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Adiabatic Slope ?

  1. Apr 7, 2008 #1
    1. The problem statement, all variables and given/known data

    The slope of an adiabatic curve will be equal to r times the slope of an isothermal curve only when we use the ideal gas equation

    2. Relevant equations

    3. The attempt at a solution

    Adibatic process, P(V)^g = constant
    Differentiating w.r.t V


    dp/dv= -[g(V)^g-1]P/v^g

    dp/dv= -g P/V

    Slope of an isothermal curve using P=nRT/V is -nRT/V^2 = -P/V

    Therefore slope Adiabatic = g times slope isothermal

    Am i Correct ??

    Can we use vander waal eqn also ?
  2. jcsd
  3. Apr 7, 2008 #2
    You've proven what you put it --- the formula pV^g=constant for adiabatic expansion uses the ideal gas law for its derivation. You have to start more generally.
  4. Apr 7, 2008 #3
    how do i do that ?
  5. Apr 8, 2008 #4
    Start with a general equation of state, i.e. p is some unknown function of V and T. Relate the gradients of isothermal and adiabatic curves to its values or derivatives. Set them to be proportional to each other, with a constant -g, and then solve the resulting differential equations to show that you end up with p=GVT where G is some (unknown) constant.
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