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Heat process

  1. Nov 15, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider an ideal gas whose volume is initially [tex]V_0=1cm^3[/tex] that is initially at [tex]P_0=1[/tex] atmosphere. It goes through a cycle a-b-c-a where b-c is at constant temperature.
    1)According to the sketch, does the system absorb or release heat?
    2)Calculate the heat exchange in a complete cycle.
    3)What's the total variation of entropy in a complete cycle?

    2. Relevant equations
    None given.


    3. The attempt at a solution

    Since it's an ideal gas, PV=nRT holds.
    1)So when the system goes through a-b, its temperature duplicates and as [tex]Q=mc\Delta T[/tex], we have [tex]Q>0[/tex].
    For b-c Q=0 since it's an isothermal.
    For c-a, heat is released I believe, in the same amount heat has been gained from the environment for the part a-b.
    2)Q=0.
    3)[tex]\Delta S=\int \frac{dQ}{T}=0[/tex] but I'm unsure.
     

    Attached Files:

  2. jcsd
  3. Nov 15, 2009 #2

    Andrew Mason

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    Apply the first law. In a complete cycle, positive work is done but is there a change in internal energy? What does this tell you about heat flow?
    Again, apply the first law. All you have to do is determine the amount of work done in a cycle to work this out.

    If you think of entropy as a state function, this easy to answer.

    AM
     
  4. Nov 15, 2009 #3

    fluidistic

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    Thanks for helping. I must say I'm a rusty on this part of into physics.
    For part 1), the first law states [tex]\Delta E =Q-W[/tex] where [tex]W[/tex] is the work done by the system, which is equal to [tex]P\Delta V[/tex] in this case since it's a (an ideal) gas.
    You pointed out that the work done in a cycle is positive, but I don't see it.
    I realize that the internal energy of the gas changes within the cycle, but after a complete cylcle its internal energy is the same as the start of the cycle. So [tex]\Delta E=0[/tex], hence [tex]Q=W=P\Delta V[/tex]. But after a complete cycle, doesn't [tex]\Delta V=0[/tex]?
    As you can see, I still need some guidance.
     
  5. Nov 15, 2009 #4

    Andrew Mason

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    Positive work is done from b to c. Negative work (ie work is done on the gas in compressing it) is done from c to a. No work is done from a to b. The total work is the area under b-c less the area under c-a, which is the area inside the graph.

    Since U is unchanged over one cycle, you know that net heat flow, Q = W = area inside the graph.

    Since positive work is done in one cycle, the first law tells you that over a complete cycle the system absorbs more heat than it releases.

    To work out the heat flow, determine the heat flow for each of the three parts of the cycle using the first law and the ideal gas equation. For part b to c, since it is isothermal (ie. internal energy does not change) AND work is being done, what can you say about the heat flow? Can you calculate it? Hint: use PV = nRT ; P = nRT/V in calculating:

    [tex]\int_b^c PdV[/tex]

    For part a to c, no work is being done on the gas. What can you say about the heat flow? Can you calculate it?

    Since you know the total heat flow over the full cycle and you now know the heat flow into the gas, you can calculate the heat flow out of the gas from c to a.

    AM
     
  6. Nov 16, 2009 #5

    fluidistic

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    Thank you very much for the detailed reply. I'm currently studying from my textbook and will come back to this problem later when I'll remember what I should.
     
  7. Nov 17, 2009 #6

    fluidistic

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    I can say that [tex]\Delta E=Q-W=0 \Rightarrow Q=W[/tex] Using the integral you provided, I get that [tex]W=nRT \ln 2[/tex].
    a to b I guess you meant. I can say that the heat flows equal the change in internal energy of the gas. I don't really know how to calculate it. Since the pressure duplicate I guess the internal energy also duplicate because the temperature has to duplicate. Isn't it [tex]E=\frac{f}{2}nRT[/tex] where T is the temperature at point a and f is the number of freedom's degree of the gas' molecules? I find strange that the exercise doesn't say anything whether the gas is monoatomic or not.

    Not sure if this is it, but [tex]\Delta E_{a-b}=-\Delta E_{c-a}=-P_0V_0[/tex] but f and T are unknowns.
    By the way I should be giving the final results with 5 significant digits, so I can't stay with any unknowns.
    I appreciate your help.
     
  8. Nov 24, 2009 #7

    fluidistic

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    I'm still working on this problem. Now I understand some things much better thanks to you. I'm redoing it, except part a which is now obvious.
    For a-b, [tex]W=0\Rightarrow \Delta E=Q=\frac{3nRT_0}{2}[/tex].
    For b-c, [tex]\Delta E=0\Rightarrow Q=W=\int P dV=nRT\int _{V_0}^{2V_0}\frac{dV}{V}=nRT\ln 2[/tex].
    For c-a, [tex]W=-P_0V_0 \Rightarrow \Delta E=Q_{c-a}P_0V_0[/tex].
    I'd like to work out [tex]Q_{c-a}[/tex]. I know that [tex]\Delta E[/tex] after a cycle is worth [tex]0\Rightarrow Q_{c-a}=-\frac{3nRT_0}{2P_0V_0}[/tex] in which the negative sign makes sense since heat is released contrarily to the other 2 processes.

    Now I try to find out the total heat transfer after a cycle, adding them up I reach [tex]Q_{\text{total}}=\frac{3nRT_0}{2}+nRT\ln 2-\frac{3nRT_0}{2P_0V_0}[/tex], I know it's positive but I'm not able to demonstrate it.

    For the last question, [tex]\Delta S = \int \frac{dQ}{T}[/tex]. I think I have to calculate the entropy in each part of the cycle and sum them up. Am I right?
    My intuition tells me the change of entropy of the gas must be 0 since the gas return to its initial state, is that right? If this is right then [tex]Q_{a-b}=-Q_{c-a}[/tex] but this is not what I've found...
    Where did I go wrong?
     
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