{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EO_440_PS5_Sol_F11

# EO_440_PS5_Sol_F11 - 39.948 = 4.31 x 10 4 cms 1 c mp = 2RT...

This preview shows pages 1–3. Sign up to view the full content.

Chemistry 440 Problem Set 5 – Solutions Fall 2011 1. ! = RT 2 N A " p At constant V, p varies directly with T p = nRT V and ! = RT 2 N A " x V nRT = V 2 N A " n λ is independent of temperature. 2. The Maxwell distribution of speeds is ƒ(s) = 4 ! M 2 ! RT " # \$ % & ' 3/2 s 2 e ( Ms 2 /2RT The factor, M 2RT , can be evaluated as M 2RT = 28.02 x 10 ! 3 kg mol ! 1 2 x 8.314J K ! 1 mol ! 1 ( ) x 500K ( ) = 3.37 x 10 ! 6 m ! 2 s 2 Though ƒ(s) varies over the range 290 to 300 m s -1 , the variation is small for this small range and its value at the center of the range can be used. ƒ 295ms ! 1 ( ) = 4 " ( ) x 3.37 x 10 ! 6 m ! 2 s 2 " # \$ % & ' ( 3/2 x 295ms ! 1 ( ) 2 x e ! 3.37 x 10 ! 6 ( ) x 295 ( ) 2 = 9.06 x 10 ! 4 m ! 1 s Therefore, the fraction of molecules in the specified range is ƒ x ν = (9.06 x 10 -4 m -1 s) x (10 m s -1 ) = 9.06 x 10 -3 corresponding to 0.91 percent.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chemistry 440 Problem Set 5 – Solutions Fall 2011 Page 2 3. The formula for determining the mean free path is ! = RT 2 N A " p solving for p, p = RT 2 N A !" with λ = 10 cm p = 8.3145J K ! 1 mol ! 1 ( ) x 298.15K ( ) 2 x 6.03 x 10 23 mol ! 1 ( ) x 0.36 x 10 ! 18 m 2 ( ) x 0.10m ( ) = 0.081 Pa 4. c rms = 3RT M = 3 8.314 x 10 7 ( ) 298 ( ) 39.948 = 4.31 x 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 39.948 = 4.31 x 10 4 cms ! 1 c mp = 2RT M = 2 8.314 x 10 7 ( ) 298 ( ) 39.948 = 3.52 x 10 4 cms ! 1 c = 8RT ! M = 8 8.314 x 10 7 ( ) 298 ( ) 39.948 ! ( ) = 3.97 x 10 4 cms " 1 5. c rms = 3RT M T = c rms ( ) 2 M 3R = 1.50 x 10 3 ms ! 1 ( ) 2 47.998 x 10 ! 3 kgmol ! 1 ( ) 3 8.314J K ! 1 mol ! 1 ( ) = 4.33 x 10 3 K Chemistry 440 Problem Set 5 – Solutions Fall 2011 Page 3 6. E trans = N 1 2 mc 2 ! " # \$ % & = 3 2 NkT T = mc 2 3k = Mc 2 3R = 4.0026 ( ) 2.74 x 10 4 ( ) 2 3 8.314 x 10 7 ( ) = 12.1 K 7. ! = kT 2 1/2 " p [19] = 1.381 x 10 # 23 J K # 1 ( ) x 298.15K ( ) 2 1/2 ( ) x 0.43 x 10 # 18 m 2 ( ) x p ( ) = 6.8 x 10 # 3 m p / Pa ( ) = 6.7 x 10 # 8 m p / atm a) When p = 10 atm, λ = 6.7 x 10-9 m, or 6.7 nm . b) When p = 1 atm, λ = 67 nm. c) When p = 10-6 atm, λ = 6.7 cm . The mean free path is inversely proportional to p and to z....
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern