How do you derive the pV Work formula, W= "integral"pdV?

[KingDaniel]

This is a Thermodynamics question. I need help in how to derive the Work done in changing from State 1 and State 2.
ie: How do you derive the pV Work formula, W= "integral"dW= "integral"pdV

 

[calculo2718]

Defining work to be [;W = \int F dx;] and pressure to be [;P = \frac{F}{A};]

We can substitute F in the in the work equation (assuming constant force) and obtain [;W = \int PA dx;] which is effectively [;W = \int PdV;].

 

[KingDaniel]

@calculo2718 , what you've shared made sense at first read but after thinking about it a bit, I think with your "W = \int PA dx", you should get:

W = PA * \int dx , since PA is constant,
and finally W = PAx , since "\int dx"=x,
and not W = \int PdV

I thought that maybe Ax could be considered as dV, but I'm not sure.

Please try to explain further as there might be something I'm not considering

 

[RedDelicious]

Substitute PA=F into your equation for W and when you integrate A with respect to x, you treat P as a constant. The integral of area gives you volume.

 

[Qwertywerty]

Work done is defined as ( ∫F.dx ) . Force is not necessarily a constant and may vary at different points .

It would , however , be a constant ( approximately ) over a length dx . So we write F as P.A at and write the work as ( ∫PA.dx ) .

A.dx is the small volume dV that is crossed by an acting force . So we write the work done as ( ∫P.dV ) ,where P is not necessarily a constant .

 

[KingDaniel]

@RedDelicious , if I do that, I'll end up with at the same solution of W = pAx = pV I believe

@Qwertywerty , what you've said seems to makes sense and I want to understand it this way but I need to clear a few things first:

If you integrate F with respect to x, you get F*x (which is what we were taught in high school physics, without the integral). So if you integrate, pA with respect to x, shouldn't you get pAx = pV, and not integral of pdV? If F is not constant, don't we just treat it as average of all the F's, and still treat it as a constant?

 

[calculo2718]

@RedDelicious , if I do that, I'll end up with at the same solution of W = pAx = pV I believe

@Qwertywerty , what you've said seems to makes sense and I want to understand it this way but I need to clear a few things first:

If you integrate F with respect to x, you get F*x (which is what we were taught in high school physics, without the integral). So if you integrate, pA with respect to x, shouldn't you get pAx = pV, and not integral of pdV? If F is not constant, don't we just treat it as average of all the F's, and still treat it as a constant?

Yes, you learned W = Fx in high school but that was assuming F was constant, I shouldn't have said assuming F is constant, but in my defense it was nearly 5 in the morning when I posted.

Think about it like this, force is a function of position(not necessarily constant), area is also a function of position so its more like W = [;\int F(x)dx;] and [;W = \int P(x)A(x)dx;], [;A(x)dx;] is essentially [;dV;].

 

[Qwertywerty]

Here - ∫dx = x + c
∫kx.dx = k(x∧2)/2 + c
∫k(x∧2).dx = k(x∧3)/3 + c

and so on , where k and c are some random constants .

Now suppose F = kx , could you remove F from the integral of ∫F.dx ?
F = P.A will however be true at any instant .

Thus we can write W =∫F.dx or W= ∫PA.dx .

In the second case it is not necessary that P or A is a constant over some large distance . They may simply be functions as P(x) and A(x) . So can you now remove either of them from your integral ?

However we can write A(x).dx as the small volume that that the particle , on which the force acts , moves through at any x .
Thus work can be written as W = ∫P.dV .

I hope this helps .