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Mixtures of Gases Dalton’s Law of Partial Pressures All gases respond in the same way to changes in pressure, volume and temperature. Therefore for our calculations it is unimportant whether all the molecules in the sample are the same. A mixture of gases that do not react with one another behaves like a single pure gas. John Dalton showed how to calculate the pressure of a mixture of gases (1801). His reasoning was something like this: The total pressure of the mixture of gases was the sum of the pressures of the gases A and B if they were alone in the container. Dalton summarised his observations in his Law of Partial Pressures: The total pressure of a mixture of gases is the sum of the partial pressures of its components: P = PA + PB + … He called the partial pressure of each gas: The pressure it would exert if it occupied the container alone. According to kinetic theory, molecules of gas A have the same average kinetic energy the molecules of gas B, since they have the same temperature. Furthermore, they have no physical or chemical interaction on each other. Consequently, the act of mixing two or more gases does not change the average kinetic energy of any of the gases, so each gas will exert the same pressure that it would exert if it were the only gas in the container. but The amount of each gas in the mixture (the number of its moles) will affect its participation in the total pressure. If nA mol of gas A and nB mole of gas B are mixed, the total number of moles in the mixture is (nA+ nB). The ratio of the number of moles of A to the total number of moles present is called the mole fraction of A, XA: nA nA XA= = nA + nB ntotal The fraction of the total pressure that is due to gas A is given by the mole fraction of A. The partial pressure of A, therefore is nA PA= ﴾ ﴿ Ptotal = XA Ptotal nA + nB The partial pressure for gas B nB PB= ﴾ ﴿ Ptotal = XB Ptotal nA + nB Notice that the sum of the mole fraction is: XA + XB = 1 nA nB nA + nB + = = 1 nA + nB nA + nB nA + nB Example: A mixture of 40.0 g of oxygen and 40.0 g of helium has a total pressure of 0.900 atm. What is the partial pressure of oxygen? Mw of O2 = 32.0 g.mol-1 Mw of He = 4.00 g.mol-1 If it is not very soluble in water, a gas evolved in in the course of a reaction is frequently collected over water. It is conducted into an inverted bottle that has been filled with water. The gas displaces the water, and the collected gas is mixed with water vapour. The total pressure of the mixture is the sum of the partial pressure of the gas and the partial pressure of the water vapour. Example: A 370.0 mL sample of oxygen is collected over water at 23º and a barometric pressure of 0.992 atm. What volume would this sample occupy dry and at STP? Molecular Speeds Derivation of the Ideal Gas Law from the Kinetic Theory Consider a gas sample that contains N molecules, each having a mass of m. If this sample is enclosed in a cube a cm on a side, the total volume of the gas is: V = a3 cm3 The pressure of the gas on any wall is due to the impacts of the molecules on that wall. The force of each impact can be calculated from the change in momentum per unit time. a cm Momentum before the collision : mv Momentum after the collision : - mv (negative because the direction is changed) Change in the momentum in each collision: 2mv Consider the shaded wall and take into account only those molecules moving in the direction of x axis. A molecule moving in this direction will strike this wall every 2a cm of its path . If its velocity is v cm/s, the number of collisions with the wall in every second: v/2a The total change in momentum per molecule in one second: the number of collisions . the change in momentum mv2 (v/2a﴿ 2mv = a The total changes in momentum (force) for all the molecules striking the wall in one of the three possible directions, x in one second: N mv2 Nmv2 ﴾ ﴿.﴾ ﴿= 3 a 3a Pressure is force per unit area. The area of the shaded wall is a2. The pressure on the wall: force Nmv2 Nmv2 pressure = = / a2 = area 3a 3a3 Nmv2 1 Since V = a3 P= or PV = Nmv2 3V 3 It can be written: 2 1 PV = ﴾ N﴿﴾ mv2﴿ 3 2 Since KE = 1/2mv2 therefore: PV= 2/3N(KE) The average molecular kinetic energy, KE, is directly proportional to the absolute temperature, T, and the number of the molecules is directly proportional to the number of moles, n. Since N(KE) α nT we can write the equation as: 2 PV α ﴾ ﴿ nT 3 Using a constant R the proportionality will give an equation: PV = nRT … Molecular Speeds Previously we derived the expression 1 PV = ﴾ ﴿ N mv2 3 For one mole of gas, the number of molecules, N, is Avogadro’s number; and N times the mass of a single molecule, m, is the molecular weight, M: 1 PV = ﴾ ﴿ Mv2 3 For one mole, PV = RT; thus, 1 RT = ﴾ ﴿ Mv2 3 Rearranging and solving for the molecular speed, we obtain v = √ 3RT/M The speed, v, in this equation, is the root-mean-square speed. It is the speed of a molecule that has the average kinetic energy of a collection of molecules at the temperature under consideration. To obtain v in m/s, R must be expressed in appropriate units. If M is expressed in g/mol, the appropriate value of R is 8.3143. 103 g.m2/(s2. K.mol). The root- mean- square speed of H2 (g) molecule at 0 °C is: v = √ 3RT/M v = √ 3[8.3143. 103 g.m2/(s2. K.mol)].273 K/2.016 g.mol-1 = 1.84 . 103 m.s-1 and at 100 °C : v = √ 3[8.3143. 103 g.m2/(s2. K.mol)].373 K/2.016 g.mol-1 = 2.15 . 103 m.s-1 This speeds are high 4.12 . 103 mile.hr-1 and 4.81 . 103 miles.hr-1. Although a given molecule travels at a high speed, its direction is continually being changed through collisions with other molecules. At STP conditions a H2 molecule, on average, undergoes about 1.4 . 1010 collisions in one second. The average distance travelled between the collisions is 1.3 . 10-5 cm; this value is called mean free path of hydrogen. Example: At what temperature the root- mean- square speed of N2O (g) equals to the root- mean- square speed of N2 (g) at 300 K? Answer: 471 K The Maxwell Distribution of Speeds Not all of the molecules of a gas have the same kinetic energy and speed. The speed and the direction of molecules in a sample of gas change continually. So the molecular speeds and kinetic energy are distributed over a range. Distribution of molecular speeds of different gases in different temperatures are investigated by Scottish scientist James Maxwell. His conclusions are summarised in some graphs: Heavy molecules like CO2 travel with speeds close to their average values. Light molecules such as H2 not only have a higher average speed in the same temperature, but the speed of many of them are very different from their average speed. Why hydrogen and helium are very rare in the earth’s atmosphere, but are abundant on massive planets like Jupiter? The spread of speeds widens as the temperature increases. At low temperatures, most molecules have speeds close to their average speed. At high temperatures, a high proportion have speeds widely different from their average speed. The molecules of all gases have a wide range of speeds. As the temperature increases, the average speed and the range of speeds increase. The average Molecular Kinetic Energy We previously obtained this equation: 2 PV = N(KE) 3 So we have: 3 PV KE= 2N For one mole of gas, PV = RT and N is Avogadro’s number: 3 RT KE= 2N To obtain KE in Joules R = 8.314 J.K-1.mol-1 Example: What is the average Kinetic Energy of a H2 molecule at 0 °C? Distribution of molecular kinetic energies over a range gives typical curve like that for molecular speeds. Molecular Motion Diffusion and Effusion The gradual dispersal of one substance through another is called diffusion like the spread of pheromones and perfume through air (fig 05, a). It helps to keep the the composition of the atmosphere approximately constant, because abnormally high concentrations of one gas diffuse away and disperse. The escape of one substance (particularly a gas) through a small hole into vacuum is called effusion (fig 05, b) . Effusion occurs whenever a gas is separated from a vacuum by a porous barrier or a single pin hole. A gas escapes through the hole from the high concentration to the low concentration region. Graham’s Law of Effusion Suppose that samples of two gases, A and B, are confined separately in identical containers under the same conditions of temperature and pressure. According the kinetic theory they have the same average kinetic energy: KEA = KEB 1/2 mAvA2 = 1/2 mBvB2 mAvA2 = mBvB2 vA2 mB = vB2 mA vA mB =√ vB mA The ratio of the molecular masses, mB / mA, is the same as the ratio of the molecular weights MB / MA, therfore: vA MB =√ vB MA Imagine that each container has an identical, extremely small opening ( an orifice) in it. Gas molecules will escape through these orifices; molecular effusion. The rate of effusion, r, is equal to the rate at which molecules strike the orifice, which in turn is proportional to molecular speed, v. Molecules that move rapidly will effuse at a faster rate than slower moving molecules. So the ratio vA / vB is the same as the ratio of effusion rates, rA / rB : rA MB =√ rB MA It can be expressed in terms of gas densities. Since the desity of a gas, d, is proportional to the molecular weight of the gas, M: rA dB =√ rB dA Hydrogen will effuse four times more rapidly than oxygen, because its lighter than oxygen: rH2 MO2 32 =√ =√ = √ 16 = 4 rO2 MH2 2 In the figure below, the plug on the left is soaked in HCl and that on the right in aqueous ammonia. Formation of ammounium chloride occurs where the two gases meet. The reaction occurs closer to the HCl plug because this gas has the greater molar mass and thus diffuses more slowly. Graham’s law of effusion has been used for separation of isotopes. Uranium occurs in nature as 0.72% 92235U and 99.28% 92238U. Of the two isotopes, only 92235U will undergo nuclear fission. It is necessary to separate 92235U from 92238U. By converting uranium to uranium hexafloride, which boils at 56 °C. The uranium hexafloride actually is a mixtur of 235UF (A) and 238UF (B). This mixture as a gas and at low temperature is allowed to 92 6 92 6 effuse through a porous barrier. The lighter 92235UF6 effuses 1.004 times faster than the heavier 92238UF6 . Hence the emerging gas has a higher 92235UF6 content than the original mixture. By repeating this effusion procedure for thousands of times a significant separation will accomplish. rA MB 352 =√ =√ = 1.004 rB MA 349 Liquefaction of Gases Liquefaction of a real gas is the consequence of intermolecular attractions. If the pressure is high, the molecules are close together, and intermolecular forces are appreciable. This forces are opposed by the motion of the molecules; thus; liquefaction is favoured by low temperatures, where the average kinetic energy of the molecules is low. The behaviour of a gas deviates more and more from ideality as the temperature is lowered and the pressure is raised. At extremes of these conditions, gases liquefy. The higher the temperature of the gas, the more difficult it is to liquefy and the higher pressure that must be employed. For each gas there is a temperature above which it is impossible to liquefy the gas no matter how high the applied pressure. This temperature is called the critical temperature. The critical temperature of a gas gives an indication of the strength of the intermolecular forces. The weaker the intermolecular forces, the lower the critical temperature. As can be seen in Table 10.6, helium with the weakest intermolecular forces, liquefies below 5.3 K. Water with strong forces is liquid. The minimum pressure needed to liquefy a gas at its critical temperature is called the critical pressure. The Joule-Thomson Effect It is necessary to cool many gases below room temperature (295 K) before these substances can be liquefied. When most compressed gases are allowed to expand to a lower pressure, they cool. In the expansion work is done against the intermolecular attractive forces. The energy used in performing this work must be taken from the kinetic energy of the gas molecules themselves; hence; the temperature of the gas decreases. This effect was studied by James Joule and William Thompson (Lord Kelvin) during the years 1852-1862. A Linde refrigerator for liquefying gases. The compressed gas gives up heat to the surroundings in the heat exchanger and is cooled further as it passes through the coil past which cooled gas circulates. The gas is cooled still further by the Joule-Thompson effect as it emerges through the throttle. This gas cools the incoming gas and is recirculated through the system. Eventually, the temperature of the incoming gas is so low that it condenses to a liquid.

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