Starting with the virial theorem for an equilibrium configuration show that:

(a) the total kinetic energy of a finite gaseous configuration is equal to the total internal energy if $\gamma=C_{p} / C_{v}=5 / 3, \text { where } C_{p} \text { and } C_{v}$ are the molar specific heats of the gas at constant pressure and at constant volume, respectively,

(b) the finite gaseous configuration can be in Newtonian gravitational equilibrium only if $C_{p} / C_{v}>4 / 3$ .

Question

Starting with the virial theorem for an equilibrium configuration show that:

(a) the total kinetic energy of a finite gaseous configuration is equal to the total internal energy if [latex]\gamma=C_{p} / C_{v}=5 / 3, ext { where } C_{p} ext { and } C_{v}[/latex] are the molar specific heats of the gas at constant pressure and at constant volume, respectively,

(b) the finite gaseous configuration can be in Newtonian gravitational equilibrium only if [latex]C_{p} / C_{v}>4 / 3[/latex].

Accepted Answer

For a finite gaseous configuration, the virial theorem gives

[latex]2 \bar{K}+\overline{\sum_{i} r _{i} \cdot F _{i}}=0[/latex].

[latex]\bar{K} ext { is the average total kinetic energy, } F _{i}[/latex] is the total force acting on molecule i by all the other molecules of the gas. If the interactions are Newtonian gravitational of potentials [latex]V\left(r_{i j} ight) \sim \frac{1}{r_{i j}}[/latex] we have

[latex]\sum_{i} r _{i} \cdot F _{i}=-\sum_{j eq i} r_{i j} \frac{\partial V\left(r_{i j} ight)}{\partial r_{i j}}=\sum_{j

[latex]r_{i j}=\left| r _{i}- r _{j} ight|[/latex].

[latex] ext { Hence } 2 \bar{K}+\bar{V}=0 ext {, where } \bar{V}[/latex] is the average total potential energy.
We can consider the gas in each small region of the configuration as ideal, for which the internal energy density [latex]\bar{u}( r )[/latex] and the kinetic energy [latex] ext { density } \bar{K}( r ) ext { satisfy }[/latex]

[latex]\bar{u}( r )=\frac{2}{3} \frac{1}{\gamma-1} \bar{K}( r ) \quad ext { with } \bar{K}( r )=\frac{3}{2} k T( r )[/latex].

[latex] ext { Hence the total internal energy is } \bar{U}=2 \bar{K} / 3(\gamma-1) [/latex].

[latex] ext { When } \gamma=\frac{5}{3}, \bar{U}=\bar{K} ext {. In general the Virial theorem gives } 3(\gamma-1) \bar{U}+[/latex] [latex]\bar{V}=0[/latex] , so the total energy of the system is

[latex]E=\bar{U}+\bar{V}=(4-3 \gamma) \bar{U}[/latex].

For the system to be in stable equilibrium and not to diverge infinitely, we require [latex]E<0 . ext { Since } \bar{U}>0[/latex], we must have

[latex]\gamma<\frac{4}{3}[/latex].

Question 2.162: Starting with the virial theorem for an equilibrium configur...

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