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Thermodynamics Interesting Entropy Problem

  1. Sep 25, 2014 #1
    1. The problem statement, all variables and given/known data
    A piston‐cylinder initially contains 0.5 m3 of an ideal gas at 150 kPa and 20 degrees. The gas is suddenly subjected to a constant external pressure of 400 kPa and it is compressed in a manner that the final temperature is also 20oC. Assume the surroundings are also at 20 degrees. Take Cp=2.5 Ro. Determine:
    (i) What is the entropychange of (a) the system,
    (b) the surroundings and
    (c) the universe or total?
    (ii) Is the process reversible,irreversible or impossible?

    2. Relevant equations

    3. The attempt at a solution
    For the system, because the temperature didn't change after compression.
    Hence U=0;
    W can be obtained by:
    W=PV = 400KPa(V2-V1);
    V2 can be obtained by using ideal gas equation;
    PV= nRT;

    However, I encounter difficulties in doing so because, if so, there is only one Q value, Q leaking out from the system = Q entering the surrounding. Hence the entropy will be zero instead, which is not right by the answer given.
    The solution is given as follow and I dont really understand how they calculate the entropy change for system and equate it with isothermal process. And I doubt why the two Q values are different.
  2. jcsd
  3. Sep 25, 2014 #2
    You are confused because the explanation they provided you is very poor.

    You calculated the work done by the system on the surroundings correctly, and from this, it follows that the heat transferred from the surroundings to the system Q is equal to this work. So, both W and Q are negative, and you correctly determined them for this irreversible process.

    But, the change in entropy for the system is not equal to Q/T for this irreversible process. To determine the change in entropy, you always need to consider a reversible process path between the initial and final states of the system. This differs from the actual irreversible process path that the system was subjected to. The explanation they provided you considered a reversible path between the initial and final states, and ended up getting the correct result for the entropy change of the system. But, for this reversible path, the Q is different from the real path.

    Now, for the surroundings. The analysis assumes that the deformation of the surroundings takes place reversibly, and that the heat transfer to the surroundings all takes place at 20 C (because the surroundings also have a huge reservoir at 20 C). So, the change in entropy for the surroundings (which experienced a reversible process) is -Q/293, where Q is the actual heat transferred in the real process (which is reversible for the surroundings, but not for the system). The sum of the entropy change for the surroundings plus the entropy change for the system should add up to a number greater than zero.

    I hope that this makes some kind of sense.

  4. Sep 25, 2014 #3
    Thanks for your reply!

    I am trying to grasp what you typed. However, I dont understand why entropy is determined by always considering a reversible process. And why, the actual process taken for the system is irreversible?

    Really thankful for your help and I am still a bit confused :(((
  5. Sep 25, 2014 #4
    I'm going to answer this question first, before discussing why the actual process taken for the system is irreversible.
    Below is the text from a blog that I had had on Physics Forums before the blogs were removed in early September. It explains how the concept of Entropy was discovered and evolved.


    Suppose that we have a closed system that at initial time ti is in an initial equilibrium state, with internal energy Ui, and at a later time tf, it is in a new equilibrium state with internal energy Uf. The transition from the initial equilibrium state to the final equilibrium state is brought about by imposing a time-dependent heat flow across the interface between the system and the surroundings, and a time-dependent rate of doing work at the interface between the system and the surroundings. Let [itex]\dot{q}(t)[/itex] represent the rate of heat addition across the interface between the system and the surroundings at time t, and let [itex]\dot{w}(t)[/itex] represent the rate at which the system does work on the surroundings at the interface at time t. According to the first law (basically conservation of energy),
    [tex]\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W[/tex]
    where Q is the total amount of heat added and W is the total amount of work done by the system on the surroundings at the interface.

    The time variation of [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] between the initial and final states uniquely characterizes the so-called process path. There are an infinite number of possible process paths that can take the system from the initial to the final equilibrium state. The only constraint is that Q-W must be the same for all of them.

    If a process path is irreversible, then the temperature and pressure within the system are inhomogeneous (i.e., non-uniform, varying with spatial position), and one cannot define a unique pressure or temperature for the system (except at the initial and the final equilibrium state). However, the pressure and temperature at the interface can be measured and controlled using the surroundings to impose the temperature and pressure boundary conditions that we desire. Thus, TI(t) and PI(t) can be used to impose the process path that we desire. Alternately, and even more fundamentally, we can directly control, by well established methods, the rate of heat flow and the rate of doing work at the interface [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex]).

    Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:
    where [itex]\dot{V}(t)[/itex] is the rate of change of system volume at time t. However, if the process path is reversible, the pressure P within the system is uniform, and

    [itex]P_I(t)=P(t)[/itex] (reversible process path)

    Therefore, [itex]\dot{w}(t)=P(t)\dot{V}(t)[/itex] (reversible process path)

    Another feature of reversible process paths is that they are carried out very slowly, so that [itex]\dot{q}(t)[/itex] and [itex]\dot{w}(t)[/itex] are both very close to zero over then entire process path. However, the amount of time between the initial equilibrium state and the final equilibrium state (tf-ti) becomes exceedingly large. In this way, Q-W remains constant and finite.


    In the previous section, we focused on the infinite number of process paths that are capable of taking a closed thermodynamic system from an initial equilibrium state to a final equilibrium state. Each of these process paths is uniquely determined by specifying the heat transfer rate [itex]\dot{q}(t)[/itex] and the rate of doing work [itex]\dot{w}(t)[/itex] as functions of time at the interface between the system and the surroundings. We noted that the cumulative amount of heat transfer and the cumulative amount of work done over an entire process path are given by the two integrals:
    In the present section, we will be introducing a third integral of this type (involving the heat transfer rate [itex]\dot{q}(t)[/itex]) to provide a basis for establishing a precise mathematical statement of the Second Law of Thermodynamics.

    The discovery of the Second Law came about in the 19th century, and involved contributions by many brilliant scientists. There have been many statements of the Second Law over the years, couched in complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800's.

    Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:
    where TI(t) is the temperature at the interface with the surroundings at time t. He carried out extensive calculations on many systems undergoing a variety of both reversible and irreversible paths and discovered something astonishing. He found that, for any closed system, the values calculated for the integral over all the possible reversible and irreversible paths (between the initial and final equilibrium states) was not arbitrary; instead, there was a unique upper bound (maximum) to the value of the integral. Clausius also found that this result was consistent with all the "word definitions" of the Second Law.

    Clearly, if there was an upper bound for this integral, this upper bound had to depend only on the two equilibrium states, and not on the path between them. It must therefore be regarded as a point function of state. Clausius named this point function Entropy.

    But how could the value of this point function be determined without evaluating the integral over every possible process path between the initial and final equilibrium states to find the maximum? Clausius made another discovery. He determined that, out of the infinite number of possible process paths, there existed a well-defined subset, each member of which gave the same maximum value for the integral. This subset consisted of what we call today the reversible process paths. So, to determine the change in entropy between two equilibrium states, one must first conceive of a reversible path between the states and then evaluate the integral. Any other process path will give a value for the integral lower than the entropy change.

    So, mathematically, we can now state the Second Law as follows:

    [tex]I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}\leq \Delta S=\int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}[/tex]
    where [itex]\dot{q}_{rev}(t)[/itex] is the heat transfer rate for any of the reversible paths between the initial and final equilibrium states, and T(t) is the system temperature at time t (which, for a reversible path, is equal to the temperature at the interface with the surroundings). This constitutes a precise mathematical statement of the Second Law of Thermodynamics.

    I hope that this description works for you, and gives you a better picture of why entropy change can only be calculated using a reversible path between the initial and final equilibrium states of a closed system.

    Last edited: Sep 30, 2014
  6. Sep 26, 2014 #5
    It really answer my concerns and I understand now:) So now the actual path taken for the system, is irreversible because? I am trying to think about it, pressure added, gas compressed, heat loss to the surroundings and hence remain the same temperature. To take a reversible path, one must ensure that the work is done by the gas to the surrounding, and heat is added to the system and the temperature of the system still remain the same. Which means that, Q is totally transfered to W, which violates the second law of thermodynamics. So it is an irreversible path because reversible path is not possible. Am i right? Thank you
    Last edited by a moderator: Sep 30, 2014
  7. Sep 26, 2014 #6
    To have a reversible path, the gas must be compressed gradually (quasistatically) so that there are negligible temperature or pressure variations present throughout the gas during the entire deformation, and the pressure throughout the gas comes very close to matching the imposed pressure at the interface with the surroundings at all times. If the pressure at the interface is suddenly raised from 150kPa to 400kPa, the compression will not be gradual, and there will be both temperature and pressure gradients arising within the gas during the compression (which will give rise to viscous dissipation of mechanical energy to heat, and conductive dissipation of temperature gradients). As a result, if you try to take both the system and the surroundings from the final state back to the original state, you will be unable to do so. In your system, even though Q matches W for the irreversible path (which certainly doesn't violate the 2nd law), there will be pressure and temperature non-uniformities present within the gas which will later prevent you from being able to identify any path from the final state to the initial state without causing a change in either the system or the surroundings.

  8. Sep 26, 2014 #7
    Really thank you:) It worth spending one whole day working out what is entropy! Thanks a lot!
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