How do long-range interactions between atoms affect the virial theorem?

[eme]

Homework Statement

Suppose that there are long-range interactions between atoms in a gas in the form of central forces derivable from a potential. $$V(r) = \frac k r^m $$ where r is the distance between any pair of atoms and m is a positive integer. Assume further that relative to any given atom the other atoms are distributed in space such that their volume density is given by the Boltzmann favor: $$ \rho(r) = \frac N V e^{\frac -U(r) kT},$$ where N is the total number of atoms in a volume V. Find the addition to the virial of Clausius resulting from these forces between pairs of atoms, and compute the resulting correction to Boyle's Law. Take N so large that sums may be replaced by integrals. While closed results can be found for any positive $m$, if desired, the mathematics can be simplified by taking $m = +1$

Homework Equations

$$ \overline T = -\frac 1 2 \overline{ \sum_i \mathbf{F_i \cdot r_i} }$$, where the right-hand term is the addition to the virial of Clausius.

and if $V(r) = a r^n$ then $$\overline T = -\frac 1 2 \overline V$$

The Attempt at a Solution

My idea is for the first part, to find the addition to the virial of Clausius, find the force $\mathbf F = -\nabla V$ so i can write it the first equation. For the second part I'm kind of lost, i want to use the potential energy but I'm not really sure how to find the average potential energy.

The problem is 3.12 in Goldstein third edition.

 

[mark726]

Hello,

Thank you for bringing this interesting problem to our attention. I have worked through the solution and here is my approach:

For the first part, we need to find the addition to the virial of Clausius. To do this, we need to find the force between two atoms, which is given by $$\mathbf{F} = -\nabla V = -\frac{km}{r^{m+1}}\mathbf{r}$$
where $\mathbf{r}$ is the distance between the two atoms. Now, we can rewrite this as $$\mathbf{F} = -\frac{km}{r^{m+1}}\frac{\mathbf{r}}{r} = -\frac{km}{r^{m+2}}\mathbf{r}$$
Note that this is a central force, which means it only depends on the distance between the two atoms. Now, we can use this force to find the addition to the virial of Clausius, which is given by $$\overline{T} = -\frac{1}{2}\overline{\sum_i \mathbf{F_i \cdot r_i}}$$
where $\mathbf{F_i}$ is the force on the $i^{th}$ atom and $\mathbf{r_i}$ is the position of the $i^{th}$ atom. Since the force is central, we can rewrite this as $$\overline{T} = -\frac{1}{2}\overline{\sum_i \mathbf{F_i \cdot r_i}} = -\frac{1}{2}\overline{\sum_i F_i r_i} = -\frac{1}{2}\overline{\sum_i F_i}\overline{\sum_i r_i}$$
Note that the average of the position of the atoms is zero, since they are distributed evenly in space. Therefore, we can simplify this to $$\overline{T} = -\frac{1}{2}\overline{\sum_i F_i}\overline{\sum_i r_i} = -\frac{1}{2}\overline{\sum_i F_i}\cdot 0 = 0$$
Hence, we see that there is no addition to the virial of Clausius due to these long-range interactions between atoms.

For the second part, we need to compute the resulting correction to Boyle