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Thermodyanmics, basic Integration and total differential.

  1. Oct 23, 2013 #1
    Example 1.

    A non ideal gas undergoes isothermal doubling of the pressure and you are asked to calculate the total work done; where the equation of state V=v(T,p), has been specified. Where V is the volume, T the temperature.

    Attempt:


    - The only way around this that I can see is using W=-∫pdV. ( But I can only justify its use if the reaction is reversible, and we are not told this....?)
    - Anyway, to continue with this method, I first of all find the total differential dV=∂V/∂pdp+∂V/∂T dT*
    - The 2nd term will =0 as it is isothermal.
    - Now my main quesstion is, before intergrating - dp - on the first term - surely I need to express T in terms of p before integrating? (whilst in the partial derivative it is treated as a constant, before I integrate it cant be correct to treat is as a constant as there is known information on the relationship between p and V)
    E.g say I end up with 2po∫po k/T p^3 dP, where k is a constant, do I need to express T in terms of P before integrated?

    On second thoughts, this seems okay to do this time around, as T is constant, but explicitly, where/how does this step in , such that I could generalise this for cases when T is not constant, see example below:


    2nd Example

    - To calculate the work done in a carnot cycle, ideal gas, let the volumes be Va,Vb,Vc,Vd, where Va-Vb is a isothermal expansion and it operates between reservoirs at Th and Tl.
    - Okay, now if im to consider the work done in the adiabatic expansion:

    Attempt

    1) Use of W=-∫pdv is THIS TIME justified as the carnot cycle consists of reversible processes by definition.
    - I attain the same chain rule as above *, this time used to attain the total differential of V=V(p,T) as related by the ideal gas law: pV=nRT
    -W=∫=nR dT + ∫nRTP^-1dp
    - the first term looks okay to deal with , ΔT must =Th-Tl (my justification being adiabatic expansion so the gas expands without any head input, it does work on the environment, so its temperature must decrease) so that's my limits; and 'dT' looks fine to carry out as nR is a constant.
    - But now I look at the second term, T is not constant. Surely it must be expressed in terms of p before carrying out the integration. Well I can use pV=nRT , so attain T=T(P,V), so now I want to eliminate the V to get everything solely in terms of p, well I can turn to pV^γ =constant. (as the process is adiabatic)*2

    - BUT, the question did not specify any information regarding gamma - Cp/Cv or whether it is monoatomic etc.

    *2 Bit of a long shot, but could it be in someway that because the processes are reversible- equilibrium at all stages- that only the final and starting values of T are required.



    Many Thanks in advance to anyone who can shed some light on this :) !!
     
  2. jcsd
  3. Oct 24, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This problem is more physics than mathematics but:
    No. That equation is always true, not just for reversible reactions.

    As you add later, you are given that T is a constant.
    [tex]\frac{k}{T}\int_{p_0}^{2p_0} p^3 dp= \frac{k}{4T}\left[p^4\right]_{p_0}^{2p_0}= \frac{k}{4t}(16p_0^4- p_0^4)= \frac{15k}{4T}p_0^4[/tex]
     
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