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This paradox, is a paradox to me only, probably because of my incomplete knowledge.

Assume a non-quasi-static process done on a gas for instance.

Internal energy of a system is a function of two of it's coordinates (if it has three). So it is defined, and exists, as long as our system is in equilibrium. So we have two value for initial and final internal energies in the process.

Work, however isn't defined during a non-quasi-static process since pressure is undefined when system isn't in equilibrium.

Now assume two conditions. If heat is defined during a non-quasi-static process, then work must also be defined during the non-quasi-static process, like this:

$$W=\Delta U-Q$$

If heat is undefined during a non-quasi-static process, then internal energy, as the sum of two undefined quantities, must also be undefined.

$$\Delta U=W+Q$$

What's wrong?

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  • $\begingroup$ Why would the sum of two undefined quantities be undefined? "Undefined" here is not like when you divide by zero, it just means you literally haven't defined it. $\endgroup$
    – ZachMcDargh
    Commented Jan 17, 2017 at 18:50
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    $\begingroup$ They are both defined, even if P has not a defined global value the gas still makes a defined force on the piston (even if this force is unpredictable). Q is also defined, why do you think it is not? $\endgroup$
    – user126422
    Commented Jan 17, 2017 at 19:03
  • $\begingroup$ @ZachMcDargh so they are both undefined? $\endgroup$
    – AHB
    Commented Jan 17, 2017 at 19:17
  • $\begingroup$ @AlbertAspect So would it be correct to say that the infinitesimal work in undefined? This is what Zemansky is stating. That during a very short time, P is not defined, to the infinitesimal work is also undefined. $\endgroup$
    – AHB
    Commented Jan 17, 2017 at 19:18
  • $\begingroup$ @AHB I would not say that either. I am not really sure why Zemansky would say that. My guess is that he means that dW=PdV is undefined because P is undefined, so you should rather go back and use the definition: dW=Fdx=F/AdV instead. $\endgroup$
    – user126422
    Commented Jan 18, 2017 at 1:48

2 Answers 2

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Both "work done" on the system and *heat transferred" to the system are completely and unambiguously definable and measurable irrespective of whether the process is or is not reversible and/or quasi-static, and the same goes for the internal energy of the system. These are boundary conditions, in effect. Hence the equation $\Delta U = \Delta W + \Delta Q$ has meaning if the letters are properly defined as being (1) change in the internal energy of the system: $\Delta U$, (2) work done on the system by an external agent: $\Delta W$, (3) heat transferred to the system by an external agent: $\Delta Q$.

What is not true in general but is true for reversible processes is that $\Delta Q$ and $\Delta W$ can also be written as a first order differential form of the system's internal parameters, such as $\Delta Q = TdS$ and $\Delta W= - pdV + qdE + mdB + ...$.

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In a non-equilibrium process, the amount of work done to go through the process varies each time you perform it. Therefore one can only speak of an expectation value of the work done, as in Jarzynski's equality. However, "work" and "heat" are simply a convenient way to partition the total change in energy. The energy $U$ is always defined: it is simply the total kinetic and potential energy of all of the particles in the system. The fact that partitioning it into heat and work is no longer convenient does not invalidate its definition or existence.

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