A tall closed column contains pure methane (molecular mass )

PHYSICS · Multi-part · Fermi University Physics · Thermal, Statistical & Complex Systems · Statistical foundations and the equilibrium ensembles · Statistical foundations and the equilibrium ensembles

Thermal Physics, Ch The grand canonical ensemble, ensemble equivalence and fluctuations

Problem

A tall closed column contains pure methane (molecular mass 16 u16\,\mathrm{u}), treated as an ideal gas held at the single temperature T=250 KT = 250\,\mathrm{K} throughout and in diffusive equilibrium in the uniform gravitational field g=9.81 m s−2g = 9.81\,\mathrm{m\,s^{-2}}. At the base of the column the pressure is p0=3.0 Pap_0 = 3.0\,\mathrm{Pa}. Identical adsorption sites are fixed to the inner wall at various heights zz above the base. Each site is either empty (energy 00, no molecule) or holds exactly one methane molecule, whose energy is then −ε0-\varepsilon_0 with ε0=0.55 eV\varepsilon_0 = 0.55\,\mathrm{eV}, measured from a free gas molecule at the same height. Sites do not interact with one another and are weakly coupled to the gas, which acts as the reservoir fixing TT and the local chemical potential. ((a)) Write the grand partition function Z\mathcal Z of one site and show that its mean occupation is θ=p/(p+p1/2)\theta = p/(p + p_{1/2}), where pp is the local pressure and p1/2=(kBT/λT3) e−ε0/kBTp_{1/2} = (k_BT/\lambda_T^3)\,e^{-\varepsilon_0/k_BT}. Evaluate p1/2p_{1/2}. ((b)) Find the coverage θ\theta of a site at the base. ((c)) Find the coverage of a site z=8.0 kmz = 8.0\,\mathrm{km} above the base. You may use the ideal-gas results of Chapter 14, μ=kBTln⁡(nλT3)\mu = k_BT\ln(n\lambda_T^3) with λT=h/2πmkBT\lambda_T = h/\sqrt{2\pi m k_BT}, and Chapter 8's statement that in a uniform field it is the total chemical potential, internal plus mgzmgz, that is uniform at equilibrium. Constants: kB=1.381×10−23 J K−1k_B = 1.381\times10^{-23}\,\mathrm{J\,K^{-1}}, h=6.626×10−34 J sh = 6.626\times10^{-34}\,\mathrm{J\,s}, 1 u=1.661×10−27 kg1\,\mathrm{u} = 1.661\times10^{-27}\,\mathrm{kg}, 1 eV=1.602×10−19 J1\,\mathrm{eV} = 1.602\times10^{-19}\,\mathrm{J}.

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