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In an undergraduate thermodynamics book, the authors use the case of free expansion of an ideal gas to argue that the internal energy $U$ of the gas depends only on its temperature $T$. I'm not sure if the authors mean to say that $\biggl(\frac{\partial U}{\partial P}\biggr)_V = \biggl(\frac{\partial U}{\partial V}\biggr)_P = 0$ or if $V = V(T)$ and $P = P(T)$ - in other words, all changes in internal energy are ultimately attributable to changes in temperature. I assume it must be the latter.

Supposing it is the former - i.e. $\biggl(\frac{\partial U}{\partial P}\biggr)_V = \biggl(\frac{\partial U}{\partial V}\biggr)_P = 0$, then let us consider the adiabatic expansion of a gas from volume $V_1$ to $V_2$. The First Law tells us that

\begin{eqnarray} \Delta U &=& Q – W \\ &=& 0 -P(V_2 – V_1) = - P\Delta V \end{eqnarray}

What am I to conclude from this?

Mauricio
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Zachary Candelaria
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  • You must indicate in your partial derivatives what’s being held constant; otherwise, the interpretation is ambiguous. 2. The heat Q disappeared in your derivation. Please consider editing your question to clarify.
    • – Chemomechanics Apr 07 '23 at 16:25
  • @Chemomechanics Actually, there is no ambiguity. If $f(x_1,x_2,...,x_n)$ is differentiable in all of its variables, then the definition of the partial $\frac{\partial f}{\partial x_i}$ ensures that all variables save $x_i$ are held constant. The notational convention of explicitly showing which variables are held constant is unnecessary. Secondly, $Q$ vanishes because an adiabatic process is one in which there is no heat transfer. – Zachary Candelaria Apr 07 '23 at 17:15
  • That's not how partial derivatives work in thermodynamics contexts; see this explanation, for example. – Chemomechanics Apr 07 '23 at 17:29
  • Ah, I see! Now it makes sense. Thanks – Zachary Candelaria Apr 07 '23 at 17:38
  • So $nC_v\Delta T=-P\Delta V$ – Chet Miller Apr 07 '23 at 20:16