# How does volume affect microstates?

Mar 24, 2018

Entropy is a function of the volume. As the volume increases in a system, the entropy increases.

As we know,

$S = {k}_{B} \ln t$

where $t$ is the most probable distribution of microstates in a system.

Hence, the number of microstates increases due to increasing volume.

To show this, we verify that ${\left(\frac{\partial S}{\partial V}\right)}_{T}$ is nonzero (and in fact, must be positive). This would show that $S = S \left(V\right)$ (and in fact, $S = S \left(T , V\right)$).

From the Helmholtz Maxwell Relation,

$\mathrm{dA} = - S \mathrm{dT} - P \mathrm{dV}$

Since $A$ is a state function, the cross-derivatives are equal:

${\left(\frac{\partial S}{\partial V}\right)}_{T} = {\left(\frac{\partial P}{\partial T}\right)}_{V}$

Volume must depend on pressure and temperature. If we write the total differential of the volume as a function of $T$ and $P$:

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

By definition, the isothermal compressibility factor is

$\kappa = - \frac{1}{V} {\left(\frac{\partial V}{\partial P}\right)}_{T}$

and the coefficient of thermal expansion is:

$\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P}$

As a result, we replace those terms to get:

$\mathrm{dV} = V \alpha \mathrm{dT} - V \kappa \mathrm{dP}$

Now we divide by $\mathrm{dT}$ at constant volume:

${\cancel{{\left(\frac{\partial V}{\partial T}\right)}_{V}}}^{0} = V \alpha {\cancel{{\left(\frac{\partial T}{\partial T}\right)}_{V}}}^{1} - V \kappa {\left(\frac{\partial P}{\partial T}\right)}_{V}$

Therefore:

$\textcolor{b l u e}{\overline{\underline{|}}} \stackrel{\text{ ")(" "((delP)/(delT))_V = alpha/kappa = ((delS)/(delV))_T" }}{|}$

Since $\alpha$ and $\kappa$ are positive physical quantities, ${\left(\frac{\partial S}{\partial V}\right)}_{T}$ must be positive for liquids and solids.

For gases, we consider merely that increasing temperature increases particle motion, and thus the force colliding with the rigid container walls. By definition, that leads to increasing gas pressure.

Therefore, ${\left(\frac{\partial S}{\partial V}\right)}_{T} = {\left(\frac{\partial P}{\partial T}\right)}_{V}$ is positive for gases as well.