0
$\begingroup$

Consider a box with substance inside. It has a fixed, constant volume and is at constant temperature. The box is split into two halves by a piston.

How does the Helmholtz free energy change if the piston is moved to the right so that the volume to right of the piston decreases by a small amount $\delta V$?

In the question, I'm meant to derive an expression for this using a Taylor series. I would appreciate some help with that as well (I already know what a Taylor series is, I just don't understand how to use it here), but if that is too 'homework-like' for this site, I would also very much appreciate just a theoretical explanation on what happens here (i.e. how the Helmholtz free energy changes).

One of my main problems is that the box's volume and temperature are fixed, so I'd assume the $F(V,T)$ Helmholtz function would be fixed as well, yet evidently it is not.

Help would be much appreciated.

Improve this question
$\endgroup$
4
  • $\begingroup$ hint: at constant temperature the change in free energy is equal to the reversible work done. (you only need 1st order expansion: $\delta M(x) = M(x+\delta x) - M(x) \approx \delta x \frac{dM}{dx}$, no need for anything more complicated Taylor "thing") $\endgroup$
    – hyportnex
    Commented May 10, 2023 at 0:26
  • $\begingroup$ What are $M$ and $x$ here? One of my issues with finding the Taylor expansion (even just its first order part like you did) is that I don't know what $x$ and $x_0$ are meant to be in $f(x) = f \left(x_0\right) + f' \left(x_0\right) \left(x-x_0\right)$ because volume is constant. Is $x$ meant to be $V$? Is it $V/2$? Is $x-x_0 = \delta V$? $\endgroup$
    – Trisztan
    Commented May 10, 2023 at 1:07
  • $\begingroup$ your problem states that a piston dividing the fixed volume can be moved by a small amount to the right, that is the $\pm \delta V$, so that one side will be $V+\delta V$ and the other $V - \delta V$. $\endgroup$
    – hyportnex
    Commented May 10, 2023 at 1:12
  • $\begingroup$ So I'm only using the Taylor expansion for one of the directions (either $\delta V$ or $-\delta V$)? And then is $x= V+\delta V$ with $x_0 = V$? Or the other way around? $\endgroup$
    – Trisztan
    Commented May 10, 2023 at 2:29

1 Answer 1

Reset to default
1
$\begingroup$

Fixing the total volume does not fix the free energy of a compound system of two subvolumes. The free energy of a compound system is the sum of the component's free energies. If the two subsystems (say, 1 and 2) have a fixed common temperature $T$, and a fixed number of particles ($N_1$, and $N_2$), the condition of a fixed total volume $V=V_1+V_2$ implies only that variations of the two subvolumes must compensate ($\delta V_1 = -\delta V_2$), i.e., their values are not independent The total free energy $$ F(T,V,N,V_1) = F(T,V_1,N_1) + F(T,V-V_1,N_2) $$ in general is a (non-constant) function of $V_1$ (at fixed $N,V,T$). It can usually be expanded in the Taylor series around a fixed value of $V_1$.

Improve this answer
$\endgroup$

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