Calculating Work with Pressure in a System

[georg gill]

The work in a system is given by

$$F\cdot\Delta h$$

$$P\cdot A\cdot\Delta h$$

$$P\cdot \Delta V$$

when one calculate work one always uses pressure from surroundings. Why is it so. I get that work becomes linear then because p is constant but I thought that if one use Newtons third law that a system is standing still if forces are equal then work as a physical change would be done if p is larger in either system or surroundings. But if one look at force from system when p is larger in system why can't one use that to calculate work on surroundings?

 

[Studiot]

Is this a thermodynamics question?

 

[georg gill]

it is from my book online here:

http://books.google.no/books?id=2Ox...ei=hWAtT5acBafE4gSQ-oGrDg#v=onepage&q&f=false

chapter 9 page 364 just above example 9.1 is where they say it

(I have bought it :) )

 

[Studiot]

OK, I though it might be about chemical thermodynamics.

Now we know where we are going.

Thermodynamics allows two kinds of work - reversible work and irreversible work.

Your book was describing reversible work, whilst your example was about irreversible work.

We can always calculate the reversible work but usually calculation of irreversible work presents difficulty.

Chemical reactions are usually carried out at atmospheric pressure, often in open reaction vessels.

This allows us to make the statement of constant pressure (of the atmosphere). Work on a gas at constant pressure is reversible work and can be expressed by the formula

w = pdV

Which is what your book says.

On the other hand the system will be solid and/or liquid and the work irreversible.

An expanding system at higher pressure than atmospheric and where the pressure is gradually changing with the expansion is not in equilibrium.

Reversible systems are equilibrium systems
Irreversible systems are non equilibrium systems.

Luckily we can say that the ( reversible) work done against the atmosphere = the (irreversible) work done by the expaning system.
So we can calculate the process work that way.

 

[georg gill]

Luckily we can say that the ( reversible) work done against the atmosphere = the (irreversible) work done by the expaning system.
So we can calculate the process work that way.

It is just so that there is one thing I find really confusing. When system is expanding it works against the pressure of atmosphere so it has to move on that so use of pressure of atmosphere would make sense for me to calculate work. But when atmosphere is working on system I would think that the work it has to do to compress system is determinded by pressure in system therefore it is less per unit of volume then if system is pressing on atmosphere. How come one can use same pressure for both cases?

And I don't seem to get why my attempt of given an example was an irreversible example?