DIE HARD Thin uniform massive rod is enclosed in large hollow shell filled with ideal gas?

Question

The rod has lenght L=1, occupies segment from (-1/2,0,0) to (+1/2,0,0) and its mass GM = 1 pulls the gas by gravitation.
As a result pressure at point (1,0) is Po =1.
How many moles of gas is under pressure greater than Po?

The rest of the data are mostly ones too:
length of the rod is L = 1,
mass of the rod is GM = 1,
temperature of gas is RT = 1,
molar mass of the gas is μ=1/2.

I will tell the person, who produces _exact answer in closed form

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Answer 1

If the gas is in equlibrium, then its density n has the Boltzmann distribution: n = n0 exp(-m φ/kT). Here, m is the mass of one molecule of gas. Multipying and dividing the exponential factor on the Avogadro number N_A, we transform it to m φ/kT = μ φ/RT = φ/2, and

n = n0 exp(-φ/2).

The rod is located along the x-axis between points L/2
φ = -ln |(x2+r2)/(x1+r1)|.

Here, x1=x-L/2, x2=x+L/2, r1=√( (x-L/2)^2+y^2+z^2), r2=√( (x+L/2)^2+y^2+z^2). Since x2-x1=L=1, and r2^2-r1^2 = x2^2-x1^2, we can write r1+r2 + x1+x2 = (x2-x1) (r1+r2) + x2^2 - x1^2 = (r1+r2) (x2-x1 + r2-r1), or

r1+r2 = (C+1)/(C-1), C = exp(-φ).

This is the equation for a prolate spheroid - the shape obtained by rotating an ellipse with foci at the rod ends about its major axis. The major a and minor b semi-axis of the ellipse are

a=(1/2)(C+1)/(C-1), b=√ [C/(C-1)^2].

The volume of the ellipsoid is

V = (4 π/3) a b^2.

The particle number N contained within a given volume is

N = ∫ n dV = (n V)| + (n0/2) ∫ V C^(-1/2) dC.

The first term is obtained by integration by parts. "|" means that one should substitute the limits. In the second integral we take into account n=n0 C^(1/2). The limits of the integration are from C =∞ (this corresponds to V=0) to C=3 (surface passing through the point (1,0,0), corresponding to maximum volume). The first integral is equal to √3 π n0. The second integtral is proportional to ∫ dC √C (C+1)/(C-1)^3 = -(1/4) ln| (√C + 1)/ (√C - 1)| -3√C/(2(C-1))-√C/(C-1)^2. Substituting the limits and combining with the first integral one gets N = (πn0/3) (√3-(1/2)ln|(√3 + 1) /(√3 - 1)|). The gas pressure the surface C=3 is n0 kT √3 = 1, or n0 = N_A/√3. If I did not make errors in algebra, then the number of moles of gas under pressure greater than 1 is

N_μ = (π/3) (1- (1/(2√3)) ln|(√3 + 1) /(√3 - 1)|) ≈ 0.65.

Answer 2

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