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Irreversible vs Reversible process

  1. Dec 22, 2014 #1
    I know that to calculate the entropy change in a process, I just need to calculate the entropy change in a process that has the initial and final states of the process and is reversible. I just don't understand what the actual difference between the irreversible versus the reversible process is!
    To clarify:
    I'm thinking about an isothermal reversible versus an isothermal irreversible process with the same initial and final states. Aren't both of them basically the same if I draw them on a PV diagram because then I'd just use the equation PV=constant? In that case, why would the entropy change be any different between the two?

    Thanks.
     
  2. jcsd
  3. Dec 22, 2014 #2

    ShayanJ

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    As you may know, we need three variables to specify the state of the system. We may choose P,V and T. Now if you consider an isothermal process, you're limiting things to a constant T plane in the PVT space. Now when you write PV=constant, you're specifying a particular path between two points in that plane and so there is no freedom here. I suspect this process is either reversible or not, and you can't have "an isothermal reversible versus an isothermal irreversible process"! For such a comparison, you should just specify the end points and find a reversible and an irreversible path between them and compare the two different paths.
     
  4. Dec 22, 2014 #3
    Ok, that makes sense, but I think PV = constant because of PV=nRT in ideal gases?
     
  5. Dec 23, 2014 #4
    In what we call an "isothermal irreversible process," you can control the temperature to be constant at the boundary of the deforming gas (e.g., the temperature of the cylinder surface), but you have no control over the temperature variations inside the gas. And the temperature inside the gas can vary with both time and spatial position. So what we are calling isothermal is not really isothermal throughout the gas and the process. Only at the beginning and at the end is the gas temperature uniform and at the same value that you started with. If you have such an isothermal irreversible process, you can not identify an isothermal reversible path that takes you between the same initial and final equilibrium states with the same amount of heat flow and work. If you could, then the entropy change would be the same for the irreversible path as for the reversible path. If you can think of an irreversible isothermal process for which you think this is possible, please describe it in detail.

    Chet
     
  6. Dec 24, 2014 #5

    Andrew Mason

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    Just to add to what Chester has said, for the reversible process the system is in equilibrium at all times with the surroundings during the process. This is not the case with the irreversible process.

    To calculate the entropy change in any process you must do two calculations:

    1. calculate [itex]\int dQ/T[/itex] over a reversible path between beginning and end states for the system and

    2. calculate [itex]\int dQ/T[/itex] over a reversible path between beginning and end states for the surroundings.

    The paths in 1. and 2. will be the same if the actual process that occurred was reversible. But they will not be the same if the actual process was irreversible.

    Let's take an irreversible expansion of a gas where the gas is put in thermal contact with a reservoir at 600 K and, due to expansion, the beginning and end states of the gas is 300 K.

    The actual heat flow relative to the gas is [itex]+\Delta Q[/itex]. The total heat flow relative to the reservoir is the same magnitude but negative ie. [itex]-\Delta Q[/itex].

    For the system, we are not concerned about the actual heat flow when determining entropy change. We are concerned only about the heat flow along a reversible path between the initial and final states.
    The reversible heat flow will always be greater than the actual heat flow. For example, in a free expansion of an ideal gas where the gas does not experience any change in internal energy and does no work, there is no actual heat flow into the gas. But a reversible path is a slow process in which work is done by the gas. So to end up at the same temperature, heat flow into the gas is required. That is the heat flow one uses to determine the entropy change of the gas: [itex]\Delta S_{gas} = \int dQ_{rev}/300[/itex]

    The change in entropy for the system is:

    [itex]\Delta Q_{rev}/300[/itex] over an isothermal reversible path (ie. it is connected to a heat reservoir at an infinitesimally higher temperature 300K + dT and expands very slowly)

    The change of entropy of the surroundings is:

    [itex]-\Delta Q_{act}/600[/itex] over an isothermal reversible path (ie. it is connected to a heat reservoir at an infinitesimally lower temperature 600K - dT and the heat flows out of the reservoir until it reaches [itex]\Delta Q_{act}[/itex].

    Total change in entropy is [itex]\Delta S_{sys} + \Delta S_{surr} = \Delta Q_{rev}/300 - \Delta Q_{act}/600 > 0[/itex]

    AM
     
    Last edited: Dec 25, 2014
  7. Dec 30, 2014 #6
    I have some more to say in this thread, and then I'm going to introduce a specific problem to illustrate in detail the difference between what is obtained in an irreversible "isothermal" process and a reversible isothermal process between the same initial and final equilibrium states. I'm hoping that the OP (Starter) or at least someone will be interested in working this problem with me. I like to use modeling to help solidify my understanding of complicated and confusing scientific concepts.

    First of all, regarding irreversible processes. In an irreversible process, the temperature within the system will typically be non-uniform, and will be varying with spatial location as a result of transient heat conduction. Often, the same can be said for the temperature variations within the surroundings. However, at the interface between the system and the surroundings, the temperature variation must be continuous, and the temperature of the system will match the temperature of the surroundings. There will not be two different temperatures at the interface, one for the system and the other for the surroundings. Even if the surroundings is an ideal heat reservoir so that its temperature is uniform, the temperature of the system at the interface must then match the temperature of the reservoir at all times during the irreversible process. If the process is reversible, the temperature of the system will match that of the reservoir not only at the interface, but throughout the system; the temperature within the system will be uniform.

    Another point I'd like to make is that, in an irreversible "isothermal" process, the change in internal energy for the system must be the same as for a reversible path between the same initial and final equilibrium states, namely zero. However the amount of work and the amount of heat flow for the irreversible process will both be different from that for reversible process. The only requirement is that the difference between work and the heat flow must be equal to zero for the system. If we calculate the integral of the heat flow divided by the interface temperature for the irreversible path, it will not be the same as for the reversible path. According to the Clausius inequality, the integral for the irreversible path will be less than that for the reversible path (where the latter is equal to the change in entropy of the system). This is why an irreversible path cannot be used to determine the change in entropy.

    The final point I'd like to make is that, when trying to get an understanding of the comparison between an irreversible process and a reversible process between the same initial and final equilibrium states, it is not imperative the we include consideration of the changes that occur in the surroundings. The total focus can be exclusively on the system. This can be done by demonstrating that the change in entropy of the system (i.e., the integral of dQ/T for the reversible path) is greater than the integral of dQ/TI for the irreversible path, where TI is the temperature at the interface with the surrounding along the path. This will be illustrated in the problem I am about to introduce.

    Problem Statement:

    Consider an ideal gas contained within a cylinder that is oriented vertically. There is a frictionless piston of mass M sitting on top of the gas, and a pile of small masses of total mass m sitting on top of the piston. The area of the cylinder and piston are A. The walls of the cylinder and the piston face in contact with the gas is always maintained constant at temperature T throughout our process. There is a perfect vacuum above the piston and masses. The system is initially at equilibrium. For our irreversible path, at time zero, all the small masses m are removed from the top of the piston instantaneously and simultaneously, and the gas is allowed to expand rapidly. Eventually, the piston will stop oscillating, and the final equilibrium state will be established. For the reversible path, the small masses are removed very gradually until they have all been removed, and the same final equilibrium state is established.

    We are going to solve for the work and heat for both the irreversible path and for the reversible path, and we are going to calculate Q/T for both paths (and compare them).

    Is there anyone out there in interested working this problem with me?

    Chet
     
    Last edited: Dec 30, 2014
  8. Dec 30, 2014 #7

    ShayanJ

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    I tried to calculate the work of the reversible process. We know that the pressure is [itex] P=\frac{M+m}{A}g \Rightarrow dP=\frac{g}{A}dm[/itex] and from ideal gas law, we have [itex] V=\frac{NkT}{P} \Rightarrow dV=-\frac{NkT}{P^2}dP [/itex]. So we have:
    [itex]
    W=-\int P \ dV=\int P \ \frac{NkT}{P^2}dP=\int \frac{NkT}{P}dP=\\ \int_{M+m}^M NkT \frac{A}{(M+m')g} \frac{g}{A} dm'=NkT \ln{(\frac{M}{M+m})}
    [/itex]
    So [itex] Q=-NkT \ln{(\frac{M}{M+m})} \Rightarrow \Delta S=-Nk \ln{(\frac{M}{M+m})} [/itex].
    But entropy is a state variable and so it doesn't matter what path the system takes(reversible or irreversible), the change in entropy is the same as the one I derived.
    There is also another point. When you say out of the cylinder is only vacuum, then the environment for the gas becomes only the cylinder and piston and masses on the piston. Then the work that the gas does, should warm up the piston and cylinder and masses through friction and so the process won't be isothermal! So we should introduce a heat bath to keep the temperature constant.
     
    Last edited: Dec 30, 2014
  9. Dec 31, 2014 #8
    Very nicely done!!!

    Regarding the need for a heat bath, I noted that the temperature of the cylinder and piston are maintained at temperature T throughout the process, but didn't specify how this would be done. One way of doing this is with an ideal heat bath. Also, since the gas is expanding, it is doing work on the piston, and this would tend to cool the gas, so heat would have to be added (as your equation shows). Rewriting your equations for the heat Q and for the entropy change ΔS,
    [tex]Q=NkT\ln{(\frac{M+m}{M})}>0[/tex]
    [tex]\Delta S=Nk\ln{(\frac{M+m}{M})}>0[/tex]

    Are you interested in next proceeding to an analysis of the irreversible path?

    Chet
     
  10. Dec 31, 2014 #9

    ShayanJ

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    Yeah, Of course. But I think there is a peculiarity here. As I said in post #2, the two conditions [itex] PV=NkT [/itex] and [itex] T=constant [/itex] uniquely determine the path. So I think we don't have an irreversible isothermal path here. So we should choose one of the many(maybe infinite) irreversible paths between the two points. But what should be our criteria?
    Also, as I said in my last post, entropy is an state variable, so whatever path we choose and calculate the change in entropy using it, we'll get nothing new.
    But actually I'm really interested to see how are we going to choose the path and calculate things when we consider an irreversible process. I can't help here, that's why I contributed in the reversible case so that you do the irreversible case! :D
     
  11. Dec 31, 2014 #10
    I love your enthusiasm. This is great.

    Before we begin, I first want to articulate what's happening physically in the irreversible case. The gas is expanding very rapidly, doing work against the piston. As a result, there is a tendency for the gas to cool. But this cooling will not take place uniformly throughout the gas because the temperature at the boundary will always be maintained at temperature T. So the boundary temperature will be hotter than the part of the deforming gas further from the boundary, and heat conduction will be occurring from the outside to the inside. This heat conduction will be driven by the temperature gradient that develops in the gas. In addition to this, the gas will be deforming very rapidly, and viscous stresses will develop within the gas. These viscous stresses will be accompanied by viscous dissipation of mechanical energy. (In the reversible process, the deformation is very slow, and the viscous stresses are negligible). The force of the gas on the piston face F(t) will be the result of a combination of the viscous stresses in the gas at that location with the local pressure (The pressure varies with spatial position within the gas). So it is not just gas pressure acting at the piston face; part of the force is the result of viscous stresses. Initially, the piston will oscillate back and forth within the cylinder, but gradually, the magnitude of the oscillation will decrease with time. This damping of the oscillation will not be the result of friction between the piston and the cylinder, since we already said that the piston is frictionless. The damping will be the result of the viscous dissipation of mechanical energy (associated with the viscous stresses) within the gas. In the end, at very long times, the piston will stop oscillating, and the final equilibrium state of the system will be attained.

    In analyzing the reversible case, you already showed that the force of the gas acting on the piston face in the initial equilibrium state is given by:
    [tex]F(0)=(M+m)g[/tex]
    and the initial equilibrium gas pressure is uniform, and given by:
    [tex]P(0)=\frac{(M+m)g}{A}[/tex]
    Similarly, the force and pressure of the gas acting on the piston face in the final equilibrium state is given by:
    [tex]F(∞)=Mg[/tex]
    [tex]P(∞)=\frac{Mg}{A}[/tex]
    These results, of course, apply to the irreversible case as well.

    OK. So now, let's get started. Let F(t) represent the force of the gas acting on the piston at time t in the reversible case, and let z(t) represent the vertical displacement of the piston relative to its initial equilibrium position. Please write down the Newton's second law force balance on the piston at arbitrary time t. (Believe it or not, this is an important first step in analyzing the irreversible case).

    Chet
     
  12. Dec 31, 2014 #11

    ShayanJ

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    At first, we were considering an ideal gas. There is no interaction between the particles of an ideal gas(they only collide elastically) and so there should be no viscousity. I'd rather consider the friction between the cylinder and the piston instead.(It seems to me that it will have the same effect you want. The gas does work on the piston that converts to heat through friction and is again given to the gas through conduction.)
    Another point is, I think the force that the gas exerts on the piston is not simply [itex] [M+m'(t)]g [/itex] because this will be simply cancelled by the gravitational force on the piston. But I don't know what is it? Does the fact that the pressure isn't constant through out the surface of the piston affect this? Maybe its actually true that this is cancelled by gravitational force because this happens in the reversible case and causes everything to be slow. So maybe in the irreversible case, this is cancelled but there is an additional term in the force which causes things to go on fast.(This seems plausible to me.)
    So we may write [itex] F(t)=[M+m'(t)]g+f(z,\dot z,t) \ [/itex], then the first part will be cancelled by the gravitational field and only [itex] f(z,\dot z,t) [/itex] remains. But what is this force? The only thing that comes into my mind, is the friction between the cylinder and the piston(which we may write as [itex] f(\dot z)=-\gamma \dot z [/itex].) This way we have:
    [itex]
    [M+m'(t)]\ddot z=-\gamma \dot z
    [/itex]
    But this means we should know how we're decreasing the masses on the piston. Different ways may give different answers(this is itself a problem) I tried [itex] m'(t)=m(1-\frac t T) [/itex](which seems to be the simplest thing we can use!) and used wolframalpha to solve the resulting differential equation but the solution was not in terms of elementary functions.
    But if we say that the part of force due to gas's pressure isn't cancelled by the gravitational force on the piston, then we should find an expression for the force due to gas's pressure which is different from [itex] [M+m'(t)]g [/itex]. But how?
    Also whatever we choose, will just make the DE more complicated but this DE is already too complicated to give us a solution we can use.
    EDIT: I tried constant friction too, but again no solution in terms of elementary functions.
     
    Last edited: Dec 31, 2014
  13. Dec 31, 2014 #12
    Actually, even though the molecules of an ideal gas collide elastically, ideal gases nonetheless exhibit viscosity. For an explanation of the molecular mechanism involved, see Transport Phenomena by Bird, Stewart, and Lightfoot, section 1.4 Molecular Theory of the Viscosity of Gases at Low Density.

    Actually, it's much simpler than this. In the problem statement, I said that in the irreversible case at time zero, the masses m are all removed at once. So, the force balance on the piston during the irreversible expansion is just given by:
    [tex]F(t) = Mg+M\frac{d^2z}{dt^2}[/tex]
    If we're able to agree on this, I'll continue with the next step.

    Chet
     
  14. Dec 31, 2014 #13
    No! The equation PV=nRT for ideal gas true if the variables in it (P, V & T) are all correspond to (thermodynamic) equilibrium. If you bring your gas having n moles irreversibly in a process the value of P at a time when Volume is V and temperature is T, can not be obtained by gas equation.


    Different! Entropy change is not a state function. Change in Entropy depends on the path(not just initial and final States. And the change in internal energy between any two particular States can be different for different process and is minimum for reversible processes.
     
    Last edited: Dec 31, 2014
  15. Dec 31, 2014 #14

    ShayanJ

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    Yeah...I forgot that we're removing them at once!!!
    But...why the force is proportional to acceleration? What is it? Also if you keep that, we'll be in trouble because it cancels the Ma term!!!
     
  16. Dec 31, 2014 #15
    The force is not proportional to the acceleration. The equation I wrote was just Newton's second law of motion for the piston.
    No. Please be patient. We're almost there.

    With the information we have, we are not going to be able to determine the detailed variation of the force F with time for our irreversible process. But, for our present purposes, we don't need to know that. All we really need to determine is the amount of work done by the gas on the piston, and this is the integral of Fdz.

    Let W(t) represent the cumulative amount of work done by the gas on the piston up to time t. Now take the force balance equation that we wrote, multiply both sides by dz/dt, and integrate from 0 to t. For the left side of the equation, I get:

    [tex]W(t)=\int_0^t{F(ξ)\frac{dz(ξ)}{dξ}dξ}=\int_0^{z(t)}{Fdz}[/tex]

    What do you get for the right-hand side of the equation?

    Chet
     
  17. Dec 31, 2014 #16

    ShayanJ

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    Oh my god. I thought you're giving an explicit form for F(t). So you just keep it unknown and wanna use it to find the work done on the piston.
    OK. For the right hand side, we have:
    [itex]
    M(g\int_0^t \frac{d z(\xi)}{d \xi} d\xi+\int_0^t \frac{d z(\xi)}{d \xi} \frac{d^2 z(\xi)}{d \xi ^2} d\xi)=M(g\int_0^{z(t)} d z(\xi)+\int_0^t \frac{d z(\xi)}{d \xi} \frac{d^2 z(\xi)}{d \xi ^2} d\xi)=\\ M(g \ z(t)+\frac 1 2 (\frac{d z(\xi)}{d\xi})^2 |_0^t )
    [/itex]
    I don't understand how its going to work out. We neither know how z(t) changes with time, nor how F(t) changes with time(or even with z!). We have one equation and two unknown functions!!!
     
  18. Dec 31, 2014 #17
    Great question. So far we have:
    [tex]W(t)=Mgz(t) + \frac{1}{2}Mv(t)^2[/tex]
    where v(t) is the velocity of the piston. The first term on the right is the potential energy of the piston at time t, and the second term is its kinetic energy. Now, if the piston oscillation is gradually damped out by viscous stresses, what is the velocity of the piston at infinite time?

    Chet
     
  19. Dec 31, 2014 #18

    ShayanJ

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    Ohhh...I couldn't see [itex] \frac{d z(\xi)}{d\xi} [/itex] is actually the velocity!!!
    Well, as [itex] t\to \infty [/itex], velocity goes to zero and so we'll have [itex] W=Mgz(t\to\infty) [/itex]. But what is [itex]z(t\to\infty)[/itex]? I know it. Its the same as the change in height in the reversible case. So we can calculate:
    [itex] \Delta V=\int dV=-\int \frac{NkT}{P^2}dP=-NkT \int_{m}^{0}\frac{A^2}{(M+m')^2g^2}\frac{g}{A} dm'=\frac{NkTA}{g}\int_0^{m} \frac{dm'}{(M+m')^2}=- \frac{NkTA}{g} \frac{1}{M+m'}|_0^{m}=\\ \frac{NkTA}{g} (\frac{1}{M}-\frac{1}{M+m})[/itex].
    So [itex] z(t\to\infty)=\frac{\Delta V}{A}=\frac{NkT}{g} (\frac{1}{M}-\frac{1}{M+m})[/itex] which gives:
    [itex]
    W=MNkT(\frac{1}{M}-\frac{1}{M+m})=\frac{mNkT}{M+m}
    [/itex].
    It was a nice trip. thanks man!
    But now I wanna see how you want to calculate the entropy. We need to know the amount of heat exchange but now we have the amount of work.
     
  20. Dec 31, 2014 #19
    Nice job. Rewriting your equation for the work a little differently, we have:
    [tex]W=NkT\frac{m}{M+m}[/tex]
    This is the work for the irreversible process. It is also the heat for the irreversible process. So,
    [tex]Q=NkT\frac{m}{M+m}[/tex]

    You already calculated the entropy change when you considered the reversible path. Since the entropy is a function of state, that is also the entropy change for the irreversible path. But now lets consider how your results stack up against the Clausius inequality that says that, for an irreversible process:
    [tex]ΔS>\frac{Q}{T_I}[/tex]
    In our problem, we are calling TI the interface temperature T. So, for the irreversible path,
    [tex]\frac{Q}{T_I}=Nk\frac{m}{M+m}[/tex]
    You found earlier that
    [tex]ΔS=Nk\ln{\frac{M+m}{M}}[/tex]
    So, if your results are to be consistent with the Clausius inequality (or visa versa), we must have that:
    [tex]\ln{\frac{M+m}{M}}>\frac{m}{M+m}[/tex]
    Is it?

    Chet

    P.S., Happy New Year
     
  21. Dec 31, 2014 #20

    ShayanJ

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    Wait! In the reversible case, we used the convention that when energy goes into the ideal gas, its sign is positive. But for the irreversible case, we focused on the piston and so our convention was changed. Now if we again go back to the first convention, we should have [itex] Q=-W=NkT\frac{m}{M+m} [/itex]. The other reason for this is that the internal energy didn't change in the reversible case and so shouldn't change here too.
    And about the inequality. We can write both sides in terms of [itex] x=\frac m M [/itex] so we have [itex] \ln(1+x) [/itex] and [itex] \frac{x}{1+x} [/itex]. Both are zero at x=0. But the derivative of the first is larger than the derivative of the second for x>0 and so the first one is larger than the second one for all x>0.
    Nice discussion. You make good as a teacher Chet!;)
    Happy New Year
     
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