Chap3prac - Chapter 3 Practice Problems(P3.1(a the number...

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

View Full Document Right Arrow Icon
Chapter 3 Practice Problems To accompany Introductory Chemical Engineering Thermodynamics © J.R. Elliott, C.T. Lira, 2001, all rights reserved. (2/3/2009) 1 of 7 (P3.1) (a) the number of microstates is N 2 (pg 91, typo in given answer, printings 1-3) (b) 3 particles total 3 ! 1 !* 2 ! 3 } 1 , 2 { = = T H p number microstates of specific arrangement (macrostate) probability = (# microstates of specific arrangement)/(total # of microstates) 8 3 2 3 3 = = prob (c ) # microstates. 20 ! 3 !* 3 ! 6 15 ! 2 !* 4 ! 6 10 ! 2 !* 3 ! 5 6 ! 2 !* 2 ! 4 } 3 , 3 { } 2 , 4 { } 2 , 3 { } 2 , 2 { = = = = = = = = T H T H T H T H p p p p (d) macrostate H T # of microstates* 0 8 1 1 7 8 2 6 28 3 5 56 4 4 70 5 3 56 6 2 28 7 1 8 8 0 1 * number of microstates = )! 8 ( ! ! 8 m m total number of microstates is 2 8 = 256, which is the same as the sum from the table. portion of microstates (probability) for requested configurations: {5:3} = 56/256 = 0.219 = 22% {4:4} = 70/256 = 0.273 = 27% {3:5) = 22% like {5:3} probability of any one of the three most evenly distributed states = 22% + 27% + 22% = 71% (e) for 8 particle system, Stirling’s approx will not apply Δ S /k = ln(p{4:4}/p{5:3}) = ln (70/56) = 0.223
Image of page 1

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

View Full Document Right Arrow Icon
Chapter 3 Practice Problems To accompany Introductory Chemical Engineering Thermodynamics © J.R. Elliott, C.T. Lira, 2001, all rights reserved. (2/3/2009) 2 of 7 (P3.3) 15 molecules in 3 boxes, molecules are identical ! ! 1 ij i j m N p = = ……………………………………… .... Eqn. 3.4 75075 ! 2 ! 4 ! 9 ! 15 1 = = p 756756 ) ! 5 ( ! 15 3 2 = = p 31 . 2 ln 1 2 = = Δ p p k S (P3.4) two dice. ?? = Δ k S for going from double sixes to a four and three. for double sixes, we have probability of 1/6 for each dice. ( ) ( ) ! 6 1 !* 6 1 ! 2 1 = p for one four and one three probability applied for 1/6 for each one in each dice, ( ) ( ) 2 * ! 6 1 !* 6 1 ! 2 2 = p 693 . 0 2 ln ln 1 2 = = = Δ p p k S (P3.5) Δ S = ??
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern