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

In summary, the Thermodynamic Identity (∂s/∂T)_x>0 holds for all processes where x is an arbitrary intensive or extensive variable of the system. This is based on the definition of specific heat at constant x, the definition of entropy, the Second Law of Thermodynamics, and the properties of cyclic processes.
  • #1
neelakash
511
1

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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  • #2
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
 

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

What is the meaning of (∂s/∂T)_x?

The symbol (∂s/∂T)_x represents the partial derivative of entropy (s) with respect to temperature (T) at constant pressure and composition.

Why is (∂s/∂T)_x>0 important in thermodynamics?

This expression represents the rate of change of entropy with respect to temperature, and when it is greater than zero, it indicates that the system is undergoing an irreversible process. This is important because irreversible processes lead to an increase in entropy, which is a fundamental concept in thermodynamics.

How is (∂s/∂T)_x calculated?

The value of (∂s/∂T)_x can be calculated using the Maxwell relation, which is a mathematical expression that relates partial derivatives of thermodynamic properties.

What does it mean when (∂s/∂T)_x>0 for a system?

If (∂s/∂T)_x is greater than zero, it means that the system is undergoing an irreversible process and the entropy of the system is increasing. This could be due to factors such as heat transfer, chemical reactions, or other irreversible processes.

Can (∂s/∂T)_x ever be negative?

No, (∂s/∂T)_x cannot be negative. This is because entropy is a measure of the disorder or randomness of a system, and it can only increase or remain constant, but never decrease. A negative value for (∂s/∂T)_x would imply a decrease in entropy, which goes against the laws of thermodynamics.

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