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Entropy change of van der Waals gas expansion

  1. Apr 19, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider ##n## moles of gas, initially confined within a volume ##V## and held at temperature ##T##. The gas is expanded to a total volume ##\alpha V##, where ##\alpha## is a constant, by a reversible isothermal expansion. Assume that the gas obeys the van der Waals equation of state $$\left ( p + \frac{n^2a}{V^2} \right )(V - nb) = nRT$$. Derive an expression for the change of entropy of the gas.

    Show further that removing a partition and allowing a free expansion into the vacuum results in the temperature of the van der Waals gas falling by an amount proportional to
    ##( \alpha-1)/ \alpha ##.

    2. Relevant equations
    ##dU = TdS - pdV ##
    ##dU = dQ + dW##

    3. The attempt at a solution
    Since the total energy for a van der Waals gas is a function of ##T## as well as ##V##, ##dU## is not 0 in the first process.

    I need an extra piece of information but I'm not sure where to look.
     
  2. jcsd
  3. Apr 19, 2015 #2
    You need to use the Maxwell relationship approach to determine the partial derivative of entropy with respect to volume at constant temperature. Start with dA = -SdT-PdV.

    Chet
     
  4. Apr 19, 2015 #3
    Thanks for the help.

    Using this method and particularly $$ \left (\frac{ \partial S}{ \partial V} \right)_T = \left (\frac{ \partial p}{ \partial T} \right )_V,$$ the formula for ##dS## becomes $$dS = \frac{nR}{(V-nb)}dV,$$ which I then integrated. I have to mention that Maxwell's relations aren't introduced until later on in the book, though.

    For the second part of the question I know ##dU## is zero but it seems to be difficult to solve the problem without knowing anything about U.
     
  5. Apr 19, 2015 #4
    Part 2 is a very interesting problem. You correctly reasoned that, for the process described, ΔU is equal to zero. One hint I can give you is that you are going to need to take into account the effect of temperature on U. The only conditions under which you know that are in the ideal gas state, where dU = CvdT. So part of your path from the initial equilibrium state of the system to its final equilibrium state is going to involve a journey through ideal gas country. Think Hess' law for getting from the initial equilibrium state to the final equilibrium state using constant T and constant V segments, as parts of the overall journey.

    Another hint: S = S(T,V), so express dS using the chain rule for differentiation, and then substitute that into the equation for dU = TdS-PdV. This will give you what you need to integrate the constant temperature segments.

    Chet
     
  6. Apr 20, 2015 #5
    I'm not entirely sure what you mean but I think I have found the solution ( which may be precisely your method).

    Writing ## U = U(V,T)##, we have

    $$ dU = 0 = \left ( \frac{ \partial U}{ \partial T} \right)_V dT + \left ( \frac{ \partial U}{ \partial V} \right)_T dV = C_v dT + \left ( T \left ( \frac{ \partial p}{ \partial T} \right)_V - p \right )dV $$

    Substituting with the van der Waals gas gives

    $$ C_vdT = -\frac{n^2a}{V^2}dV$$

    Integrating gives

    $$T_f - T_i = \frac{n^2a}{VC_v}\frac{(1-\alpha)}{\alpha}$$

    In fact everything I've done matches your hints. Thank you again.
     
    Last edited: Apr 20, 2015
  7. Apr 20, 2015 #6
    This method only works because ##\frac{n^2a}{V^2}## is independent of T, so Cv is independent of V. Otherwise, you would have had to use the more elaborate procedure that I outlined in my previous post. I didn't know in advance that Cv is independent of V for a van der waals gas.

    Also, please check on the n's in your analysis. The equation you wrote for dU is supposed to be per mole. You might need an additional n on the left hand side to cancel one of the n's on the right hand side.

    Chet
     
  8. Apr 20, 2015 #7
    If I'm to have a constant temperature sections, doesn't that invalidate dU = 0 for that section?
     
  9. Apr 20, 2015 #8
    Sure. dU is not zero over any of the parts of the path that I described. But, over the entire path, ∫dU = ΔU is equal to zero. The only constraint of the problem statement is that ΔU=0, not that dU is zero over every increment of every path.

    Chet
     
  10. Apr 20, 2015 #9
    I said before I didn't fully understand your hints. If we had two sections, the first of which involves isothermal expansion to the required volume. This increases U by a certain amount. Then the next section is cooling at constant volume, for which Cv is constant. We do this until U decreases by the same amount as before, so the final state is the same.
     
  11. Apr 20, 2015 #10
    You have the right idea now, but you generally won't know Cv at the final volume (or the initial volume for that matter). Usually, the only place we know Cv is in the ideal gas state. So to do the problem if all you know Cv(T) in the ideal gas state and the EOS for the gas, you require a 3-step process:

    Step 1: Increase the volume at constant temperature until the volume is large enough, say V*, for the ideal gas state to be achieved.

    Step 2: Change the temperature at constant V* until the final temperature is achieved (temporarily unknown).

    Step 3: Decrease the volume from V* to αV at the constant final temperature.

    To find the final temperature, algebraically set the sum of the ΔU's for these three steps equal to zero. V* should cancel out when you do this.

    Actually, V* should be infinity, so you integrate from V to infinity, or you integrate from αV to infinity.

    Chet
     
    Last edited: Apr 20, 2015
  12. Apr 21, 2015 #11
    What if you reversed the steps I gave previously. Doesn't that imply that Cv is independent of the volume ( because in both cases they must gives the same value)?

    Edit: sorry I seemed to assume that Cv was independent of temperature.
     
    Last edited: Apr 21, 2015
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