The equation of state of an ideal elastic substance is:

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

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

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

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


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

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

\begin{equation} \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} \end{equation}

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

\begin{equation} \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} \end{equation}


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

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:

\begin{equation} \mathcal{F} = \frac{KTL}{L_0} \end{equation}


\begin{equation} \boxed{ \frac{\mathcal{F}}{A} + \frac{2KTL_0^2}{AL^2} \neq \frac{\mathcal{F}}{A} + \frac{3KTL_0^2}{AL^2} } \end{equation}

  • $\begingroup$ Just perform the differentiation! $\endgroup$ Commented Aug 26, 2022 at 22:04
  • $\begingroup$ @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. $\endgroup$
    – efirvida
    Commented Aug 27, 2022 at 0:51
  • $\begingroup$ Nobody can identify the error until you show your attempt. $\endgroup$ Commented Aug 27, 2022 at 1:57
  • $\begingroup$ @Chemomechanics, I update the question with the results of my derivative, thanks for your help $\endgroup$
    – efirvida
    Commented Aug 27, 2022 at 11:49
  • $\begingroup$ (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. $\endgroup$ Commented Aug 27, 2022 at 19:40

1 Answer 1


Chemomechanics answered your question, I am merely doing the algebra:

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 $$\boxed{\boxed{ \frac{\mathcal F}{A} + \frac{3 k T L_0^2}{A L^2} =\frac{kT L}{AL_0}+ \frac{2 k T L_0^2}{A L^2} }} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.