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## Gases

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**Gases**Chapter 14**Sections**• The Gas Laws • The Combined Gas Laws and Avogadro’s Principle • Ideal Gas Law • Gas Stoichiometry**The Gas Laws**Chapter 14.1**The Gas Laws**• Boyles Law • Charles Law • Gay-Lussac’s Law • Each law relates two variables to the behavior of gasses. • Pressure • Temperature • Volume • amount**Kinetic Theory Review**• K-M theory suggests gas particles behave differently than liquids and solids. • K-M assumes the following are true about gasses: • Gas particles do not attract or repel each other • Gas particles are much smaller than the distances between them • Gas particles are in constant, random motion • No kinetic energy is lost when gas particles collide with each other or with the walls of the container • All gasses have the same average kinetic energy at a given temperature**The Nature of Gasses**• Actual gasses don’t obey all the assumptions made by K-M theory • But their behavior approximates the theory • Notice that all assumptions of K-M theory are based on the four factors mentioned about the gas laws: • Number of particles present • Temperature • Pressure • Volume of gas**Boyle’s Law**• Robert Boyle (1627-1691), Irish, studied relationship between pressure and volume. • An inverse relationship • Boyle’s Law: • The volume of a given amount of gas at constant temperature varies inversely with the pressure • So at any two different times, for the same gas, the product of pressure and volume will be the same • P1V1 = P2V2**Boyle’s Law**• http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/gaslaw/boyles_law_graph.html**Boyle’s Law Problem**• The volume of a gas at 99.0 kPa is 300.0 mL. If the pressure is increased to 188 kPa, what is the new volume? • P1V1 = P2V2 • 99.0kPa x 300.0mL = 188kPa x ? 99.0kPa x 300.0mL = ? 188kPa 158.0mL**Charles’s Law**• Jacque Charles (1746-1823) French • Studied relationship between volume and temperature • As temperature increases , so does volume of gas • MUST BE IN KELVIN • Tk = Tc + 273 • K-M Theory says as temperature increases, particles move faster, so must be further away. • Relationship between temperature and volume is a straight line with positive slope V1 T1 V2 T2 =**Charles’s Law**• Law: The volume of a gas is directly proportional to the temperature, expressed in kelvin, at a constant pressure**Charles’s Law Problem**• A gas at 89°C occupies a volume of 0.67L. At what Celsius temperature will the volume increase to 1.12L? V1 T1 V2 T2 = 1.12L ? 0.67L 89°C + 273 1.12L ? x0.67L 362 °K = = 1.12L x 362°K 0.67L ? = = 605.1 K 605.1°K – 273K = 332°C**Gay-Lussac’s Law**• Joseph Gay-Lussac studied relationship between temperature and pressure of a constained gas at fixed volume: • Found direct proportion exists between temperature (KELVIN) and pressure • Mathematically: P1 T1 P2 T2 =**Gay-Lussac’s Law Problem**• A gas in a sealed container has a pressure of 125kPa at a temperature of 30°C. If the pressure of the container is increased to 201kPa, what is the new temperature? • Convert temp to K: 30+273 = 303K P1 T1 P2 T2 125kPa 303K 201kPa T2 = = 201kPa x 303K 125kPa T2 = = 487K 487K – 273 = 214°C**Combined Gas Law and Avogadro’s Principle**Chapter 14, Section 2**Combined Gas Law**• Combine Boyle’s Law, Charles’s Law, and Gay-Lussac’s Law into one: • P1V1/T1 = P2V2/T2 • Let’s you work out problems involving more variables • Use known variables under one set of conditions to find a value of a missing variable.**Avogadro’s Principle**• 1811, Avogadro suggested that equal volumes of gasses at the same temperature and pressure contain the same number of particles: • K-M theory says the same thing • Particles are so far apart that size doesn’t matter • 1 mole of particles is 6.02 x 1023 particles • 1 mole of gas particles is 22.4 L at 0.00° C and 1.00 atm pressure • Standard Temperature and Pressure (STP)**Example Problems**• Determine the volume of a container that holds 2.4 mole of gas at STP: • 2.4 mol x 22.4L/1mol = 53.76 L = 54L • What volume will 1.02 mol of carbon monoxide gas occupy at STP? • 1.02 mol x 22.4L/1 mol = 22.8 L • A balloon will rise off the ground when it contains 0.0226 mol of He in a volume of 0.460L. How many moles of He are needed to make the balloon rise when volume is 0.885L, assuming temperature and pressure are constant? • 0.0226 mol x 0.865L/0.460L = 0.0425 mol**Ideal Gas Law**Chapter 14, Section 3**Ideal Gas Law**• P1V1/T1 = P2V2/T2 can be interpreted as saying that PV/T is a constant, k • k is based on the amount of gas present, n • Experiments have shown that k = nR, where n is number of moles and R is a constant. • Ideal Gas Law is therefore: • PV = nRT • Ideal Gas Constant (R), is: • 0.0821 L*atm/mol-K (most often used) • 8.314 L*kPa/mol K • 62.4 L*mm Hg/mol K**Real vs. Ideal**• Ideal Gas Assumptions • Particles take up no space • No intermolecular forces • Follows ideal gas law under all temperatures and pressures • Real Gas • All gas particles take up some space • All gas particles are subject to intermolecular forces • Most gasses act like ideal gasses at many/most temperatures and pressures • Deviations occur at extremely high pressures and extremely low temperatures because intermolecular forces start to show effects.**Applying Ideal Gas Law**• If you know any three variables, you can solve for the fourth: • You can solve for n (number of moles) • Combined gas law, you cannot. • You can use ideal gas law allows you to solve for molar mass and density, if mass is known • n (number of moles) = m (mass) / M (molar mass) • Sustitute m/M for n • PV = mRT/M**Solving for Density**• Density is m (mass) / V (volume) • Substitute D (Density) for m/V • M = mRT/PV = DRT/P • Or D = MP/RT**Example Problem**• Calculate the number of moles of gas contained in a 3.0L vessel at 3.00x102K with a pressure of 1.50 atm • Find: moles • Known: V = 3.0L • Known: T = 3.00x102 • Known: P = 1.50 atm • Known: R = 0.0821 L*Atm/mol*K • PV=nRT • Solve for n • N = PV/RT**Example Problem**• n = PV/RT – substitute in known information: • n = (1.50atm)(3.0L) • (0.0821 L*atm/mol*K)(3.00 x 102 K) n = 0.18 mole Evaluation: 1 mole of gas occupies 22.4L at STP. The volume is much less than 1 mole. Temperature and pressures are not dramatically different than STP. Slightly higher pressure means more gas. Slightly higher temperature means a little less gas. Units of the answer is moles – every unit cancels except 1/ 1/mol.**Example Problem – Using Molar Mass**• What is the molar mass of a pure gas that has a density of 1.40g/L at STP? • Find: molar mass (g/mole) • Known: D = 1.40g/L • Known: T=0.00°C • Known: P = 1.00atm • Known: R = 0.0821 L*atm/mol*K • Step1: Convert T to Kelvin • Tk = Tc+273 • Tk = 273**Molar Mass Problem**• Use density form of ideal gas law • M = DRT/P • Substitute known values: M = (1.40 g/L)(0.0821 L*atm/mol*K)(273K) 1 atm M = 31.4 g/mol**Gas Stoichiometry**Chapter 14, Section 4**Gas Stoichiometry – Volume only**• Example: • 2C4H10(g) + 13 O2(g)→ 8 CO2(g) + 10 H2O(g) • Remember: Avogadro’s principle states that 1 mole of gas is 22.4 L • For volume to volume calculations, you only need to know mole ratios • 2 L of butane reacts, it involves • 13 L of O2 (2L x 13 O2/2 C4H10) • 8 L of CO2 (2L x 8CO2/2C4H10) • 10 L of H2O**Gas Stoichiometry – Volume & Mass**• Ammonia is synthesized fro hydrogen and nitrogen • N2(g) + 3 H2(g)→ 2 NH3(g) • If 5.00 L of nitrogen reacts completely by this reaction at 3.0 atm and 298K, how many grams of ammonia are produced? • Analyze: you are given L, pressure, temperature. You are asked to find grams of NH3. • Known: • VN = 5.00L, P = 3.00 atm, T = 298K**Gas Stoichiometry (Cont)**• Determine the mole ratio needed • 1 volume N2 / 2 volumes NH3 • Determine volume of ammonia produced: 1 volume N2 2 volumes NH3 = 10L NH3 5.00L N2 x 3. Rearrange Gas Law to solve for n (moles): PV RT PV = nRT n = 4. Substitute values and solve for n**Gas Stoichiometry (Cont).**4. Substitute values and solve for n (3.00 atm)(10.00L) (0.0821 L*atm/mol*k)(298K) = 1.23 mol NH3 n = 5. Find the mass (M) of NH3 by finding the molecular mass and converting the moles in grams Molecular mass = 1N(14.01amu) + 3 H(1.01amu) = 17.04amu 1.23 mol NH3 x 17.04 g/mol = 21.0 g NH3 Done