The van der Waals pressure of argon, P(V,T)= RT V-b - a V^2

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Computational Physics · Scientific programming for physicists

Scientific Computing for Physicists, Ch The numerical toolkit closed: backward error and the audit of a library routine

Problem

The van der Waals pressure of argon, P(V,T)=RTV−b−aV2,P(V,T)=\frac{RT}{V-b}-\frac{a}{V^2}, with a=0.1355 Pa m6 mol−2a=0.1355\,\mathrm{Pa\,m^6\,mol^{-2}}, b=3.201×10−5 m3 mol−1b=3.201\times10^{-5}\,\mathrm{m^3\,mol^{-1}} and R=8.314 J mol−1 K−1R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}, is to be tabulated on a grid of temperature spacing ΔT\Delta T and molar-volume spacing ΔV\Delta V, away from the coexistence region, and read by bilinear interpolation. The bilinear interpolation error obeys ε≤18(ΔT2 ∣∂T2P∣+ΔV2 ∣∂V2P∣).\varepsilon \le \frac18\left(\Delta T^2\,|\partial_T^2 P| + \Delta V^2\,|\partial_V^2 P|\right). ((a)) Show that ∂T2P=0\partial_T^2 P=0 everywhere, so the bound reduces to a single term. ((b)) At T=300 KT=300\,\mathrm{K}, V=5.00×10−4 m3 mol−1V=5.00\times10^{-4}\,\mathrm{m^3\,mol^{-1}}, evaluate ∂V2P\partial_V^2P and hence the error bound for a table with ΔV=2.0×10−5 m3 mol−1\Delta V=2.0\times10^{-5}\,\mathrm{m^3\,mol^{-1}}. ((c)) Evaluate the bound again after halving ΔV\Delta V to 1.0×10−5 m3 mol−11.0\times10^{-5}\,\mathrm{m^3\,mol^{-1}}, and report the ratio of the two bounds.

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