Thermodynamics Problem: Exploiting Identity (∂s/∂T)_x>0

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SUMMARY

The discussion focuses on the thermodynamic identity (∂s/∂T)_x>0, which holds true for all processes where x is an arbitrary intensive or extensive variable. Participants clarify that this identity can be derived from the definition of specific heat at constant x, expressed as c_X = (1/m)*(dQ/dT)_X. By applying the Second Law of Thermodynamics and manipulating the equations, it is established that (∂s/∂T)_X = (c_X)/T > 0, confirming the positive nature of the entropy change with respect to temperature.

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Homework Statement



Can anyone help me to start with the Thermodynamic Identity:

(∂s/∂T)_x>0 for all processes where x is an arbitrary intensive or extensive variable of of the system.

Homework Equations


The Attempt at a Solution



I do understand that we have to exploit some property of the properties of the system.Say all of them are exact differential or so...But cannot understand how to do this Identity...
Can anyone please help me to start with it?
 
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Hope this makes sense...Please check it.


First, we recall the definition of "Specific Heat At Constant X", where
"X" is an arbitrary intensive variable (like pressure) or extensive
variable (like volume):

{Specific Heat At Constant X} = c_X = (1/m)*(dQ/dT)X

----> dQ = m*(c_X)*dT ... for constant X

where "m" is system mass. Thus, from the definition of entropy &
specific entropy (s = S/m), and the Second Law Of Thermodynamics
(for cyclic process), we have:

dS = dQ/T > 0

----> dS = {m*(c_X)*dT}/T > 0 ... for constant X

----> ds = {(c_X)*dT}/T > 0 (<--- dividing by m, & s = S/m)

----> (∂s/∂T)_X = (c_X)/T > 0 (<--- dividing by dT & at const X)

----> (∂s/∂T)_X > 0
 

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