Thermodynamics - Calorific Capacity

AI Thread Summary
The discussion focuses on deriving the total differential of entropy (dS) and volume (dV) in a thermodynamic system with constant particle numbers. The user correctly expresses dS in terms of temperature and volume derivatives, incorporating the calorific capacity at constant volume (CV). However, confusion arises regarding the interpretation of partial derivatives, particularly in equating terms involving constant pressure assumptions. The key issue is the inconsistency in applying the definitions of the partial derivatives, which affects the final relationship between CP and CV. Clarification on the proper use of these derivatives is necessary for accurate results.
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Homework Statement



Consider a system with a constant number of particles.
Write the total differential dS in terms of the derivarives \frac{∂S}{∂T} and \frac{∂S}{∂V}. Introduce CV (calorific capacity at constant volume).
Next write the total differential of the volume dV in terms of the parcial derivatives \frac{∂V}{∂T} and \frac{∂V}{∂P}. Assume that the pressure is constant. Show that the result comes in the form of:

CP-CV= Expression

Homework Equations



CP=T\frac{∂S}{∂T} , P and N Constant
CV=T\frac{∂S}{∂T} , V and N Constant


The Attempt at a Solution



First I wrote the differential of the entropy as asked:

dS = \frac{∂S}{∂T}dT + \frac{∂S}{∂V}dV

I know that \frac{∂S}{∂V} = CV/T. Substituting I get:

dS = CV/T dT + \frac{∂S}{∂V}dV

Next I found the differential of the volume:

dV = \frac{∂V}{∂T}dT + \frac{∂V}{∂P}dP

Since the pressure is constant it reduces to the form

dV = \frac{∂V}{∂T}dT

Substituting in our dS expression we get:

dS = CV/T dT + \frac{∂S}{∂V}\frac{∂V}{∂T}dT =
= CV/T dT + \frac{∂S}{∂T}dT =
= CV/T dT + CP/T dT ⇔
⇔ T\frac{dS}{dT} = CV + CP

I'm making some mistake. If anyone could point me in the right direction I'd appreciate.

Thanks!
 
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I think your problem is a confusion over the meanings of different partial derivatives. Your use of ##\frac{∂S}{∂V}## refers to changing volume, keeping temperature constant.
That's inconsistent with later equating ##\frac{∂S}{∂V}\frac{∂V}{∂T} = \frac{∂S}{∂T}##, which, to be valid, assumes pressure constant in all terms.
 
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