che359_sp10_hw4_soln - ChE 359 Homework #4 Solutions...

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Page 1 of 5 ChE 359 Homework #4 Solutions prepared by Yandi Hu 1. (25 points) We are going to look at gas bubbles expanding to fill all available space. a. (10 points) Initially, the 10 indistinguishable gas bubbles are confined to a space that only consists of 10 boxes, each of which is large enough to hold only one gas bubble. Calculate the number of ways that the bubbles can be partitioned among the 10 boxes, and use this information about the multiplicity to calculate the entropy of the gas bubble system. Since there are 10 indistinguishable gas bubbles and 10 boxes that can hold one gas bubble each, there is only one way that the bubbles can be partitioned among the ten boxes. We can also do this mathematically if we consider that there are 10 indistinguishable boxes categorized into two types – those with bubbles, and those without bubbles. In this case, the multiplicity is given by: W 10 = N ! n 1 ! n 2 ! = 10! 10!0! = 1 . The reason we cannot simply use Bose-Einstein statistics is because the Bose-Einstein model as shown in class may have more than one bubble in a box, which is not there case here. The multiplicity of the system is thus: S 10 = k ln W 10 = k ln1 = 0 b. (10 points) The space is then expanded to 50 boxes. Calculate the number of ways that the bubbles can be partitioned among the 50 boxes. Again, we just use the same formula as above. W 50 = N ! n 1 ! n 2 ! = 50! 10!40! = 10272278170 . The multiplicity of the system is thus: S 50 = k ln W 50 = k ln10272278170 = 3.18 × 10 22 J K c. (5 points) Justify why the entropy changes the way it does when the available space is increased from 10 boxes to 50 boxes. We see that the entropy increases when the number of boxes is increased, due to an increase in the number of possible arrangements of the gas bubbles in
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che359_sp10_hw4_soln - ChE 359 Homework #4 Solutions...

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