1 / 16

Characteristics of Gases

Characteristics of Gases. Vapor = term for gases of substances that are often liquids/solids under ordinary conditions Unique gas properties Highly compressible Inverse pressure-volume relationship Form homogeneous mixtures with other gases. Pressures of Enclosed Gases and Manometers.

hong
Download Presentation

Characteristics of Gases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Characteristics of Gases • Vapor = term for gases of substances that are often liquids/solids under ordinary conditions • Unique gas properties • Highly compressible • Inverse pressure-volume relationship • Form homogeneous mixtures with other gases Mullis

  2. Pressures of Enclosed Gases and Manometers • Barometer: Used to measure atmospheric pressure* • Manometer: Used to measure pressures of gases not open to the atmosphere • Manometer is a bulb of gas attached to a U-tube containing Hg. • If U-tube is closed, pressure of gas is the difference in height of the liquid. • If U-tube is open, add correction term: • If Pgas < Patm then Pgas + Ph = Patm • If Pgas > Patm then Pgas = Ph + Patm * Alternative unit for atmospheric pressure is 1 bar = 105 Pa Mullis

  3. Gas Densities and Molar Mass • Need units of mass over volume for density (d) • Let M = molar mass (g/mol, or mass/mol) PV = nRT MPV = MnRT MP/RT = nM/V MP/RT = mol(mass/mol)/V MP/RT = density M = dRT P Mullis

  4. Sample Problem: Density 1.00 mole of gas occupies 27.0 L with a density of 1.41 g/L at a particular temperature and pressure. What is its molecular weight and what is its density at STP? M.W. = 1.41 g |27.0 L = 38.1 g___ L |1.0 mol mol M = dRT d= M P = 38.1 g (1 atm)______________ = 1.70 g/L P RT mol (0.0821 L-atm )(273K) ( mol-K ) OR…AT STP: 38.1 g | 1 mol = 1.70 g/L mol | 22.4 L Mullis

  5. Example: Molecular Weight A 0.371 g sample of a pure gaseous compound occupies 310. mL at 100. º C and 750. torr. What is this compound’s molecular weight? n=PV = (750 torr)(.360L) = 0.0116 mole RT 62.4 L-torr(373 K) mole-K MW = x g_= 0.371 g = 32.0 g/mol mol 0.0116 mol Mullis

  6. Partial Pressures • Gas molecules are far apart, so assume they behave independently. • Dalton: Total pressure of a mixture of gases is sum of the pressures that each exerts if it is present alone. Pt = P1 + P2 + P3 + …. + Pn Pt = (n1 + n2 + n3 +…)RT/V = ni RT/V • Let ni = number of moles of gas 1 exerting partial pressure P1: P1 = X1P1 where X1 is the mole fraction (n1/nt) Mullis

  7. Collecting Gases Over Water • It is common to synthesize gases and collect them by displacing a volume of water. • To calculate the amount of a gas produced, correct for the partial pressure of water: • Ptotal = Pgas + Pwater • The vapor pressure of water varies with temperature. Use a reference table to find. Mullis

  8. Kinetic Molecular Theory • Accounts for behavior of atoms and molecules • Based on idea that particles are always moving • Provides model for an ideal gas • Ideal Gas = Imaginary: Fits all assumptions of the K.M. theory • Real gas = Does not fit all these assumptions Mullis

  9. 5 assumptions of Kinetic-molecular Theory • Gases = large numbers of tiny particles that are far apart. • Collisions between gas particles and container walls are elastic collisions (no net loss in kinetic energy). • Gas particles are always moving rapidly and randomly. • There are no forces of repulsion or attraction between gas particles. • The average kinetic energy of gas particles depends on temperature. Mullis

  10. Kinetic energy • The absolute temperature of a gas is a measure of the average* kinetic energy. • As temperature increases, the average kinetic energy of the gas molecules increases. • As kinetic energy increases, the velocity of the gas molecules increases. • Root-mean square (rms) speed of a gas molecule is u. • Average kinetic energy, ε ,is related to rms speed: ε = ½ mu 2 where m = mass of molecule *Average is of the energies of individual gas molecules. Mullis

  11. Maxwell-Boltzmann Distribution • Shows molecular speed vs. fraction of molecules at a given speed • No molecules at zero energy • Few molecules at high energy • No maximum energy value (graph is slightly misleading: curves approach zero as velocity increases) • At higher temperatures, many more molecules are moving at higher speeds than at lower temperatures (but you already guessed that) Just for fun: Link to mathematical details: http://user.mc.net/~buckeroo/MXDF.html Source: http://www.tannerm.com/maxwell_boltzmann.htm Mullis

  12. Molecular Effusion and Diffusion • Kinetic energy ε = ½ mu 2 • u = 3RT Lower molar mass M, higher rms speed u M Lighter gases have higher speeds than heavier ones, so diffusion and effusion are faster for lighter gases. Mullis

  13. Graham’s Law of Effusion • To quantify effusion rate for two gases with molar masses M1 and M2: r1 = M2 r2M1 • Only those molecules that hit the small hole will escape thru it. • Higher speed, more likely to hit hole, so r1/r2 = u1/u2 Mullis

  14. Sample Problem: Molecular Speed Find the root-mean square speed of hydrogen molecules in m/s at 20º C. 1 J = 1 kg-m2/s2 R = 8.314 J/mol-K R = 8.314 kg-m2/mol-K-s2 u2= 3RT = 3(8.314 kg-m2/mol-K-s2)293K M2.016 g |1 kg___ mol |1000g u2= 3.62 x 106 m2/s2 u = 1.90 x 103 m/s Mullis

  15. Example: Using Graham’s Law An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of O2 at the same temperature. What is the unknown gas? rx = MO2 0.355 = 32.0 g/mol rO2Mx 1 Mx Square both sides: 0.3552 = 32.0 g/mol Mx Mx = 32.0 g/mol = 254 g/mol  Each atom is 127 g, 0.3552 so gas is I2 Mullis

  16. The van der Waals equation • Add 2 terms to the ideal-gas equation to correct for • The volume of molecules (V-nb) • Molecular attractions (n2a/V2) Where a and b are empirical constants. P + n2a (V – nb) = nRT V2 • The effect of these forces—If a striking gas molecule is attracted to its neighbors, its impact on the wall of its container is lessened. Mullis

More Related