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Entropy for irreversible process

  1. Nov 12, 2004 #1
    Question:
    2 moles of a monatomic ideal gas at an initial pressure of 20 atm and a temperature of 450K undergo a rapid (adiabatic) free expansion from a small vessel into a vessel of 40 times greater volume. Find
    i) the change in temperature and
    ii) the increase in entropy

    The first part I know how to do. But for the second part the dQ is 0 for the adiabatic expansion. How am I going to find dS. I know it's not zero. I know adiabatic is not a reversible process. But I don't understand the approach to solve the irreversible problem. Somebody please help.
     
  2. jcsd
  3. Nov 13, 2004 #2

    Clausius2

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    Don't worry. Such a fast expansion enhances a conservation of the Temperature, because the particles haven't got enough time to propagate kinetic energy and vary it. So that, T remains constant.

    In addition, the process is adiabatic (neglecting the viscosity and fluid dissipation effects) because the characterstic time of heat transfer to the surroundings is usually much larger than the characteristic time of vessel filling.

    So that, T=0, and [tex] \delta Q=0[/tex] the process is irreversible. From the "TdS" equations of ideal gas, and dropping the rest of the terms:

    [tex] dS=-nR\frac{dV}{V}[/tex]

    or to point that from another way:

    [tex] \delta Q=-\delta W=-PdV=-nRT\frac{dV}{V}[/tex]

    and [tex]dS=\frac{\delta Q}{T}[/tex]

    as it can be derived from the 1st principle and 2nd principle.
     
  4. Nov 14, 2004 #3
    What is the dV? is it final volume - initial volume? or the other way round?
    From the equation [tex] dS=-nR\frac{dV}{V}[/tex]
    is it [tex] dS=-(2)(8.314)\frac{V_i-V_f}{V_i}[/tex]?
    Thanks.
     
    Last edited: Nov 14, 2004
  5. Nov 15, 2004 #4

    Clausius2

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    Sorry. It is written in differential form. dV is the differential of volume. You have to integrate those equations between the limits V_f and V_i. But the integration is left to you. The only hint I'm going to give you is:

    [tex] \int_{x_1}^{x_2}\frac{dx}{x}=ln\Big(\frac{x_2}{x_1}\Big)[/tex]
     
  6. Nov 15, 2004 #5
    Thank you very much. Now, I think I have understood it already.
     
  7. May 17, 2011 #6
    1. I don't think temperature remains constant in fast adiabatic expansion. In fact, temperature decreases.
    2. If you are still interested in a knowing better approach in calculating the change if entropy for a fast adiabatic process, let me know
    3. Sorry for the delay, I just discovered this forum today!
    4. Good luck
     
  8. May 17, 2011 #7

    Andrew Mason

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    The key here is to find the reversible path between the beginning and end states. The entropy is simply the integral of dQ/T over that path.

    Since no work is done, internal energy does not decrease (T is constant). So the reversible path is a quasi-static isothermal expansion.

    In a quasi-static isothermal expansion, work is done. Since it is isothermal, dU = 0 so:

    [tex]dQ_{rev} = dU + dW = 0 + PdV = nRTdV/V[/tex]


    AM
     
  9. May 17, 2011 #8
    1. Since it is a rapid process, it is adiabatic, which means dQ = 0 by definition. So it is not possible to calculate dS by dQ/T.
    2. Because of the First Law, dW = -dU = -nCvdT. (dQ = 0)This is an indication that in an adiabatic process, work is done at the expense of internal energy. Consequently, temperature reduces with the reduction in internal energy. There is an easy experiment we can do to better illustrate this, and most students and kids know about: Take an empty plastic bottle of water and squeeze it like you want to cut it in two pieces. After you squeezed it enough it will take little effort to remove the cap that usually comes out like a bullet, and this is a rapid, adiabatic expansion. When you look at the bottle you notice condensation, which is an indication that the gas cooled down during expansion and condensed.
    2. Since dQ = 0, you need to change variables:
    dU = nCvdT; and du = dQ - dW
    dS = dQ/T
    So dU = TdS - pdV
    Therefore, TdS - pdV = nCvdT; p = nRt/V
    Finally dS = R(dV/V) + nCv(dT/T), which you can integrate to find what you are looking for...
    I hope this helps
    Have a great day!
    Burke
     
    Last edited: May 17, 2011
  10. May 17, 2011 #9

    Andrew Mason

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    This is not correct. Entropy is a state function: it does not depend on the path. You have to first find the reversible path between the beginning and end states. The reversible path, in this case, does require heat flow: dS = dQrev/T. The expression for change in entropy is:

    [tex]\Delta S = \int_A^B dS = \int_A^B dQ_{rev}/T[/tex]

    The rest of your post is incorrect because you are not using the reversible path to determine the change in entropy.

    A rapid escape of air from a bottle into the atmosphere is not a free expansion. It is a rapid expansion and may be close to adiabatic BUT it does work on the surroundings since there is external pressure. In a free expansion into a vacuum, which is the case here, there is no work done by the escaping gas so dQ = dU = dW = 0 and dT = 0.

    AM
     
  11. May 17, 2011 #10
    AM,
    There is not such a thing as heat flow in an adiabatic process. That would deny the very initial hypothesis and definition of an adiabatic process. When an air balloon explodes you feel a blast of cool air, not because of any heat interchange, for the process is so rapid that there is no time for any heat exchange to happen; but because of temperature reduces as a result of internal energy going down during the expansion. Since energy U = (3/2)kT, then T goes down as well. I know it sounds contradicting that temperature may change not because of a heat exchange; but that's just the way it is.
    Rapid adiabatic processes are highly irreversible, and the equation you are suggesting to use to calculate dS is an inequality in reality
    Over and out.
     
  12. May 17, 2011 #11

    Andrew Mason

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    This is correct. But the adiabatic free expansion of a gas (ie. into a vacuum) is not a reversible process. A quasi-static isothermal expansion of a gas ends up in the same state and is reversible. A quasi-static isothermal expansion requires heat. It is this path that determines the change in entropy - ie. the reversible path.

    This is also correct. But that is not the situation given in this problem. No work is done by the gas in a free expansion. In your scenario the temperature (internal energy) decreases because there is work being done. Since dQ = 0, dU = -PdV
    This is a rapid adiabatic FREE expansion - ie. into a vacuum. A balloon burst is NOT a FREE expansion. It does work on the surroundings.

    You are certainly welcome to take whatever view you wish and argue in support of it. All we ask is that you not confuse others with incorrect information.

    AM
     
  13. May 18, 2011 #12
    AM,
    1. Since I have been using the "Post Quickly Reply" service, my statements did not come up complete. I am attaching a document with an approach to the solution that does not violate the hypothesis of the problem.
    2. If this is about being judgmental, I will tell you then that
    a. You can consider yourself lucky that I am not grading your work. If in a problem solving involving an adiabatic expansion you start by calculating a flow of heat, then I do not need to read anything else. The use of calculus may look elegant and impressive, but it is pointless, and the result will reflect a reality other than the one depicted in the problem.
    b. If you are not ready to accept criticism, then go for a second opinion or just do your homework before using a public network to make a statement that may affect your reputation
    c. I will close by saying what I tell my students: If you want to do physics, humble yourself up, accept and learn from your mistakes, and move on
     

    Attached Files:

  14. May 18, 2011 #13

    Andrew Mason

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    Your solution is correct except that you do not wish to recognize that ln(T2/T1) = 0 for a free expansion of an ideal gas (ie. T2 = T1). If you disagree, perhaps you could explain in terms of the first law how dU can be anything but 0 if dQ=0 and dW=0. If you do not agree that dW = 0 in a free expansion into a vacuum, perhaps you could explain how work is done and what it is done on.

    You also seem not to recognize that dS = dQrev/T. I am NOT saying that there is heat flow in an adiabatic free expansion of a gas. I AM SAYING that in order to determine the change in entropy you must calculate the integral of dQ/T over the reversible path between the beginning and end states. The reversible path DOES involve heat flow.
    I suggest you follow your own advice. I would add: one should carefully READ what the person is saying before one grades it. I have made it clear that I am NOT saying that an adiabatic expansion involves heat flow. I am saying that the reversible process between the beginning and end states of an ideal gas undergoing adiabatic free expansion DOES involve heat flow. There is a big difference. You don't seem to acknowledge that difference. Nor do you seem to understand why the difference is material to the calculation of entropy.

    AM
     
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