0
$\begingroup$

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 $C_V$ $$ \left( \frac{\partial U}{\partial p} \right)_V = \left( \frac{\partial U}{\partial T} \right)_V \left( \frac{\partial T}{\partial p} \right)_V = C_V \frac{K_T}{\alpha} = \frac{C_V K_T}{\alpha} $$

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.