# Understanding pV Diagrams

1. Nov 30, 2008

2. Dec 1, 2008

### horatio89

W = p*deltaV only holds for the case of constant p, that is true.

What is the general form of the work eqn. that holds for all types of paths? (Hint: It also invloves p and V, and it must reduce to the W = p*deltaV, when p is constant)

3. Dec 1, 2008

### doggieslover

Is it W = p1V1 - p2V2?

4. Dec 1, 2008

### horatio89

No, recall that work done is given by W = F.x, hence dW = F.dx, where dx is a infinitesimal displacement in the direction of the force.

For ideal gases, F = pA, hence dW = pA.dx, or dW = p.dV.

Work is then found from integrating p.dV, with respect to V from Vfinal to Vinitial, which can be interpreted as the area under the pV graph

5. Dec 1, 2008

### doggieslover

So pV from Vi to Vf, but isn't p still constant in this case?

The graph indicates that p is NOT constant.

6. Dec 1, 2008

### horatio89

no, p as indicated in this equation of the general case, is a function of V.

7. Dec 1, 2008

### doggieslover

okay let's say from 1 to 3, pressure is decreasing from 3p to 2p, how would I apply that to pV| from Vi to Vf?

p(3V) - p(2V)?

8. Dec 1, 2008

### horatio89

Integration between two points of a function can be simplified by taking the area under the curve. Because, the graph is linear, it is very easy to find the value of the integral from 1 - 3 - 6, because the area under the curve is a trapezium (or trapezoid).