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Ideal Gas Under A (Complex) Reversible Process

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data
    An ideal gas undergoes a reversible process 1-2-3-4-5. The stage from 1-2 as well as isobaric 2-3 and isochoric 4-5 stages are represented by linear segments. The Stage 2-3 is isothermal. The following relations are true:


    The initial states are V_1=3m^3 and p_1 = 2.5 atm.
    Assuming that the internal energy of an ideal gas depends only on temperature, find the total heat recieved by the gas during the entire process also find change in enthalpy (H_1-H_5).

    2. Relevant equations



    dw=pdv (isobaric)

    dw=nRT*ln(v_f-v_i) isothermic



    U=n*c_v*T+n*u_0 *****

    3. The attempt at a solution

    So far I have calculated all the pressures and volumes for all points (in atm and m^3):

    After that I started trying to find either n or T, but I am guessing the point of the problem is to solve it without those. So moving on to find Q total, I noticed that without T, this problem involves at least 4 unknowns. So my troubles come in calculating the work from 1-2 since pressure is non constant and not any special type of process, and from eliminating n and T from the equation for q total.

    ****This is also the chapter in the book before we talk about the energy of an ideal gas, so I am assuming that we are not to use that equation, but just in case there it is.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 25, 2011 #2

    rude man

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    You need to clarify what stages are isobaric, isochoric, isothermal, adiabatic, etc. better. I'm royally confused.

    Linear segments on what type of plot?
    Last edited: Sep 25, 2011
  4. Sep 26, 2011 #3
    Sorry about that just tried to copy the problem word for word. But looks like I made a mistake typing, sorry about that.

    Isochoric- 4-5
    Isobaric- 3-4
    Isothermal- 2-3
    No specified adiabatic processes.

    The plot is a P-V diagram.
  5. Sep 26, 2011 #4

    rude man

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    By the end of your last chapter, I assume it discussed Cv and Cp. Because unless it avoided the adiabatic equation pV^γ = constant, it had to mention that γ = Cp/Cv.

    By definition, Cv =∂U/∂T|V but since U = U(T only), then Cv = dU/dT.

    My first suggestion is to plot all your p's and V's on your p-V diagram.
    Then: what is ∆W on a p-V diagram when any two states are plotted?
    Can you readily see what ∫pdV and ∫Vdp are on your diagram?

    Combine the plot with dU = dQ - dW and pV = nRT and you can proceed step-by-step thru all 5 states & arrive at ∆U.

    Then of course ∆H is immediately available since you now know p5 and V5.
    Last edited: Sep 26, 2011
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