# Is entropy a state function? How? Prove it?

Jan 16, 2017

Essentially, this shows a derivation of entropy and that a state function can be written as a total derivative, $\mathrm{dF} \left(x , y\right) = {\left(\frac{\partial F}{\partial x}\right)}_{y} \mathrm{dx} + {\left(\frac{\partial F}{\partial y}\right)}_{x} \mathrm{dy}$.

From the first law of thermodynamics:

$\mathrm{dU} = \delta {q}_{\text{rev" + deltaw_"rev}}$,

where $q$ is the heat flow, $w$ is the work (which we define as $- \int P \mathrm{dV}$), and $\delta$ indicates that heat flow and work are inexact differentials (path functions).

Solving for $\delta {q}_{\text{rev}}$ gives:

$\delta {q}_{\text{rev" = dU - delw_"rev}} = {C}_{V} \left(T\right) \mathrm{dT} + P \mathrm{dV}$,

since ${\left(\frac{\partial U}{\partial T}\right)}_{V} = {C}_{V}$, the constant-volume heat capacity. For an ideal gas, we'd get:

$\delta {q}_{\text{rev}} \left(T , V\right) = {C}_{V} \left(T\right) \mathrm{dT} + \frac{n R T}{V} \mathrm{dV}$

It can be shown that this is an inexact total derivative, indicative of a path function. Euler's reciprocity relation states that for the total derivative

$\boldsymbol{\mathrm{dF} \left(x , y\right) = M \left(x\right) \mathrm{dx} + N \left(y\right) \mathrm{dy}}$,

where $M \left(x\right) = {\left(\frac{\partial F}{\partial x}\right)}_{y}$ and $N \left(y\right) = {\left(\frac{\partial F}{\partial y}\right)}_{x}$,

a differential is exact if ${\left(\frac{\partial M}{\partial y}\right)}_{x} = {\left(\frac{\partial N}{\partial x}\right)}_{y}$. If this is the case, this would indicate that we have a state function.

Let $M \left(T\right) = {\left(\frac{\partial {q}_{\text{rev}}}{\partial T}\right)}_{V} = {C}_{V} \left(T\right)$, $N \left(V\right) = {\left(\frac{\partial {q}_{\text{rev}}}{\partial V}\right)}_{T} = \frac{n R T}{V}$, $x = T$, and $y = V$. If we use our current expression for $\delta {q}_{\text{rev}}$, we obtain:

((delC_V(T))/(delV))_T stackrel(?" ")(=) ((del(nRT"/"V))/(delT))_V

But since ${C}_{V} \left(T\right)$ is only a function of $T$ for an ideal gas, we have:

$0 \ne \frac{n R}{V}$

However, if we multiply through by $\frac{1}{T}$, called an integrating factor, we would get a new function of $T$ and $V$ which is an exact differential:

$\textcolor{g r e e n}{\frac{\delta {q}_{\text{rev}} \left(T , V\right)}{T} = \frac{{C}_{V} \left(T\right)}{T} \mathrm{dT} + \frac{n R}{V} \mathrm{dV}}$

Now, Euler's reciprocity relation works:

${\left(\frac{\partial \left[{C}_{V} \left(T\right) \text{/"T])/(delV))_T stackrel(?" ")(=) ((del(nR"/} V\right)}{\partial T}\right)}_{V}$

$0 = 0$ color(blue)(sqrt"")

Therefore, this new function, $\frac{{q}_{\text{rev}} \left(T , V\right)}{T}$ can be defined as the state function $S$, entropy, which in this case is a function of $T$ and $V$:

$\textcolor{b l u e}{\mathrm{dS} \left(T , V\right) = \frac{\delta {q}_{\text{rev}}}{T}}$

and it can be shown that for the definition of the total derivative of $S$:

$\mathrm{dS} = {\left(\frac{\partial S}{\partial T}\right)}_{V} \mathrm{dT} + {\left(\frac{\partial S}{\partial V}\right)}_{T} \mathrm{dV}$

$= {\left(\frac{\partial S}{\partial T}\right)}_{V} \mathrm{dT} + {\left(\frac{\partial P}{\partial T}\right)}_{V} \mathrm{dV}$
(where we've used a cyclic relation in the Helmholtz free energy Maxwell relation)

which for an ideal gas is:

$= \frac{{C}_{V}}{T} \mathrm{dV} + \frac{n R}{V} \mathrm{dV}$