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Don't reversible processes always do the most possible work?

  1. Jan 19, 2013 #1
    I am trying to teach myself the basics of thermodynamics using a book I bought from a university book store.

    My question relates to the amount of work done by reversible vs irreversible processes. An example in the book (Chemical Principles by Atkins & Jones) reads the following:

    "A cylinder of volume 2.00 L contains 0.100 mol He (g) at 30°C. Which process does more work on the system, compressing the gas isothermally to 1.00L with a constant external pressure of 5.00 atm or compressing it reversibly and isothermally to the same final volume?"

    I know that reversible processes should always do the most work because the system is doing work against the maximum opposing force under the given conditions, but when I actually do the calculations here, I find that the irreversible process does more work.

    To find the amount of work done by the irreversible process, I used P[itex]\Delta[/itex]V and convert the quantity to Joules using 101.325 J/(L*atm). I end up with 506.6J of work being done ON the system.

    To find the amount of work done by the reversible process, I used -nRT * ln(Vfinal/Vinitial). I end up with 174.6 J of work being done on the system.

    This is seemingly inconsistent with the fact that reversible processes always do more work than irreversible processes. Could someone please explain where my logic has gone askew?

    Thank you so much!
    Last edited: Jan 19, 2013
  2. jcsd
  3. Jan 19, 2013 #2
    I have not checked the arithmetic, but this is the correct formula for an isothermal reversible compression.
  4. Jan 19, 2013 #3
    Sorry. Typo on my part; I meant reversible. I will edit when I'm back on my computer.
  5. Jan 19, 2013 #4
    I calculated the work done by both processes and found that the irreversible process does more work than the reversible process. Doesn't the reversible process always do more work?
  6. Jan 19, 2013 #5
    OK if

    [tex]\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}[/tex]

    and T1 = T2

    Then when you change V1 to V2

    Does the pressure remain constant or change?

    That is does P1 = P2 ?
  7. Jan 19, 2013 #6
    The pressure will change, right?
  8. Jan 19, 2013 #7
    So you can't use w = pΔV since the pressure is not constant.

    You are given the initial conditions so can calculate P1.

    You are also given V2 so you can calculate P2
  9. Jan 19, 2013 #8
    Ah! So is the reason you can use w = pΔV for expansions that the external pressure will not change due to the immensity of the surroundings?
  10. Jan 19, 2013 #9
    Yes, the volume change of the surroundings = volume change of the system, but if the pressure of the surroundings is say the atmosphere it is sensibly constant. This is the situation for many chemical processes. In this case expansion work done = pressure of surroundings times vol change.
  11. Jan 19, 2013 #10
    Thank you so much! I am still, however, somewhat confused by why work equals the external pressure times volume change. If the pressure being exerted by the system is only equal to the external pressure, wouldn't there not be a volume change in the first place because the two forces would cancel each other out?
  12. Jan 19, 2013 #11
    Well I make P1 = nRT/V = {(0.1)*(0.082)*(303)} / 2 = 1.24 Atm.

    Thus since the gas is compressed to half its volume at P2, P2 = 2.48 Atm

    So the compression to 5 Atm is incomplete at this point.

    There must be more to this question.
  13. Jan 19, 2013 #12
    Actually that was it. The answer they have was that the reversible process does more work. I'm guessing they didn't actually want you to calculate anything, but instead just wanted me to remember that reversible processes always do the most work.

    My confusion stems from the more general question of why work is equal to the product of external pressure and volume change, as that intimates that the force the system is exerting is merely equal but opposite to the force the surroundings exert on the system. If this were the case, it would seem that there wouldn't be any volume chance.
  14. Jan 19, 2013 #13
    The answer to this question regarding the external pressure was addressed in a recent thread https://www.physicsforums.com/showthread.php?t=663999 no more than a week ago. Check out the thread, particularly the later postings by Andrew Mason and myself. Together, he and I nailed down the answer to this question (which has puzzled students over the decades).
  15. Jan 20, 2013 #14
    Simply because it isn't in all case, so you have to know what the process is doing.
    P[itex]\Delta[/itex] is the maximum possible work in or out of a system.

    As an example suppose you have two very large containers so that any small change in volume does not affect the pressure in either container. Both containers are connected by a cylinder and massless piston ( iinertial effects of the piston can be neglected ).
    The container on the left at 1 atm is your system and the one on the right at 5 atm is your surroundings. The piston is initially on the right of the cylinder and held by a pin. There is a stop on the left of the cylinder.
    If you remove the pin, the piston travels to the left and hits the stop.

    Now, do you use the 5 atm pressure , or the 1 atm pressure to calculate the work on your system at the left? Both cannot be correct.

    The correct answer is you use the 1 atm pressure.
    Using 5 atm you can calculate the external work, so where did the missing work go?

    It might be easier to grasp if we include friction.
    With an adequate friction force between the piston and cylinder to make the process quasistatic, you can see that
    W 5atm = Wf + W 1atm
    Wf is considered lost work, or work you can never recover.

    Just be sure to consider how your system is reacting before you follow any blanket rule.
  16. Jan 20, 2013 #15
    Sorry to revive what I thought was a settled question, but in light of my new confusion regarding pressureexternal vs pressureapplied, I wanted to point out another dilemma I am having.

    If work does indeed equal the product of applied pressure (which I take to be roughly analogous to the force applied by the surroundings) and volume compressed, I find that the pressure does indeed remain constant for the very reasons the pressure does not change in irreversible gaseous expansions.

    That is to say, if we take the piston to be frictionless and massless (for sake of simplification), then the work done by the surroundings = -work done on the system.

    So, the work done by the surroundings (from w=Fd) should be Pex*ΔV. The surrounding volume seemingly increases by 1L and the pressure applied is a constant 5atm. Therefore, the work done by the surroundings = (5atm) * (1L) * 101.325 J * atm-1 *L-1.

    The work done on the system would be the opposite of this. This value, 506.6 J, is greater than the value obtained when the system is compressed reversibly to the same final volume under the same conditions.

    Could you please clarify why this is? Thank you so much for your time and help.
  17. Jan 21, 2013 #16
    I think part of the problem is the very slick way in which the 'formula' for case (1) the reversible isothermal compression is derived in Atkins.

    I do not have the particular book, since he has written several (some quite good) but he has a good model in later editions of his Physical Chemistry.

    I should tell you that it is impossible to calculate the work for the second, irreversible, example from just the information given.

    To understand why we need to work up a model of exactly how you could apply 5 atm to a container of gas at 1.24 atm.

    Say we had a cylinder and a piston. If there was 1.24 atm on one side of the piston and 5 on the other there would be considerable force on that piston, which would suffer infinite acceleration if it were massless. So not only would we need a peg or stop to release it at a time t=0 but we would also need another stop to bring it to rest at the designated second volume.

    Calculating the work for reversible isothermal compression does not involve a PV product because P is not constant. P slowly increases from zero to (in this case2.48atm) and we must integrate along the path. There is never more than in infinitesimal pressure difference across the piston.

    Incidentally the first law tells us that since the process is isothermal all the work put into the gas must be extracted as heat - the internal energy does not change.

    The difference between the PΔV that you calculate goes into (is dissipated in) whatever damping and piston control you use.
  18. Jan 21, 2013 #17
    I would like to discuss a (hopefully) simplified version of the problem introduced by 256bits in response #14. Suppose you have a cylinder closed at both ends, with a frictionless, massless piston situated at the half-way point of the cylinder. The volume of cylinder to the right is V, and the volume of cylinder to the left is also V. The piston is held in place initially by a pin (to the left of the piston). The initial gas pressure in the left half of the cylinder is 1 atm, and the initial gas pressure in the right half of the cylinder is 5 atm. Now, the pin holding the piston is released. At all times after the piston has been released, what will be the relationship between the pressures on the two faces of the piston? Well, lets do a force balance on the piston to find out. The force on the left face is pLA, and the force on the right face is pRA. If the piston had mass m, then the force balance would read:
    where v is the piston velocity to the right. But, since the mass of the piston is zero, unless the piston acceleration is infinite (which it can't be, since that gases adjacent to the piston can't accelerate that fast), the gas pressures on the right- and left faces of the piston must be equal to one another after the piston has been released:
    This does not mean that the gas pressure throughout the entire left chamber is equal to the gas pressure throughout the entire right chamber. In an irreversible process, the gas pressures in the two chambers will vary with distance from the piston. Only at the piston faces are the two gas pressures equal.

    But which is the correct pressure to assume at the piston faces, 5 atm. or 1 atm. The answer is neither. The gas pressures on both faces of the piston will lie somewhere between 5 atm. and 1 atm. A reasonable first approximation might be the geometric mean, 2.24 atm.

    This situation is somewhat analogous to a heat transfer problem in which you have two identical cylindrical metal bars, with one bar initially at 100C, and the other bar initially at 0C. At time t = 0, the two bars are brought into perfect contact with one another at one of their respective ends. The question is, what is the temperature at the contact surface between the bars at times after the two bars are brought into contact. The answer is 50C for all time. This hints at the possibility that at least for a short time after the piston is released in our piston-cylinder problem, the pressures on the two faces of the piston will not only lie between 5 atm. and 1 atm., but will also remain relatively constant (for a while). So the gas on the high pressure side would be expanding against a constant pressure of ~2.24 atm., and the gas on the low pressure side would be getting compressed by a constant pressure of ~2.24 atm.

    This is all I wanted to discuss at this time. I would like to receive some feedback on the rationale I have presented so far.
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