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Ideal gases: irreversible transformation

  1. Sep 22, 2007 #1
    A balloon featured with a negligible thermal capacity contains [tex]V_{l}=3l[/tex] of ideal gas and is immersed in a lake (thermal bath) at the depth of [tex]h_{l}=10m[/tex] beneath the lake surface. If it is brought to the depth of [tex]h_{l}=3m[/tex], how much is the heat exchange?
  2. jcsd
  3. Sep 22, 2007 #2
    I did 2 typos. The first [tex]h_{l}[/tex] has to be replaced with [tex]h_{1}[/tex] and the second with [tex]h_{2}[/tex]... anyway, could you help me please? Thanks
  4. Sep 22, 2007 #3
    Sounds like a case of isothermal expansion.
  5. Sep 22, 2007 #4
    yes, it is. But I don't know how to calculate that heat...
  6. Sep 22, 2007 #5
    Remember that for an isothermal expansion, the internal energy doesn't change, so work done is equal (up to a sign) to the heat transfer. Now, what's the work done by the gas in the expansion?
  7. Sep 22, 2007 #6
    I was just arrived to this same conclusion. But now, about work integral [tex]W=\integral_{V_{1}}^{V_{2}}p_{ext}dV[/tex], what is the correct expression I have to use for external pressure? I supposed it is a constant because of fast rising. But what's the value?
  8. Sep 22, 2007 #7
  9. Sep 22, 2007 #8
    I'd expect that the internal pressure equals the external pressure. As it's isothermal, that also allows you to work out the volume. You should get an integral involving 1/V, and so some logs out the end.
  10. Sep 22, 2007 #9
    Surely you'd be right if transformation could be thought as a reversible transformation (a nearly static one). I learned that heat exchange and work done by the system are not function of state and are strictly dependent upon the particular path the transformation has gone across.
  11. Sep 22, 2007 #10
    Isothermal expansion is reversible
  12. Sep 22, 2007 #11
    The equation of state [tex]PV=nRT[/tex] keeps always its validity, but the reversibility is not a prerequisite at all. If transformation takes place with rapidity, alas, your last sentence is not satisfactory anymore.
  13. Sep 22, 2007 #12
    Okay, that is true. However, assuming that in moving the balloon doesn't put significant amounts of energy into the system, the answer will be the same as if the entire process happened quasi-statically.
  14. Sep 23, 2007 #13
    I was told that the correct calculation is given by using as [tex]p_{ext}=const[/tex]. Nevertheless, this constant is the external pressure corresponding to [tex]h_{2}[/tex]! This is the incomprehensible enigma I'm stuck in!
  15. Sep 23, 2007 #14
    I don't see why the external pressure is constant -- pressure of the water would be depth dependent.
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