Question

How does entropy change with pressure ?

Answer 1

Starting from the first law of thermodynamics and the relationship of enthalpy H to internal energy U:
$\Delta U=q_{rev}+w_{rev}$
$\Delta H=\Delta U+\Delta \left(PV\right)=q_{rev}+w_{rev}+\Delta \left(PV\right)$
where:
$q_{rev}$ and $w_{rev}$ are the most efficient (reversible) heat flow and work, respectively. $w_{rev}=-PdV.$
P is pressure and V is volume.
THE ENTHALPY MAXWELL RELATION
Using the differential form, we get:
$dH=\partial q_{rev}+\partial w_{rev}+d\left(PV\right)$
Another relationship that relates with heat flow is the one for entropy and reversible heat flow:
$dS=\frac{\partial q_{rev}}{T}$
Thus, utilizing this relationship and invoking the Product Rule on $d\left(PV\right),$ we get:
$dH=TdS-PdV+PdV+VdP$
$dH=TdS+VdP$
which is what you would get for the Maxwell relation.
ENTHALPY VS ENTROPY
When we relate pressure then, to entropy, with $S=S(T,P):$
$dS=\frac{dH}{T}-\frac{V}{T}dP$
For an ideal monoatomic gas, $PV=nRT,$ so:
$dS=\frac{dH}{T}-\frac{nR}{P}dP$
Finally, when we integrate this, we get:
${\int }_{S_1}^{S_2}dS={\int }_{H_1}^{H_2}\frac{dH}{T}-nR{\int }_{P_1}^{P_2}\frac{1}{P}dP$
Enthalpy is $\Delta H={\int }_{H_1}^{H_2}dH={\int }_{T_1}^{T_2}{C}_{p}dT,$ where $C_p$ is the heat capacity at constant pressure in $J/K.$ Thus:
$\Rightarrow {\int }_{T_1}^{T_2}\frac{C_p}{T}dT-nR{\int }_{P_1}^{P_2}\frac{1}{P}dP$
Since $C_p$ for a monoatomic ideal gas is $C_V+nR=\frac{3}{2}nR+nR=\frac{5}{2}nR.$ with $C_V$ as the constant-volume heat capacity and the $\frac{3}{2}$ coming from the three linear degrees of freedom $(x,y,z)$ this becomes:
$\Delta S_{sys}=\frac{5}{2}nRln|\frac{T_2}{T_1}|-nRln|\frac{P_2}{P_1}|$
PRESSURE VS ENTROPY

Therefore, if pressure increases, a negative contribution is made on the change in entropy of an ideal gas, but depending on the change in temperature, the actual change in entropy for the system might be positive or negative.

(Regardless, the entropy of the universe is $\ge 0$.)
On the other hand, the change in volume of a liquid is appreciably low upon small increases in pressure that should substantially compress a gas, so the change in pressure of a liquid makes a smaller negative contribution to the change in entropy.
With solids, I would not expect pressure to significantly alter any entropy patterns they already have for small values for pressure that would otherwise be significant for gases.
For di/polyatomicsolids, we can consider either the complexity or the bond strength, as it relates to the number of "ways" it can exist. One of the most popular thermodynamics equations is:
$\Delta S=k_{b}ln\Omega$
where:
$k_b$ is the Boltzmann constant, $1.38\times {10}^{-23}J/K.$
$\Omega$ is the number of microstates consistent with a chosen macrostate, which is proportiona; to the number of observable "snapshots" of molecular motion.
The stronger the bond, the lower the magnitude of the entropy because the lower the number of microstates available to the solid.