Finding equation of state from thermal expansion coefficient and isothermal compressibility

Question

I'm stuck on a problem that I found in a book (Modern Thermodynamic with Statistical Mechanics, Helrich S., problem 5.2).

The text of the problem is that:

Consider a solid material for which:

$$ \frac{1}{\kappa_T} = \frac{\varepsilon}{2V_0}\left[\frac{2\Gamma c_v T}{\varepsilon}\,\frac{V_0}{V} - 3\left(\frac{V_0}{V}\right)^3\right] $$ $$ \beta = \frac{1}{T}\left[1 + 3\,\frac{\varepsilon}{2\Gamma c_v T}\left(\frac{V_0}{V}\right)^2\right] $$

Where $\varepsilon$ is a constant with the units of energy, $\Gamma$ is a dimensionless constant and $V_0$ is a reference volume less than $V$. The temperature range is such that we may assume that the specific heat at constant volume $c_v$ is independent of temperature. Find the thermal equation of state.

In this book the convention defines $\kappa_T$ as the isothermal compressibility and $\beta$ as the thermal expansion coefficient.

$$ \beta = \frac 1V \left(\frac{\partial V}{\partial T}\right)_P $$

$$ \kappa = - \frac 1V \left(\frac{\partial V}{\partial P}\right)_T $$ The answer key for this problem says:

$$ P(V,T) = \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2V_0}\left[\left(\frac{V_0}{V}\right)^5 - \left(\frac{V_0}{V}\right)^3\right] $$

Here's the procedure I tried to apply. I wanted to do a contour integration of this equation because I can express the partial derivatives as a function of $\kappa _T$ and $\beta$: $$ dP(T,V) = \left(\frac{\partial P}{\partial V}\right)_TdV + \left(\frac{\partial P}{\partial T}\right)_VdT $$ given that: $$ \left(\frac{\partial P}{\partial T}\right)_V = \frac{\beta}{\kappa_T} $$ $$ \left(\frac{\partial P}{\partial V}\right)_T = - \frac{1}{V \kappa_T} $$

But by using this procedure I have a problem with the integration of $c_v$ with respect to the volume, and even if I assume this constant I still don't get the result.

Which approach would you suggest?

Answer 1

Your approach is all right but the solution given by the textbook is wrong :), at least if no approximation is to be made.

Let's go the other way around: start from $$P(V,T) = \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2V_0}\left[\left(\frac{V_0}{V}\right)^5 - \left(\frac{V_0}{V}\right)^3\right]$$ and then derive the values of $\kappa_T$ and $\beta$ (by the way, shouldn't it rather be $\chi_T$ (see here) and $\alpha$ (here)?). As you mentioned, to do this, we need to compute partial derivatives of $P$:

$$ \left(\frac{\partial P}{\partial T}\right)_V = \frac{\Gamma c_v}{V} $$

$$ \left(\frac{\partial P}{\partial V}\right)_T = -\left(\frac{\Gamma c_vT}{V^2} + \frac{\varepsilon}{2{V_0}^2}\left[5\left(\frac{V_0}{V}\right)^6 - 3\left(\frac{V_0}{V}\right)^4\right]\right) $$

which give

$$ \frac{1}{\kappa_T} = -V \left(\frac{\partial P}{\partial V}\right)_T = \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2{V_0}}\left[5\left(\frac{V_0}{V}\right)^5 - 3\left(\frac{V_0}{V}\right)^3\right] $$ and

$$ \beta = \kappa_T \left(\frac{\partial P}{\partial T}\right)_V = \frac1T \left(1 + \frac{\varepsilon }{2\Gamma c_V T}\left[5\left(\frac{V_0}{V}\right)^4 - 3\left(\frac{V_0}{V}\right)^2\right] \right)^{-1} $$

Clearly, that's not what the textbook gives in the first place, but it's close enough to understand what they did: a Taylor expansion at order 3 in $V_0/V$ in booth cases, which gives back the expressions

$$ \frac{1}{\kappa_T} \approx \frac{\Gamma c_vT}{V} + \frac{\varepsilon}{2{V_0}}\left[ - 3\left(\frac{V_0}{V}\right)^3\right] $$

and

$$ \beta \approx \frac1T \left(1 + \frac{\varepsilon }{2\Gamma c_V T}\left[ - 3\left(\frac{V_0}{V}\right)^2\right] \right)^{-1} \approx \frac1T \left(1 + \frac{\varepsilon }{2\Gamma c_V T}\left[ 3\left(\frac{V_0}{V}\right)^2\right] \right) $$

It really should have been clearer in the text that you could suppose $V\gg V_0$ (and not only $V>V_0$) and I don't see how you could derive the term in $\left(\frac{V_0}{V}\right)^5$ from this as it is completely neglected to find back $\kappa_T$ and $\beta$