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Homework Help: Physical Chemistry

  1. Sep 5, 2010 #1
    The problem statement, all variables and given/known data
    Cloud moves from alt. 2000m (P = 0.802atm) to 3500m (P = 0.602) when it encounters a mountain. It expands adiabatically. The initial temp is 288K, CP,m for the air is 28.86 J/Kmol (assume ideal). What is the final temp and will it drop it's moisture?

    The attempt at a solution
    This is what I did. CV,m = CP,m - R
    CV,mdT = -[tex]\frac{RT}{V}[/tex]dV
    [tex]\int[/tex]CV,mdT = [tex]\int[/tex]-[tex]\frac{RT}{V}[/tex]dV divide by T
    CV,m[tex]\int[/tex][tex]\frac{1}{T}[/tex]dT = -R[tex]\int[/tex][tex]\frac{1}{V}[/tex]dV
    CV,mln([tex]\frac{T2}{T1}[/tex]) = R ln([tex]\frac{V1}{V2}[/tex]) rearrange
    T2 = T1(V1/V2)R/CV,m this to solve for T2, but 2 variables, so...

    P1V1[tex]\gamma[/tex] = P2V2[tex]\gamma[/tex] solve for the ratio V1/V2
    [tex]\frac{V1}{V2}[/tex] = ([tex]\frac{P2}{P1}[/tex])1/[tex]\gamma[/tex]

    T2 = T1 (([tex]\frac{P2}{P1}[/tex])1/[tex]\gamma[/tex])R/CV,m

    CV,m = CP,m - R = 20.546 [tex]\gamma[/tex] = CP,m/CV,m = 1.405

    T2 = 288 (([tex]\frac{0.602}{0.802}[/tex])1/1.405)8.314/20.546 = 265K

    Someone please tell me if I did this right, if not, what did I do wrong..........and how do I know if it will drop moisture?

    This is my first HW question in Pchem I and only the 2nd week of the semester, so please explain anything so I can understand it.
    Last edited: Sep 5, 2010
  2. jcsd
  3. Sep 8, 2010 #2


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    You should reverse the order of your first and second steps. Divide by T before integrating. Once the integral signs are there, you can't just divide to move variables from one side of the equation to the other. Your solution otherwise looks okay, but you could have done it more directly by simply using just the ideal gas law and the relationship [itex]PV^\gamma=c[/itex], where c is a constant.

    I don't know about the moisture. Perhaps someone else can chime in on that part of the problem.
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