The First Law of Thermodynamics for an adiabatic process is: $\Delta U=W$. The internal energy is extensive. Thermodynamic work is neither extensive nor intensive in general, but is it extensive, if the process is adiabatic ?
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2 Answers
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Work is extensive in the sense that if you a system $\lambda$ times as big, work will be $\lambda$ times bigger.
However, the distinction between intensive and extensive is typically restricted to state functions and work is not a state function.
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$\begingroup$ Work is always path-dependent. Saying that process is adiabatic is specifying the path. You're essentially saying: "On a single path, work is path-independent" it is kind of true but doesn't carry much information. $\endgroup$ Commented Feb 2, 2022 at 6:23
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$\begingroup$ Also, path dependence is not linked to extensivity/intensivity. $\endgroup$ Commented Feb 2, 2022 at 6:24
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$\begingroup$ The question remains: If the work is path independent for an adiabatic process, should I conclude that the system is extensive for this specific path/process ? $\endgroup$– FrostCommented Feb 2, 2022 at 7:23
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$\begingroup$ I don't think I understand the question. The fundamental principle of thermodynamics is that an equilibrium state of a system in question is parametrized by a set of extensive functions (e.g. U, V, N). $\endgroup$ Commented Feb 2, 2022 at 7:30
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First of all, work is not a property of a system, it depends on the path. So in the sense that the terms intensive and extensive normally refer to intensive and extensive properties, work is neither an intensive nor extensive property.
Work is path dependent. It only equals the change in internal energy for an adiabatic path. For a non adiabatic processes there are an infinite number of possible paths connecting the same two equilibrium states and thus an infinite number of possible amounts of work connecting the two states. That work may be considered to be path independent for an adiabatic process does not make work a property. That's because a property is path independent between two states for any process.
That said, work can be divided by mass to obtain an amount of work done per unit mass. This is the case for work done for open systems where mass flows into and out of the system. But that doesn't make work an intensive property as when one divides an extensive property like internal energy by mass to obtain the intensive property of internal energy per unit mass.
Hope this helps.
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$\begingroup$ Ref: M. Zemansky and R. Dittman, Heat and Thermodynamics, 7th ed., p. 76:"The importance of (4.1) is that thermodynamic work which is generally path-dependent, becomes path-independent for an adiabatic process". Here (4.1) refers to $W=\Delta U$. $\endgroup$– FrostCommented Feb 2, 2022 at 16:32
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$\begingroup$ I like Zemansky's book and often quote it. My argument is just because work may equal the change in a property for an adiabatic process does not make work a property. The Zemansky quote does not say that since work is path independent for an adiabatic process that makes work a property. It would be illogical to say that. A property is something that is path independent FOR ANY PROCESS. Work is only path independent for an adiabatic process. IMO to say that work is a property, but only for an adiabatic process, contradicts the definition of a property. $\endgroup$– Bob DCommented Feb 2, 2022 at 16:45
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$\begingroup$ @Frost I revised my answer adding two sentences to the end of the second paragraph to clarify this point. $\endgroup$– Bob DCommented Feb 2, 2022 at 16:49