Is there a clear analogy here between volume and entropy? Looking at: $ \frac{dQ}{T} = dS, \frac{dW}{P} = dV$

Question

The first law of thermodynamics states: $$ dE = TdS + PdV $$

And entropy is defined as:

$$ \int \frac{dQ}{T} = S \qquad(1)$$

If we make an analogous relation using the second term in the first law, it looks like this:

$$ \int \frac{dW}{P} = V \qquad(2) $$

And what does this exactly say? $(1)$ says that the heat added to a system at temperature $T$ results in an increase of entropy $S$. $(2)$ says (if it makes any sense) that the work on a system at certain pressure $P$ results in a change of volume $V$. Like pushing in a piston $dW$ at pressure $P$ and thus a volume change of $ W/P = - V$.

Is there a clear analogy here between volume and entropy? Is there any way equation $(2)$ is useful? My goal is to get a better picture of entropy, and since volume is easily visualised it seems that this would be a nice way of visualising entropy. Also the fact that volume and entropy both have a lowerbound $0$.

Answer 1

I’d prefer to write your equations as $$\int\frac{q_\text{rev}}{T}=\Delta S;$$$$\int\frac{w_\text{rev}}{P}=\Delta V;$$ where $q_\text{rev}$ and $w_\text{rev}$ correspond to infinitesimal reversible heat and work, respectively (the presence of “d” might mislead us into thinking that there’s a heat or work state variable that can be differentiated).

Note that we’re describing a change in the extensive state variable (entropy and volume), not its absolute value.

These equations hold for a closed system in which only mechanical work is considered; the existence of other types of work would violate the second equation. Under these constraints (and the constraint of reversibility), yes, entropy and volume are analogous. Entropy is the “stuff” that’s driven to shift by a temperature imbalance, just as volumes are exchanged due to pressure imbalances. (Note however that for real processes, which are all irreversible, entropy is generated. No analogous behavior exists for volume—at least in classical thermodynamics.)

Answer 2

The general expression for the first law (using the OP sign convention) is: $dE = \delta q + \delta w$. When the process is reversible it is possible to write: $dE = TdS + PdV$. This expression is the diferential of a function E(S,V): $$dE = \frac{\partial E}{\partial S}dS + \frac{\partial E}{\partial V}dV$$ The similarity between V and S in the expression is that both are independent functions of the internal energy, while temperature and pressure are partial derivatives: $$T = \frac{\partial E}{\partial S}$$$$P = \frac{\partial E}{\partial V}$$