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Gases Chapter 12 Properties of Gases Compressibility Mass Volume (fill container) Exert Pressure Diffuse through other gases Low Density Kinetic Molecular Theory (KMT) 1. 2. 3. 4. 5. 6. All gases are matter (have mass and take up space). These particles are in constant, rapid and random motion. All collisions are perfectly elastic. (no energy is lost in the collision) The force of gas particle collisions on the walls of the container creates pressure; however, gas particles do not exert force on each other. At a given temperature, all gas particles have the same amount of kinetic energy (temperature). The distance between gas particles is very large. Gas Pressure Gas pressure results from the force of colliding particles on a given area. The metric unit of force is the newton (similar to the ounce in the English system). A newton of force acting on a square meter is called a pascal. Normal atmospheric pressure at sea level is 101.3 kPa. Measuring Gases In order to describe a gas, we use four variables to make predictions about the behavior of the particles. Amount of a gas (n) expressed in moles 2. Amount of space the gas takes up, volume (V), as measured in liters. *note: this is the volume of the container 3. Measure of kinetic energy, temperature (T), in degrees Kelvin 1. 1. 4. *note: K = C + 273 Measure of the force exerted by the gas particles on the walls of the container, pressure (P) Kinetic Energy and Temperature Temperature is an indication of kinetic energy. T1 T2 T2 > T1 Speed of Particles The Kelvin Temperature Scale Kinetic Energy of Molecules 0 (Absolute Zero Scale) K = oC + 273 -273oC -300 0 -200 73 -100 173 0 100 oCelcius scale 273 373 Kelvin scale Pressure Pressure is the measure of the amount of force of an object per unit area. Newton / m2 = Pascal Atmospheric Pressure is the amount of pressure exerted by the entire atmosphere on the surface of the Earth. A barometer is an instrument that measures atmospheric pressure A manometer is an instrument that measures the pressure of an enclosed gas Mercury Barometer Manometer STP Pressure and temperature changes affect gas volume. To compare experimental results, scientist convert their results to standard temperature and pressure (STP) Standard Conditions: Standard Temperature = 0oC (273 K) Standard Pressure = 101.3 kPa (1atm) {760 torr} Pressure Units Kilopascals (kPa) Atmospheres (atm) 1 atm = 101.3 kPa Torricellis (torr) 1 torr = 1 mm Hg 760 torr = 101.3 kPa (other units are listed on page 420) Pressure Units Atmosphere (atm) Millimeter Hg (mmHg) Pascal (Pa) Pounds per square inch (psi) Kilopascal (kPa) 1 atm = Standard Pressure 1 atm = 760 mmHg 1 atm = 101,325 Pa 1 atm = 14.7 psi 1 atm = 101.3 kPa Conversions Covert using Dimensional Analysis Information you have x conversion factor (unknown/known) Convert 665 mmHg to kPa 665 mmHg x 101.3 kPa = 88.6 kPa 760 mmHg Classwork: page 421 #1-4 Homework: page 422 #1-12 Boyle’s Law Pressure and Volume are inversely proportional at a given temperature and number of molecules. VP = k (a constant value) Pressure Effects on Volume V 1/P Inverse Proportionality Boyle’s Law continued V 1P 1 = k 1 and V 2P 2 = k 2 since k1 = k2 (that’s why its called a constant) V 1P 1 = V 2P 2 This is the mathematical expression for Boyle’s Law A Boyle’s Law Problem A gas in a 242 cm3 container exerts a pressure of 87.6 kPa. What volume would the gas occupy at standard atmospheric pressure? Identify Variables V1 = 242 cm3 V2 = ? P1 = 87.6 kPa P2 = 101.3 kPa Two Methods of Solving 1. Algebraic: V1P1 = V2P2 V2 = V1P1 P2 V2 = 242 cm3 x 87.6 kPa 101.3 kPa V2 = 209 cm3 Another Method 2. Logic and reason: The new volume (V2) depends upon the old volume (V1) multiplied by a pressure change. 87.6 V2 = 242 cm3 x _______ 101.3 Since the pressure is increasing, the volume should decrease. Multiply by a fraction less than 1. Charles’s Law At a given pressure and number of molecules, Volume and Temperature are directly proportional. V = k T Two Temperature Scales b Volume y = mx + b V = kT + b V = kT + 0 V = kT 0 b -300 0 -200 73 -100 173 0 100 oCelcius scale 273 373 Kelvin scale Charles’s Law Continued V1 = k1 T1 V2 = k2 T2 V 1 = V2 T1 T2 and since k1 = k2 This is the mathematical expression for Charles’s Law A Charles’s Law Problem A 225 cm3 volume of gas is collected at 58oC. What volume would the gas occupy at standard temperature? Identify Variables V1 = V2 = 225 cm3 ? T1 = 58oC + 273 = 331 K T2 = 0oC + 273 = 273 K Two Methods of Solving 1. Algebraic: V1 / T1 = V2 / T2 V2 = V1T2 T1 V2 = 225 cm3 x 273 K 331 K V2 = 186 cm3 The Other Method 2. Logic and reason: The new volume (V2) depends upon the old volume (V1) multiplied by a temperature change. 273 V2 = 225 cm3 x _______ 331 Since the temperature is decreasing, the volume should decrease. Multiply by a fraction less than 1. Gay-Lussac’s Law P1 = k1 T1 and P2 = k2 T2 since k1 = k2 P1 = P2 T1 T2 This is the mathematical expression for Gay-Lussac’s Law The Combined Gas Law In most laboratory situations, both temperature and pressure change. The Combined Gas Law is used to calculate the affects of these changes. A Combination of Gas Laws Boyle’s Law: V 1/P Charles’s Law: V T Gay-Lussac’s Law: PT V T/P To change a proportionality to an equality… multiply by a proportionality constant. V = kT P V1P1 = V2P2 VP = k T1 T2 T V 1P1 = V 2P 2 T1 T2 Solve for each variable V2 = V1P1T2 / T1P2 T2 = V2P2T1 / V1P1 P2 = V1P1T2 / T1V2 Sample Problem What volume would 955 cm3 of a gas measured at 58 oC and 108.0 kPa occupy at 76 oC and 123.0 kPa? 3 955 cm V1 = V2 = ? T1 = 331 K T2 = 349 K P1 = P2 = 108.0 kPa 123.0 kPa Method #1 V1P1 / T1 = V2P2 / T2 V2 = V1P1T2 / (T1P2) V2 = (955 cm3) (108.0 kPa) (349 K) (331 K) (123.0 kPa) V2 = 884 cm3 Method #2 108 349 x --------V2 = 955 cm3 x --------331 temperature increases volume increases pressure increases volume decreases V2 = 884 cm3 123 Gas Law Variables P V T When T is constant P V When P is constant T V When V is constant T P Dalton’s Law of Partial Pressures Gases are often collected by water displacement. These gases contain water vapor. To find the pressure of the dry gas alone, Dalton’s Law of Partial Pressures is used. The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each gas in the mixture. Dalton’s Law of Partial Pressures Gases are often collected by water displacement. These gases contain water vapor. To find the pressure of the dry gas alone, Dalton’s Law of Partial Pressures is used. The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each gas in the mixture. The Mathematical Expression of Dalton’s Law PT = P1 + P2 + …. Pn Example: Gas A = 1.0 atm Gas B = 1.5 atm Gas C = 0.5 atm PT = 3.0 atm The Pressure of a “Dry” Gas When a gas is collected over water, the total pressure is the result of the dry gas plus the water vapor pressure. PT = Pgas + Pvapor To determine the pressure of the dry gas alone, subtract the water vapor pressure from the total pressure. ( P. 401 ) Pgas = PT - Pvapor Diffusion If a bottle of ammonia were opened at the front of the room, the odor could soon be detected at the back of the room. Diffusion is the random movement of particles through another substance. Effusion is gas under pressure escaping through a small opening. Diffusion Kinetic Energy The equation for finding the kinetic energy of a particle: K.E. = ½ mv2 m = mass of the particle v = velocity (speed) of the particle Kinetic Energy K.E. = ½ mv2 #@*~!!* ~##~@!& Graham’s Law If gases A and B are at the same temperature….. ____ ____ K.E.A = K.E.B ½ mAv2A = ½ mBv2B mAv2A = mBv2B Collecting terms… v2A = mB v 2B mA vA = vB mB mA This is the mathematical expression for Graham’s Law The Meaning of Graham’s Law “The ratio of the velocities of two gases is equal to the square root of the inverse ratios of their masses.” mAv2A = mBv2B if: mA > mB then: vB > vA (The lightest gas always travels faster) See practice problems on page 438 A Graham’s Law Experiment HCl(g) + NH3(g) NH4Cl(s) HCl v = d/t vA = dA/tA vB = dB/tB tA = t B vA / vB = dA / dB (experimental) vA / vB = √mB / mA (accepted) % error: | exp. – accepted | x 100 accepted NH3 Ideal Gases The gas laws we have studied only apply to ideal gases. There are no ideal gases. Properties of Ideal Gases Ideal gases have: 1. Point mass (Particles have mass but occupy no volume) 2. No mutual attractions (No matter how close or how slow the particles are, they do not attract each other) Ideal Gas Laws Work Real gas particles occupy negligible volume compared to the total volume occupied by the gas… (unless under very high pressures) Real Gases Ideal Gas Laws Work •Real gas particles occupy negligible volume compared to the total volume occupied by the gas… (unless under very high pressures) •Real gas particles have such weak intermolecular attraction forces that they do not affectively attract each other… (unless at very low temperatures) Real Gas Laws Ideal gas laws work for real gases except at very high pressures or at very low temperatures. Why not use real gas laws that work all the time? Ideal gas law: PV = KT Solve for V Real gas law: [(P + n2a)/V2] [V – nb] = KT Solve for V The Real Gas Law Solution V = -(P + n2a) (P + n2a)2 – 4(KT)(P +n2a)(nb) 2KT Manometer Problems A Closed Arm Manometer = ? kPa 165 torr x 101.3 kPa 760 torr = 22.0 kPa Manometer Problems An Open Arm Manometer The gas exerts 120 torr more pressure than the atmosphere. Gas pressure = 872 torr 872 torr x 101.3 kPa = 760 torr 116 kPa