Chapter 10 Gases10.1 Characteristics of Gases10.2 Pressure10.3 The Gas Laws10.4 The Ideal-Gas Equation10.5 Further Applications of the Ideal-Gas Equation10.6 Gas Mixtures and Partial Pressure10.7 Kinetic-Molecular Theory10.8 Molecular Effusion and Diffusion10.9 Real Gases: Deviations from Ideal Behavior1. Characteristics of Gases•Gases are highly compressible and occupy the full volume of their containers.•When a gas is subjected to pressure, its volume decreases.•Gases always form homogeneous mixtures with other gases.•The total volume of the molecules of a gas sample is only about 0.1 % of the volume of their containers.12. Pressure•Pressure is the force acting on an object per unit area:P=FA•Gravity exerts a force on the earth’s atmosphere.•A column of air 1 m2in cross section exerts a force of 105N.•The pressure of a 1 m2column of air is 100 kPa.Atmospheric Pressure and the Barometer•SI Units: 1 N = 1 kg⋅m/s2; 1 Pa = 1 N/m2.•Atmospheric pressure is measured with a barometer.•If a tube is inserted into a container of mercury open to the atmosphere, the mercury will rise 760 mm up the tube.•Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column.•Units: 1 atm= 760 mmHg = 760 torr= 1.01325 ×105Pa = 101.325 kPa.2Mercury BarometerMercury Manometer•The pressures of gases not open to the atmosphere are measured in a manometer. •A manometer consists of a bulb of gas attached to a U-tube containing Hg:–If Pgas< Patmthen Pgas+ Ph2= Patm.–If Pgas> Patmthen Pgas= Patm+ Ph2.3. The Gas LawsThe Pressure-Volume RelationshipBoyle’s Law-the volume, V, of a fixed quantity of gas at constant temperature is inversely proportional to its pressure, P.Mathematically:V = constant x 1/PPV = constant3•A plot of Vversus Pis a hyperbola.•Similarly, a plot of Vversus 1/Pmust be a straight line passing through the origin.•Boyle used a manometer to carry out experiments to establish this law.The Temperature-Volume Relationship•Charles’s Law: The Vof a gas is directly proportional to the T. •We know that hot air balloons expand when they are heated.•Mathematically:V=constant×TVT=constant4•A plot of Vversus Tis a straight line.•When Tis measured in °C, the intercept on the temperature axis is -273.15°C. •We define absolute zero, 0 K = -273.15°C.The Quantity-Volume Relationships•Gay-Lussac’s Law of combining volumes: at a given Tand P, the Vof gases that react are ratios of small whole numbers.•Avogadro’s Law•Avogadro’s Hypothesis: equal Vof gas at the same Tand Pwill contain the same number of molecules.•Avogadro’s Law: the Vof gas at a given Tand Pis directly proportional to the number of moles of gas, n.•Mathematically:V=constant×n•We can show that 22.4 L of any gas at 0°C contain 6.02 ×1023gas molecules.54. The Ideal Gas Equation•Consider the three gas laws.•Boyle’s Law:V∝1 (constant n,T•Charles’s Law:V∝P)T (constant n,P)•Avogadro’s Law:V∝n (constant P,T)•We can combine these into a general gas law:V∝nTP•If Ris the constant of proportionality (called the gas constant), then:V=R⎛⎜nT⎞⎝P⎟⎠•The ideal gas equation is:•R= 0.08206 L·atm/mol·K= 8.314 J/mol·KPV=nRT•We define STP (standard Tand P) = 0°C, 273.15 K, 1 atm.•Vof 1 mol of gas at STP is:V=nRT(1 mol)(0.08206P= L·atm/mol·K)(273.15 K)1.000 atm=22.41 L6Exercise:SF6is a colorless, odorless, very unreactivegas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5 oC.Numerical Values of the Gas Constant, R, in Various UnitsUnits ValueL⋅atm/mol⋅K0.08206J/mol⋅K8.314cal/mol⋅K1.987m3⋅Pa/mol⋅K*8.314L_____________________________⋅torr/mol⋅K62.36 *SI unit.Relating the Ideal-Gas Equation and The Gas Laws•If PV= nRTand nand T are constant, then PV = constant and we have Boyle’s law.•Other laws can be generated similarly.•In general, if we have a gas under two sets of conditions, then:P1V1P2V2nT=11n2T27Exercise:A small bubble rises from the bottom of a lake, where thetemperature and pressure are 8 oCand 6.4 atm, to the surfaceof the water, where the temperature is 25 oCand pressure is1.0 atm. Calculate the final volume (in mL) of the bubble if itsinitial volume was 2.1 mL.P1V1n=R=P2V21T1n2T2P1V1n=P2V21T1n2T2P1V1=P2V2T1T25.Further Applications of the Ideal-Gas Equation•Density has units of mass over volume. •Rearranging the ideal-gas equation with Mas molar mass gives:PV=nRTnV=PRTnMV=d=PMRTGas Densities and Molar Mass•The molar mass, M,of a gas is given by:M=dRTPVolumes of Gases in Chemical Reactions•The ideal-gas equation relates P, V, and Tto number of moles, n, of gas.•The ncan then be used in stoichiometric calculations.8Exercise:Calculate the densityof hydrogen bromide (HBr) in grams perliter (g/L) at 733 mmHg and 46 oC.6. Gas Mixtures and Partial Pressures•Because gas molecules are so far apart, it is assumed that they behave independently.•Dalton’s Law: in a gas mixture the total Pis given by the sum of partial Pof each component:Ptotal=P1+P2+P3+L•Each gas obeys the ideal gas equation:Pi=ni⎛⎜RT⎞⎝V⎟⎠•Combining the equations:Ptotal=(n1+n2+n3+L)⎛⎜RT⎞⎝V⎟⎠Partial Pressures and Mole Fraction•Let nibe the number of moles of gas ithat exert a partial pressure Pi, then:Pi=ΧiPtotalwhere Χiis the mole fraction(ni/nt).9Exercise:Ethanol burns in air:C2H5OH(l) + O2(g) →CO2(g) + H2O(l)Balance the equation and determine the volume of air in litersat 35.0 oCand 790. mmHg required to burn 227 g of ethanol.Assume the air is 21.0 % O2by volume.Collecting Gases over Water•It is common to synthesize gases and collect them by displacing a volume of water.•To calculate the amount of gas produced, it is necessary to correct for the partial pressure of the water vapor:Ptotal=Pgas+Pwater7. Kinetic Molecular Theory•Theory developed to explain gas behavior.•Theory of moving molecules.•Assumptions:–Gases consist of a large number of molecules in constant random motion.–Volume of individual molecules negligible compared to Vof container.–Intermolecular forces (forces between gas molecules) are negligible.10•Assumptions (cont’d):–Energy can be transferred between molecules, but total kinetic energy is constant at constant T.–Average kinetic energy of molecules is proportional to T.•Kinetic molecular theory permits an understanding of Pand Ton the molecular level.•Pof a gas results from the number of collisions per unit time on the walls of container.•Magnitude of Pgiven by how often and how hard the molecules strike the walls of the container. •Gas molecules have an average kinetic energy.•Not all molecules have the same energy.Distribution of Molecular Speed•There is a distribution of individual energies of gas molecules in any sample of gas.•As the Tincreases, the average kinetic energy of the gas molecules increases.•As kinetic energy increases, the velocity of the gas molecules increases.•Root mean square speed, u, is the speed of a gas molecule having average kinetic energy.•Average kinetic energy, ε, is related to root mean square speed, rms, u:ε=1mu2211Application to Gas Laws•As Vincreases at constant T, the average kinetic energy of the gas remains constant. Therefore, uis constant. However, Vincreases so the gas molecules have to travel further to hit the walls of the container. Therefore, Pdecreases.•If Tincreases at constant V, the average kinetic energy of the gas molecules increases. Therefore, there are more collisions with the container walls and the Pincreases.8. Molecular Effusion and Diffusion•As kinetic energy increases, the velocity of the gas molecules increases.•Average kinetic energy of a gas is related to its mass:ε=12mu2•Consider two gases at the same T: the lighter gas has a higher uthan the heavier gas.•Mathematically:u=3RTM12•The lower the molar mass, M, the higher the u.Exercise:The temperature in the stratosphere is –23oC. Calculate theroot-mean-square speeds of Nin this region.2, O2, and O3moleculesGraham’s Law of Effusion•As kinetic energy increases, the velocity of the gas molecules increases.•Effusion is the escape of a gas through a tiny hole (a balloon will deflate over time due to effusion).•The rate of effusion can be quantified.13•Consider two gases with molar masses M1and M2. The relative rate of effusion is given by:ru3RT11M1Mr===22u23RTMM21•Only those molecules that hit the small hole will escape through it.•Therefore, the higher the uthe greater the probability of a gas molecule hitting the hole.Diffusion and Mean Free Path•Diffusion of a gas is the spread of the gas through space.•Diffusion is faster for light gas molecules.•Diffusion is significantly slower than uspeed (consider someone opening a perfume bottle: it takes while to detect the odor but rmsspeed at 25°C is about 1150 mi/hr).•Diffusion is slowed by gas molecules colliding with each other.•Average distance a gas molecule travels between collisions is called mean free path.•At sea level, mean free path is about 6 ×10-6cm. (~ 100-300 times the dimension of atoms)149.Real Gases: Deviations from Ideal Behavior From the ideal gas equation, we have:PVRT=n•For 1 mol of gas, PV/RT= 1 for all P.•In a real gas, PV/RTvaries from 1 significantly.•The higher the Pthe more the deviation from ideal behavior.•From the ideal gas equation, we have:PVRT=n•For 1 mol of gas, PV/RT = 1 for all T.•As Tincreases, gases behave more ideally.15•The assumptions in kinetic molecular theory show where ideal gas behavior breaks down.•Recall the assumptions:-the molecules of a gas have finite volume;-molecules of a gas do attract each other.How do changes in P, V,and Trelate to changes at the molecular level? PRESSURE•As the Pon a gas increases, the molecules are forced closer together.•As the molecules get closer together, the Vof the container gets smaller.•The smaller the container, the higher is the fraction of the space occupied by the gas molecules.•Therefore, the higher the P, the less the gas resembles an ideal gas.VOLUME•As Vdecreases at constant P, or Pincreases at constant T, the intermolecular distance decreases.•The smaller the distance between gas molecules, the more likely attractive forces will develop between the molecules.•Therefore, the less the gas resembles an ideal gas.16TEMPERATURE•With increasing T, the gas molecules move faster andfurther apart.•Also, higher Tmeans more energy available to break intermolecular forces.•Therefore, the higher the T, the more ideal the gas.The van derWaals Equation•Two terms are added to the ideal gas equation. One to correct for the actual volume of molecules and the other to correct for intermolecular attractions. •The correction terms generate the van der Waals equation:2P=nRTnaV−nb−V2where aand bare empirical constants. 17The van der Waals EquationnRT2P=V−nb−naV2Corrects for Corrects for molecular molecular volumeattraction•General form of the van der Waals equation:⎛⎜2⎜P+na⎞⎟(V−nb)=nRT⎝V2⎟⎠18