How can power be defined as the derivative of work with respect to time if work is not a function?

Question

In thermodynamics, it is frequently said that work and heat are path functions with inexact differentials $\delta Q$, $\delta W$, so one can not talk about a change in work or a change in heat since they are not state properties of the system. If so, why do I usually find power defined as the derivative of work with respect to time if work is not a function and we can not speak about changes in work ?

And in general if work is $W = \int_C{\textbf{F}}^{}\cdot d\textbf{x}$, how can we evaluate $P(t)=\frac{d W}{dt}$?

Answer 1

If so, why do I usually find power defined as the derivative of work with respect to time if work is not a function

Just because work is shown as an inexact differential in the differential form of the first law doesn't mean work is not a function. It just means the work function (path) needs to be specified.

In the case of boundary work, it means that pressure as a function of volume needs to be specified in order to evaluate the equation

$$W=\int_{V1}^{V2} PdV$$

This is no different than saying force as a function of displacement has to be specified in order to evaluate the equation

$$W=\int_{x1}^{x2} \vec F \cdot d\vec x$$

Work in both equations is a path function. In thermodynamics it's pressure as a function of volume. In mechanics it's force as a function of displacement.

Hope this helps.

Answer 2

My two cents. I think that's because sometimes it's possible to convert inexact differential to the exact one using integrating factors. For example, if the process is quasi-static adiabatic , then there is no heat exchange in the system, namely $\delta Q = 0$. In this case from the first law of thermodynamics follows that, $$ \delta W = -dU \tag 1,$$

Now in that type of process using (1) we can define perfectly valid thermodynamic power, which would be :

$$ P = \frac {\delta W}{dt} = -\frac {dU}{dt} \tag 2$$

and this does not depend on path taken, because U is state function too.

Let's take a quasi-static adiabatic process of an expanding gas which moves a piston. Since in this process $$dU = - pdV \tag 3,$$

we can re-write (2) as :

$$ P = p \frac {dV}{dt} = p \frac {d(A \cdot x)}{dt} = p A \frac {dx}{dt} = pAv, \tag 4$$

where $p$ gas pressure, $A$ - piston cross section and $v$,- piston moving speed.