Chapter 11: Gases PDF

Chapter 11: Gases 1  Outline Properties of gases Kinetic Molecular Theory (KMT) Pressure and units, standard conditions Ideal Gas Law Gas Laws Partial Pressures 2 ...

Chapter 11: Gases 1  Outline Properties of gases Kinetic Molecular Theory (KMT) Pressure and units, standard conditions Ideal Gas Law Gas Laws Partial Pressures 2  Skills Needed Prior Knowledge: Stoichiometric relationships in a chemical reaction Intensive properties – per mole Physical states, including both atomic and both macroscopic definitions Force Temperature, including units and conversions Math Skills: conversions, solving an equation for an unknown variable, relationships in ratios, stoichiometry, labeling and reading a graph 3  The Physical States of Matter Temperature Most substances can exist as solid, liquid, or gas under appropriate pressure and temperature conditions. 4  Gases vs Liquids and Solids Differences: 1) Gas volume changes significantly with pressure 2) Gas volume changes significantly with temperature 3) Gases flow very freely 4) Gases have relatively low densities 5) Gas can form a homogenous solution in any proportion with each other. 5  Kinetic-Molecular Theory (KMT) Explains behavior of gases at a molecular level, instead of macroscopic level Questions: How do gas particles create pressure? What happens to gas particles under pressure? Why aren’t liquids and solids compressible? Why does each gas contribute to the total pressure, proportional to its number of particles present? What effect does higher temperature have on gas particles? What does temperature measure on the molecular scale? Why doesn’t 1 mol of a heavier gas exert more pressure than 1 mol of a lighter gas? 6 Kinetic-Molecular Theory (KMT) Each gas particle colliding with the wall exerts a force. The greater the number of particles, the more frequently they collide with the walls, the higher the registered force (or pressure). 7  Kinetic-Molecular Theory (KMT) Postulates (Main Ideas): Particle volume: volume of each particle is so small in comparison to the total volume that it is assumed to be zero. Particle motion: constant, random, straight-line motion, except when they collide with the walls, or each other Particle collisions; collisions are elastic; total kinetic energy is constant Describes an ideal gas 8  Kinetic Energy and Gas Behavior Kinetic energy – energy associated with motion Why don’t heavier molecules exert more pressure and take up more volume than lighter molecules at the same temperature? 1 Ek = mass  speed 2 Relative number of molecules O2 (32) 2 Temperature – measure of with a given speed N2 (28) H2O (18) average kinetic energy of He (4) particles H2 (2) Molecular speed at a given T 9  Pressure Gravity of Earth attracts atmospheric gases, so exert a uniform force on everything (about 14.7 lb/in2, or psi) force pressure ( p ) = area Measuring Atmospheric Pressure: Barometer – instrument that measures atmospheric pressure mm Hg – common unit for pressure; sea level is 760 mm Hg If H2O was used, column would be 34 ft tall! 10  Units of Pressure m SI unit for force: newton (N) 1 N = 1 kg  2 s m kg  2 SI unit for pressure: pascal (Pa) 1 Pa =1 N2 = 1 2s = 1 kg 2 m m ms Standard atmosphere (atm): 1 atm = 101,325 Pa = 101.325 kPa mm Hg = renamed torr: 1 mm Hg = 1 torr 760 mm Hg = 760 torr = 1 atm GIVEN ON EXAMS: 1 bar = 100,000 Pa = 100 kPa atm, Pa, kPa, torr, mmHg, bar 11  “Standard Conditions” To better understand factors that influence the behavior of a gas, we have a set of “standard conditions” or standard temperature and pressure (STP) STP = 0°C (or 273.15 K) and 1 atm (or 760 torr) At STP, one mole of an ideal gas will occupy 22.4 L 12  The Gas Laws Physical behavior of gas can be described completely by 4 variables: Pressure (P) Volume (V) Temperature (T) Amount, or moles, (n) Interdependent, so any one of these can be determined from measuring the other 3. 13  Memorize equation Ideal Gas Law pV = nRT Becomes any of the other laws when 1 or more variables are held constant. L  atm = 0.0821 =R Value of R mol  K will be given Gas constant Other values of R, with other units: kPa  L 8.314 J 8.314 mol  K mol  K 14  Remaining Gas Laws pV = nRT ALL can be determined using the Ideal Gas Law: Boyle’s Law pV = constant p V = p V i i f f Vi V f Charles’s Law V = constant = T Ti T f Gay-Lussac’s Law P = constant Pi Pf = T Ti T f Avogadro’s Law 𝑉𝑖 𝑉𝑓 V = constant = n 𝑛𝑖 𝑛𝑓 Combined Gas Law pV = constant PiVi Pf V f = T Ti Tf 15  The Gas Laws – Summary Table pV = nRT Constant Variable What changes? Name of Law; Equation Properties Properties Relationship Temperature (T) Pressure (P) As we increase volume, we Boyle’s Law; PiVi = PfVf Amount (n) Volume (V) see a decrease in pressure Indirect relationship (and vice versa) (inversely related) Pressure (P) Volume (V) As we increase volume, we Charles’ Law; Vi V f Amount (n) Temperature (T) see an increase in temperature Direct relationship = (and vice versa) Ti T f As we increase temperature, Gay-Lussac’s Law; Pi Pf Volume (V) Amount (n) Pressure (P) Temperature (T) we see an increase in pressure Direct relationship = (and vice versa) Ti T f Temperature (T) Volume (V) As we increase amount, we see Avogadro’s Law; Pressure (P) Amount (n) an increase in volume Direct relationship (and vice versa) Amount (n) Pressure (P) Combined Gas Law PiVi Pf V f Volume (V) = Temperature (T) Ti Tf Gas Laws PhET Simulation: https://phet.colorado.edu/sims/html/gas-properties/latest/gas-properties_en.html 16  Types of Gas Law Problems pV = nRT 2 main types of gas law problems: 1) A change in one of the 4 variables (P, V, n, T), causes a change in the other(s), remaining variables are held constant Ideal Gas Law reduces to one of the others Use one of the other gas laws! R is not used, units must be consistent, T must be in K 2) One variable is unknown, but 3 are known and no change occurs Use pV = nRT 17  Practice with Gas Laws pV = nRT R = 0.0821 L∙atm/mol∙K 1. A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 21°C. 18  Practice with Gas Laws pV = nRT R = 0.0821 L∙atm/mol∙K 2. A sample of methane exerts 1.5 atm of pressure and occupies a volume of 10.0 L. What is the volume if the pressure is increased to 6.0 atm? 19  iClicker pV = nRT R = 0.0821 L∙atm/mol∙K 3. A sample of gas initially at 298 K occupies a volume of 25 L. When the gas is heated to a temperature of 500. K, what is the volume occupied by the gas? 20  iClicker pV = nRT R = 0.0821 L∙atm/mol∙K 4. A gas is filled in a container at 5.60 atm and 23.0°C. For safety reasons, the pressure must not exceed 9.50 atm. What is the maximum temperature (in K) to which the gas can be heated? 21  iClicker pV = nRT R = 0.0821 L∙atm/mol∙K 5. A sample of gas at 350 K exerts a pressure of 3.0 atm. When the gas is cooled to 250 K, it exerts a pressure of 0.75 atm and occupies a volume of 3.0 L. What is the volume initially occupied by the gas? 22 760 torr = 1 atm iClicker pV = nRT R = 0.0821 L∙atm/mol∙K 6. A steel tank used for fuel delivery is fitted with a safety valve that opens if the internal pressure exceeds 1000. torr. It is initially filled with methane at 23°C and 0.991 atm, and then placed in boiling water, at 100°C. Will the safety valve open? A) Yes B) No 23  Density, Molar Mass, Partial Pressures pV = nRT We can also rearrange the ideal gas law to find: gas density molar mass partial pressure of each gas in a mixture amount of gaseous reactant or product in a mixture 24  Determining the Gas Density pV = nRT Density = mass in grams Molar Mass = grams n= mass in grams Volume mol Molar Mass (g/mol) n m pV = nRT → pV = RT M m Mp density = (pressure)(molar mass) Rearrange to isolate m and V: = V RT RT d = P(MM) RT 25  Determining the Gas Density – Example Problem What is the density (in g/L) of CO2 gas at 273.15 K and 1 atm? MM(CO2) = 44.01 g/mol R = 0.0821 L∙atm/mol∙K 26  Determining the Molar Mass pV = nRT Molar Mass = grams n= mass in grams mol Molar Mass (g/mol) n pV m pV pV = nRT → n= → = RT M RT mRT molar mass = (mass)(RT) so M = pV pV 27  Determining the Molar Mass – Example Problem An organic chemist isolates a colorless liquid from a petroleum sample. She collects the liquid in a flask, vaporizes it completely, and collects the following data: V = 213 mL mass of flask + gas = 78.416 g T = 100.0 °C mass of empty flask = 77.834 g R = 0.0821 L∙atm/mol∙K p = 754 torr (760 torr = 1atm) Determine the molar mass of the liquid. pV = nRT 28  Ideal Gas Law and Stoichiometry – Example Problem Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ideal molar ratio from ideal gas law balanced equation gas law P,V,T AMOUNT (mol) AMOUNT (mol) P,V,T of gas A of gas A of gas B of gas B Copper reacts with any oxygen present as an impurity in ethylene used to make polyethylene. The copper is regenerated when hot H2 reduces the copper(II) oxide, forming the pure metal and H2O. What volume of H2 at 765 torr and 225 °C is needed to reduce 35.5 g of copper(II) oxide? R = 0.0821 L∙atm/mol∙K 29  Partial Pressures Ideal Gas Law holds for any gas in ordinary conditions, whether pure or a mixture Gases mix homogeneously in any proportions Each gas in a mixture behaves like it’s the only gas present (i.e., assume no chemical interactions between gas molecules) Discovered by John Dalton, 1801, after a lifetime of studying humidity 34  Dalton’s Law of Partial Pressures When water vapor is added to dry air, the total air pressure increases by the pressure of the water vapor Phumid air = Pdry air + Padded vapor So each gas in a mixture exerts a partial pressure equal to the pressure it would exert by itself. Dalton’s Law of Partial Pressures Ptot = P1 + P2 + P3 +... 31  Dalton’s Law of Partial Pressures Mole Fraction (χ) Each component in a gas mixture contributes a fraction of the total moles Sum of mol fractions of all components must be 1, like with mass fraction 𝑛𝐴 𝑛𝐴 𝑛𝐵 𝑛𝐵 χ𝐴 = = χ𝐵 = = 𝑛𝑡𝑜𝑡 𝑛𝐴 + 𝑛𝐵 𝑛𝑡𝑜𝑡 𝑛𝐴 + 𝑛𝐵 Mole Percent (%): Multiply XA above by 100 Partial Pressure = Mole Fraction x Ptot 𝑃𝐴 = χ𝐴 × 𝑃𝑡𝑜𝑡 𝑃𝐵 = χ𝐵 × 𝑃𝑡𝑜𝑡 32  Dalton’s Law of Partial Pressures – Example Problem A sample of a gas mixture was prepared using 79 mol % N2, 17 mol % O2, and 4 mol % H2. The total pressure of the sample is 1.25 atm. What is the mol fraction and partial pressure of each gas in the mixture? Mole Fraction A: Partial Pressure A: nA nA A = = PA =  A  Ptot ntot n A + nB 33  Summary You should now be able to: Discuss the differences between gases and condensed phases. Explain the postulated of kinetic molecular theory. Use KMT to explain and predict the behavior of gases. Identify relative speeds of gases based on KMT and molecular mass. Interconvert between different pressure units. Explain how gases exert pressure, using KMT. Describe the relationships between the 4 variables of state (pressure, volume, temperature, amount). Calculate changes in state, given changes in the variables. Given three variables, calculate the 4th using the Ideal Gas Law. Use the Ideal Gas Law in stoichiometric calculations. Use the Ideal Gas Law to calculate the density or molar mass of a gas. Use Dalton’s Law of Partial Pressure to calculate the pressure exerted by a single gas in a mixture; to calculate the mole fraction of a gas. 34  Extra information about: Boyle’s Law, Charles’ Law, Gay-Lussac’s Law, and Dalton’s Law of Partial Pressures  Boyle’s Law When you increase the Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Pext pressure, the volume decreases, and vice versa. Pgas Pext Pext increases, V (mL) P (torr) T and n fixed 20.0 780 1.56x104 15.0 1038 1.56x104 d2 Pgas d1 10.0 1560 1.56x104 5.0 3112 1.56x104 Higher Pext causes lower V, which results in At any T, Pgas = Pext as particles hit the more collisions, because particles hit the walls walls from an average distance, d1. from a shorter average distance (d2 < d1). As a result, Pgas = Pext again.  Boyle’s Law 1) Volume is inversely proportional to pressure. 2) The product of P∙V is a constant. P∙V = const. [at constant T and n] How does changing the volume affect the pressure of a sample of gas? PiVi = PfVf  3.0 Volume and Temperature Boyle’s Law is true only at constant temperature, but 2.0 they didn’t know why until the early 1800s. Volume (L) n = 0.04 mol P = 1 atm French scientists Charles and Gay-Lussac determined n = 0.02 mol P = 1 atm relationship between volume and temperature 1.0 When sample of gas is heated, the volume increases, and n = 0.04 mol P = 4 atm vice versa. –273 –200 –100 0 100 200 300 400 500 (C) 0 73 173 273 373 473 573 673 773 (K) C Temperature When sample of gas is heated, the volume increases, and vice versa, regardless of composition of gas. Extrapolate from experimental measurements, all lines converge to 1 point: -273.15 °C (absolute 0) The point at which the volume for the gas becomes 0. Because of this, all temperatures with gas laws must be in K.  Charles’s Law Volume is directly proportional to temperature (T↑, V↑). Ratio of volume to temperature is constant. V = const. [at constant P and n] T Like with Boyle’s Law, can use Charles’s Law to determine how the volume will change as the temperature is varied. Vi V f = Ti T f  Molecular View of Charles’s Law Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Patm Pgas Patm Patm T2 T2 T1 Pgas T increases V increases Pgas fixed n At T1, Pgas = Patm. Higher T increases Thus, V increases until collision frequency, Pgas = Patm at T2. so Pgas > Patm.  Temperature and Pressure You air up your car tires in the afternoon (high of 89°F). The tire gauge says that you have inflated them to just the right amount, but when you get in your car the next morning (after a low that night of 45°F, because Texas in April is weird) your fancy tire sensor in your car warns you that one of the tires has a low pressure and needs air. What happened? Amonton’s Law: Pressure and temperature are directly proportional Ratio of pressure and temperature is constant. P Pi Pf = const. [at constant V and n] = T Ti T f  Molecular View of Dalton’s Law Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Gas A Gas B Mixture When gas A is mixed with of A and B gas B, Ptotal = PA + PB and When gases A and B are Stopcock the numbers of collisions separate, each exerts the opened, of particles of each gas total pressure in its own piston with the container walls container. depressed are in proportion to the at fixed T amount (mol) of that gas. Closed Open PA = Ptotal PB = Ptotal Ptotal = PA + PB = 1.5 atm = 1.0 atm = 0.50 atm ntotal = 0.90 mol nA = 0.60 mol nB = 0.30 mol XA = 0.67 mol XB = 0.33 mol