What is the relation between entropy and pressure in strict thermodynamic terms? I read on the net that it is a general consensus that entropy in a system decreases as pressure increases and vice versa. 


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*How can one reach such conclusion using the characteristic equation in terms of entropy?
$$dS=(1/T)dU+(p/T)dV-(μ_B/T)dn_b$$
and lets assume there is no change in molecular mass, so we use only the first 2 clauses.

*Can we prove, without making any other assumptions like ideal gas or other simplifications, that the strict thermodynamic term $(\dfrac{\partial S}{\partial p})_T$ is, as the general consensus is, negative, thus increasing pressure, decreases entropy?
 A: From a Maxwell relation we known that:
$$\left(\frac{\partial S}{\partial p}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{p}$$
For gases and most liquids, $\left(\frac{\partial V}{\partial T}\right)_{p}>0$, so indeed the entropy decreases as pressure increases.
Derivation: Starting with the Gibbs free energy relation: 
$$dG=VdP-SdT$$
$$dG=\left(\frac{\partial G}{\partial P}\right)_{T}dP+\left(\frac{\partial G}{\partial T}\right)_{P}dT$$
we find the thermodynamic definition of volume and entropy:
$$V=\left(\frac{\partial G}{\partial P}\right)_{T}\quad S=-\left(\frac{\partial G}{\partial T}\right)_{P}$$
since we know from calculus:
$$\frac{\partial}{\partial T}\left(\frac{\partial G}{\partial P}\right)_{T}=\frac{\partial}{\partial P}\left(\frac{\partial G}{\partial T}\right)_{P}
 $$
It follows that:
$$\left(\frac{\partial V}{\partial T}\right)_{P}=-\left(\frac{\partial S}{\partial P}\right)_{T}$$
A: You are right: further specifications i.e. constraints have to be provided in order to provide such affirmation. 
Your expression of entropy as a thermodynamical potential in terms of internal energy, volume and particle number, does not provide a unique relation between two thermodynamic variables. Instead it provides a dependence of the entropy, as the main function describing a system, as a function of the parameters which are free and independent of each other in this case.
Choosing for example, the ideal gas as the system, then you are constraining further the system to the ideal gas relation $PV=nRT$, which can in turn give the desired relation as was shown already in the other answer. 
A: Let's start with
$dU = T\,dS - p\,dV$.
The enthalpy is defined as
$H = U + p\,V$,
so
$dH = dU + p\,dV + V\,dp = T\,dS + V\,dp$.
Remember that the enthalpy is the heat at constant pressure, and that
$dH = C_p\,dT$.
This is used to define the heat capacity at constant pressure $C_p$.
From these last equations,
$\displaystyle dS = \frac{C_p}{T} dT - \frac{V}{T}dp$,
and from the definition of exact differential,
$\displaystyle dS = \left.\frac{\partial S}{\partial T}\right)_p dT +
                    \left.\frac{\partial S}{\partial p}\right)_T dp$.
Therefore,
$\displaystyle \left.\frac{\partial S}{\partial p}\right)_T = -\frac{V}{T}$,
which is always negative because both $V$ and $T$ can only be positive.
Notice that you don't need to assume any particular equation of state for the gas, i.e. you don't need to assume that the gas is ideal.
