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When all other thermodynamic variables pertaining to a system are kept constant, how does the entropy $S$ of a system change with the increase in temperature? Does it strictly increase with the increase in temperature? Whatever is the behaviour i.e., whether entropy increase or decrease with the increase in temperature be worked out from the thermodynamic relations?

One way to find the answer is to find the sign of the derivative $\Big(\frac{\partial S}{\partial T}\Big)_{V,N}$. If it's positive, then I can say for sure that $S$ increases with $T$.

Please do not answer by citing examples of the temperature dependence of entropy of some specific systems.

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    $\begingroup$ Rigorous proof according to what axioms? $\endgroup$
    – AccidentalFourierTransform
    Commented May 20, 2018 at 16:25
  • $\begingroup$ @AccidentalFourierTransform Basic thermodynamic relations for PVT system (laws of thermodynamics), Maxwell's relations etc. But I do not want examples such as ideal gas entropy. $\endgroup$
    – Solidification
    Commented May 20, 2018 at 16:38
  • $\begingroup$ Related: physics.stackexchange.com/questions/232990/… $\endgroup$
    – Chemomechanics
    Commented May 20, 2018 at 17:11
  • $\begingroup$ A fully rigorous proof (given in an explicit system of axioms) is provided in Section V of this well-known paper by Lieb and Yngvason. Note that it requires dozens of pages of development to get to this point... $\endgroup$
    – Yvan Velenik
    Commented May 20, 2018 at 17:17

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If it is OK to take the following as a starting point, then I will proceed further: $$dU=TdS-PdV+\mu dN$$ If I use this equation to determine the partial derivative of U with respect to T at constant V and N, I obtain:$$\left(\frac{\partial U}{\partial T}\right)_{V,N}=T\left(\frac{\partial S}{\partial T}\right)_{V,N}$$But, the left hand side of this equation is equal to $NC_V$. Therefore, $$\left(\frac{\partial S}{\partial T}\right)_{V,N}=N\frac{C_V}{T}$$The sign of this is always positive.

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