You will need to enter two vectors into Matlab, P_TP and P_s. Then you will fit P_s/P_TP versus P_TP to a straight line and determine the temperature of the sample by extrapolating to zero pressure. The Matlab commands, and the response to those commands inside Matlab, will be printed here in
this font.
Start Matlab at the UNIX shell prompt by typing matlab.
ptp=[100;200;300;400;500] ptp = 100 200 300 400 500 ps=[233.9;471.7;714.77;962.7;1215.4] ps = 1.0e+03 * 0.2339 0.4717 0.7148 0.9627 1.2154 ps_div_ptp=ps./ptp ps_div_ptp = 2.3390 2.3585 2.3826 2.4068 2.4308 help polyfit POLYFIT Polynomial curve fitting. POLYFIT(x,y,n) finds the coefficients of a polynomial p(x) of degree n that fits the data, p(x(i)) ~= y(i), in a least-squares sense. [p,S] = POLYFIT(x,y,n) returns the polynomial coefficients p and a matrix S for use with POLYVAL to produce error estimates on predictions. If the errors in the data, y, are independent normal with constant variance, POLYVAL will produce error bounds which contain at least 50% of the predictions. See also POLY, POLYVAL, ROOTS. [p,S]=polyfit(ptp,ps_div_ptp,1) p = 0.0002 2.3140 S = -741.6198 -2.0226 0 -0.9535 3.0000 0 0.0029 0 format short e p p = 2.3185e-04 2.3140e+00 help polyval POLYVAL Polynomial evaluation. If p is a vector of length d+1 whose elements are the coefficients of a polynomial, then y = POLYVAL(p,x) is the value of the polynomial evaluated at x. y = p(1)*x^d + p(2)*x^(d-1) + ... + p(d)*x + p(d+1) If X is a matrix or vector, the polynomial is evaluated at all points in X. [y,delta] = POLYVAL(p,x,S) uses the optional output generated by POLYFIT to generate error estimates, y +/- delta. If the errors in the data input to POLYFIT are independent normal with constant variance, y +/- delta contains at least 50% of the predictions. See POLYVALM for evaluation in a matrix sense. ptp_calc=0:10:500; [ps_div_ptp_calc,delta]=polyval(p,ptp_calc,S); plot(ptp_calc,ps_div_ptp_calc) hold Current plot held plot(ptp,ps_div_ptp,'+g') xlabel('P_TP (torr)') ylabel('P_s/P_TP') t=273.16*p(2) t = 6.3208e+02 delta(1)*273.16 ans = 6.6743e-01
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