4.6.3 Special Cases

There are two special cases of mechanical work.

Free Expansion

The first is free expansion. Intuitively it seems that when something pushes, it does work. Intuition is wrong. While it is true that you can't do mechanical work without a "push", you must be pushing against something and that something must move. Both a "push" and movement are necessary. Here there is nothing to push against and so there is no work. One can see this by substituting 0 for p_ext in the formula for differential work. The result is:

$$dw = 0$$ (4.6.14)

which is what I claimed.

Expansion against a Constant Pressure

Actually, we did this case before developing the formula for differential work. But it is useful to see how we can go the other way. Here's an example:

A cylinder containing a gas with an initial volume of V₁ expands to a final volume of V₂ against a constant pressure of p_ext. How much work is done?

The formula for differential work is:

$$dw = - p_{ext} dV$$ (4.6.15)

We need to integrate this. Note the limits of the integration:

$$\int_{0}^{w} dw = - p_{ext} \int_{V_1}^{V_2}$$ (4.6.16)

$$w = - p_{ext} ( V_2 - V_1 ) = - p_{ext} \Delta V$$ (4.6.17)

Which is certainly simple enough.