Presentation on theme: "EMPIRICAL GAS LAWS Boyles Law P 1 V 1 = P 2 V 2 Charles LawV 1 / T 1 = V 2 / T 2 Guy-Lussacs LawP 1 / T 1 = P 2 / T 2 Avogadros LawV 1 / n 1 = V 2 / n."— Presentation transcript:
EMPIRICAL GAS LAWS Boyles Law P 1 V 1 = P 2 V 2 Charles LawV 1 / T 1 = V 2 / T 2 Guy-Lussacs LawP 1 / T 1 = P 2 / T 2 Avogadros LawV 1 / n 1 = V 2 / n 2 Combined Gas Law P 1 V 1 / T 1 = P 2 V 2 / T 2 Ideal Gas Law PV = nRT P = pressure (atm)V = volume (L) n = chemical amount (mol)T = Temperature (K) R = ideal gas constant = L-atm / mol-K
Avogadros 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 Avogadros Hypothesis, the balloon filled with one mole of He will occupy that same volume as a balloon filled with one mole of Xe. V nSo for a gas, the volume and the moles are directly related. V n
Avogadros Hypothesis A sample of N 2 gas at 3.0 atm and 20.0 o C is known to occupy a volume of 1.43 L. What volume would a mole sample of NH 3 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 = L-atm/mol-K, and T = 20.0 o C = 293K n = PV / RT = (3.0 atm x 1.43L) / (0.082 L-atm/mol-K x 293K) = moles of N2N2 So since the moles of N2 N2 is mol and the moles of ammonia is mol according to Avogadros hypothesis the volume of NH 3 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 & T = 273 K
EMPIRICAL GAS LAWS & STP 1. The gases in a rigid Helium filled ball at 25 o C 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 21 o C & 760 mmHg had a volume of mL. What would be the volume of the gas under STP conditions? 4. How many atoms are present in 1.0 L of SO 3 gas at STP?
IDEAL GAS LAW Q. What is the pressure inside a gas balloon if it filled with 852 g of Xe gas at 25.0 o C and occupies a volume of 7.00 L? P = ? 852 g Xe ( 1 mol / 131 g) = 6.50 mol V = 7.00 L T = 25 o C = 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 atm. ·How much would a balloon weigh if it contained 40.0 L of O 2 gas at 987 mmHg and 45.3 o C? T 2 = 1914 o C mass O 2 = 63.7 g mass O 2 = 63.7 g
DENSITY OF A GAS The density of a gas at STP can be calculated by d STP = 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 g/L at STP. d (STP) = (82 g/mol) / 22.4 L/mol) = 3.66 g/L Cl 2
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 o C. ·Identify an unknown homonuclear diatomic gas that was found to have a density of g/L at 2.57 atm and 177 o C. d = (82 g/mol*0.772 atm) / (0.082 L-atm/mol-K*330K) = 2.34 g/L N2N2N2N2
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 lighter the gas, the faster it moves. The General Rule is: The lighter the gas, the faster it moves. Grahams 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.
DALTONS 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. P total = P gas A + P gas B + P gas 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 O 2 and 1.48 atm of N 2 ? P T - P O2 - P N2 = P CO 2.5 atm atm atm = 0.62 atm
STOICHIOMETRY & THE GAS LAWS. Write a balanced chemical equation 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 O 2 is needed to combust L of C 3 H 8 ? C 3 H O 2 3 CO H 2 O Due to Avogadros 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 L C 3 H 8 (5 mol O 2 / 1 mol C 3 H 8 ) = 1740 L O 2
STOICHIOMETRY & THE GAS LAWS 2. How many grams of CO 2 is produced from L of C 3 H 8 if the temperature is 40.0 o C and the pressure is 654 torr? C 3 H O 2 3 CO H 2 O P = 654 torr (1 atm / 760 torr) = atm T = 40 o C = 313 K PV / RT = n = (0.861 atm) (348.0 L) /(0.082 L-atm/mol-K) (313 K) = mol of C 3 H mol C 3 H 8 (3 mol CO 2 / 1 mol C 3 H 8 ) = mol CO g of CO mol CO 2 (44 g / 1 mol) = 1541 g of CO 2
STOICHIOMETRY & THE GAS LAWS 3. In lab, you decomposed potassium chlorate into oxygen and potassium chloride. What volume of O 2 at STP can be formed from 3.65 g of potassium chlorate? 2 KClO 3 3 O KCl 3.65 g (1 mol / 122.6g) = mol KClO g (1 mol / 122.6g) = mol KClO mol KClO 3 (3 mol O 2 / 2 mol KClO 3 ) = mol O mol KClO 3 (3 mol O 2 / 2 mol KClO 3 ) = mol O mol O 2 ( 22.4 L / 1 mol) = 1.00 L mol O 2 ( 22.4 L / 1 mol) = 1.00 L
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 98 o C 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 23 o C. 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 CaCO 3 was required to neutralize the spill. CaCO 3 (s) + 2HCl (aq) CaCl 2 (aq) + H 2 O (l) + CO 2 (g) What volume of CO 2 gas was released by the neutralization at 735 mmHg and 20 o C? Hydrogen effuses first by a factor of g/mol SO g 745 L
Group Study Problem # 20b 1. Measured at 65 o C and torr, the mass of 3.21 L of a gas is 3.5 g. What is the molar mass of this gas 2. A mL sample of air is analyzed and found to contain g N 2, g CO 2 and g O 2 at 35 o C. What is the total pressure of the sample and the partial pressure of each component? 3. What volume would 5.30 L of H 2 gas at STP occupy if the temperature was increased to 70 o F and the pressure to 830 torr? 4. Carbon monoxide is produced by the reaction of coke with oxygen from preheated air. 2 C + O 2 2 CO. How many liters of atmospheric oxygen at an effective pressure of 182 torr and a temperature of 29.0 o C are required to produce 895 L of carbon monoxide at 846 torr and 1700 o C? 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 o C and the pressure is atm?
Group Study Problem Answers: g/mol 2. P N 2 =7.53 atm, P O 2 =1.55 atm, P CO 2 =0.380 atm; P T = 9.46 atm L L L