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Adiabatic irreversible expansion

  1. Feb 15, 2015 #1
    This example is taken from the wikipedia page describing irreversible processes.

    I just want to make sure I understand correctly why the initial state can't be reached anymore. I assume the transitions to be quasi-static, but there is friction between the piston and the cylinder. If so, during the transition from state 1 to state 2, when the piston rises, some of the work is transformed into heat, which increases the internal energy of the gas. During the compression, again, some of the work is turned into heat, so the pressure from state 1 is now achieved at a higher volume (because some work was converted into internal energy). So the entropy of the system (or universe) has increased.
    Is this correct?
    Last edited: Feb 15, 2015
  2. jcsd
  3. Feb 15, 2015 #2
    It would help if you could provide a link to the Wiki page you are referring to.

  4. Feb 15, 2015 #3
  5. Feb 15, 2015 #4
    No. The description implicitly assumes that the piston is frictionless. Also, if the deformation were quasi-static, then the process would be reversible, and the system and surroundings could actually be returned to their initial states. What is happening here is that the mass m1 is being replaced by a much smaller mass m2, and the gas is then being allowed to expand rapidly and irreversibly. The irreversibility comes in because the expansion will be very rapid, and the gas will experience viscous dissipation of mechanical energy to heat. This is essentially the same thing that happens in a spring-mass-damper system, with the damper being analogous to the viscous dissipation in the gas. If the spring-mass-damper system is made to deform quasi-statically, then the damper will not dissipate significant mechanical energy (because its effect is proportional to the rate of deformation), and the deformation will be reversible. However, if the system is allowed to deform rapidly, then the damper will dissipate significant mechanical energy to heat.

    When the smaller mass m2 is replaced by the larger mass m1 and the gas is allowed to compress rapidly and irreversibly, the same thing happens again. Here again viscous dissipation comes in.

  6. Feb 15, 2015 #5
    Thank you for your answer. But the overall effect of letting the gas "expand freely" (finite difference between the outside and the inside pressure) or assuming that the piston is not frictionless and the process is done quasi-statically would be the same: adding heat to the gas when it could be avoided, correct? So when the gas is back to the pressure from state 1, its volume (and implicitly temperature) would be higher than in the initial state.
  7. Feb 15, 2015 #6
    No. Please forget about friction. What they are trying to explain happens without friction. You can still expand the gas quasistatically even without friction, by applying an additional external force to allow the piston to rise gradually. Here are the two cases:

    1. Additional external force applied to piston so that gas expands gradually (rather than rapidly). Gas expansion takes place adiabatically and reversibly. Negligible viscous heating occurs in gas.

    2. No supplemental external force applied to piston so that gas expands rapidly. Gas expansion takes place adiabatically, but irreversibly. Significant viscous heating occurs in gas. Final pressure is the same as in case 1.

    In both cases, the gas cools when it expands, but the final temperature in case 1 (the reversible case) is lower than in case 2 (the irreversible case).

    Now for compression:

    1. Compression force on piston increased gradually (rather than rapidly). Gas compression takes place adiabatically and reversibly. Negligible viscous heating occurs in gas.

    2. Compression force on piston applied suddenly. Gas compression takes place adiablatically, but irreversibly. Significant viscous heating occurs in gas. Final pressure is the same as in case 1 (and the same as pressure before expansion).

    In both cases, the gas heats when it compresses, but the final temperature in case 1 (the reversible case) is lower than in case 2 (the irreversible case). The final temperature in case 1 is the same as before the original expansion. The final temperature in case 2 is higher.

  8. Feb 15, 2015 #7
    Hi Razvan. If you're interested, we can solve the problem quantitatively both ways (reversible and irreversible), and compare the results. Any interest?

  9. Feb 16, 2015 #8
    The only reason I decided to assume that the piston is not frictionless is because I know exactly how much of the work done turns into heat that way.
    So, the differential equation that models the process is
    $$dU = -PdV + \frac{F}{A} dV$$
    $$\frac{C_v}{R}(PdV+VdP) = -PdV + \frac{F}{A} dV$$
    Which results in (if I didn't make any mistakes)
    $$(P-\frac{FR}{AC_p})V^{\gamma} - constant$$

    If, instead, we let the gas expand "freely", by replacing the mass ##m1## with ##m2##, instead of the term ##\frac{F}{A} dV## there would be something different, but the overall effect would be the same. Is this so wrong to assume?
    Thank you.
  10. Feb 16, 2015 #9
    There's a fundamental difference. In the frictional case you have articulated, the irreversibility is directly related to the behavior of the piston, not the gas. In the rapid irreversible expansion, the irreversibility is directly related to the gas, not the piston. The latter is what they had in mind in the Wiki article. Aren't you curious about how the irreversibility of the gas expansion plays out.

  11. Feb 16, 2015 #10
    I would like to see how it could be modeled, but I am afraid I do not have enough knowledge in this area. Despite that, if you are willing to explain it, I would be more than happy to follow it.
    All that I can think about that situation is that the parameters of the gas would no longer be defined. So different parts of the gas will have different pressure and temperature, which I think would be responsible for an increase in the kinetic energy of some molecules, but that (organised) energy finally will transform into heat.
  12. Feb 16, 2015 #11

    Yes. That's correct, but we can still do the problem for the irreversible case (not exactly the same way), and then see how the results compare. I'll get back with you a little later and define the problem more precisely so that we are both totally comfortable with the problem statement.

  13. Feb 16, 2015 #12
    Problem Statement

    We have a gas in a vertical cylinder of cross sectional area A. There is a massless, frictionless piston sitting on top of the gas, and a large mass m1 sitting on top of the piston. The cylinder is contained in a vacuum chamber so that the absolute pressure in the room surrounding the cylinder is zero (this makes solving the problem a little easier). The cylinder/piston combination is perfectly insulated, so that any expansion or contraction of the gas occurs adiabatically. The system is initially at thermodynamic equilibrium at temperature T0 and volume V0. The initial pressure of the gas in the cylinder is m1g/A.

    We are going to consider the reversible expansion of the gas and the irreversible expansion of the gas, both to a new pressure m2g/A.

    Reversible Expansion: In the reversible expansion, we replace the mass m1 with a new mass m2, and start out by holding the piston down manually with an additional downward force m1g - m2g. We then very gradually ease off on the additional downward force until it is zero.

    Irreversible Expansion: In the irreverislbe expansion, we again replace the mass m1 with a new mass m2, and start out by holding the piston down manually with an additional downward force m1g - m2g. However, in this case, rather than very gradually easing off on the additional downward force, we remove it suddenly (i.e., drop it to zero) and let nature take its course until the system finally equilibrates on its own.

    Our goal is to determine the final temperature and volume of the gas in each of these two cases.

    Is this problem statement satisfactory to you?

  14. Feb 16, 2015 #13
  15. Feb 16, 2015 #14
    An adiabatic reversible expansion respects $$PV^{\gamma} - const.$$
    So the final volume is
    $$V_f = (\frac{m_1}{m_2})^\frac{1}{\gamma}V_0.$$
  16. Feb 16, 2015 #15
    And the final temperature?

  17. Feb 16, 2015 #16
    I will write the steps, so other readers can benefit from this as well.
    Because ##PV^{\gamma}## is constant, from the ideal gas law (##PV=nRT##), we get that
    ##TV^{\gamma-1}## is constant.
    Finally, we have
    $$T_0V_0^{\gamma-1} = T_fV_f^{\gamma-1}$$
    $$T_f = T_0 (\frac{V_0}{V_f})^{\gamma-1} = T_0 (\frac{m_2}{m_1})^{\frac{\gamma-1}{\gamma}}$$
  18. Feb 16, 2015 #17
    Very nice. I'll have to get back with you later to get us started on the irreversible. I'm out with my grandchildren right now, and I need to be able to type on my computer. This is from my IPhone.

  19. Feb 16, 2015 #18
    Ok. Please do not let me take up your time. I wish you a wonderful day.
  20. Feb 16, 2015 #19
    Irreversible Case

    The key to doing the irreversible case is to figure out how to get the work done by the gas on the surroundings (i.e., the piston and the mass m2).

    We're going to assume that the mass m2 is glued to the piston, so it can't separate from the piston if oscillation is involved.

    As you mentioned, in the irreversible case, the pressure of the gas and the temperature of the gas will not be uniform within the cylinder during the expansion, so we can't use the ideal gas law (except at the initial and final states). But we have to ask ourselves, "what really determines the amount of work that the gas does on the surroundings?" Well, the work done on the surroundings is determined soley by the force that the gas exerts at the base of the piston, FI, where the subscript I refers to the interface between the system (the gas) and the surroundings (the piston and mass). This interface is the only location where the gas does work on the surroundings. The work is equal to the integral of FIdx, where dx is the upward displacement of the piston. Thus,
    But, how can we get the work on the surroundings if we can't use the ideal gas law to get the force on the interface? Well, to get that for the irreversible case, we need to turn attention to what is happening on the other side of the interface (the piston and mass combo).

    Even though this is a course in thermodynamics, there is no reason why we are not allowed to use what we learned in freshman physics (mechanics) to help us. Suppose I write a Newton's second law force balance on the piston and mass. Then we have:
    Now, suppose we multiply this equation by the velocity v = dx/dt and integrate from time t = 0 when the piston is released to some arbitrary time t. We obtain:
    Do you see how I obtained this equation?

    Now, the left hand side of this equation integrates to:
    where KE(t) is the kinetic energy of the mass m2 at time t.

    The term ##\int_0^t{F\frac{dx}{dt}dt}## integrates to:
    where W(t) is the cumulative amount of work that the gas does on the piston/mass combo up to time t.

    Therefore, if we combine these results, we have that:
    Now, we expect that, in the irreversible case, the mass will start out by overshooting its final equilibrium position, and will oscillate about this location. However, as time progresses, as a result of viscous dissipation in the gas, the oscillation will gradually decay to zero. In the final equilibrium state at infinite time, the kinetic energy of the mass will be zero. Therefore, the work that the gas does on the surroundings between the initial and final equilibrium states will simply be given by:
    This is the result that we have been looking for.

    I'm going to stop here and give you a chance to digest all this.

  21. Feb 17, 2015 #20
    From $$m_2\frac{d^2x}{dt^2}=F_I-m_2g \vert \cdot \frac{dx}{dt}$$
    On the left side there is
    $$m_2va = m_2\frac{d}{dt}(\frac{1}{2}v^2)$$
    On the right side there is no major change.
    The work done on the piston and mass has decreased the internal energy of the gas.
    $$\frac{C_v}{R}(P_fV_f-P_0V_0) = - \frac{m_2g}{A}(V_f-V_0)$$
    which results in
    $$V_f = V_0\frac{m_2+\frac{C_v}{R}m_1}{m_2+\frac{C_v}{R}m_2} = V_0(\frac{2}{5}+\frac{3m_1}{5m_2})$$
    Is this correct?
    Thank you very much for your help.
    Last edited: Feb 17, 2015
  22. Feb 17, 2015 #21
    Nice work!!!

    Yes, it's correct for γ=5/3. But, how about expressing it in terms of γ? And the final temperature in terms of γ is?

    Have you done any calculations to see how these results compare quantitatively with those for the reversible case? I think you'll find that, when the two masses differ from one another by only a little bit, the two sets of results will be very similar. But when m2 is much smaller than m1, there will be a very significant difference.


    PS, I wanted to point out the, for the irreversible case, the work is just equal to the change in potential energy of m2, but, for the reversible case, because you apply the supplemental force, the work is much larger than just the potential energy change.
  23. Feb 17, 2015 #22
    From ##C_p = C_v + R## we can express ##V_f## as
    $$V_f = V_0(1+\frac{1}{\gamma}\frac{m_1-m_2}{m_2})$$
    And from
    $$\frac{T_f}{T_0} = \frac{P_fV_f}{P_0V_0} = \frac{m_2}{m_1}(1+\frac{1}{\gamma}\frac{m_1-m_2}{m_2})$$
    $$T_f = T_0\frac{m_2}{m_1}(1+\frac{1}{\gamma}\frac{m_1-m_2}{m_2})$$
  24. Feb 17, 2015 #23
    Very nice. I really like your style, particularly the way you manipulated the results into their final forms.

    If you still have any energy left, you can compare the final results for the temperatures in the two cases pretty easily by expanding each of them in a power series in ε = (m1-m2)/m1. The linear terms in the two series will match exactly, but the quadratic terms will differ (in the way that you would expect).

  25. Feb 17, 2015 #24
    In the irreversible case, ##\frac{T_f}{T_0}## is linear in ##\frac{m_1-m_2}{m_1}##.
    $$\frac{T_f}{T_0} = 1+ (\frac{m_1-m_2}{m_1})\frac{1-\gamma}{\gamma}$$
    In the reversible case, if we use
    $$(1+x)^a = 1+\frac{a}{1!}x + \frac{a(a-1)}{2!}x^2 + \cdots$$
    we get
    $$\frac{T_f}{T_0} = (1+\frac{m_2-m_1}{m_1})^{\frac{\gamma-1}{\gamma}} = 1+ (\frac{m_1-m_2}{m_1})\frac{1-\gamma}{\gamma} - (\frac{m_1-m_2}{m_1})^2\frac{\gamma-1}{2\gamma^2} + \cdots$$
    I used minus for the last term to show that it is negative.
  26. Feb 17, 2015 #25
    Very nice indeed. So the conclusion is that, as expected, the amount of cooling when expanding with the reversible case is more than when expanding with the irreversible case. These same equations also apply to compression. The amount of compressional heating with the reversible case is less than with the irreversible case.

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