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Explaining Adiabatic Process taking Van Der Waals Equations?

  1. Jul 7, 2007 #1
    To Forum Physicists / Administrators...
    Equations for an ideal gas undergoing an adiabatic process are 3 viz
    1) P(v^g)=const
    2)T(V^(g-1)) =constant and
    3)T(P^((1-g)/g))= constant

    where g=specific heat for const pressure/specific heat for const volume.These equations suits the Ideal gas since they are based on the assumption that PV=RT.

    There is a derivation of van der waals equation according to which a modification of the ideal gas law was suggested taking into account the vander waals forces and nonzero size of molecules .(it also approximates the behavior of real fluids, taking into account the nonzero size of molecules and the attraction between them). The equation is (p + a / v2)(v − b) = kT.How can we modify the three equations cited above for gas(or fluid) undergoing an adiabatic process if van der waals correction is taken into account? How to derive and solve them?. I searched web pages and didn't find any.I'm a new comer
     
  2. jcsd
  3. Jul 7, 2007 #2

    olgranpappy

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    Is this a homework question? If so, I believe that the rule is that you should post in the homework forums. Anyways. You should review the derivation of equations 1 thru 3 starting from the ideal gas equation PV=RT. Then simply work thru the same steps but starting with the van der waals equation rather than the ideal gas equation.
     
  4. Jul 9, 2007 #3
    NO!, its NOT AT ALL a homework question.I'm a post garduate in COMMERCE not student in a high school. Actually as a hobby I search & learn physics thru web pages and learnt special relativity myself (from the scratch in just 3 months). Gas laws were well understood until this puzzled me since I saw about the van der waals equation and I ALREADY did that "simply work thru the same steps but starting with the van der waals equation rather than the ideal gas equation" part even before searching web pages.The way the calculus was applied to original Eqns gets far complex as it takes a form of dy/dx=f(x,y). One cannot find any web pages that deals with this problem.i learnt that after spending Terabytes of browsing quota.HENCE I NEED HELP!!..AND I EXPECT AN ANSWER.....
     
    Last edited: Jul 9, 2007
  5. Jul 9, 2007 #4

    olgranpappy

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    So, what did you arrive at after working through the same steps? An equation of the form:
    [tex]
    \frac{dP}{dV}+Pf(V)+g(V)=0\;?
    [/tex]

    What expressions did you find for the functions [tex]f(V)[/tex] and [tex]g(V)[/tex]?
     
  6. Jul 11, 2007 #5

    Andrew Mason

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    Try following this derivation for the Ideal adiabatic condition but use the Van der Waals equation:

    [tex]\left(P + a\left(\frac{n}{V}\right)^2\right)\left(\frac{V}{n} - b\right) = RT[/tex]

    rather than the Ideal Gas law: PV/n=RT

    AM
     
  7. Jul 14, 2007 #6
    Here Is A Try


    I tried , but as I asked earlier, the final result was not in any explicit form.Its not even clear how
    to get the relationships between temperature and volume or pressuer-temperature using this.

    For Ideal gas:
    PV=RT;Replacing R and T with R= Cp-Cv; T= -PdV/Cv (-ve since energy is expended for volume increase)
    and differentiating PV gives PdV+VdP=(Cp-Cv)(-PdV/Cv)
    Now Cp/Cv=Y ('Y' resembles gamma in plain txt format)
    hence, PdV+VdP=PdV-YPdV => VdP=-YPdV
    dividing throught by PV => Log(P)=-YLog(V)+C
    Antilog(Log(P)+YLog(V))=K => P(V^Y)=K
    This equation can be expressed EXPLICITLY in terms of P and V; thus:
    P=K(V^-Y) and V=(K/P)^1/Y

    With Van der waals correction:

    (P+a/V^2)(V-b)=RT. Putting this as
    PV+a/V-Pb-ab/V^2=RT.Again RT=(Cp-Cv)(-PdV/Cv) for the same reason
    Differentiating LHS
    PdV+VdP-a/V^2-bdP+2ab/V^3=PdV-YPdV
    Dividing throught by PV
    Log(P)-a/(V^3*P)-b/V(Log(P))+2ab/P*V^4+Y*Log(V)=C
    Solving(or further complicating ) to
    (P^[1-b/V])( V^Y )(EXP(a[2b-v]/P*V^4])=K
    Given the Inputs P1, V1, a and b ;and P2, other than guess work how would we get V2?
    So this Bold one not Solvable in terms of one of the variables
     
  8. Jul 14, 2007 #7
    I tried , but as I asked earlier, the final result was not in any explicit form.Its not even clear how
    to get the relationships between temperature and volume or pressuer-temperature using this.

    For Ideal gas:
    PV=RT;Replacing R and T with R= Cp-Cv; T= -PdV/Cv (-ve since energy is expended for volume increase)
    and differentiating PV gives PdV+VdP=(Cp-Cv)(-PdV/Cv)
    Now Cp/Cv=Y ('Y' resembles gamma in plain txt format)
    hence, PdV+VdP=PdV-YPdV => VdP=-YPdV
    dividing throught by PV => Log(P)=-YLog(V)+C
    Antilog(Log(P)+YLog(V))=K => P(V^Y)=K
    This equation can be expressed EXPLICITLY in terms of P and V; thus:
    P=K(V^-Y) and V=(K/P)^1/Y

    With Van der waals correction:

    (P+a/V^2)(V-b)=RT. Putting this as
    PV+a/V-Pb-ab/V^2=RT.Again RT=(Cp-Cv)(-PdV/Cv) for the same reason
    Differentiating LHS
    PdV+VdP-a/V^2-bdP+2ab/V^3=PdV-YPdV
    Dividing throught by PV
    Log(P)-a/(V^3*P)-b/V(Log(P))+2ab/P*V^4+Y*Log(V)=C
    Solving(or further complicating ) to
    (P^[1-b/V])( V^Y )(EXP(a[2b-v]/P*V^4])=K
    Given the Inputs P1, V1, a and b ;and P2, other than guess work how would we get V2?
    So this not Solvable in terms of one of the variables
     
  9. Aug 29, 2008 #8
    I don't think that your hint will actually work. In the derivation you cite, they use a relationship:

    [tex] dU = n\, c_v\, dT [/tex]

    For the ideal gas, the internal energy is only a function of the temperature, so the above relationship holds for all processes, not only constant volume ones. However, in general the above relationship is only true along an isochore, and we're moving on an adiabat.

    To substantiate this further, you can look at the expression for the internal energy of the van der Waals gas as a function of (T,V,N) given in Wikipedia. There I find that:

    [tex] U = \frac{3}{2} N k T - \frac{a N^2}{V} [/tex]

    Along an adiabat in general T and V will change, and so we can't just use the specific heat at constant volume to get the energy change.
     
  10. Sep 19, 2008 #9
    From the above expression for the internal energy of a van der Waals gas, one can derive an expression of form f(T,V)=constant in the following way:

    From dU=dQ+dW and dW=p dV, we have dQ=dU + p dV = 0 (since the process is adiabatic)

    Then expand dU in terms of dT and dV to get

    (dU/dT) dT + (dU/dV) dV + p dV = 0

    Calculate the derivatives from the expression for U above, and substitute for p from the original van der Waals equation of state. This will give you a separable ODE in T and V; from the ODE solution you can derive the relation f(T,V)=constant. From this, you should then be able to work out the expressions f(P,V)=constant and f(T,P)=constant .
     
  11. Sep 19, 2008 #10
    oops -- that should have been dW=-p dV
     
  12. Sep 22, 2008 #11
    Your above calculation is fine. However, if the question is about a system where we don't have an expression for the internal energy, U, then I think we are stuck. The first answer suggested that one use dU= cv dT. My reply was that the internal energy can be a function of volume, and I gave the vdW gas expression for U, not supplied by the original posting. You are absolutely right that if we use this expression for U(T,V), we can solve for the adiabat. But if we don't have such an expression, I don't think that we can (in general) solve for adiabats.

    How does one calculate U for the vdW gas from the vdW equation of state? On Wikipedia, they work from Statistical Mechanics. Callen derives an expression for S(U,V,N) for the van der Waals fluid from the vdW equation, but he does have to make some assumptions. If all one has is the vdW equation and you can't make any other assumptions, I think you may be stuck.

    However, I would like to use your derivation as a homework problem in my next assignment on the vdW gas. :-)
     
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