Computing Young's modulus of an ideal elastic substance using it's thermodinamic equation of state

The equation of state of an ideal elastic substance is:

$$$$\mathcal{F} = KT \left[\left(\frac{L}{L_0}\right) - \left(\frac{L}{L_0}\right)^{-2}\right] \tag{1}$$$$

Where $$K$$ is a constant and $$L_0$$ (the value of $$L$$ at zero tension) is a function of temperature only ($$L_0(T)$$).

Show that the isothermal Young's modulus is given by

$$$$Y = \frac{\mathcal{F}}{A} + \frac{3KTL_0^2}{AL^2} \tag{2}$$$$

Exersice 2.7.a of: Heat and Thermodynamics 7th Revised edition. Mark W. Zemansky; Richard H. Dittman

If:

$$$$Y = \frac{L}{A}\left(\frac{\partial \mathcal{F}}{\partial L}\right)_T \tag{3}$$$$

Can perform the derivative of equation 3 with equation 1 as:

\begin{aligned} \mathcal{F} & = KT \left[\left(\frac{L}{L_0}\right)-\left(\frac{L}{L_0}\right)^{-2}\right] \\ & = \frac{KTL}{L_0} - \frac{KTL_0^2}{L^2} \end{aligned} \tag{4}

substituting 4 in 3, solvig the derivarives and reducing terms:

\begin{aligned} Y & = \frac{L}{A} \left(\frac{\partial}{\partial L}\right)_T \left[\frac{KTL}{L_0} - \frac{KTL_0^2}{L^2} \right] \\ & = \frac{L}{A} \left[ \left(\frac{\partial}{\partial L} \frac{KTL}{L_0}\right)_T - \left(\frac{\partial}{\partial L}\frac{KTL_0^2}{L^2}\right)_T \right] \\ & = \frac{L}{A} \left[\frac{KT}{L_0} + \frac{2KL_0^2T}{L^3} \right] \\ \\ & = \frac{L}{A} \left[\frac{KTL^3 + 2KL_0^3T}{L^3 L_0} \right] \\ & = \frac{KTL^3 + 2KL_0^3T}{AL^2 L_0} \\ & = KT\frac{L^3 +2L_0^3}{AL^2 L_0} \\ & = KT\left[ \frac{L}{AL_0} + \frac{2L_0^2}{AL^2} \right] \\ & = \frac{KTL}{AL_0} + \frac{2KTL_0^2}{AL^2} \\ \end{aligned}

Which:

$$$$\boxed{ \frac{KTL}{AL_0} + \frac{2KTL_0^2}{AL^2} \neq \frac{\mathcal{F}}{A} + \frac{3KTL_0^2}{AL^2} }$$$$

Then just by solving the derivative $$\partial \mathcal{F}/\partial L$$ of eq.3 in eq 1. leadsme to a path where I miss the term $$\mathcal{F}/A$$ in eq. 2. That's where I think I'm missing something in the theory, which is what I'm looking for.

Even if I can demostrate that:

$$$$\mathcal{F} = \frac{KTL}{L_0}$$$$

then:

$$$$\boxed{ \frac{\mathcal{F}}{A} + \frac{2KTL_0^2}{AL^2} \neq \frac{\mathcal{F}}{A} + \frac{3KTL_0^2}{AL^2} }$$$$

• Just perform the differentiation! Commented Aug 26, 2022 at 22:04
• @Chemomechanics, that's what I tried but, If I perform the derivative $\partial \mathcal{F}/\partial L$ of eq.1. I will miss the term $\mathcal{F}/A$ in eq 2. That's why I think that I'm missing something. Or doing the derivative in wrong way. Commented Aug 27, 2022 at 0:51
• Nobody can identify the error until you show your attempt. Commented Aug 27, 2022 at 1:57
• @Chemomechanics, I update the question with the results of my derivative, thanks for your help Commented Aug 27, 2022 at 11:49
• (Edited to correct typo.) I don’t understand the inequality you wrote twice. If you plug in $\mathcal{F}$, you can verify that it’s an equality. Commented Aug 27, 2022 at 19:40

You say $$\boxed{ \frac{k T L}{A L_0} + \frac{2 k T L_0^2}{A L^2} \neq \frac{\mathcal F}{A}+\frac{3 k T L_0^2}{A L^2} }$$ but this is not true: From your Eq (1) $$\frac{\mathcal F}{A} =\frac{kT L}{AL_0}-\frac{k T L_0^2}{L^2}$$ Then $$\boxed{\boxed{ \frac{\mathcal F}{A} + \frac{3 k T L_0^2}{A L^2} =\frac{kT L}{AL_0}+ \frac{2 k T L_0^2}{A L^2} }}$$