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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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2 Answers2

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$\left(\frac{\partial U}{\partial V}\right)_{T,N}=\left(\frac{\partial U}{\partial P}\right)_{T,N}=0$ for the ideal gas. Thus, there's no contradiction with $\Delta U\neq 0$ for pressure–volume work on an closed adiabatic ideal gas system, as the temperature varies.

Chemomechanics
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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 $T$.

It not strictly true that internal energy depends only on the temperature. It also depends on $N$, the number of particles. But, let's assume that $N$ is fixed. (Or, hold it fixed in all partial derivatives).

I'm not sure if the authors mean to say that $\biggl(\frac{\partial U}{\partial P}\biggr)_V = \biggl(\frac{\partial U}{\partial V}\biggr)_P = 0$

No, neither of the above are true.

Remember, for an ideal gas: $$ U = \frac{3Nk}{2}T = \frac{3}{2}PV\;. $$

So, for example $$ \left(\frac{\partial U}{\partial P}\right)_V = \frac{3V}{2}\;. $$

or if $V = V(T)$ and $P = P(T)$

Well, not really. For example, for an ideal gas we have: $$ V = \frac{2U}{3P} = \frac{NkT}{P} = V(T,P,N)\;. $$

What am I to conclude from this?

That you have made and error in reasoning, as explained above.

hft
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