Gases – Topic Outlines Gas Pressure Gas Laws & Ideal Gas Equation Density of Gases Stoichiometry involving gas reactions Kinetic Molecular Theory of Gases Root-mean-square speed Rates of Gas Diffusion and Effusion Deviation of Real Gases Atmospheric Chemistry

SI units = Newton/meter 2 = 1 Pascal (Pa) 1 standard atmosphere = 101,325 Pa 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr Pressure

Gas and Liquid Pressure Gas pressure is due to force exerted by gaseous molecules colliding with the surface of container wall. Liquid or hydraulic pressure is due to force exerted by liquid molecules colliding against the surface of the container walls. Liquid Pressure = g.d.h (g= gravitational constant; d = liquid density; h = height of liquid column)

Barometer Device used to measure atmospheric pressure. Mercury flows out of the tube until the pressure of the column of mercury standing on the surface of the mercury in the dish is equal to the pressure of the air on the rest of the surface of the mercury in the dish.

Device used for measuring the pressure of a gas in a container. Manometer or Pressure Gauge

Opened-end Manometer

The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals. Pressure Conversions: An Example

Ideal Gas Laws Boyle’s Law: –For a given quantity of gas at constant temperature, volume is inversely proportional to the pressure; –V = b/P (b is a proportionality constant); P 1 V 1 = P 2 V 2 Charles’s Law: –For a given quantity of gas at constant pressure, volume is directly proportional to the temperature in Kelvin; –V = cT (c is a proportionality constant); V 1 /T 1 = V 2 /T 2 Avogadro’s Law: –At fixed temperature and pressure, volume is directly proportional to the moles of gas; –V = an (a is a proportionality constant); V 1 /n 1 = V 2 /n 2

Boyle’s Law

Exercise A sample of helium gas occupies 12.4 L at 23°C and atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant? 9.88 L

Volume and Temperature (in Kelvin) are directly related (constant P and n). V=bT (b is a proportionality constant) K = °C K is called absolute zero. Charles’s Law

Exercise Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)? L

Volume and number of moles are directly related (constant T and P). V = an (a is a proportionality constant) Avogadro’s Law

Exercise If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure? 76.3 L

Ideal Gas Law We can bring all of these laws together into one comprehensive law:  V = bT (constant P and n)  V = an (constant T and P)  V = k/P (constant T and n) PV = nRT (where R = L·atm/mol·K, the universal gas constant)

Exercise An automobile tire at 23°C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire mol

Exercise What is the pressure in a L tank that contains kg of helium at 25°C? 114 atm

Exercise At what temperature (in °C) does 121 mL of CO 2 at 27°C and 1.05 atm occupy a volume of 293 mL at a pressure of 1.40 atm? 696°C

For 1 mole of an ideal gas at 0°C and 1 atm, the volume of the gas is L. STP = standard temperature and pressure  0°C and 1 atm  Therefore, the molar volume is L at STP. Molar Volume of an Ideal Gas

Standard Temperature and Pressure STP = Standard Temperature and Pressure –Standard Temperature = 0 o C = K –Standard Pressure = atm (760.0 torr) At STP, mole of ideal gas = L; Ideal gas constant, R = L.atm / K.mol = L.torr / K.mol ; Other value of R is x 10 4 mL.torr / K.mol ;

Exercise A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O 2 are present? 3.57 g

Density-Molar Mass Relationships Density = Mass / Volume = grams / L Mass (in grams) = mole x molar mass = nM

 d = density of gas  T = temperature in Kelvin  P = pressure of gas  R = universal gas constant Molar Mass of a Gas

Exercise What is the density of F 2 at STP (in g/L)? 1.70 g/L

Dalton’s Law of Partial Pressure In a mixture of gases, the total pressure is equal to the sum of the partial pressures of individual gases. If the partial pressures of gases A, B, C,…are P A, P B, P C,.., then the total pressure is P Total = P A + P B + P C + … Since P A = n A RT/V; P B = n B RT/V, and P C = n C RT/V, P T = (n A RT/V) + (n B RT/V) + (n C RT/V) = (n A + n B + n C + …)(RT/V) = (n total )RT/V

Exercise Consider the following apparatus containing helium in both sides at 45°C. Initially the valve is closed. –After the valve is opened, what is the pressure of the helium gas? 3.00 atm 3.00 L 2.00 atm 9.00 L

Exercise 27.4 L of oxygen gas at 25.0°C and 1.30 atm, and 8.50 L of helium gas at 25.0°C and 2.00 atm were pumped into a 5.81-liter tank at 25°C. Calculate the new partial pressure of oxygen atm Calculate the new partial pressure of helium atm Calculate the new total pressure of both gases atm

The Need for a Model/Theory -Kinetic Molecular Theory (KMT) So far we have considered “what happens,” but not “why.” In science, “what” always comes before “why.”

1)The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero). Postulates of the Kinetic Molecular Theory

2)The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. Postulates of the Kinetic Molecular Theory

3)The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other. Postulates of the Kinetic Molecular Theory

4)The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. Postulates of the Kinetic Molecular Theory

Kinetic Molecular Theory Gas is composed of tiny particles with finite masses but negligible molecular volume; Gas particles are in constant random motion, and colliding with the with the container walls; Molecular collisions are completely elastic – energy is conserved during collisions; Intermolecular attractions between gas particles are nonexistent; The average molecular kinetic energy for gas depends only on its temperature in Kelvin.

Kinetic Molecular Theory Two gases at the same temperature have the same average molecular kinetic energy; At the same temperature, the lighter gaseous molecules have higher average velocities;

Kinetic Molecular Theory

Concept Check You are holding two balloons of the same volume. One contains helium, and one contains hydrogen. Complete each of the following statements with “different” or “the same” and be prepared to justify your answer. He H2H2

Concept Check Complete the following statement with “different” or “the same” and be prepared to justify your answer. The pressures of the gas in the two balloons are ________________. He H2H2

Concept Check Complete the following statement with “different” or “the same” and be prepared to justify your answer. The temperatures of the gas in the two balloons are _____________. He H2H2

Concept Check Complete the following statement with “different” or “the same” and be prepared to justify your answer. The numbers of moles of the gas in the two balloons are __________. He H2H2

Concept Check Complete the following statement with “different” or “the same” and be prepared to justify your answer. The densities of the gas in the two balloons are __________. He H2H2

KMT Explanation of Boyle’s Law Pressure-Volume Relationship At constant temperature, gaseous particles travel with the same average speed. If volume is compressed, the average distance traveled by particles decreases; This leads to a higher frequency of molecular collisions with walls, which leads to higher gas pressure.

Concept Check Sketch a graph of: I.Pressure versus volume at constant temperature and moles.

Molecular View of Boyle’s Law

Concept Check Sketch a graph of: II.Volume vs. temperature (  C) at constant pressure and moles.

Concept Check Sketch a graph of: III.Volume vs. temperature (K) at constant pressure and moles.

Molecular View of Charles’s Law

Concept Check Sketch a graph of: IV.Volume vs. moles at constant temperature and pressure.

Molecular View of The Ideal Gas Law

KMT Explanation of Charles’s Law Volume–Temperature Relationship The volume of a gas will increase proportionally as the temperature increases at constant pressure. Increasing the temperature causes: –molecules to travel faster; –rate of molecular collision with container walls increases. To maintain a constant pressure when temperature increases: –the volume of the gas must increase to reduce the frequency of collisions.

Concept Check Which of the following best represents the mass ratio of Ne:Ar in the balloons? 1:1 1:2 2:1 1:3 3:1 Ne Ar V Ne = 2V Ar

Concept Check You have a sample of nitrogen gas (N 2 ) in a container fitted with a piston that maintains a pressure of 6.00 atm. Initially, the gas is at 45  C in a volume of 6.00 L. You then cool the gas sample.

Concept Check Which best explains the final result that occurs once the gas sample has cooled? a)The pressure of the gas increases. b)The volume of the gas increases. c)The pressure of the gas decreases. d)The volume of the gas decreases. e)Both volume and pressure change.

Concept Check The gas sample is then cooled to a temperature of 15  C. Find the new gas volume. (Hint: A moveable piston keeps the pressure constant overall, so what condition will change?) 5.43 L

KMT Explanation of Charles’s Law Pressure-Temperature Relationship Gas pressure will increase proportionally as the temperature is increased at constant volume. As the temperature increases, average molecular speed increases; This causes frequency of collisions to increase, which leads to an increase in pressure.

KMT Explanation of Avogadro’s Law Volume – Mole Relationship At constant temperature, increasing the number of gas molecules leads to higher rates of collisions and higher pressure; To maintain a constant pressure, volume of gas must expand. Avogadro’s law: at constant temperature and pressure, gas volume increases proportionally with the number of moles.

R = J/K·mol (J = joule = kg·m 2 /s 2 ) T = temperature of gas (in K) M = mass of a mole of gas in kg Final units are in m/s. Root Mean Square Velocity

Diffusion and Effusion Diffusion – the mixing of gases. Effusion – describes the passage of a gas through a tiny orifice into an evacuated chamber. Rate of effusion measures the speed at which the gas escapes from the chamber.

Effusion

Diffusion

M 1 and M 2 represent the molar masses of the gases. Graham’s Law of Effusion

Real Gases We must correct for non-ideal gas behavior when: –Pressure of the gas is high. –Temperature is low. Under these conditions: –Concentration of gas particles is high. –Attractive forces become important.

Plots of PV/nRT Versus P for Several Gases (200 K)

Plots of PV/nRT Versus P for Nitrogen Gas at Three Temperatures

Equation for Real Gases van der Waals equation for real gases includes two factors: one correct the effect of intermolecular forces (n 2 a/V 2 ) and the other correct the effect of molecular volume (nb), such that, {P + (n 2 a/V 2 )}(V – nb) = nRT For 1 mole of gas (n = 1) (P + a/V 2 )(V – b) = RT This yields the expression for pressure, P = {RT/(V – b)} – a/V 2

Real Gases (van der Waals Equation) corrected pressure P ideal  corrected volume V ideal

Values of the van der Waals Constants for Some Gases The value of a reflects how much of a correction must be made to adjust the observed pressure up to the expected ideal pressure. A low value for a reflects weak intermolecular forces among the gas molecules.

Characteristics Real Gases The actual pressure for a real gas is lower than that expected for an ideal gas. The existence of intermolecular (attractive) forces lowers the observed pressure in real gases.

Atmospheric Chemistry Earth atmosphere is divided into 5 regions: –Troposphere (ca. 0 – 12 km) – climate changes occur –Stratosphere (ca. 12 – 50 km) – where ozone layer is –Mesosphere (ca. 50 – 80 km, –Thermosphere (ca. 80 – 1000 km, and –Exosphere (ca. >1000 km) Principal components of atmosphere: –N 2, O 2, Ar, H 2 O, and CO 2 –H 2 O and CO 2 make the Earth warm and livable

The Ozone Layer The formation of ozone in the Stratosphere: Photo-dissociation of O 2 : O 2  2O; Reaction of O with O 2 to form ozone molecule: O 2 + O  O 3 Absorption of uv radiation by ozone molecule: O 3 + uv  O 2 + O

Destruction of Ozone Layer Photo-dissociation of CFCs to produce a free-radical Cl: CCl 2 F 2 + uv  CClF 2 + Cl Reaction of Cl with O 3 : Cl + O 3  ClO + O 2 Reaction of ClO with O and re-generation of Cl: ClO + O  O 2 + Cl Overall reaction: O 3 + O  2O 2

Two main sources:  Transportation  Production of electricity Combustion of petroleum produces CO, CO 2, NO, and NO 2, along with unburned molecules (hydrocarbons) from gasoline. Air Pollution

Atmospheric Pollutants Formation of nitric oxide, NO, in internal combustion engines and subsequent oxidation to NO 2 : N 2 + O 2  2NO 2NO + O 2  2NO 2 Photo-dissociation of NO 2 : NO 2 + uv  NO + O Reaction of reactive O atom with O 2 to form O 3 O 2 + O + M  O 3 + M* (M – any particle that can absorb excess heat)

Atmospheric Pollutants Photo-dissociation of O 3 that produces O 2 and excited O*: O 3 + uv  O 2 + O* Reaction of O* with H 2 O to produce excited hydroxyl radical OH: H 2 O + O*  2OH Reaction of OH radical with NO 2 forms HNO 3 : OH + NO 2  HNO 3 HNO 3 causes acid rain.

Photochemical Smog Reaction of hydroxyl radical with HC, such as methane, and other subsequent reactions involving free-radicals: CH 4 + OH  H 2 O + CH 3 CH 3 + O 2  CH 3 O 2 CH 3 O 2 + NO  CH 3 O + NO 2 CH 3 O + O 2  HCHO + HO 2 HO 2 + NO  NO 2 + OH Overall reaction: CH 4 + 2O 2 + 2NO  HCHO + 2NO 2 + H 2 O

Concentration for Some Smog Components vs. Time of Day

Other Reactions in Photochemical Smog Reactions of OH radical lead to formation of atmospheric pollutant PAN: CH 3 CH 3 + OH  CH 3 CH 2 + H 2 O CH 3 CH 2 + O 2  CH 3 CH 2 O 2 CH 3 CH 2 O 2 + NO  CH 3 CH 2 O + NO 2 CH 3 CH 2 O + O 2  CH 3 CHO + HO 2 HO 2 + NO  NO 2 + OH CH 3 CHO + OH  CH 3 CO + H 2 O CH 3 CO + O 2  CH 3 COO 2 CH 3 COO 2 + NO 2  CH 3 COO 2 NO 2 (PAN) (PAN = peroxyacetylnitrate is an eye irritant)

Sulfur forms SO 2 gas when burned. SO 2 oxidizes into SO 3, which combines with water droplets in the air to form sulfuric acid, H 2 SO 4. Sulfur Oxides (from Burning Coal for Electricity)

The Formation of Acid Rain, HNO 3 The formation of NO, NO 2, and HNO 3 : N 2 + O 2  2NO 2NO + O 2  2NO 2 3NO 2 + H 2 O  2HNO 3 + NO NO 2 + OH  HNO 3 ; Action of HNO 3 on marble structures: CaCO 3 (s) + 2HNO 3  Ca(NO 3 ) 2 + H 2 O + CO 2

The Formation of Acid Rain, H 2 SO 4 Combustion of sulfur-containing coal produces SO 2 gas, which is further oxidized to SO 3 : S + O 2  SO 2 2SO 2 + O 2  2SO 3 Reaction of SO 3 with rain water produces H 2 SO 4 : SO 3 + H 2 O  H 2 SO 4 (aq) Reaction of H 2 SO 4 with marble structures: CaCO 3 (s) + H 2 SO 4 (aq)  CaSO 4 (aq) + H 2 O (l) + CO 2 (g)

Removal of SO 2 From Flu-Gas Decomposition of CaCO 3 to produce quicklime: CaCO 3 (s)  CaO (s) + CO 2 (g) Reaction of CaO with SO 2 to form CaSO 3 : CaO (s) + SO 2 (g)  CaSO 3 (s) (CaSO 3 is disposed in landfills)

A Schematic Diagram of a Scrubber