Is work a state function for an isobaric process?

Question

For an isobaric process, work is $P(V_2-V_1)$. This quantity seems to be depending only on initial and final volume. Is work a state function for an isobaric process?

This post imported from StackExchange Physics at 2025-01-22 11:00 (UTC), posted by SE-user Shikhar Chamoli

Comments

Hint: Doesn't the path depend on what $p$ you have chosen?

This post imported from StackExchange Physics at 2025-01-22 11:00 (UTC), posted by SE-user Physiker

Answer 1

Good question.

Once you have specified a path, the thermodynamic differentials become exact differentials. The reason they are 'inexact' to begin with is that different paths end up in different changes in the state variables.

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Note : All statements made for reveriblse processes

This post imported from StackExchange Physics at 2025-01-22 11:00 (UTC), posted by SE-user Brian

Comments

And in this particular case the work is an exact differential as $dW =d(PV)$? With $P$ constant. Correct?

This post imported from StackExchange Physics at 2025-01-22 11:00 (UTC), posted by SE-user Kashmiri

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yes, but the property of it being exact can onyl be explicitly understood when we integrate it $W(V)= P(V - V_1)$ if you have an reference volume, as long as you are on isobaric path, the work is function of volume

This post imported from StackExchange Physics at 2025-01-22 11:00 (UTC), posted by SE-user Brian

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Since we've $dW =d(PV)$ P being constant, isn't it obvious that the integral will be independent of path and depend only only volume change?

This post imported from StackExchange Physics at 2025-01-22 11:00 (UTC), posted by SE-user Kashmiri

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I can't figure out teh symbol for inexact differential with 'd' having a line on it, but if you are writing dW= - P dV then it's difficult to understand it's exact because whatever process you use, the infinitesimal work is the same expression

This post imported from StackExchange Physics at 2025-01-22 11:00 (UTC), posted by SE-user Brian

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Like suppoe adiabatic $ P = \frac{C}{V^{\gamma} }$ , now this path will also have the same kind of work $dW = - P dV$ but you can't differentiate it from this explicitly unless you know what the state function for work is by integrating it

This post imported from StackExchange Physics at 2025-01-22 11:00 (UTC), posted by SE-user Brian

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I deliberately wrote $dW =d(PV)$ with $P$ inside braces, I believe $d(PV)$ quaifies to be evidently an exact differential and hence a state function.

This post imported from StackExchange Physics at 2025-01-22 11:00 (UTC), posted by SE-user Kashmiri

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I disagree. The concept of exact differential depends on the coefficients of the differential form. It has nothing to do with the choice of a specific path. In that case, the dependence of the result on the initial and final point is a trivial consequence of the fundamental theorem of Calculus and it would be true for any differential form.

This post imported from StackExchange Physics at 2025-01-22 11:00 (UTC), posted by SE-user GiorgioP-DoomsdayClockIsAt-90