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I A proper Thermodynamic system

  1. Nov 11, 2016 #1

    MathematicalPhysicist

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    I have this question that in it there's this expression "A proper Thermodynamic system".

    I thought that means that the first law (or is it the second) that ##dS \ge 0##, i.e. the change of entropy always increase.

    But I am not sure, can someone clear to me this matter?

    Thanks in advance.
     
  2. jcsd
  3. Nov 11, 2016 #2
    Have you tried looking it up? Your question can be answered in a second of searching on the internet.
     
  4. Nov 11, 2016 #3

    MathematicalPhysicist

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    I tried searching google, and didn't find the exact phrasing:"A proper Thermodynamic system" then what is it?
     
  5. Nov 11, 2016 #4
    I guess I didn't understand your question. I thought you were confused by what the first and second laws of thermodynamics are, since you got them mixed up above. You still should look them up for that reason.
    I think you are expecting a technical definition for "proper thermodynamic system", when it just means proper "thermodynamic system", where proper is just a non-technical English word, and thermodynamic system is just a system that obeys thermodynamics, which is basically anything that isn't some make-believe model.
     
  6. Nov 11, 2016 #5

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    I have this question which I don't know how to start with and the teacher assistant isn't well responsive in the email.

    Maybe you know better than me:

    We have: the equations of system: ##u =3/2 pv , p=av T^n##, I first was ask to find ##T(u,v), p(u,v)## which I found (##n## is an integer and ##a## some constant).
    Now I am asked to find for which values of ##n## is the system a proper theormodynamic system?

    I thought of using ##dS \ge 0## in the end I got that: ##n \ge constant/ (av^3 \ln T)##, I thought that since should be valid for every ##T>0## then when ##T\ to \infty## we get: ##n \ge 0##, but I am not sure.

    Do you know of this problem before?
    I searched my reference with no help.
     
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