 Download Download Presentation EMPIRICAL GAS LAWS Boyle’s Law P 1 V 1 = P 2 V 2 Charles’ Law V 1 / T 1 = V 2 / T 2 Guy-Lussac’s Law P 1

# EMPIRICAL GAS LAWS Boyle’s Law P 1 V 1 = P 2 V 2 Charles’ Law V 1 / T 1 = V 2 / T 2 Guy-Lussac’s Law P 1

Download Presentation ## EMPIRICAL GAS LAWS Boyle’s Law P 1 V 1 = P 2 V 2 Charles’ Law V 1 / T 1 = V 2 / T 2 Guy-Lussac’s Law P 1

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1. EMPIRICAL GAS LAWS Boyle’s Law P1V1 = P2V2 Charles’ Law V1 / T1 = V2 / T2 Guy-Lussac’s Law P1 / T1 = P2 / T2 Avogadro’s Law V1 / n1 = V2 / n2 Combined Gas Law P1V1 / T1 = P2 V2 / T2 Ideal Gas Law PV = nRT P = pressure (atm) V = volume (L) n = chemical amount (mol) T = Temperature (K) R = ideal gas constant = 0.08206 L-atm / mol-K

2. 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

3. 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!!!

4. At STP, gas molecules are so far apart that for 1 mole of gas, the overall volume does not change.STP : P = 1 atm & T = 273 K

5. EMPIRICAL GAS LAWS & STP 1. The gases in a rigid Helium filled ball at 25oC exerts a pressure of 1.2 atm. If the ball is placed in a freezer and the pressure decreases to 1/8 of its original value, what is the temperature inside the ball? 2. A balloon containing 2.50 moles of He has a volume of 850 mL at a certain temperature and pressure. How many grams of He would have to be removed from the balloon in order for the volume to decrease to 250 mL under the same conditions? 3. A sample of Ne gas at 21oC & 760 mmHg had a volume of 250.0 mL. What would be the volume of the gas under STP conditions? 4. How many atoms are present in 1.0 L of SO3 gas at STP?

6. 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

7. 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

8. DENSITY OF A GAS 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

9. DENSITY OF A GAS The density of a gas not at STP can be calculated by d = (MM) P / R T • Calculate the density of hydrogen sulfite gas at 587 torr and 56.9 oC. • Identify an unknown homonuclear diatomic gas that was found to have a density of 1.950 g/L at 2.57 atm and 177oC. d = (82 g/mol*0.772 atm) / (0.082 L-atm/mol-K*330K) = 2.34 g/L N2

10. Properties of Gases • DIFFUSION Diffusion is the ability of two or more gases to mix spontaneously until a uniform mixture is formed. Example: A person wearing a lot of perfume walks into an enclosed room, eventually in time, the entire room will smell like the perfume. • EFFUSION • Effusion is the ability of gas particles to pass through a small opening or membrane from a container of higher pressure to a container of lower pressure. • The General Rule is: The lighter the gas, the faster it moves. • Graham’s Law of Effusion: • Rate of effusion of gas A = √(molar mass B / molar mass A) • Rate of effusion of gas B • The rate of effusion of a gas is inversely proportional to the square root of the molar mass of that gas.

11. 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 = PgasA + P gasB + PgasC … 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

12. 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

13. 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

14. 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

15. PRACTICE PROBLEM # 20b 1. Both hydrogen and helium have been used as buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor? 2. At STP, 560 mL of a gas has a mass of 1.08 g. What is the molecular weight of the gas? 3. 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? 4. 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? 5. 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? Hydrogen effuses first by a factor of 1.41 43.2 g/mol SO3 986 g 745 L

16. Group Study Problem # 20b 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?

17. 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