[galpinj]

Okay guys,

So I watched the video series on thermodynamics by Khan academy months ago, but always found one thing he mentioned quite unsettling. I spent a couple hours trying to figure it out online, to no avail. In the end I decided it was best to just memorize the equation and move on. But, time and again, this question pops up in my studying and vexes me. I have resolved to find the answer, and as such appreciate any input from the community!

In Khan's videos (and a plethora of other videos) we can see that work is defined as PΔV for an isobaric reaction; however, in all circumstances that I have seen there is always a change in pressure accompanying the change in volume! No matter how tiny the change in volume (dV), there always seems to be an accompanying tiny change in pressure (dP). Why is it okay to just ignore this change in pressure while we drool over the changing value of volume? Why can we just assume pressure is constant and volume is changing?

I've taken a screen shot to show exactly what I mean below; you can see a simple volume and pressure graph, in which both state values are changing:
link: https://www.khanacademy.org/science/physics/thermodynamics/laws-of-thermodynamics/v/pv-diagrams-and-expansion-work (around the 5 minute mark)

Furthermore, why can we not have changes to work in an isochoric reaction? I know that, if the volume is held constant, then no actual work is being done by/on the system, but clearly we are changing the pressure, and that should count for something, right?? I've seen videos that state, in this circumstance, the change in pressure should be considered part of internal energy (ΔU), but I don't see how pressure can factor into ΔU. I can kind of see how changing pressure factors into Q (pressure will go up if more heat is added), but I still find the concept confusing.

[Borek]

To be precise

[tex]W = \int P dV[/tex]

but as long as P doesn't change too much with the change in volume we can approximate work quite well by PΔV.

[galpinj]

To be precise

[tex]W = \int P dV[/tex]

but as long as P doesn't change too much with the change in volume we can approximate work quite well by PΔV.

Would we be able to find work if both pressure and volume were changing, and we were accounting for both changes? Put another way, why is it that work requires pressure to be constant?

Thank you

[Borek]

Would we be able to find work if both pressure and volume were changing, and we were accounting for both changes?

Sure thing.

Put another way, why is it that work requires pressure to be constant?

It doesn't.

Integral form never states that P=const, quite the opposite - P can be a function of V.

We can simplify calculations assuming P=const, then we get PΔV - but in general it is not necessary, it just in most cases makes life easier and is accurate enough.

[Vidya]

To be precise

[tex]W = \int P dV[/tex]

but as long as P doesn't change too much with the change in volume we can approximate work quite well by PΔV.

Thank you

If you integrate this function mathematically
 W= PdV + VdP

[Enthalpy]

I understand it more as a matter of math than physics.

That is, as dV is taken arbitrarily small (and calculus gives methods for that), the change in P gets arbitrarily small too, so that P and P+dP can legitimately be taken as identical. Or better: the error is < dV×dP, and as dV gets small, dV×dP becomes negligible as compared with P×dV.

Physicists stop the reasoning here because physical functions use to be continuous, differentiable, analytical and so on. Mathematicians are more petty-minded (or rather, they treat more general and exotic objects) and would remember that some conditions must be met for such methods to work.

You know, some functions are not analytic. For instance exp[-1/(1-x²)] extended by 0 outside [-1; +1] can't be computed from its series expansion despite it's indefinitely differentiable. Other functions are continuous nowhere, for instance 1_Q whose value is 1 on rational numbers and 0 on irrational ones. With that in mind, a mathematician would add a few words of conditions.

[galpinj]

Thank you guys!

This question was driving me crazy, so I really appreciate everyone's help.