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Thermodynamic Adiabats

  1. Nov 5, 2016 #1
    1. The problem statement, all variables and given/known data
    The internal Energy of a system is ##U=aP^2V## with a positive constant a. Find the adiabats of this system in the P-V plane.
    The solution is $$P=\frac 1 a\left( \sqrt{\frac {V_0} V}-1\right)$$

    2. The attempt at a solution
    the first law with the given internal energy:
    $$a(2PVdP+P^2dV)=-PdV$$
    Integration:
    $$\frac {2adP} {1+aP}=- \frac{dV} V$$
    $$2 ln(1+aP) = -lnV+const.$$
    and
    $$(1+aP)^2=\frac 1 V+const.$$
    $$P=\frac 1 a\left( \sqrt{\frac 1 V+const}-1\right)$$
     
  2. jcsd
  3. Nov 5, 2016 #2

    TSny

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    You have a small but crucial error in going from the first equation above to the second.
     
  4. Nov 12, 2016 #3
    Thanks for your Tip. Should it be like that?
    ##ln((1+aP)^2)=ln\frac 1 V+lnV_0=ln\frac {V_0} V## with ##V_0=e^{const.}##
     
  5. Nov 12, 2016 #4

    TSny

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    Yes, that's a nice way to do it.

    The important thing is that if you have something like ##\ln y = \ln x + const##, it does not follow that ##y = x + const##.

    If you have ##\ln y = \ln x + k## where ##k## is a constant, then "taking exponentials of both sides" yields
    ##e^{\ln y} = e^{\ln x + k} = e^k e^{\ln x}##.

    So, ##\ln y = \ln x + k## implies ##y = C x ## where ##C = e^k## is a new constant.
     
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