRT V P = (6.50 mol) (0.082 L-atm / mol-K) (298 K) 7.00 L P = 22.7 atm

IDEAL GAS LAW What would be the temperature of 100 g of Ar gas contained in a 500 L sealed container at 0.8976 atm. How much would a balloon weigh if it contained 40.0 L of O2 gas at 987 mmHg and 45.3 oC? T2 = 1914 oC mass O2 = 63.7 g

The density of a gas at STP can be calculated by dSTP = molar mass/molar volume Calculate the density of hydrogen sulfite gas at STP. Identify an unknown homonuclear diatomic gas that was found to have a density of 3.165 g/L at STP. d (STP) = (82 g/mol) / 22.4 L/mol) = 3.66 g/L Cl2

DALTON’S LAW OF PARTIAL PRESSURES If there is more than one gas present in a container, each gas contributes to the total pressure of the mixture. Ptotal = Pgas A + P gas B + Pgas C … If the total pressure of a system was 2.5 atm, what is the partial pressure of carbon monoxide if the gas mixture also contained 0.4 atm O2 and 1.48 atm of N2? PT - PO2 - PN2 = PCO 2.5 atm - 0.4 atm - 1.48 atm = 0.62 atm

Implications of DALTON’S LAW OF PARTIAL PRESSURES If a gas is collected by the downward displacement of water, the gas collected is a mixture of the gas you want and water vapour (small amount)…it is a mixture Ptotal = Pgas A + P water vapour If the atmospheric conditions were 101 kPa, then the pressure of the gas is 101kPa minus the pressure of the water vapour…consult a chart for values…it is usually small like 2 or 3 kPa PT – PH20 = Pgas 101kPa – 2 kPa = 99 kPa

Implications of DALTON’S LAW OF PARTIAL PRESSURES If a mixture of gases is found in one container, they are all under the same conditions of V and T Ptotal = Pgas A + P gas B… ntotal = ngas A + n gas B… Pgas A = (n gas A /n total) x P total Pgas B = (n gas B /n total) x P total The mole fraction (moles of A/ total moles) can be used to find the partial pressure of each gas.

Implications of DALTON’S LAW OF PARTIAL PRESSURES Example: 32 g of oxygen gas (mm is 32g/mol) and 32 g of hydrogen gas (mm is 2.0 g/mol) are mixed together in a 25.0 L container. The pressure is found to be 100.0 kPa. Find the pressure due to oxygen gas and the pressure due to hydrogen gas. PO2 = (1.00 moles /(1.00 moles + 16moles)) x 100kPa PH2 = (16.0 moles /(1.00 moles + 16moles)) x 100kPa Notice that even if equal GRAMS of each substance are mixed, it does not mean 50 kPa of pressure for each. The key in chemistry is moles and not mass.

STOICHIOMETRY & THE GAS LAWS 1. Write a balanced chemical equation 2. Convert to moles (if gas, use PV=nRT or Molar Volume) 3. Use the mole ratio to convert from moles of “A” to moles of “B”. 4. Convert moles of “B” to desired measurement, if a gas use PV=nRT. 1. What volume of O2 is needed to combust 348.0 L of C3H8? C3H8 + 5 O2  3 CO2 + 4 H2O Due to Avogadro’s Hypothesis, the moles of a gas are directly related to the volume of a gas therefore it is possible to use the mole ratio on volumes of gas. 348.0 L C3H8 (5 mol O2 / 1 mol C3H8) = 1740 L O2

STOICHIOMETRY & THE GAS LAWS 2. How many grams of CO2 is produced from 348.0 L of C3H8 if the temperature is 40.0oC and the pressure is 654 torr? C3H8 + 5 O2  3 CO2 + 4 H2O P = 654 torr (1 atm / 760 torr) = 0.861 atm T = 40oC + 273 = 313 K PV / RT = n = (0.861 atm) (348.0 L) /(0.082 L-atm/mol-K) (313 K) = 11.67 mol of C3H8 11.67 mol C3H8 (3 mol CO2 / 1 mol C3H8) = 35.02 mol CO2 35.02 mol CO2 (44 g / 1 mol) = 1541 g of CO2

STOICHIOMETRY & THE GAS LAWS 3. In lab, you decomposed potassium chlorate into oxygen and potassium chloride. What volume of O2 at STP can be formed from 3.65 g of potassium chlorate? 2 KClO3  3 O2 + 2 KCl 3.65 g (1 mol / 122.6g) = 0.02977 mol KClO3 0.02977 mol KClO3 (3 mol O2 / 2 mol KClO3) = 0.04466 mol O2 0.04466 mol O2 ( 22.4 L / 1 mol) = 1.00 L

CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) + H2O (l) + CO2 (g) PRACTICE PROBLEMS 1. At STP, 560 mL of a gas has a mass of 1.08 g. What is the molecular weight of the gas? 2. A gas is known to be either sulfur dioxide or sulfur trioxide. Its density at 98oC and 1.08 atm is 2.84 g/L. Which gas is it? 3. A 50.0 L cylinder of nitrogen has a pressure of 17.1 atm at 23oC. What is the mass of nitrogen in the cylinder? 4. When a 2.0 L bottle of concentrated HCl was spilled, 3.0 kg of CaCO3 was required to neutralize the spill. CaCO3 (s) + 2HCl (aq)  CaCl2 (aq) + H2O (l) + CO2 (g) What volume of CO2 gas was released by the neutralization at 735 mmHg and 20 oC? 43.2 g/mol SO3 986 g 745 L

Group Study Problems 1. Measured at 65 oC and 500.0 torr, the mass of 3.21 L of a gas is 3.5 g. What is the molar mass of this gas 2. A 100.0 mL sample of air is analyzed and found to contain 0.835 g N2, 0.0640 g CO2 and 0.197 g O2 at 35 oC. What is the total pressure of the sample and the partial pressure of each component? 3. What volume would 5.30 L of H2 gas at STP occupy if the temperature was increased to 70oF and the pressure to 830 torr? 4. Carbon monoxide is produced by the reaction of coke with oxygen from preheated air. 2 C + O2  2 CO . How many liters of atmospheric oxygen at an effective pressure of 182 torr and a temperature of 29.0oC are required to produce 895 L of carbon monoxide at 846 torr and 1700oC? 5. Hydrogen gas is produced by the complete reaction of 8.34 g of aluminum metal with an excess of gaseous hydrogen sulfate. How many liters of hydrogen will be produced if the temperature is 50.0 oC and the pressure is 0.950 atm?

Group Study Problem Answers: 1. 45.9 g/mol 2. PN2=7.53 atm, PO2=1.55 atm, PCO2=0.380 atm; PT = 9.46 atm 3. 5.23 L 4. 320 L 5. 12.9 L