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Gases A Review Elements that exist as gases at 250C and 1 atmosphere Physical Characteristics of Gases • Gases assume the volume and shape of their containers. • Gases are the most compressible state of matter. • Gases will mix evenly and completely when confined to the same container. • Gases have much lower densities than liquids and solids. Properties of Gases Gas properties defined using only 4 varibles • V = volume of the gas (L) • T = Kelvin temperature (K) • n = number of moles (mol) • P = pressure in either (atm), (mm Hg) or (kPa) Force Pressure = Area (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa Barometer 10 miles 4 miles Sea level 0.2 atm 0.5 atm 1 atm Atmospheric pressure Barometer Manometer Boyle’s 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 (1627-1691). Son of Early of Cork, Ireland. Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire. Boyle’s Law P a 1/V P x V = constant P1 x V1 = P2 x V2 Constant temperature Constant amount of gas A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P2 = P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL = = 4460 mmHg 154 mL Charles’s Law If n and P are constant, then V = (nR/P)T = kT V and T are directly related. Jacques Charles (17461823). Isolated boron and studied gases. Balloonist. Charles’s original balloon Modern long-distance balloon As T increases V increases Variation of gas volume with temperature at constant pressure. Charles’ & Gay-Lussac’s Law VaT V = constant x T V1/T1 = V2/T2 Temperature must be in Kelvin T (K) = t (0C) + 273.15 A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? T1 = 125 (0C) + 273.15 (K) = 398.15 K T2 = V2 x T1 V1 = 1.54 L x 398.15 K 3.20 L = 192 K Avogadro’s Law 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 Avogadro’s Law V a number of moles (n) V = constant x n V1/n1 = V2/n2 Constant temperature Constant pressure 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 1 mole NH3 4NO + 6H2O 1 mole NO At constant T and P 1 volume NH3 1 volume NO Ideal Gas Equation Boyle’s law: V a 1 (at constant n and T) P Charles’ law: V a T (at constant n and P) Avogadro’s law: V a n (at constant P and T) Va V = constant x nT P =R nT P nT P R is the gas constant PV = nRT The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT (1 atm)(22.414L) PV R= = nT (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K) What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm PV = nRT nRT V= P 1 mol HCl n = 49.8 g x = 1.37 mol 36.45 g HCl 1.37 mol x 0.0821 V= V = 30.6 L L•atm mol•K 1 atm x 273.15 K What is the volume (in liters) occupied by 49.8 g of HCl at STP? grams of HCl moles of HCl liters of HCl at STP 1 mol HCl 22.41 L V 49.8 g HCl 30.6 L 36.45 g HCl 1 mol HCl Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR P = = constant T V P1 P2 = T1 T2 P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K T2 = 1.20 atm x 358 K = 1.48 atm P2 = P1 x 291 K T1 Density (d) Calculations PM m d= = V RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT M= P d is the density of the gas in g/L A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.00C. What is the molar mass of the gas? dRT M= P M= g 2.21 L M = 54.6 g/mol 4.65 g m = = 2.21 d= 2.10 L V x 0.0821 L•atm mol•K 1 atm x 300.15 K g L Gas Stoichiometry • You can apply the discoveries of Gay -Lussac and Avogadro to calculate the stoichiometry of chemical reaction. • Step 1 Write a balanced chemical equation. • Step 2 Convert all units to common units • Step 3 Start with given and Use factor-label skills (Via MOLES) to get to desired units. • Step 4 If gas laws are needed use appropriately at this point or prior to factor- label step. Gas Stoichiometry What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) g C6H12O6 mol C6H12O6 5.60 g C6H12O6 x 6 mol CO2 1 mol C6H12O6 x = 0.187 mol CO2 180 g C6H12O6 1 mol C6H12O6 V= nRT = P mol CO2 V CO2 L•atm x 310.15 K mol•K 1.00 atm 0.187 mol x 0.0821 = 4.76 L Gas Stoichiometry problems • 1) Ammonia can react with Oxygen to produce nitrogen and water. If 1.78 L of O2 reacts, what volume of nitrogen will be produce? (Temp and pressure remain constant) Ans.= 1.19 L of N2 • 2) Ethylene gas( C2H4) combust in Air to form H2O and CO2. If 13.8 L of C2H4 measured at 21°C and 1.038 atm burns completely with O2, calculate the volume of CO2 produced assuming the CO2 is measured at 44°C and 0.989 atm. Ans.= 32.6 L of CO2 • 3) When Arsenic (III) sulfide is roasted in air, it reacts with oxygen to produce arsenic (III) oxide and sulfur dioxide. When 89.5 g As2S3 is roasted with excess O2, what volume of SO2 is produced? The gaseous product is measured at 20°C and 98.0 kPa. Ans.= 27.1 L of SO2 Dalton’s Law of Partial Pressures V and T are constant P1 P2 Ptotal = P1 + P2 Consider a case in which two gases, A and B, are in a container of volume V. nART PA = V nA is the number of moles of A nBRT PB = V nB is the number of moles of B PT = PA + PB PA = XA PT nA XA = nA + nB nB XB = nA + nB PB = XB PT Pi = Xi PT mole fraction (Xi) = ni nT A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 Xpropane = 8.24 + 0.421 + 0.116 = 0.0132 Ppropane = 0.0132 x 1.37 atm = 0.0181 atm Chemistry in Action: Scuba Diving and the Gas Laws P Depth (ft) Pressure (atm) 0 1 33 2 66 3 V PV = nRT Kinetic Molecular Theory of Ideal Gases 1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. 3. Gas molecules exert neither attractive nor repulsive forces on one another. 4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy Kinetic theory of gases and … • Compressibility of Gases • Boyle’s Law P a collision rate with wall Collision rate a number density Number density a 1/V P a 1/V “12m10an3.mov” • Charles’ Law P a collision rate with wall Collision rate a average kinetic energy of gas molecules Average kinetic energy a T PaT “12m10an2.mov” Kinetic theory of gases and … • Avogadro’s Law P a collision rate with wall Collision rate a number density Number density a n Pan “12m10an1.mov” • Dalton’s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas Ptotal = SPi Deviations from Ideal Gas Law • Real molecules have volume. • There are intermolecular forces. –Otherwise a gas could not become a liquid. Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT PV = 1.0 n= RT Repulsive Forces Attractive Forces Deviations from Ideal Gas Law Account for volume of molecules and intermolecular forces with Van der Waal’s Equation. Measured V = V(ideal) Measured P ( P n2 a + ----V2 ) V - nb nRT vol. correction intermol. forces J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910. Deviations from Ideal Gas Law Cl2 gas has a = 6.49, b = 0.0562 For 8.0 mol Cl2 in a 4.0 L tank at 27 oC. P (ideal) = nRT/V = 49.3 atm P (van der Waals) = 29.5 atm Van der Waals equation nonideal gas 2a n ( P + V2 ) (V – nb) = nRT } } corrected pressure corrected volume During Calculations plug in a and b values into the equation and solve algebraically