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Thermal Physics Introduction- solving for entropy and temperature change

  1. Sep 27, 2012 #1
    1. The problem statement, all variables and given/known data

    A container is divided into two parts by a thermally conducting wall. There are N atoms of a monatomic ideal gas on the left side, 2N on the right. The gas on the left is initially at absolute temperature 200K, the gas on the right at 500K.
    a. After thermal equilibrium is reached, what is the temperature in the box (on either side)?
    b. How much does the entropy of the whole system change in reaching thermal equilibrium?

    T_L=N
    T_R=2N
    T_f=final temp


    2. Relevant equations


    E=(3/2)N*k_b*T
    deltaE=deltaQ
    dS=(k_b*T)^-1*dQ


    3. The attempt at a solution

    a. For the first part, I used the equation for E to get (3/2)N*k_b*(T_f-T_L)=deltaQ and (3)N*k_b*(T_f-T_R)=-deltaQ, because deltaE_L=-deltaE_R

    (3/2)N*k_b*(T_f-T_L)=-(3)N*k_b*(T_f-T_R), then crossing out k_b and N and solving:
    T_f-T_L=2(T_f-T_R)
    so T_f=2T_R-T_L = 1000K-200K = 800K
    I'm not sure if i did this right or not

    b. dS=(k_b*T)^-1*dQ and dQ=dE, so
    dS=(k_b*T)^-1*(3/2)N*k_b*dT for the left side
    dS=(3/2)N*T^-1*dT
    S=(3/2)N*ln(T) from T_L to T_f, so
    S=(3/2)N*ln(T_f/T_L)
    and similarly, S=3N*ln(T_f/T_R)

    I know that I'm missing a k_b in the end. The entropy on either side of the wall is not equivalent. I'm not sure where I went wrong.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 27, 2012 #2

    rude man

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    Well, you know that T_f - T-L has to be positive and T_f - T_R has to be negative, right? So how can those two differences have the same sign?
    You need to get the right T_f first.

    Well, that would be a neat trick; you're starting with 200K and 500K and end up with 800K without doing any work or adding heat to the system?
    Your S is the change in entropy, not the entropy. So there's a change in S on the left = ΔS_L and a change in the right ΔS_R, and the total entropy change for your system is ΔS_total = ΔS_L + ΔS_R.

    So now recompute those entropy changes and then sum. Just don't get a negative result for the net change in S, or Clausius et al will roll over in their graves! :biggrin:
     
  4. Sep 27, 2012 #3
    Okay so I recomputed the temperature using -(Tf-Tr) and got 400. From there i got left S= (3/2)N*ln(2) (Tf/Tl = 2) which gave me 1.04N J/K. On the right I got S=3N*ln(4/5) which was -.669N J/K. I summed them, but I think I'm still missing a k_b somewhere. I crossed them out in the same step i.e.
    dS=(k_b*T)^-1*(3/2)N*k_b*dT for the left side
    dS=(3/2)N*T^-1*dT

    but the answer when multiplied by N without a k_b is too large, isn't it?
     
  5. Sep 27, 2012 #4

    rude man

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    I'm not sure what the b is in your equations, but somewhere you dropped something all right.You have the right expressions for the two entropy changes now, except for a constant coefficient. If you just keep track of your coefficients (the N, k, b etc.) you will get the right answer.

    Hint: one form of that coefficient is: divide N by some (big) number, then multiply by another number involving R, the universal gas constant. It doesn't involve k or b in that form.

    I have my answer waiting if you want to compare w/ yours.
     
  6. Sep 28, 2012 #5
    I got .370N*k_b where k_b is the Boltzmann constant 1.38E-23, does that look like yours?
     
  7. Sep 28, 2012 #6

    rude man

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    Must be absolutely correct! We agree!
     
  8. Sep 28, 2012 #7
    Thank you!
     
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