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Chapter 12 Gases and Their Properties

Chapter 12 Gases and Their Properties. Three phases of matter:. Gas. Liquid. Solids. The phase of a material is determined by balance between the kinetic energy in the particles and their interactions.

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Chapter 12 Gases and Their Properties

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  1. Chapter 12 Gases and Their Properties Three phases of matter: Gas Liquid Solids The phase of a material is determined by balance between the kinetic energy in the particles and their interactions Condensed phases: intermolecular interaction overcome the kinetic energy holding the molecules together Gas phase: intermolecular interactions are too weak to overcome the interaction and the molecules are separated and move around freely

  2. Properties of Gasses To completely describe the state of a gaseous system, we need to know: How many particles? Moles How much space they occupy? Volume On average how fast are they moving? Temperature Pressure V K.E K.E αT P Conservation of momentum

  3. Pressure Pressure is defined as a force experienced over a given surface area P = F/A Units N/m2 = Pa (Pascal) J/m3 = Pa (Pascal) Energy/volume !!! PV = energy Other Units 1 atm = 101.3 kPa Recall 1 kPa = 1000 Pa 760 mm Hg = 1 atm 1 mm Hg = 1 Torr 1 atm = 101.3 kPa = 760 mm Hg = 760 Torr

  4. Gas Laws In the 17th and 18th century many observations were made on gasses relating T, V, P and n (number of particles). Robert Boyle (1662) demonstrated that the pressure of a gas varies inversely proportional to the volume of a gas (using air) For example, the gas-tight syringe is placed in a sealed flask, then compressed air is applied to the flask. As the applied pressure increases, the volume of the syringe decreases Boyle’s Law

  5. Gas Laws Jacques Charles (~1787) demonstrated that at constant pressure the volume of a gas is a direct linear function of its temperature Charles’ law Charles “Extrapolation” By graphing the volume of the gas against temperature, 0C had no significance, but all measured gases seem to converge to a common zero point: absolute zero We must always use absolute temperatures (Kelvin scale) with the gas laws!

  6. Gas Laws Amedeo Avogadro (1811) demonstrated that the volume of a gas depends directly on the quantity of gas present. Consider a gas forming reaction: Ex) HOOH(l)→ H2O(l) + ½ O2(g) If you use twice as much reactants you get twice the volume of gas, under constant pressure and temperature His work with gases lead directly to the development of the molecular hypothesis and to the determination of Avogadro’s Number

  7. Ideal Gas Law The modern gas law is a combination of all three (Boyle, Charles, Avogadro) into one combined equation, the ideal gas law P = pressure V = volume n = moles (quantity) T = absolute temperature (K) Why is it called the ideal gas law? Serious deviations occur at (a) very low temperature or (b) very high pressure Air obeys the ideal gas law to within 1% under “normal” conditions R is the ideal gas constant which is unit dependent. • P(Pascal), V(m3), n(mol): 8.3145 Pa m3 mol-1 K-1 = 8.3145 J mol-1 K-1 • P(atm), V(Litre), n(mol): 0.0820574 L atm mol-1 K-1

  8. Ideal Gas Law Ex) Compute the number of moles of 1.00 L an ideal gas under standard temperature and pressure (STP). Standard temperature is room temperature at 0.00 oC Standard pressure is ambient pressure of 1.000 atm P = 1.000 atm = 101.3 kPa = 101300 Pa T = 0.00 oC = 273.15 K m (H2) = 1.0 g; m (Ar) = 8.0 g = 0.00100 m3 V = 1.00 L = (1.00 dm3)(0.1 m/dm)3 n =(101300 Pa)( 0.00100 m3)/(8.314 J/mol K)(273.15 K) = 0.0446 mol Ex) Compute the volume of 1.000 mol of gas at STP V = (1.000 mol)(8.314 J/molK)(273.15 K)/(101300 Pa) = 0.02242 m3 = (0.002242 m3)(10 dm/m)3 = 22.42 dm3= L

  9. Ideal Gas Law Divide by V 1) Boyle’s Law If n and T is constant Divide by P 2) Charles’ law If n and P is constant Avogadro’s law If T and P is constant

  10. Ideal Gas Law 3) Devide by nT Constant Therefore we can directly compare any two (different) systems

  11. Ideal Gas Law We can compare a system to itself which has undergone a change. Closed System If the number of particles remains the same: We can change any one of P, V and T, keep one constant and predict the remaining variable Change T, with fixed V, compute P Change P, with fixed V, compute T Change V, with fixed T, compute P Change P, with fixed T, compute V Change T, with fixed P, compute V Change V, with fixed P, compute T We can change any two of P, V and T and predict the remaining variable

  12. Changing Temperature With Constant Volume. Ex) A gas is in a closed 3.00 L container at 100. oC with a pressure of 5.00 atm. Predict the pressure in Pa when the system has cooled to 50. oC P1 = 5.00 atm = (5.00 atm)*(101.3 kPa/atm) = 507000 Pa T1 = 100. oC = 100. + 273.15 K V1 = 0.00300 m3 = 373 K T2 = 50. oC = 100. + 273.15 K = 333 K Decreasing T increases K.E. If the V does not increase, since the amount of K.E. per unit volume is decreased, P must decrease. Volume remains the same Solve for P2 P2 = (507000 Pa)(333 K)/(373 K) = 463000 Pa

  13. Changing Volume Under Constant Temperature. Ex) A gas is in a closed 3.00 L container at 100. oC with a pressure of 5.00 atm. Predict the pressure in Pa when the system has expanded to 50.0 L. P1 = 5.00 atm = (5.00 atm)*(101300 Pa/atm) = 507000 Pa T1 = 100. oC = 100. + 273.15 K = 373 K = T2 V1 = 0.00300 m3 V2 = 0.0500 m3 Increasing the volume with out changing the temperature, means that the gas has to less energy per unit volume to generate pressure. Therefore P decreases. Temperature Is constant Solve for P2 P2 = (507000 Pa)(0.00300 m3)/(0.0500 m3) = 30400 Pa

  14. Changing Volume Under Constant Pressure. Ex) A gas in a closed 3.00 L container at 100. oC with a pressure of 5.00 atm. Predict the temperature in K when the system has increased in volume to 50.0 L. P1 = 5.00 atm = P2 = (5.00 atm)*(101.3 kPa/atm) = 507000 Pa T1 = 100. oC = 100. + 273.15 K = 373 K V1 = 0.00300 m3 V2 = 0.0500 m3 Increasing the volume, while the pressure is maintained, requires the gas to have much more energy to maintain the pressure. Therefore T increases. Pressure constant Solve for T2 T2 = (373 K)(0.0500 m3)/(0.00300 m3) = 6220 K

  15. Changing Volume and Pressure Simultaneously Ex) A gas in a closed 3.00 L container at 100. oC with a pressure of 5.00 atm. Predict the temperature in K when the system has decreased in volume to 0.300 L and increased in pressure to 10.00 atm. P1 = 5.00 atm = (5.00 atm)*(101.3 kPa/atm) = 507000 Pa P2 = 10.00 atm = (10.00 atm)*(101.3 kPa/atm) = 1013000 Pa T1 = 100. oC = 100. + 273.15 K = 373 K V1 = 0.00300 m3 V2 = 0.000300 m3 Solve for T2 P has been doubled, but V has been decreased to 10 %. The increase in P does not compensate for the decrease in V, therefore, T should decrease to 2*10% = 20 % its original value = 66.3 K T2 = (1013000 Pa)(0.000300 m3)(373 K)/(507000 Pa)(0.00300 m3)

  16. Densityof ideal gases Densities of solids and liquids are ~1 g mL-1 (up to a high of about 15 g ml-1 Hg) Densities of gases at “normal” conditions are about 1000 times lower Hence are reported in units of g L-1 Solids and liquids are very difficult to compress: density almost does not vary, and hence are regarded as a fundamental property Gases are compressible and thus gas density is variable and not regarded as a fundamental property as with solids and liquids Ex) Compute the density of an N2(g) under STP. r = (101300 Pa)(28.02 g/mol)/(8.314 J/mol K)(273.15 K) = 1249 g/m3 = 1.249 g/L

  17. Density of ideal gases From the density of a gas (or vapor) one can determine the molecular mass. Chloroform is a common liquid used in the laboratory. It readily vaporizes. If the pressure of CHCl3 is 195 mmHg in a flask at 25C and r = 1.25 g L–1, what is its molar mass? P = [(195 mm Hg)/(760 mm Hg/atm)](101300 Pa/atm) = 26000 Pa T = 25 + 273 = 298 K r = (1.25 g/L)(1000 L/m3) = 1250 g/m3 M = (1250 g/m3)(8.314 J/molK)(298 K)/(26000 Pa) = 119 g/mol

  18. Dealing with Gas Mixtures Avogadro’s Law states that the number of particles is the key to the behaviour of gases The mole fraction expresses the relative number of particles (atoms, molecules) of specific kinds of gases in a mixture of gases For a two component mixture: A + B: For a three component mixture: A + B + C Mole fraction is a new unit of concentration

  19. Dalton’s Law of Partial Pressures The total pressure of a gas mixture equals the sum of the partial pressures of the individual gases (when the Ideal Gas Law is valid) Consequences of Dalton’s Law: consider two gases A and B in the same container Now combine the equation for the total mixture and that of one component A Rearranging gives: In general:

  20. Illustrating Dalton’s Law Consider the two flasks, one containing nitrogen, the other containing half as many molecules of oxygen Since the flasks are the same size and the temperature is the same, the pressure in the oxygen flask is half that of the nitrogen If we combine the oxygen with the nitrogen (making a sort of synthetic air), we find that the pressure obtained is the sum of the individual pressures

  21. Dalton’s Law What is the total pressure in atmospheres of a gas mixture that contains 1.00 g of H2 and 8.00 g of Ar in a 3.0 L container at 27C? What are the partial pressures of the two gases? V = 3.0 L = 0.0030 m3 T = 27+273.15 K = 300 K m (H2) = 1.00 g; m (Ar) = 8.00 g We need nH2 and nAr nAr = (8.00 g/39.95 g/mol) = 0.200 mol nH2 = (1.00 g/2.016 g/mol) = 0.496 mol = 412000 Pa = 166000 Pa

  22. Dalton’s Law Silane, SiH4, reacts with O2 to give silicon dioxide and water SiH4(g) + 2 O2(g)  SiO2(s) + 2 H2O(g) If you mix SiH4 with O2 in the correct stoichiometric ratio, and if the total pressure of the mixture is 120 mm Hg, what are the partial pressures of SiH4 and O2? When the reactants have been completely consumed, what is the total pressure in the flask Ratio is 1:2 After reaction (e.g. by sparking mixture) Notice that the number of H20 molecules produced is equal to the number of O2 consumed. Because SiO2 is solid it does not contribute to gas pressure

  23. Kinetic Molecular Theory of Gases Gases are the best understood among the four phases of matter Their behaviour has been predicted by a mathematical theory developed by Clausius, Maxwell and Boltzmann. The theory is based on some simplifying assumptions: 1. A pure gas consists of a large number of identical molecules separated by distances that are large compared with their size 2. The molecules of a gas are constantly moving in random directions with a distribution of speeds 3. The molecules of a gas exert no forces on one another except during collisions, so that between collisions they move in straight lines with constant velocities 4. The collisions of the molecules with each other and with the walls of the container are elastic; no energy is lost during a collision From these postulates, it is possible to calculate a model that leads mathematically to the ideal gas law

  24. Pressure and kinetic energy Consider the consequences of this theory on collisions with the walls of the container (A) Pressure is created by the impact of the gas molecules on a solid surface, e.g. the container walls (B) If each particle doubles in mass, the impulses are twice as large (C) If the gas density is doubled, then twice as many collisions occur with the walls (D) If the average molecule speed is doubled, there are twice as many collisions per time interval and each impulse has twice the momentum

  25. Kinetic energy When these postulates are expressed mathematically, the kinetic energy of each gas molecule is calculated, using the usual definition random motion it gas moleculeshave different velocities and kinetic energy The distribution of velocities Increases with increasing temperature The Maxwell-Boltzmanndistribution indicates a statisticaldistribution of molecule velocities Even describing the average speedcauses difficulties, and there arethree common ways to do so:v(mp) – most probable speedv(av) – average speed v(rms) – root mean square speed

  26. Maxwell-Boltzmann Statistical Mechanics The mathematical model that takes all of the above factors into account results in the following prediction: Recall that P  V = energy (in units: N / m2  m3 = N m = J) Compare the M-B equation to the ideal gas law (R in Joules): We now combine and rearrange the terms, and convert from number of particles to moles of particles. (NA = Avogadro’s number) In practice we will always use the root-mean-square velocity, for which by rearrangement: (m*NA = M = molar mass) R = 8.3145 J mol-1 K-1 because J = kg2m2/s2, M must be in kg/mol!!!!!

  27. Maxwell-Boltzmann Statistical Mechanics You have a sample of helium gas at –33C, and you want to increase the average speed of helium atoms by 10.0%. To what temperature should the gas be heated to accomplish this? T1 = -33 + 273.15 = 240 K; M (He) = 0.0040026 kg/mol; v2 = 1.1v1

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