Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chem 300 - Ch 27/#2 Today’s To Do List l KMT (Continued) Bimolecular Collision Frequencies Mean Free Path.

Similar presentations


Presentation on theme: "Chem 300 - Ch 27/#2 Today’s To Do List l KMT (Continued) Bimolecular Collision Frequencies Mean Free Path."— Presentation transcript:

1 Chem 300 - Ch 27/#2 Today’s To Do List l KMT (Continued) Bimolecular Collision Frequencies Mean Free Path

2 Collision Frequency with Wall l Find the # molecules that collide with an area from a specific direction with speed u per unit time. l Integrate over all speeds & angles: Z wall =  /4 = (8RT/  M) 1/2  = N A n/V = N a P/RT Z wall = (N a P/RT)(RT/2  M) 1/2

3 Examples Z wall = (N a P/RT)(RT/2  M) 1/2 l 1. At 25 o C & 1 bar, Z wall = 2.88x10 23 s -1 -cm -2 l 2. Outside surface of Space Shuttle facing away from the Sun: T~1K, P=10 -10 bar Z = (6.02x10 23 )(10 -10 bar)/(0.0831L-bar/mol)(1K) x[(8.31J/mK)(1K)/(2  )(.028Kg)] 1/2 = 4.8x10 14 /s-cm 2 l 3. Avg speed: = (8RT/  M) 1/2 = 27.5 m/s l 4. u rms = (3RT/M) 1/2 = 29.8 m/s l 5. u pp = (2RT/M) 1/2 = 24.4 m/s

4 Collision with Molecules l With an identical Stationery Molecule l With identical Moving Molecules l With different Moving Molecules l Total Collision Frequency of All Molecules with a Minimum Energy (E a )

5 Stationery Molecule l To collide: Must come within a target area occupied by another molecule. Collision Cross Section (  ) =  d 2 l Z A =  =  (8kT/  m) 1/2  =N A n/V

6 Moving Target l Consider: A Relative Speed: = (8kT/   ) 1/2 A Relative Mass (  ) = m 1 m 2 /(m 1 + m 2 ) If Identical Molecules: m 1 = m 2  m ===>  = m/2 = 2 1/2 l Z A =  = 2 1/2 

7 Total Collision Freq. (Z AA ) l All Identical Molecules: Z AA =(½)  Z A = (1/2 1/2 )  2  l A & B Molecules: Z AB =  A  B  AB  AB = average of A & B cross sections

8 Minimum KE Collisions l Collision freq./unit volume for molec. A & B with relative speed between u r and u r + du r – l dZ AB = l  A  B  AB (  /kT) 3/2 (2/  ) 1/2 exp(-  u r 2 /2kT)u r 3 du r l u r 3 favors more collisions among high-speed molecules.

9

10 Mean Free Path l Average distance travelled between collisions (l): l l = /Z a = RT/(2 1/2 N a  P) l Example: N 2 @ T=1K, P=10 -10 bar l l=(0.082)(1K)/[(2 1/2 x6.02x10 23 x 0.45x10 -18 m 2 x10 -10 bar)] = 2.1 m

11 Example: He in upper atm T=1000K, P=10 -12 bar l  = 0.218x10 -18 m 2 l l = 7x10 5 m = 467 mi l = 2x10 3 m/s l Avg time between collisions: t avg = l/ = 400 sec = 0.1 hr

12 Perspective l Typical gas (N 2 ) @ 1 atm & 25 o C: l Collection of molec. Traveling with 350 m/s l Each molec. Collides with another every 1 ns l Between collisions it travels 100-10 3 molecular diameters. l Since d< { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/12/3369769/slides/slide_12.jpg", "name": "Perspective l Typical gas (N 2 ) @ 1 atm & 25 o C: l Collection of molec.", "description": "Traveling with 350 m/s l Each molec. Collides with another every 1 ns l Between collisions it travels 100-10 3 molecular diameters. l Since d<

13 Next Time l Debye-Huckel Theory l Odds and Ends


Download ppt "Chem 300 - Ch 27/#2 Today’s To Do List l KMT (Continued) Bimolecular Collision Frequencies Mean Free Path."

Similar presentations


Ads by Google