Do you work well under pressure?

In thermodynamics, Work done is given as dw= -P (external).dv. Why do we consider the pressure against which the work is being done and why not the internal pressure?

• And how is energy transferred through pressure - volume work?

• Answer:

  Because we are talking about work done by external agent (Throughout the answer I've assumed only mechanical work can be done.) First of all, we should realize that this definition is used in context of first law of thermodynamics. In the first law of thermodynamics, we are basically talking about energy conservation. So we are interested in energy transferred to the system which is heat input and work done by external agent. This net input (dq and dw) is equal to net change in internal energy of the system. If question is about negative sign, we can easily see it coming by considering simple system of gas enclosed in a chamber along with moving piston. Work done by internal forces is taken care of in internal energy calculations. So we don't need to worry about it as far as we are talking in context of first law. But if you are interested in calculating work done by internal forces you can integrate P(internal).dv    (No negative sign here)