Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall

Air Pressure & Shallow Wells water for many homes is supplied by a well less than 30 ft. deep with a pump at the surface the pump removes air from the pipe, decreasing the air pressure in the pipe the outside air pressure then pushes the water up the pipe the maximum height the water will rise is related to the amount of pressure the air exerts Tro, Chemistry: A Molecular Approach

Atmospheric Pressure pressure is the force exerted over an area on average, the air exerts the same pressure that a column of water 10.3 m high would exert 14.7 lbs./in2 so if our pump could get a perfect vacuum, the maximum height the column could rise is 10.3 m Tro, Chemistry: A Molecular Approach

Gases Pushing gas molecules are constantly in motion as they move and strike a surface, they push on that surface push = force if we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting pressure = force per unit area Tro, Chemistry: A Molecular Approach

The Effect of Gas Pressure the pressure exerted by a gas can cause some amazing and startling effects whenever there is a pressure difference, a gas will flow from area of high pressure to low pressure the bigger the difference in pressure, the stronger the flow of the gas if there is something in the gas’s path, the gas will try to push it along as the gas flows Tro, Chemistry: A Molecular Approach

Atmospheric Pressure Effects differences in air pressure result in weather and wind patterns the higher up in the atmosphere you climb, the lower the atmospheric pressure is around you at the surface the atmospheric pressure is 14.7 psi, but at 10,000 ft it is only 10.0 psi rapid changes in atmospheric pressure may cause your ears to “pop” due to an imbalance in pressure on either side of your ear drum Tro, Chemistry: A Molecular Approach

Pressure Imbalance in Ear If there is a difference in pressure across the eardrum membrane, the membrane will be pushed out – what we commonly call a “popped eardrum.” picture from Tro Intro Chem 2nd ed. Tro, Chemistry: A Molecular Approach

The Pressure of a Gas result of the constant movement of the gas molecules and their collisions with the surfaces around them the pressure of a gas depends on several factors number of gas particles in a given volume volume of the container average speed of the gas particles Tro, Chemistry: A Molecular Approach

Measuring Air Pressure use a barometer column of mercury supported by air pressure force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury gravity Tro, Chemistry: A Molecular Approach

Common Units of Pressure Average Air Pressure at Sea Level pascal (Pa), 101,325 kilopascal (kPa) 101.325 atmosphere (atm) 1 (exactly) millimeters of mercury (mmHg) 760 (exactly) inches of mercury (inHg) 29.92 torr (torr) pounds per square inch (psi, lbs./in2) 14.7 Tro, Chemistry: A Molecular Approach

since mmHg are smaller than psi, the answer makes sense Example 5.1 – A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg? Given: Find: 132 psi mmHg Concept Plan: Relationships: 1 atm = 14.7 psi, 1 atm = 760 mmHg psi atm mmHg Solution: Check: since mmHg are smaller than psi, the answer makes sense

Manometers the pressure of a gas trapped in a container can be measured with an instrument called a manometer manometers are U-shaped tubes, partially filled with a liquid, connected to the gas sample on one side and open to the air on the other a competition is established between the pressure of the atmosphere and the gas the difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere Tro, Chemistry: A Molecular Approach

Manometer for this sample, the gas has a larger pressure than the atmosphere, so Tro, Chemistry: A Molecular Approach

Boyle’s Law pressure of a gas is inversely proportional to its volume constant T and amount of gas graph P vs V is curve graph P vs 1/V is straight line as P increases, V decreases by the same factor P x V = constant P1 x V1 = P2 x V2 Tro, Chemistry: A Molecular Approach

Boyle’s Experiment added Hg to a J-tube with air trapped inside used length of air column as a measure of volume Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

Tro, Chemistry: A Molecular Approach

Boyle’s Experiment, P x V Tro, Chemistry: A Molecular Approach

When you double the pressure on a gas, the volume is cut in half (as long as the temperature and amount of gas do not change) Tro, Chemistry: A Molecular Approach

Boyle’s Law and Diving since water is denser than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm at 20 m the total pressure is 3 atm if your tank contained air at 1 atm pressure you would not be able to inhale it into your lungs Tro, Chemistry: A Molecular Approach

Example 5. 2 – A cylinder with a movable piston has a volume of 7 Example 5.2 – A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm? Given: Find: V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm V2, L Concept Plan: Relationships: P1 ∙ V1 = P2 ∙ V2 V1, P1, P2 V2 Solution: Check: since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does

Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2780 mL, what was it originally? Tro, Chemistry: A Molecular Approach

P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly) V1, P1, P2 V2 A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2780 mL, what was it originally? Given: Find: V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm V1, mL Concept Plan: Relationships: P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly) V1, P1, P2 V2 Solution: Check: since P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does

Charles’ Law volume is directly proportional to temperature constant P and amount of gas graph of V vs T is straight line as T increases, V also increases Kelvin T = Celsius T + 273 V = constant x T if T measured in Kelvin Tro, Chemistry: A Molecular Approach

Charles’ Law – A Molecular View the pressure of gas inside and outside the balloon are the same at high temperatures, the gas molecules are moving faster, so they hit the sides of the balloon harder – causing the volume to become larger the pressure of gas inside and outside the balloon are the same at low temperatures, the gas molecules are not moving as fast, so they don’t hit the sides of the balloon as hard – therefore the volume is small Tro, Chemistry: A Molecular Approach

The data fall on a straight line. If the lines are extrapolated back to a volume of “0,” they all show the same temperature, -273.15°C, called absolute zero

Example 5. 3 – A gas has a volume of 2. 57 L at 0. 00°C Example 5.3 – A gas has a volume of 2.57 L at 0.00°C. What was the temperature at 2.80 L? Given: Find: V1 =2.57 L, V2 = 2.80 L, t2 = 0.00°C t1, K and °C Concept Plan: Relationships: T(K) = t(°C) + 273.15, V1, V2, T2 T1 Solution: Check: since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does

Practice – The temperature inside a balloon is raised from 25 Practice – The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air? Tro, Chemistry: A Molecular Approach

The temperature inside a balloon is raised from 25. 0°C to 250. 0°C The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air? Given: Find: V1 =10.0 L, t1 = 25.0°C L, t2 = 250.0°C V2, L Concept Plan: Relationships: T(K) = t(°C) + 273.15, V1, T1, T2 V2 Solution: Check: since T and V are directly proportional, when the temperature increases, the volume should increase, and it does

Avogadro’s Law volume directly proportional to the number of gas molecules V = constant x n constant P and T more gas molecules = larger volume count number of gas molecules by moles equal volumes of gases contain equal numbers of molecules the gas doesn’t matter Tro, Chemistry: A Molecular Approach

Example 5. 4 – A 0. 225 mol sample of He has a volume of 4. 65 L Example 5.4 – A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L? Given: Find: V1 =4.65 L, V2 = 6.48 L, n1 = 0.225 mol n2, and added moles Concept Plan: Relationships: mol added = n2 – n1, V1, V2, n1 n2 Solution: Check: since n and V are directly proportional, when the volume increases, the moles should increase, and it does

Ideal Gas Law By combing the gas laws we can write a general equation R is called the gas constant the value of R depends on the units of P and V we will use 0.08206 and convert P to atm and V to L the other gas laws are found in the ideal gas law if two variables are kept constant allows us to find one of the variables if we know the other 3 Tro, Chemistry: A Molecular Approach

Example 5.6 – How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C? Given: Find: V = 3.24 L, P = 24.3 psi, t = 25 °C, n, mol Concept Plan: Relationships: 1 atm = 14.7 psi T(K) = t(°C) + 273.15 P, V, T, R n Solution: Check: 1 mole at STP occupies 22.4 L, since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas

Standard Conditions since the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions STP standard pressure = 1 atm standard temperature = 273 K 0°C Tro, Chemistry: A Molecular Approach

Practice – A gas occupies 10. 0 L at 44. 1 psi and 27°C Practice – A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions? Tro, Chemistry: A Molecular Approach

A gas occupies 10. 0 L at 44. 1 psi and 27°C A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions? Given: Find: V1 = 10.0 L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0°C V2, L Concept Plan: Relationships: 1 atm = 14.7 psi T(K) = t(°C) + 273.15 P1, V1, T1, R n P2, n, T2, R V2 Solution: Check: 1 mole at STP occupies 22.4 L, since there is more than 1 mole, we expect more than 22.4 L of gas

Molar Volume solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L 6.022 x 1023 molecules of gas notice: the gas is immaterial we call the volume of 1 mole of gas at STP the molar volume it is important to recognize that one mole of different gases have different masses, even though they have the same volume Tro, Chemistry: A Molecular Approach

Molar Volume Tro, Chemistry: A Molecular Approach

Density at Standard Conditions density is the ratio of mass-to-volume density of a gas is generally given in g/L the mass of 1 mole = molar mass the volume of 1 mole at STP = 22.4 L Tro, Chemistry: A Molecular Approach

Gas Density density is directly proportional to molar mass Tro, Chemistry: A Molecular Approach

Example 5.7 – Calculate the density of N2 at 125°C and 755 mmHg Given: Find: P = 755 mmHg, t = 125 °C, dN2, g/L 1 atm = 760 mmHg, MM = 28.01 g T(K) = t(°C) + 273.15 P, MM, T, R d Concept Plan: Relationships: Solution: Check: since the density of N2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is

Molar Mass of a Gas one of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law Tro, Chemistry: A Molecular Approach

the value 31.9 g/mol is reasonable Example 5.8 – Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg Given: Find: m=0.311g, V=0.225 L, P=886 mmHg, t=55°C, molar mass, g/mol m=0.311g, V=0.225 L, P=1.1658 atm, T=328 K, molar mass, g/mol P, V, T, R n n, m MM Concept Plan: Relationships: 1 atm = 760 mmHg, T(K) = t(°C) + 273.15 Solution: Check: the value 31.9 g/mol is reasonable

Practice - Calculate the density of a gas at 775 torr and 27°C if 0 Practice - Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g Tro, Chemistry: A Molecular Approach

the value 1.65 g/L is reasonable Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g Given: Find: m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C, density, g/L m=9.988g, n=0.250 mol, P=1.0197 atm, T=300. K density, g/L P, n, T, R V V, m d Concept Plan: Relationships: 1 atm = 760 mmHg, T(K) = t(°C) + 273.15 Solution: Check: the value 1.65 g/L is reasonable

Mixtures of Gases when gases are mixed together, their molecules behave independent of each other all the gases in the mixture have the same volume all completely fill the container  each gas’s volume = the volume of the container all gases in the mixture are at the same temperature therefore they have the same average kinetic energy therefore, in certain applications, the mixture can be thought of as one gas even though air is a mixture, we can measure the pressure, volume, and temperature of air as if it were a pure substance we can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules Tro, Chemistry: A Molecular Approach

Partial Pressure the pressure of a single gas in a mixture of gases is called its partial pressure we can calculate the partial pressure of a gas if we know what fraction of the mixture it composes and the total pressure or, we know the number of moles of the gas in a container of known volume and temperature the sum of the partial pressures of all the gases in the mixture equals the total pressure Dalton’s Law of Partial Pressures because the gases behave independently Tro, Chemistry: A Molecular Approach

Composition of Dry Air Tro, Chemistry: A Molecular Approach

The partial pressure of each gas in a mixture can be calculated using the ideal gas law Tro, Chemistry: A Molecular Approach

Example 5.9 – Determine the mass of Ar in the mixture PAr = 0.275 atm, V = 1.00 L, T=298 K massAr, g PHe=341 mmHg, PNe=112 mmHg, Ptot = 662 mmHg, V = 1.00 L, T=298 K massAr, g Given: Find: Ptot, PHe, PNe PAr PAr, V, T nAr mAr Concept Plan: Relationships: PAr = Ptot – (PHe + PNe) Ptot = Pa + Pb + etc., 1 atm = 760 mmHg, MMAr = 39.95 g/mol Solution: Check: the units are correct, the value is reasonable

Practice – Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe. Tro, Chemistry: A Molecular Approach

the unit is correct, the value is reasonable Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe Given: Find: Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol PNe, atm nXe, V, T, R PXe Ptot, PXe PNe Concept Plan: Relationships: Solution: the unit is correct, the value is reasonable Check:

Mole Fraction the fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes the ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction, c for gases, = volume % / 100% the partial pressure of a gas is equal to the mole fraction of that gas times the total pressure Tro, Chemistry: A Molecular Approach

Mountain Climbing & Partial Pressure our bodies are adapted to breathe O2 at a partial pressure of 0.21 atm Sherpa, people native to the Himalaya mountains, are adapted to the much lower partial pressure of oxygen in their air partial pressures of O2 lower than 0.1 atm will lead to hypoxia unconsciousness or death climbers of Mt Everest carry O2 in cylinders to prevent hypoxia on top of Mt Everest, Pair = 0.311 atm, so PO2 = 0.065 atm Tro, Chemistry: A Molecular Approach

Deep Sea Divers & Partial Pressure its also possible to have too much O2, a condition called oxygen toxicity PO2 > 1.4 atm oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions its also possible to have too much N2, a condition called nitrogen narcosis also known as Rapture of the Deep when diving deep, the pressure of the air divers breathe increases – so the partial pressure of the oxygen increases at a depth of 55 m the partial pressure of O2 is 1.4 atm divers that go below 50 m use a mixture of He and O2 called heliox that contains a lower percentage of O2 than air Tro, Chemistry: A Molecular Approach

Partial Pressure & Diving Tro, Chemistry: A Molecular Approach

Ex 5. 10 – Find the mole fractions and partial pressures in a 12 Ex 5.10 – Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K Given: Find: mHe = 24.2 g, mO2 = 43.2 g V = 12.5 L, T = 298 K cHe, cO2, PHe, atm, PO2, atm, Ptotal, atm nHe = 6.05 mol, nO2 = 0.135 mol V = 12.5 L, T = 298 K cHe=0.97817, cO2=0.021827, PHe, atm, PO2, atm, Ptotal, atm mgas ngas cgas ntot, V, T, R Ptot Concept Plan: Relationships: cgas, Ptotal Pgas MMHe = 4.00 g/mol MMO2 = 32.00 g/mol Solution:

Collecting Gases gases are often collected by having them displace water from a container the problem is that since water evaporates, there is also water vapor in the collected gas the partial pressure of the water vapor, called the vapor pressure, depends only on the temperature so you can use a table to find out the partial pressure of the water vapor in the gas you collect if you collect a gas sample with a total pressure of 758.2 mmHg* at 25°C, the partial pressure of the water vapor will be 23.78 mmHg – so the partial pressure of the dry gas will be 734.4 mmHg Table 5.4* Tro, Chemistry: A Molecular Approach

Vapor Pressure of Water Tro, Chemistry: A Molecular Approach

Collecting Gas by Water Displacement Tro, Chemistry: A Molecular Approach

Ex 5.11 – 1.02 L of O2 collected over water at 293 K with a total pressure of 755.2 mmHg. Find mass O2. Given: Find: V=1.02 L, PO2=737.65 mmHg, T=293 K mass O2, g V=1.02 L, P=755.2 mmHg, T=293 K mass O2, g Ptot, PH2O PO2 PO2,V,T nO2 gO2 Concept Plan: Relationships: 1 atm = 760 mmHg, Ptotal = PA + PB, O2 = 32.00 g/mol Solution:

Practice – 0. 12 moles of H2 is collected over water in a 10 Practice – 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure. Tro, Chemistry: A Molecular Approach

12 moles of H2 is collected over water in a 10. 0 L container at 323 K 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure. Given: Find: V=10.0 L, nH2=0.12 mol, T=323 K Ptotal, atm Concept Plan: Relationships: nH2,V,T PH2 PH2, PH2O Ptotal 1 atm = 760 mmHg Ptotal = PA + PB, Solution:

Reactions Involving Gases the principles of reaction stoichiometry from Chapter 4 can be combined with the gas laws for reactions involving gases in reactions of gases, the amount of a gas is often given as a volume instead of moles as we’ve seen, must state pressure and temperature the ideal gas law allows us to convert from the volume of the gas to moles; then we can use the coefficients in the equation as a mole ratio when gases are at STP, use 1 mol = 22.4 L P, V, T of Gas A mole A mole B P, V, T of Gas B Tro, Chemistry: A Molecular Approach

Ex 5. 12 – What volume of H2 is needed to make 35 Ex 5.12 – What volume of H2 is needed to make 35.7 g of CH3OH at 738 mmHg and 355 K? CO(g) + 2 H2(g) → CH3OH(g) Given: Find: mCH3OH = 37.5g, P=738 mmHg, T=355 K VH2, L nH2 = 2.2284 mol, P=0.97105 atm, T=355 K VH2, L g CH3OH mol CH3OH mol H2 P, n, T, R V Concept Plan: Relationships: 1 atm = 760 mmHg, CH3OH = 32.04 g/mol 1 mol CH3OH : 2 mol H2 Solution:

Ex 5. 13 – How many grams of H2O form when 1 Ex 5.13 – How many grams of H2O form when 1.24 L H2 reacts completely with O2 at STP? O2(g) + 2 H2(g) → 2 H2O(g) Given: Find: VH2 = 1.24 L, P=1.00 atm, T=273 K massH2O, g g H2O L H2 mol H2 mol H2O Concept Plan: Relationships: H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP 2 mol H2O : 2 mol H2 Solution:

Practice – What volume of O2 at 0 Practice – What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g) (MMHgO = 216.59 g/mol) Tro, Chemistry: A Molecular Approach

What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g) Given: Find: mHgO = 10.0g, P=0.750 atm, T=313 K VO2, L nO2 = 0.023085 mol, P=0.750 atm, T=313 K VO2, L g HgO mol HgO mol O2 P, n, T, R V Concept Plan: Relationships: 1 atm = 760 mmHg, HgO = 216.59 g/mol 2 mol HgO : 1 mol O2 Solution:

Properties of Gases expand to completely fill their container take the shape of their container low density much less than solid or liquid state compressible mixtures of gases are always homogeneous fluid Tro, Chemistry: A Molecular Approach

Kinetic Molecular Theory the particles of the gas (either atoms or molecules) are constantly moving the attraction between particles is negligible when the moving particles hit another particle or the container, they do not stick; but they bounce off and continue moving in another direction like billiard balls Tro, Chemistry: A Molecular Approach

Kinetic Molecular Theory there is a lot of empty space between the particles compared to the size of the particles the average kinetic energy of the particles is directly proportional to the Kelvin temperature as you raise the temperature of the gas, the average speed of the particles increases but don’t be fooled into thinking all the particles are moving at the same speed!! Tro, Chemistry: A Molecular Approach

Gas Properties Explained – Indefinite Shape and Indefinite Volume Because the gas molecules have enough kinetic energy to overcome attractions, they keep moving around and spreading out until they fill the container. As a result, gases take the shape and the volume of the container they are in. Tro, Chemistry: A Molecular Approach

Gas Properties Explained - Compressibility Because there is a lot of unoccupied space in the structure of a gas, the gas molecules can be squeezed closer together Tro, Chemistry: A Molecular Approach

Gas Properties Explained – Low Density Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume, the result is they have low density Tro, Chemistry: A Molecular Approach

Density & Pressure result of the constant movement of the gas molecules and their collisions with the surfaces around them when more molecules are added, more molecules hit the container at any one instant, resulting in higher pressure also higher density Tro, Chemistry: A Molecular Approach

Gas Laws Explained - Boyle’s Law Boyle’s Law says that the volume of a gas is inversely proportional to the pressure decreasing the volume forces the molecules into a smaller space more molecules will collide with the container at any one instant, increasing the pressure Tro, Chemistry: A Molecular Approach

Gas Laws Explained - Charles’s Law Charles’s Law says that the volume of a gas is directly proportional to the absolute temperature increasing the temperature increases their average speed, causing them to hit the wall harder and more frequently on average in order to keep the pressure constant, the volume must then increase Tro, Chemistry: A Molecular Approach

Gas Laws Explained Avogadro’s Law Avogadro’s Law says that the volume of a gas is directly proportional to the number of gas molecules increasing the number of gas molecules causes more of them to hit the wall at the same time in order to keep the pressure constant, the volume must then increase Tro, Chemistry: A Molecular Approach

Gas Laws Explained – Dalton’s Law of Partial Pressures Dalton’s Law says that the total pressure of a mixture of gases is the sum of the partial pressures kinetic-molecular theory says that the gas molecules are negligibly small and don’t interact therefore the molecules behave independent of each other, each gas contributing its own collisions to the container with the same average kinetic energy since the average kinetic energy is the same, the total pressure of the collisions is the same Tro, Chemistry: A Molecular Approach

Dalton’s Law & Pressure since the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side Tro, Chemistry: A Molecular Approach

Deriving the Ideal Gas Law from Kinetic-Molecular Theory pressure = Forcetotal/Area Ftotal = F1 collision x number of collisions in a particular time interval F1 collision = mass x 2(velocity)/time interval no. of collisions is proportional to the number of particles within the distance (velocity x time interval) from the wall Ftotal α mass∙velocity2 x Area x no. molecules/Volume Pressure α mv2 x n/V Temperature α mv2 P α T∙n/V,  PV=nRT Tro, Chemistry: A Molecular Approach

Calculating Gas Pressure Tro, Chemistry: A Molecular Approach

Molecular Velocities all the gas molecules in a sample can travel at different speeds however, the distribution of speeds follows a pattern called a Boltzman distribution we talk about the “average velocity” of the molecules, but there are different ways to take this kind of average the method of choice for our average velocity is called the root-mean-square method, where the rms average velocity, urms, is the square root of the average of the sum of the squares of all the molecule velocities Tro, Chemistry: A Molecular Approach

Boltzman Distribution Tro, Chemistry: A Molecular Approach

Kinetic Energy and Molecular Velocities average kinetic energy of the gas molecules depends on the average mass and velocity KE = ½mv2 gases in the same container have the same temperature, the same average kinetic energy if they have different masses, the only way for them to have the same kinetic energy is to have different average velocities lighter particles will have a faster average velocity than more massive particles Tro, Chemistry: A Molecular Approach

Molecular Speed vs. Molar Mass in order to have the same average kinetic energy, heavier molecules must have a slower average speed Tro, Chemistry: A Molecular Approach

Temperature and Molecular Velocities _ KEavg = ½NAmu2 NA is Avogadro’s number KEavg = 1.5RT R is the gas constant in energy units, 8.314 J/mol∙K 1 J = 1 kg∙m2/s2 equating and solving we get: NA∙mass = molar mass in kg/mol as temperature increases, the average velocity increases Tro, Chemistry: A Molecular Approach

Temperature vs. Molecular Speed as the absolute temperature increases, the average velocity increases the distribution function “spreads out,” resulting in more molecules with faster speeds Tro, Chemistry: A Molecular Approach

Ex 5.14 – Calculate the rms velocity of O2 at 25°C Given: Find: O2, t = 25°C urms Concept Plan: Relationships: T(K) = t(°C) + 273.15, O2 = 32.00 g/mol MM, T urms Solution:

Mean Free Path molecules in a gas travel in straight lines until they collide with another molecule or the container the average distance a molecule travels between collisions is called the mean free path mean free path decreases as the pressure increases Tro, Chemistry: A Molecular Approach

Diffusion and Effusion the process of a collection of molecules spreading out from high concentration to low concentration is called diffusion the process by which a collection of molecules escapes through a small hole into a vacuum is called effusion both the rates of diffusion and effusion of a gas are related to its rms average velocity for gases at the same temperature, this means that the rate of gas movement is inversely proportional to the square root of the molar mass Tro, Chemistry: A Molecular Approach

Effusion Tro, Chemistry: A Molecular Approach

Graham’s Law of Effusion for two different gases at the same temperature, the ratio of their rates of effusion is given by the following equation: Tro, Chemistry: A Molecular Approach

Ex 5. 15 – Calculate the molar mass of a gas that effuses at a rate 0 Ex 5.15 – Calculate the molar mass of a gas that effuses at a rate 0.462 times N2 Given: Find: MM, g/mol Concept Plan: Relationships: rateA/rateB, MMN2 MMunknown N2 = 28.01 g/mol Solution:

Ideal vs. Real Gases Real gases often do not behave like ideal gases at high pressure or low temperature Ideal gas laws assume no attractions between gas molecules gas molecules do not take up space based on the kinetic-molecular theory at low temperatures and high pressures these assumptions are not valid

The Effect of Molecular Volume at high pressure, the amount of space occupied by the molecules is a significant amount of the total volume the molecular volume makes the real volume larger than the ideal gas law would predict van der Waals modified the ideal gas equation to account for the molecular volume b is called a van der Waals constant and is different for every gas because their molecules are different sizes Tro, Chemistry: A Molecular Approach

Real Gas Behavior because real molecules take up space, the molar volume of a real gas is larger than predicted by the ideal gas law at high pressures Tro, Chemistry: A Molecular Approach

The Effect of Intermolecular Attractions at low temperature, the attractions between the molecules is significant the intermolecular attractions makes the real pressure less than the ideal gas law would predict van der Waals modified the ideal gas equation to account for the intermolecular attractions a is called a van der Waals constant and is different for every gas because their molecules are different sizes Tro, Chemistry: A Molecular Approach

Real Gas Behavior because real molecules attract each other, the molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures Tro, Chemistry: A Molecular Approach

Van der Waals’ Equation combining the equations to account for molecular volume and intermolecular attractions we get the following equation used for real gases a and b are called van der Waal constants and are different for each gas Tro, Chemistry: A Molecular Approach

Real Gases a plot of PV/RT vs. P for 1 mole of a gas shows the difference between real and ideal gases it reveals a curve that shows the PV/RT ratio for a real gas is generally lower than ideality for “low” pressures – meaning the most important factor is the intermolecular attractions it reveals a curve that shows the PV/RT ratio for a real gas is generally higher than ideality for “high” pressures – meaning the most important factor is the molecular volume Tro, Chemistry: A Molecular Approach

PV/RT Plots Tro, Chemistry: A Molecular Approach

Structure of the Atmosphere the atmosphere shows several layers, each with its own characteristics the troposphere is the layer closest to the earth’s surface circular mixing due to thermal currents – weather the stratosphere is the next layer up less air mixing the boundary between the troposphere and stratosphere is called the tropopause the ozone layer is located in the stratosphere Tro, Chemistry: A Molecular Approach

Air Pollution air pollution is materials added to the atmosphere that would not be present in the air without, or are increased by, man’s activities though many of the “pollutant” gases have natural sources as well pollution added to the troposphere has a direct effect on human health and the materials we use because we come in contact with it and the air mixing in the troposphere means that we all get a smell of it! pollution added to the stratosphere may have indirect effects on human health caused by depletion of ozone and the lack of mixing and weather in the stratosphere means that pollutants last longer before “washing” out Tro, Chemistry: A Molecular Approach

Pollutant Gases, SOx SO2 and SO3, oxides of sulfur, come from coal combustion in power plants and metal refining as well as volcanoes lung and eye irritants major contributor to acid rain 2 SO2 + O2 + 2 H2O  2 H2SO4 SO3 + H2O  H2SO4 Tro, Chemistry: A Molecular Approach

Pollutant Gases, NOx NO and NO2, oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plants as well as lightning storms NO2 causes the brown haze seen in some cities lung and eye irritants strong oxidizers major contributor to acid rain 4 NO + 3 O2 + 2 H2O  4 HNO3 4 NO2 + O2 + 2 H2O  4 HNO3 Tro, Chemistry: A Molecular Approach

Pollutant Gases, CO CO comes from incomplete burning of fossil fuels in cars, trucks, and power plants adheres to hemoglobin in your red blood cells, depleting your ability to acquire O2 at high levels can cause sensory impairment, stupor, unconsciousness, or death Tro, Chemistry: A Molecular Approach

Pollutant Gases, O3 ozone pollution comes from other pollutant gases reacting in the presence of sunlight as well as lightning storms known as photochemical smog and ground-level ozone O3 is present in the brown haze seen in some cities lung and eye irritants strong oxidizer Tro, Chemistry: A Molecular Approach

Major Pollutant Levels government regulation has resulted in a decrease in the emission levels for most major pollutants Tro, Chemistry: A Molecular Approach

O3(g) + UV light  O2(g) + O(g) Stratospheric Ozone ozone occurs naturally in the stratosphere stratospheric ozone protects the surface of the earth from over-exposure to UV light from the sun O3(g) + UV light  O2(g) + O(g) normally the reverse reaction occurs quickly, but the energy is not UV light O2(g) + O(g)  O3(g) Tro, Chemistry: A Molecular Approach

CF2Cl2 + UV light  CF2Cl + Cl Ozone Depletion chlorofluorocarbons became popular as aerosol propellants and refrigerants in the 1960s CFCs pass through the tropopause into the stratosphere there CFCs can be decomposed by UV light, releasing Cl atoms CF2Cl2 + UV light  CF2Cl + Cl Cl atoms catalyze O3 decomposition and removes O atoms so that O3 cannot be regenerated NO2 also catalyzes O3 destruction Cl + O3  ClO + O2 O3 + UV light  O2 + O ClO + O  O2 + Cl Tro, Chemistry: A Molecular Approach

Ozone Holes satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions Tro, Chemistry: A Molecular Approach