Presentation on theme: "Dr. S. M. Condren Chapter 12 BEHAVIOR OF GASES Dr. S. M. Condren BEHAVIOR OF GASES."— Presentation transcript:
Dr. S. M. Condren Chapter 12 BEHAVIOR OF GASES
Dr. S. M. Condren BEHAVIOR OF GASES
Dr. S. M. Condren Importance of Gases Airbags fill with N 2 gas in an accident.Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3.Gas is generated by the decomposition of sodium azide, NaN 3. 2 NaN 3 2 Na + 3 N 22 NaN 3 2 Na + 3 N 2 if bag ruptures 2 Na + 2 H 2 O 2 NaOH + H 2if bag ruptures 2 Na + 2 H 2 O 2 NaOH + H 2
Dr. S. M. Condren THREE STATES OF MATTER
Dr. S. M. Condren THREE STATES OF MATTER
Dr. S. M. Condren General Properties of Gases There is a lot of free space in a gas.There is a lot of free space in a gas. Gases can be expanded infinitely.Gases can be expanded infinitely. Gases occupy containers uniformly and completely.Gases occupy containers uniformly and completely. Gases diffuse and mix rapidly.Gases diffuse and mix rapidly.
Dr. S. M. Condren Properties of Gases Gas properties can be modeled using math. Model depends on V = volume of the gas (L) T = temperature (K) n = amount (moles) P = pressure (atmospheres)
Dr. S. M. Condren Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643)
Dr. S. M. Condren Pressure Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). P of Hg pushing down related to Hg densityHg density column heightcolumn height
Dr. S. M. Condren Pressure Column height measures P of atmosphere 1 standard atm = 760 mm Hg1 standard atm = 760 mm Hg = 29.9 inches Hg = about 34 feet of water SI unit is PASCAL, Pa, where 1 atm = kPa
Dr. S. M. Condren Effect of Pressure Differential
Dr. S. M. Condren IDEAL GAS LAW Brings together gas properties. Can be derived from experiment and theory. P V = n R T
Dr. S. M. Condren Boyles Law If n and T are constant, then PV = (nRT) = k This means, for example, that P goes up as V goes down. Robert Boyle ( ). Son of Earl of Cork, Ireland.
Dr. S. M. Condren A bicycle pump is a good example of Boyles law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire. Boyles Law
Dr. S. M. Condren Boyles Law
Dr. S. M. Condren Charless Law If n and P are constant, then V = (nR/P)T = kT V and T are directly related. Jacques Charles ( ). Isolated boron and studied gases. Balloonist.
Dr. S. M. Condren Charless original balloon Modern long-distance balloon
Dr. S. M. Condren Charless Law Balloons immersed in liquid N 2 (at -196 ˚C) will shrink as the air cools (and is liquefied).
Dr. S. M. Condren Charless Law
Dr. S. M. Condren Hypothesis Avogadros Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related. twice as many molecules
Dr. S. M. Condren The gases in this experiment are all measured at the same T and P. 2 H 2 (g) + O 2 (g) 2 H 2 O(g) Avogadros Hypothesis
Dr. S. M. Condren Combining the Gas Laws V proportional to 1/PV proportional to 1/P V prop. to TV prop. to T V prop. to nV prop. to n Therefore, V prop. to nT/PTherefore, V prop. to nT/P V = 22.4 L for 1.00 mol whenV = 22.4 L for 1.00 mol when Standard pressure and temperature (STP) ST = 273 K SP = 1.00 atm
Dr. S. M. Condren Using PV = nRT How much N 2 is reqd to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = Latm/Kmol Solution 1. Get all data into proper units V = 27,000 L V = 27,000 L T = 25 o C = 298 K T = 25 o C = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm memorize
Dr. S. M. Condren Using PV = nRT How much N 2 is reqd to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? R = Latm/Kmol Solution 2. Now calc. n = PV / RT n = 1.1 x 10 3 mol (or about 22 kg of gas)
Dr. S. M. Condren Ideal Gas Constant R = L*atm/mol*K R has other values for other sets of units. R = mL*atm/mol*K = J/mol*K = cal/mol*K
Dr. S. M. Condren Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution Strategy: Calculate moles of H 2 O 2 and then moles of O 2 and H 2 O. Finally, calc. P from n, R, T, and V.
Dr. S. M. Condren Gases and Stoichiometry 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the pressure of O 2 at 25 o C? Of H 2 O? Solution #mol H 2 O 2 = 1.1g H 2 O 2 (1mol/ 34.0g H 2 O 2 ) = mol H 2 O 2 #mol O 2 = (0.032mol H 2 O 2 )(1mol O 2 /2mol H 2 O 2 ) = 0.016mol O 2 P of O 2 = nRT/V = (0.016mol)(0.0821L*atm/K*mol)(298K) 2.50L = 0.16 atm
Dr. S. M. Condren Gases and Stoichiometry What is P of H 2 O? Could calculate as above. But recall Avogadros hypothesis. V n at same T and P P n at same T and V There are 2 times as many moles of H 2 O as moles of O 2. P is proportional to n. Therefore, P of H 2 O is twice that of O 2. P of H 2 O = 0.32 atm 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g)
Dr. S. M. Condren Daltons Law John Dalton
Dr. S. M. Condren Daltons Law of Partial Pressures What is the total pressure in the flask? P total in gas mixture = P A + P B +... Therefore, P total = P (H 2 O) + P (O 2 ) = 0.48 atm Daltons Law: total P is sum of pressures. Daltons Law: total P is sum of PARTIAL pressures. 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) 0.32 atm 0.16 atm 0.32 atm 0.16 atm
Dr. S. M. Condren Collecting Gases over Water
Dr. S. M. Condren Example A student generates oxygen gas and collects it over water. If the volume of the gas is 245 mL and the barometric pressure is 758 torr at 25 o C, what is the volume of the dry oxygen gas at STP? P water = 23.8 torr at 25 o C P O2 = P bar - P water = ( ) torr = 734 torr P 1 = P O2 = 734 torr; P 2 = SP = 760. torr V 1 = 245mL; T 1 = 298K; T 2 = 273K; V 2 = ? (V 1 P 1 /T 1 ) = (V 2 P 2 /T 2 ) V 2 = (V 1 P 1 T 2 )/(T 1 P 2 ) = (245mL)(734torr)(273K) (298K)(760.torr) = 217mL
Dr. S. M. Condren Higher Density air Low density helium PV = nRT n = P V RT m = P MV RT Where m => mass M => molar mass and density (d) = m/V d = m/V = PM/RT d and M are proportional GAS DENSITY
Dr. S. M. Condren USING GAS DENSITY The density of air at 15 o C and 1.00 atm is 1.23 g/L. What is the molar mass of air? What is air? mass/mol = 1.23 g/ mol = 29.1 g/mol 79% N 2 ; M 28g/mol 20% O 2 ; M 32g/mol 1. Calc. moles of air. V = 1.00 LP = 1.00 atmT = 288 K n = PV/RT = mol 2. Calc. molar mass Reasonable?
Dr. S. M. Condren KINETIC MOLECULAR THEORY (KMT) Theory used to explain gas laws. KMT assumptions are Gases consist of atoms or molecules in constant, random motion.Gases consist of atoms or molecules in constant, random motion. P arises from collisions with container walls.P arises from collisions with container walls. No attractive or repulsive forces between molecules. Collisions elastic.No attractive or repulsive forces between molecules. Collisions elastic. Volume of molecules is negligible.Volume of molecules is negligible.
Dr. S. M. Condren Kinetic Molecular Theory Because we assume molecules are in motion, they have a kinetic energy. KE = (1/2)(mass)(speed) 2 At the same T, all gases have the same average KE. As T goes up for a gas, KE also increases – and so does the speed.
Dr. S. M. Condren Kinetic Molecular Theory where u is the speed and M is the molar mass. speed INCREASES with Tspeed INCREASES with T speed DECREASES with Mspeed DECREASES with M Maxwells equation
Dr. S. M. Condren Velocity of Gas Molecules Molecules of a given gas have a range of speeds.
Dr. S. M. Condren Velocity of Gas Molecules Average velocity decreases with increasing mass. All gases at the same temperature
Dr. S. M. Condren GAS DIFFUSION AND EFFUSION DIFFUSION is the gradual mixing of molecules of different gases.
Dr. S. M. Condren GAS EFFUSION is the movement of molecules through a small hole into an empty container. EFFUSION is the movement of molecules through a small hole into an empty container.
Dr. S. M. Condren GAS DIFFUSION AND EFFUSION Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to Tproportional to T inversely proportional to M.inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T. He
Dr. S. M. Condren GAS DIFFUSION AND EFFUSION Grahams law governs effusion and diffusion of gas molecules. Thomas Graham, Professor in Glasgow and London. Rate of effusion is inversely proportional to its molar mass.
Dr. S. M. Condren HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube. HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube. Gas relation of mass to rate of diffusion Gas Diffusion relation of mass to rate of diffusion
Dr. S. M. Condren Deviations from Ideal Gas Law Real molecules haveReal molecules have volume. There areThere are intermolecular forces. –Otherwise a gas could not become a liquid.
Dr. S. M. Condren Deviations from Ideal Gas Law Account for volume of molecules and intermolecular forces with Account for volume of molecules and intermolecular forces with VAN DER WAALSs EQUATION. Measured V = V(ideal) Measured P intermol. forces vol. correction J. van der Waals, , Professor of Physics, Amsterdam. Nobel Prize nRT V - nb V 2 n 2 a P )(
Dr. S. M. Condren Cl 2 gas has a = 6.49, b = For 8.0 mol Cl 2 in a 4.0 L tank at 27 o C. P (ideal) = nRT/V = 49.3 atm P (van der Waals) = 29.5 atm Deviations from Ideal Gas Law
Dr. S. M. Condren
Carbon Dioxide and Greenhouse Effect
Dr. S. M. Condren Composition of Air at Sea Level
Dr. S. M. Condren Some Oxides of Nitrogen N 2 O NO NO 2 N 2 O 4 2 NO 2 = N 2 O 4 brown colorless NO x