How does volume affect microstates?

1 Answer
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_BlntS=kBlnt

where tt 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 ((delS)/(delV))_T(SV)T is nonzero (and in fact, must be positive). This would show that S = S(V)S=S(V) (and in fact, S = S(T,V)S=S(T,V)).

From the Helmholtz Maxwell Relation,

dA = -SdT - PdVdA=SdTPdV

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

((delS)/(delV))_T = ((delP)/(delT))_V(SV)T=(PT)V

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

dV = ((delV)/(delT))_PdT + ((delV)/(delP))_TdPdV=(VT)PdT+(VP)TdP

By definition, the isothermal compressibility factor is

kappa = -1/V ((delV)/(delP))_Tκ=1V(VP)T

and the coefficient of thermal expansion is:

alpha = 1/V ((delV)/(delT))_Pα=1V(VT)P

As a result, we replace those terms to get:

dV = ValphadT - VkappadPdV=VαdTVκdP

Now we divide by dTdT at constant volume:

cancel(((delV)/(delT))_V)^(0) = Valpha cancel(((delT)/(delT))_V)^(1) - Vkappa ((delP)/(delT))_V

Therefore:

color(blue)barul|stackrel(" ")(" "((delP)/(delT))_V = alpha/kappa = ((delS)/(delV))_T" ")|

Since alpha and kappa are positive physical quantities, ((delS)/(delV))_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, ((delS)/(delV))_T = ((delP)/(delT))_V is positive for gases as well.