w pv then dw pdv vdp but why is the volume work w pdv but not vdp??

First of all, I don't know if the formula w=pv is correct, at least I can't physically understand what it is trying to say. Furthermore, we know that the volume work is PDV, which is not deduced by w=pv, and you can imagine that drawing a PV image, no matter how complicated, is to find the area below the plot line. So the subject's question, I think you must have no doubts in the isobaric situation, because dp 0;So in the isothermal and adiabatic process, according to pv nrt, both sides are differentiated, in the case of isothermal, because of dt 0, we get pdv vdp 0, but then we can't continue to calculate, so the book is calculated from another angle, and the adiabatic process, pdv vdp nrdt, pdv is volumetric work is definitely true, but vdp to be honest, I don't know what physical meaning it is, it only has abstract mathematical meaning.

I've said so much, I don't know if you understand, in short, the volume work is equal to PDV, and there is no VDP behind

Let's analyze it in the form of derivatives. w'=p'v+v'p.If v is a variable, p is a constant, p'=0, i.e. w=

PDV is generally used for gases to do work equal to the acupressure multiplied by the varying volume.

In the same way, if the two positions are reversed, there is w= vdpThen this formula should be that the work done by the gas is equal to the volume multiplied by the pressure of the change.

And then both are variables, and when it comes to multivariate functions, I don't know how to do it yet.

I'm not good at physics, pure mathematical analysis, don't look for me if I'm wrong.

Because the assumption in this case is that p is constant. The volume is changing. So your points are different. To be precise, the definition of this topic is that you can move, that is, reversible

process.In this case, dw=pdv, no vdp

You think so. reversible process.

Although v and p can be regarded as invariant, they should be counted as work dw.

If you think about it, under a certain pressure, changing the volume is a work. However, keeping the volume unchanged and changing the pressure is not work, and the volume remains the same, where does the work come from.

VDP represents the change in pressure with the same volume. And because the volumetric invariant system does not do work, vdp=0 and is rounded.

The formula for the work done by gas expansion to the outside world is w=pv. The equation of state of an ideal gas is pv=n·r·t. There is a variant of the form p=n v·r·t.

From the variable form p=n v·r·t and the formula w=pv for the work done by gas expansion to the outside world, it is shown that there are two variables for the work done by the gas when the volume is v1.

There is w = p · v1 = n v · r · t · v1. One variable is the concentration n v and the other is the temperature t. In other words, a change in concentration and temperature will cause a change in work.

The physical meaning represented by v1 is the product of the area under force and the displacement that occurs in the direction of the force. It is clear: the second law of thermodynamics describes the temperature t, and the microscopic probability orientation of matter describes the concentration nv.

Upstairs didn't say the key, the first PV=NRT, so PV is actually the energy of the system, the previous W is not the meaning of volume work, the back is volume work, the meaning of the same letter is not necessarily the same.

Because volumetric work is defined as the work that is produced along with the change in volume. When calculating, whoever is a variable is the integral object

Because the assumption in this case is that p is constant. The volume is changing. So your points are different.

To be precise, the definition of this topic is reversible processIn this case, dw=pdv, no vdp

Let's analyze it in the form of derivatives

w'=p'v+v'p.If v is a variable, dup is a constant, p'=0, i.e., w= PDV is generally used to do work on gas equal to the DAO acupressure multiplied by the weight volume of the variation.

In the same way, if the two positions are reversed, there is w= vdpThen this formula should be that the work done by the gas is equal to the volume multiplied by the pressure of the change.

And then both are variables, and when it comes to multivariate functions, I don't know how to do it yet.

I'm not good at physics, pure mathematical analysis, don't look for me if I'm wrong.

The upstairs didn't talk about the key, the key is that the existing dw=pdv w=pv is the case where the pressure is constant.

w=pv.

dw=pdv+vdp.

w ＝ pdv

The work done by the volume is 0, so there is no vdp