Why is the thermal expansion coefficient of an incompressible substance constant with pressure?

Question

Upon learning that $C_P=C_V$ for incompressible substances, the takeaway for why this is true is due to volume being constant, which comes from the thermal compressibility, $\beta_T=0$ for an incompressible substance. This information I understand.

I am confused about incompressibility's effect on enthalpy and internal energy. An example from my textbook describes the effect that incompressibility has on changes in volume, enthalpy, internal energy, and entropy with a solid that undergoes a change in pressure at constant temperature.

The volume and entropy are constant, which makes sense to me. I fail to understand the effect this situation has on enthalpy and internal energy. It starts with the equation for isothermal, pressure change of enthalpy: $dh=[v-T(\frac{\partial v}{\partial T})_P]dP=v[1-T\alpha]dP$, where $\alpha$ is the coefficient of thermal expansion. It then says that $\alpha$ is "essentially constant with pressure" with no further explanation. Every source I read says this is valid to assume, but I cannot find any reasoning.

Furthermore, I am confused on the connection with internal energy and its relation with $\alpha$. When changing pressure in isothermal process, my book asserts there's no change in entropy, since entropy can be written as $ds=\frac{C_P}{T}dT-(\frac{\partial v}{\partial T})_PdP=\frac{C_P}{T}dT-\beta_Tv(\frac{\partial P}{\partial T})_vdP$, isothermal so $dT=0$, and incompressible, so $\beta_T=0$. Thus, $ds=0$, implying constant entropy.

Using this same logic, I assumed internal energy would be constant as well, because $du=Tds-Pdv$, both entropy and volume are constant, so they are zero. Yet, my book says internal energy would be $du=-vT\alpha dP$, coming from $u=h-Pv$.

With that, my questions are: For an incompressible substance, why can we assume $\alpha$ to be constant with pressure? Second, why is internal energy non-zero?

Answer 1

(Summarizing comments as an answer.)

1. No material is perfectly incompressible. The idealization of incompressibility, or $\left(\frac{\partial V}{\partial P}\right)_T\approx 0$, doesn't necessarily imply that the volume is always constant, or $dV=0$, which is a distinct assumption. It implies only that the volume remains essentially constant as any amount of reasonable pressure is applied and nothing else is happening. But if notable thermal expansion/contraction, or $\left(\frac{\partial V}{\partial T}\right)_P\neq 0$, is happening, then the volume could be changing significantly.

2. If one wants to consider a near-incompressible material that does exhibit notable thermal expansion, one has to be careful with the triple-product rule $\left(\frac{\partial V}{\partial T}\right)_P=-\left(\frac{\partial V}{\partial P}\right)_T\left(\frac{\partial P}{\partial T}\right)_V$ because a well-defined thermal expansion coefficient $\alpha\equiv\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$ and a minuscule compressibility $\beta_T\equiv-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$ requires $\left(\frac{\partial P}{\partial T}\right)_V$ to have a very large magnitude indeed. (The physical interpretation is that we'd have a hard time ever stopping the material from thermally expanding or contracting because it's so stiff.) Contradictions can arise if $\beta_T$ or $\left(\frac{\partial V}{\partial P}\right)_T$ is taken as exactly zero.

3. The dependence of the thermal expansion coefficient $\alpha$ on pressure (at constant temperature), or $$\begin{align}\left(\frac{\partial \alpha}{\partial P}\right)_T&=\frac{\partial}{\partial P}\left[\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P\right]_T\\&=-\frac{1}{V^2}\left(\frac{\partial V}{\partial P}\right)_T\left(\frac{\partial V}{\partial T}\right)_P+\frac{1}{V}\left(\frac{\partial^2 V}{\partial P\partial T}\right)_{T,P}\\&=\beta_T\alpha-\left(\frac{\partial\beta_T}{\partial T}\right)_P,\end{align}$$ may or may not be reasonably assumed to be negligible depending on the material and range of pressures being considered; the expansion makes it clear that the assumption is bolstered if $\beta_T$ is small at all temperatures. Try looking up the $\alpha$ vs. $P$ data for your material of interest and seeing if the pressure dependence of $\alpha$ is notable. (Example for amorphous polymers. Example for alkali halides. Example for iron.)

4. Agreed that near-incompressibility implies near-constant entropy for an isothermal single phase when only pressure–volume work is being considered and thermal expansion is being ignored; thank you for clarifying the reasoning on that point. (If a phase change occurs, for example, then the entropy would be nonconstant even at constant temperature. The same applies if other types of work or effects exist that tend to change the volume, for example.)

5. Constant internal energy ($dU=0$), doesn't imply that $U$ is zero; $U=U_0+\int dU$, where $U_0$ is an arbitrary reference zero. The internal energy doesn't really have an absolute reference zero; only changes in $U$ are relevant in this context.

6. We can expand the internal energy for the system described above as a function of two variables, $T$ and $P$, and apply the assumptions of constant temperature and near-incompressibility as follows:

$$\begin{align}\require{cancel}dU&=\left(\frac{\partial U}{\partial T}\right)_P \cancelto{0}{dT}+\left(\frac{\partial U}{\partial P}\right)_TdP\\&=\left(\frac{\partial (T\,dS-P\,dV)}{\partial P}\right)_TdP\\&=\left[T\left(\frac{\partial S}{\partial P}\right)_T-P\cancelto{\approx 0}{\left(\frac{\partial V}{\partial P}\right)_T}\right]dP\\&=-T\left(\frac{\partial V}{\partial T}\right)_PdP\\&=-TV\alpha\,dP,\end{align}$$

where I've used a Maxwell relation to convert the partial derivative of entropy. This agrees with the expression in your book. Note that the material property $\alpha$ can still appear in an expression even if the temperature is assumed to be constant in this particular example. Note also that if thermal expansion is negligible, then both $\alpha\approx 0$ and $dS\approx 0$ imply that $dU\approx 0$.

I believe this reconciles the issues of the compressibility possibly being near zero, the thermal expansion possibly being nonzero, $dU$ being essentially zero upon compression only but possibly nonzero if thermal expansion/contraction is occurring, and $U$ generally being nonzero (depending on the arbitrary reference zero that has been set). The main point to emphasize is that since $\beta_T$ being exactly zero is unphysical, it may be necessary to assume only that it's very small to avoid contradictions in other derivations. I don't know what text you're using, so I can't discuss how this point is addressed there.