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1 Dalton’s Law n The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container. n The total pressure is the sum of the partial pressures. n P Total = P 1 + P 2 + P 3 + P 4 + P 5... n For each P = nRT/V

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2 Dalton's Law n P Total = n 1 RT + n 2 RT + n 3 RT +... V V V n In the same container R, T and V are the same. n P Total = (n 1 + n 2 + n 3 +...)RT V n P Total = (n Total )RT V

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3 The mole fraction n Ratio of moles of the substance to the total moles. symbol is Greek letter chi symbol is Greek letter chi = n 1 = P 1 n Total P Total = n 1 = P 1 n Total P Total

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4 Examples n The partial pressure of nitrogen in air is 592 torr. Air pressure is 752 torr, what is the mole fraction of nitrogen? n What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm?

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5 P 1 /P T = n 1 /n T = 1 592 torr /752 torr =.787 = n 592 torr /752 torr =.787 = n.787 = P n /P T =P n /5.25 atm= 4.13atm.787 = P n /P T =P n /5.25 atm= 4.13atm

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6 Examples 3.50 L O 2 1.50 L N 2 2.70 atm n When these valves are opened, what is each partial pressure and the total pressure? 4.00 L CH 4 4.58 atm 0.752 atm

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7 Find the partial pressure of each gas P 1 V 1 =P 2 V 2 2.70atm(4.00L) = P(9.00L) = 1.20 atm 4.58atm(1.5L) = P(9.00L) =.76atm.752atm(3.50L) = P(9.00L) =.292atm 1.20atm +.76atm+.292atm = 2.26atm

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8 Vapor Pressure n Water evaporates! n When that water evaporates, the vapor has a pressure. n Gases are often collected over water so the vapor pressure of water must be subtracted from the total pressure to find the pressure of the gas. n It must be given. Table of vapors pressures as different temperatures

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9 Example n N 2 O can be produced by the following reaction NH 4 NO 3 N 2 O + 2H 2 O n what volume of N 2 O collected over water at a total pressure of 785torr and 22ºC can be produced from 2.6 g of NH 4 NO 3 ? ( the vapor pressure of water at 22ºC is 21 torr)

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10 2.6gNH 4 NO 3 1mol NH 4 NO 3 1molN 2 O = 0.0325mol NO 2 80.06g 1mol NH 4 NO 3 80.06g 1mol NH 4 NO 3 PV=nRT V=nRT/P PV=nRT V=nRT/P V= 0.0325mol(62.4torr L/mol K)(295K) / (785torr – 21torr) V=.77L V=.77L

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12 Kinetic Molecular Theory n Theory tells why the things happen. n explains why ideal gases behave the way they do. n Assumptions that simplify the theory, but don’t work in real gases. The particles are so small we can ignore their volume. The particles are in constant motion and their collisions cause pressure.

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13 Kinetic Molecular Theory The particles do not affect each other, neither attracting or repelling. The average kinetic energy is proportional to the Kelvin temperature.

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14 What it tells us n (KE) avg = 3/2 RT n This the meaning of temperature. n u is the particle velocity. n u is the average particle velocity. n u 2 is the average of the squared particle velocity. the root mean square velocity is u 2 = u rms the root mean square velocity is u 2 = u rms

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15 Combine these two equations n For a mole of gas n N A is Avogadro's number n n

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16 Combine these two equations n n m is kg for one particle, so N a m is kg for a mole of particles. We will call it M n Where M is the molar mass in kg/mole, and R has the units 8.3145 J/Kmol. n The velocity will be in m/s

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17 Example n Calculate the root mean square velocity of carbon dioxide at 25ºC. n Calculate the root mean square velocity of hydrogen gas at 25ºC. n Calculate the root mean square velocity of chlorine gas at 250ºC.

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18 Solutions CO 2 u rms = (3(8.314J/molK)(298)/.04401kg/mol) 1/2 = 411 m/s H 2 U rms = (3(8.314J/molK)(298)/.00202kg/mol) 1/2 = 1918m/s Cl 2 U rms = (3(8.314J/molK)(523)/.0709kg/mol) 1/2 = 429m/s

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19 Range of velocities n The average distance a molecule travels before colliding with another is called the mean free path and is small (near 10 -7 ) n Temperature is an average. There are molecules of many speeds in the average. n Shown on a graph called a velocity distribution

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20 number of particles Molecular Velocity 273 K

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21 number of particles Molecular Velocity 273 K 1273 K

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22 Velocity n Average increases as temperature increases. n Spread increases as temperature increases.

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23 Effusion n Passage of gas through a small hole, into a vacuum. n The effusion rate measures how fast this happens. n Graham’s Law the rate of effusion is inversely proportional to the square root of the mass of its particles.

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24 Effusion n Passage of gas through a small hole, into a vacuum. n The effusion rate measures how fast this happens. n Graham’s Law the rate of effusion is inversely proportional to the square root of the mass of its particles.

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25 Deriving n The rate of effusion should be proportional to u rms n Effusion Rate 1 = u rms 1 Effusion Rate 2 = u rms 2

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26 Deriving n The rate of effusion should be proportional to u rms n Effusion Rate 1 = u rms 1 Effusion Rate 2 = u rms 2

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27 Diffusion n The spreading of a gas through a room. n Slow considering molecules move at 100’s of meters per second. n Collisions with other molecules slow down diffusions. n Best estimate is Graham’s Law.

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28 Helium effuses through a porous cylinder 3.20 times faster than a compound. What is it’s molar mass? √X / √4g = 3.2 √X = 6.4g X = 40.96g/mol

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29 If 0.00251 mol of NH 3 effuse through a hole in 2.47 min, how much HCl would effuse in the same time? √M NH3 / √M HCl = √17 / √36.5=.687 times faster 2.51 x 10 -3 mol (.687) = 1.72 x 10 -3 mol HCl

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30 A sample of N 2 effuses through a hole in 38 seconds. what must be the molecular weight of gas that effuses in 55 seconds under identical conditions? √M N2 / √X = 55/38 = 1.45 √28 / 1.45 = √X 13.32g/mol = X

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31 Real Gases n Real molecules do take up space and they do interact with each other (especially polar molecules). n Need to add correction factors to the ideal gas law to account for these.

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32 Volume Correction n The actual volume free to move in is less because of particle size. n More molecules will have more effect. n Bigger molecules have more effect n Corrected volume V’ = V - nb n b is a constant that differs for each gas. P ’ = nRT (V-nb) P ’ = nRT (V-nb)

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33 Pressure correction n Because the molecules are attracted to each other, the pressure on the container will be less than ideal gas n Depends on the type of molecule n depends on the number of molecules per liter. n since two molecules interact, the effect must be squared.

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34 Pressure correction n Because the molecules are attracted to each other, the pressure on the container will be less than ideal n depends on the number of molecules per liter. n since two molecules interact, the effect must be squared.

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35 Altogether n n Called the Van der Waal’s equation if rearranged n Corrected Corrected Pressure Volume

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36 Where does it come from n a and b are determined by experiment. n Different for each gas. n Look them up n Bigger molecules have larger b. n a depends on both size and polarity. n once given, plug and chug.

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37 Example n Calculate the pressure exerted by 0.5000 mol Cl 2 in a 1.000 L container at 25.0ºC n Using the ideal gas law. n Van der Waal’s equation –a = 6.49 atm L 2 /mol 2 –b = 0.0562 L/mol

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Chapter 5 The Gas Laws.

Chapter 5 The Gas Laws.

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