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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.

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Presentation on theme: "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."— Presentation transcript:

1 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

2 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 )RT V n P Total = (n Total )RT V

3 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

4 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?

5 5 P 1 /P T = n 1 /n T =  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

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

7 7 Find the partial pressure of each gas P 1 V 1 =P 2 V atm(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

8 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

9 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)

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

11 11

12 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.

13 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.

14 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

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

16 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 J/Kmol. n The velocity will be in m/s

17 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.

18 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

19 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 ) n Temperature is an average. There are molecules of many speeds in the average. n Shown on a graph called a velocity distribution

20 20 number of particles Molecular Velocity 273 K

21 21 number of particles Molecular Velocity 273 K 1273 K

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

23 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.

24 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.

25 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

26 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

27 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.

28 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

29 29 If 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 mol (.687) = 1.72 x mol HCl

30 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

31 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.

32 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)

33 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.

34 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.

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

36 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.

37 37 Example n Calculate the pressure exerted by mol Cl 2 in a 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 = L/mol


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