# Calculation of work (P.dV) in an isothermal system

1. Oct 25, 2012

### metalrose

Here's the textbook way of calculating the work done by an ideal gas in an isothermal case.

PV=nRT
P.dV=(nRT/V).dV
∴ ∫P.dV=nRT∫dV/V
→ W2-W1=nRT*ln(V2/V1)

My question.
Consider a cylindrical (or of any other shape) container of surface area A and a frictionless movable piston attached.
Let the coordinate x run along the length of this container, with origin at the bottom of the container.
Let the piston be at some x at some point of time.

At this x, P.V= constant = nRT
This holds for all x.
Now since V(x)=A.x
P(x)=constant/A.x=K/x .... for some contant k (=nRT/A)

Now P and V are both functions of x.
While calculating the infinitesimal work P.dV, how can we treat P(x) constant over the range dx?

differentiating both sides, d[P(x)V(x)]=0
→ V(x).dP + P(x).dV=0
∴ P(x).dV= -V(x).dP

Shouldn't this relation hold?

I ask this because I was solving a similar problem, although which wasn't isothermal, involved P and V that depended on x and the approach used there was the second one.

2. Oct 25, 2012

### nasu

The relationship holds and if you integrate -VdP instead of pdV you get the same result, don't you?
I am not sure what is the question...

3. Oct 28, 2012

### DaTario

Usually, in calculus, functions of x are taken as being constant along any differential displacement dx.

4. Oct 28, 2012

### metalrose

Any rigorous mathematical explanation you could point me to? I know that if dx is infinitesimally small, f(x) will not change much in dx, but it will change neverthless. So i am looking for a rigorous mathematical explanation of why we can treat it as constant.