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

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 $$0$$.

• Except that the equation between dQ and dS applies only to a reversible path. May 1 at 12:08

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).
• Thank you for your answer @chemomechanics. So just like a pressure differential 'pushes out' a volume $\Delta V$, a temperature differential 'pushes out' an entropy $\Delta S$. In a certain sense the differentials create a new 'space' for the particles to move. Where as pressure literal space, and temperature new 'energy' states to reach. An irreversible path creates entropy but doesn't create volume is the biggest difference if I understand correctly May 8 at 21:02
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: $$T = \frac{\partial E}{\partial S}$$$$P = \frac{\partial E}{\partial V}$$