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The closed system work PdV is path dependent but it is path independent for an adiabatic process as can be shown by applying conservation of energy on a closed system (dQ = dE + PdV),here dE is change in energy of system & dQ is change in heat, here for adiabatic dQ=0 , so PdV= -dE, since energy of a system is it's property so it is path independent and hence in this case work is also independent of path).

I have this homework question:

It is desired to bring about a certain change in the state of a system by performing work on the system under adiabatic conditions.

a) The amount of work needed is path dependent.

b) Work alone cannot bring about such a change of state.

c) The amount of work needed is independent of path.

d) More information is needed to conclude anything about the path dependence or otherwise of the work needed.

The answer given is (a). Shouldn't it be (c)? Or is there an exception to above said condition.

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  • $\begingroup$ pl.check whether your path is reversible or irreversible- in general p.dv will be path dependent. $\endgroup$ – drvrm Aug 31 '18 at 12:00
  • $\begingroup$ @shashank tyagi. Think we are going around in circles so I am withdrawing my answer. However, if the process is adiabatic then there is only one value of work for that particular process. But that work depends on it being an adiabatic process. I was only trying to prove that there is no unique value of work between the two states that you happen to connect by an adiabatic process. You said it is a closed system, but then there is no mass transfer regardless of the speed of the process. You also said it is adiabatic. Therefore there is no heat transfer regardless of the speed of the process. $\endgroup$ – Bob D Aug 31 '18 at 18:05
  • $\begingroup$ That is how adiabatic process avoids heat transfer, either by very high speed process or by heavy insulation, so no it won't be adiabatic regardless of speed(only if insulation is mentioned, then it can). I agree with you that there is no unique value of work between two states but I think you misunderstood the question, you said there is only one value of work for that particular adiabatic process, which would also be true for other processes, if you pick one & fix the path to it, like a particular isothermal process(please note a particular not any isothermal process). Continued in next... $\endgroup$ – shashank tyagi Sep 1 '18 at 3:36
  • $\begingroup$ But I think the question boil downs to the another question that can there be multiple adiabatic processes between given two states, for which the work would be same? Because the derivation in question just mentions the process is adiabatic, it doesn't mention that it is a particular unique adiabatic process. So can there be multiple adiabatic processes between two states? If there can be, then for all of them dQ=0, & according to derivation it appears for all of those different adiabatic paths, work would be same. $\endgroup$ – shashank tyagi Sep 1 '18 at 3:42
  • $\begingroup$ Maybe you define a closed system adiabatic process different than I am aware of. The definitions I know are: (1) a closed system is defined as one that does not exchange mass with its surroundings and (2) an adiabatic process is one involving no heat transfer between a system and its surroundings, owing to the nature of the boundary between them (normally a perfectly thermally insulated boundary). As far as I know, these apply regardless of the speed of the process. Can you cite a reference that heat/mass transfer for a closed system adiabatic process depends on the speed of the process? $\endgroup$ – Bob D Sep 1 '18 at 12:24
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The issue of path dependence relates to whether, for a given change in state from an initial thermodynamic equilibrium state to a final thermodynamic equilibrium state, the amount of heat and the amount of work depend on the process path. So, in general, if you have an adiabatic process, there may be other (non-adiabatic) process paths between the same two states that involve a different amount of heat (non-zero) and work.

That being said, however, the way this specific question is phrased is ambiguous. One interpretation is that it seems to be restricted to only adiabatic paths. In that case, if there were other adiabatic paths possible to get from the same initial state to the final state (which I doubt), the amount of work would be independent of path. On the other hand, if it were interpreted as not being restricted to only adiabatic paths, (a) would be the answer.

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  • $\begingroup$ Now that you have mentioned. Can there be multiple adiabatic paths for a given system between two fixed states? Also isothermal and polytropic (for a given value of polytropic index) processes, can there be multiple paths of any one type of them? As we know that the slope for adiabatic would be γ(-P/V),(on P-V plot),here γ is adiabatic index & the curve has to pass through given 2 fixed points? $\endgroup$ – shashank tyagi Aug 31 '18 at 12:44
  • $\begingroup$ Are you also considering irreversible paths? If you have an adiabatic irreversible path, there is no adiabatic reversible path at all between the initial and final states. The reversible path must involve some heat transfer along the path. By an isothermal path, are you including irreversible paths in which that initial and final temperatures are the same and in which the system is in contact with with a constant temperature reservoir at the same initial temperature? $\endgroup$ – Chet Miller Aug 31 '18 at 14:43
  • $\begingroup$ I'm going to reiterate. The term path-dependence refers only to the two thermodynamic equilibrium end points. For a given pair of end points, the work is always dependent on path. It is meaningless to ask whether work is path-dependent for an adiabatic process; for the two end points of an adiabatic process, there will always be non-adiabatic paths for which the work is different. In one of your comments, you said that an adiabatic process does not necessarily refer to a process for which the net heat transfer is zero. But this is exactly what it means. (continued in next comment) $\endgroup$ – Chet Miller Aug 31 '18 at 21:36
  • $\begingroup$ You indicated that it means that the process is fast enough that there is no heat transfer with the surroundings. But this means either that (a) there is no net heat transfer or (b) the final state is not an equilibrium state, and, if we allow more time for the system to equilibrate with its surrounding, heat transfer will occur. But, for path independence to apply, the two end states must be thermodynamic equilibrium states. $\endgroup$ – Chet Miller Aug 31 '18 at 21:41
  • $\begingroup$ I agree with you in that for an adiabatic process between 2 states there will always be non-adiabatic ones for which work is different. But question is can there be different adiabatic paths between same 2 states? If there can be, then for all of them dQ=0 & it appears from derivation above that for all of them, work would be same. So it is meaningful in this sense to ask whether work is path dependent for adiabatic process between 2 states (because I am talking just about possible adiabatic processes, not the non-adiabatic, which I know exist). Continued.... $\endgroup$ – shashank tyagi Sep 1 '18 at 3:52

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