I need to obtain the derivative of internal energy $U$ w.r.t. pressure $p$ at a constant volume $V$. Realizing that $\mathrm{d} U = T \mathrm{d} S - p \mathrm{d} V$, I rewrite $$ \left( \frac{\partial U}{\partial p} \right)_V = \left( \frac{\partial U}{\partial S} \right)_V \left( \frac{\partial S}{\partial p} \right)_V $$

The first term is equal to $T$ $$ \left( \frac{\partial U}{\partial S} \right)_V = T $$

For the second term I used $$ \mathrm{d} U = T \mathrm{d} S - p \mathrm{d} V $$ again, from which $$ \left( \frac{\partial U}{\partial S} \right)_V = T, \quad \left( \frac{\partial U}{\partial V} \right)_S = - p $$ and therefore $$ \left( \frac{\partial p}{\partial S} \right)_V = -\left( \frac{\partial T}{\partial V} \right)_S \quad \implies \quad \left( \frac{\partial S}{\partial p} \right)_V = \frac{1}{\left( \frac{\partial p}{\partial S} \right)_V} = - \frac{1}{\left( \frac{\partial T}{\partial V} \right)_S} = - \left( \frac{\partial V}{\partial T} \right)_S $$

So far, we have $$ \left( \frac{\partial U}{\partial p} \right)_V = \left( \frac{\partial U}{\partial S} \right)_V \left( \frac{\partial S}{\partial p} \right)_V = - T \left( \frac{\partial V}{\partial T} \right)_S $$

For $(\partial V/\partial T)_S$ I use $$ \left( \frac{\partial x}{\partial y} \right)_z \left( \frac{\partial y}{\partial z} \right)_x \left( \frac{\partial z}{\partial x} \right)_y = -1 $$ with $x = V$, $y = T$ and $z = S$ $$ \left( \frac{\partial V}{\partial T} \right)_S \left( \frac{\partial T}{\partial S} \right)_V \left( \frac{\partial S}{\partial V} \right)_T = -1 $$

Using the definition for the heat capacity at constant volume, we get $$ C_V = \left( \frac{\partial Q}{\partial T} \right)_V = T \left( \frac{\partial S}{\partial T} \right)_V \quad \implies \quad \left( \frac{\partial T}{\partial S} \right)_V = \frac{1}{\left( \frac{\partial S}{\partial T} \right)_V} = \frac{1}{C_V / T} = \frac{T}{C_V} $$

For $\left( \frac{\partial S}{\partial V} \right)_T$ I use Helmholtz free energy $$ \mathrm{d} F = - p \mathrm{d} V - S \mathrm{d} T $$ so $$ \left( \frac{\partial F}{\partial V} \right)_V = - p, \quad \left( \frac{\partial F}{\partial T} \right)_S = - S $$ therefore $$ \left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial p}{\partial T} \right)_V $$ and for the last time $$ \left( \frac{\partial p}{\partial T} \right)_V \left( \frac{\partial T}{\partial V} \right)_p \left( \frac{\partial V}{\partial p} \right)_T = -1 $$ with $$ \alpha = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_p \quad \implies \quad \left( \frac{\partial T}{\partial V} \right)_p = \frac{1}{\left( \frac{\partial V}{\partial T} \right)_p} = \frac{1}{\alpha V} $$ and $$ K_T = - \frac{1}{V} \left( \frac{\partial V}{\partial p} \right)_T \quad \implies \quad \left( \frac{\partial V}{\partial p} \right)_T = - K_T V $$ and therefore $$ \left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial p}{\partial T} \right)_V = - \frac{1}{\left( \frac{\partial T}{\partial V} \right)_p \left( \frac{\partial V}{\partial p} \right)_T} = \frac{\alpha}{K_T} $$ going back to $$ \left( \frac{\partial V}{\partial T} \right)_S = - \frac{1}{\left( \frac{\partial T}{\partial S} \right)_V \left( \frac{\partial S}{\partial V} \right)_T} = - \frac{1}{(T/C_V) (\alpha/K_T)} = - \frac{C_V K_T}{\alpha T} $$ and going back to the internal energy... $$ \left( \frac{\partial U}{\partial p} \right)_V = - T \left( \frac{\partial V}{\partial T} \right)_S = - T \left( - \frac{C_V K_T}{\alpha T} \right) = \frac{C_V K_T}{\alpha} $$

However, the formula should be $$ \left( \frac{\partial U}{\partial p} \right)_V = \frac{C_p K_T}{\alpha} - \alpha V T $$

My question is: how do I obtain the second formula as quickly as possible without using Mayer's relation $C_p - C_V = (\alpha^2/K_T) V T$?

P.S. I also found a quicker way to obtain the formula with $C_V$ $$ \left( \frac{\partial U}{\partial p} \right)_V = \left( \frac{\partial U}{\partial T} \right)_V \left( \frac{\partial T}{\partial p} \right)_V = C_V \frac{K_T}{\alpha} = \frac{C_V K_T}{\alpha} $$


1 Answer 1


$$dH=dU+pdV+Vdp=C_pdT+V(1-\alpha T)dP$$So, $$dU=C_pdT-pdV-V\alpha Tdp$$The rest is basically what you've already done with your second method.

  • $\begingroup$ Would you please explain how we get the first equation? I don't quite see the line of reasoning... $\endgroup$
    – user16320
    Jun 29, 2020 at 23:28
  • $\begingroup$ The first equation for dH is derived in every thermo book. The derivation starts from dH=TdS+Vdp. Then we express dS in terms of partial derivatives with respect to T and p. Then the Maxwell equation is used to express partial S with respect to p in terms of partial V with respect to T. Hope this makes sense. $\endgroup$ Jun 29, 2020 at 23:37
  • $\begingroup$ The Maxwell relation I alluded to starts from dG=-SdT+Vdp $\endgroup$ Jun 29, 2020 at 23:58
  • $\begingroup$ It does. So since you express dS in terms of dT and dp, you actually almost do a derivation of Mayer's relation (which also uses dS expressed via dT and dp instead of dT and dV)... $\endgroup$
    – user16320
    Jun 30, 2020 at 3:36
  • $\begingroup$ Actually, this relationship is more fundamental than Mayer's relation, and Mayer's relation is typically derived from it, rather than the other way around. $\endgroup$ Jun 30, 2020 at 12:29

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