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one of the main equation when it comes to Thermodynamics is the relationship between internal energy, heat and work done by the system $q = u + w$. As total derivatives we can write:

$$dQ = dU + dW$$ $$\rightarrow TdS = dU + PdV$$

I'm trying to prove that

$$P = T \left(\frac{\partial P}{\partial T}\right)_V - \left(\frac{\partial E}{\partial V}\right)_T$$ a so called reciprocity relation. I thought what if I could write $TdS$ as $T\partial S$ taking $T$ constant?

if $$P \partial V = T \partial S - \partial E$$

then

$$P = \left(\frac{T\partial S - \partial E}{\partial V}\right)_T = T \left(\frac{\partial S}{\partial V}\right)_T - \left(\frac{\partial E}{\partial V}\right)_T$$

and using one of Maxwell's thermodynamic relations: $$\left(\frac{\partial P}{\partial T}\right)_V = \left(\frac{\partial S}{\partial V}\right)_T$$

last equation could be written as

$$P = T \left(\frac{\partial P}{\partial T}\right)_V - \left(\frac{\partial E}{\partial V}\right)_T$$

I'm puzzled because I suspect that assuming $T\partial S$ is not valid or not always. How could I know under which assumptions it is?

nuwe
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1 Answers1

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The fact the way you proved things are not that correct ( and that lead you the confusion ). I would rather follow the following proof :

$$TdS=dU+pdV$$ Now suppose $S=S(V,T)$ and $U=U(V,T)$ then (I'm doing a little bit of abuse of notation to make it compact so keep track of constant during derivative) $$T\left[\partial_VS \ dV +\partial_T S \ dT\right]=\partial_V U \ dV+\partial_T U\ dT+pdV$$ $$(T\partial_VS-\partial_V U)dV+(T\partial_T S-\partial_T U)dT=pdV $$ Using Maxwell's relation $\partial_VS=\partial_TP$ and equating term of same differential $$T\partial_TP-\partial_V U=pdV$$ or $$T\left(\frac{\partial P}{\partial T}\right)_V-\left(\frac{\partial U}{\partial V}\right)_T=p$$

Himanshu
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