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If $h=h(T, P)$.

Does $ dh = c_pdT + \left[v - T\left(\frac{\partial v}{\partial T}\right)_P \right]dP \Rightarrow h_2 - h_1 = \int_{T_1}^{T_2} c_pdT + \int_{P_1}^{P_2}\left[v - T\left(\frac{\partial v}{\partial T} \right)_P\right]dP $ ?

If so, how?

I apologize for this, but I just haven't been able to find an appropiate justification for this operative behavior in any of the Calculus, Differential Equations and Thermodynamics books in my possession. I'm particular bugged by the "integration of differentials" and how it, before me, seems to damage the symmetry of the first equation in the statement.

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We can always write

$$dh=\left(\frac{\partial h}{\partial T}\right)_P dT+ \left(\frac{\partial h}{\partial P}\right)_T dP;$$

this is just expansion in $T$ and $P$. By definition,

$$c_P\equiv\left(\frac{\partial h}{\partial T}\right)_P.$$

Then, we can write $h=g+Ts$ (by definition of $g$) and $dg=-s\,dT+v\,dP$ (the fundamental relation) and thus obtain

$$\left(\frac{\partial h}{\partial P}\right)_T=\left(\frac{\partial g+Ts}{\partial P}\right)_T=v+T\left(\frac{\partial s}{\partial P}\right)_T=v-T\left(\frac{\partial v}{\partial T}\right)_P=v(1-T\alpha),$$

where we've used a Maxwell relation to go from $(\partial s/\partial P)_T$ to $-(\partial v/\partial T)_P$. This gives

$$dh=c_P\, dT+ v(1-\alpha T) dP. $$

Make sense?

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  • $\begingroup$ I had this exact question just days ago $\endgroup$ – morbo Sep 21 at 10:31
  • $\begingroup$ I was explained that concerning my question, the operation that justifies the second equation is a line integral. Still, I do appretiate you contribution; thank you. $\endgroup$ – Wachuke Oct 5 at 0:17

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