1 CHAPTER 8 8 1 Ideal gas law v RT 0 082054 375 15 38513 p 2 van der Waals equation a f v p 2 v b RT v 12 02 f v 2 2 v 0 08407 0 082054 375 v Any of the techniques in Chaps 5 or 6 can be used to determine the root as v 15 0727 L mol The Newton Raphson method would be a good choice because a the equation is relatively simple to differentiate and b the ideal gas law provides a good initial guess The Newton Raphson method can be formulated as vi 1 p vi p a vi b RT vi2 a 2a vi b 3 2 vi vi Using the ideal gas law for the initial guess results in an accurate root determination in a few iterations i 0 1 2 xi 15 38513 15 07285 15 07268 f xi 0 608865 0 000323 9 5 10 11 f xi 1 949774 1 947683 1 947682 a 2 0718 0 0011 8 2 The function to be solved is f R ln 1 R 1 X Af R 1 X Af R 1 0 R 1 R 1 X Af or substituting XAf 0 9 f R ln 1 0 1R R 1 0 R 0 1 R 1 0 1R A plot of the function indicates a root at about R 0 43 PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 2 2 0 2 0 0 2 0 4 0 6 0 8 1 4 6 Bisection with initial guesses of 0 01 and 1 can be used to determine a root of 0 4299373 after 16 iterations with a 0 004 8 3 The function to be solved is f x x 6 0 05 0 1 x 2 x A plot of the function indicates a root at about x 0 028 0 1 0 0 0 02 0 04 0 06 0 1 Because the function is so linear false position is a good choice Using initial guesses of 0 02 and 0 04 the first iteration is xr 0 04 0 021458 0 02 0 04 0 028173 0 01483 0 021458 After 3 iterations the result is 0 028249 with a 0 003 8 4 The function to be solved is f t 10 1 e 0 04t 4e 0 04t 9 3 0 A plot of the function indicates a root at about t 55 2 0 2 0 20 40 60 80 100 4 6 Bisection with initial guesses of 0 and 60 can be used to determine a root of 53 711 after 16 iterations with a 0 002 8 5 The function to be solved is f x 4 x 42 2 x 2 28 x 0 015 0 PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 3 a A plot of the function indicates a root at about x 16 0 8 0 6 0 4 0 2 0 0 2 0 5 10 15 20 b The shape of the function indicates that false position would be a poor choice recall Fig 5 14 Bisection with initial guesses of 0 and 20 can be used to determine a root of 16 01563 after 8 iterations with a 0 488 Note that false position would have required many more iterations to attain comparable accuracy i 1 2 3 4 5 6 7 8 xl f xl xr 20 10 0 01492 20 15 0 01355 20 17 5 0 0062 17 5 16 25 0 0062 16 25 15 625 0 0062 16 25 15 9375 0 00319 16 25 16 09375 0 00117 16 09375 16 01563 0 00117 xu 0 10 15 15 15 15 625 15 9375 15 9375 f xr 0 01355 0 00620 0 02010 0 001318 0 00319 0 00117 6 92E 06 0 0006 f xl f xr 0 000202 8 41E 05 0 00012 8 2E 06 1 98E 05 3 73E 06 8 1E 09 6 99E 07 a 100 000 33 333 14 286 7 692 4 000 1 961 0 971 0 488 8 6 The functions to be solved are K1 K2 c c 0 x1 x 2 c a 0 2 x1 x 2 2 cb 0 x1 c c 0 x1 x 2 c a 0 2 x1 x 2 c d 0 x 2 or f1 x1 x 2 f 2 x1 x 2 5 x1 x 2 2 50 2 x1 x 2 20 x1 4 10 4 5 x1 x 2 3 7 10 2 50 2 x1 x 2 10 x 2 Graphs can be generated by specifying values of x1 and solving for x2 using a numerical method like bisection x1 0 1 2 3 4 5 first equation x2 8 6672 6 8618 5 0649 3 2769 1 4984 0 2700 x1 0 1 2 3 4 5 second equation x2 4 4167 3 9187 3 4010 2 8630 2 3038 1 7227 These values can then be plotted to yield PROPRIETARY MATERIAL The McGraw Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw Hill for their individual course preparation If you are a student using this Manual you are using it without permission 4 12 1st eq 2nd eq 8 4 0 0 1 2 3 4 5 4 Therefore the root seems to be at about x1 3 3 and x2 2 7 Employing these values as the initial guesses for the two variable Newton Raphson method gives f1 3 3 2 7 2 36 10 6 f2 3 3 2 7 2 33 10 5 f1 9 9 10 5 x1 f 2 5 185 10 3 x1 f1 5 57 10 5 x 2 f 2 9 35 10 3 x 2 J 6 37 10 7 x1 3 3 x 2 2 7 2 36 10 6 9 35 10 3 2 33 10 5 5 57 10 5 6 37 10 7 3 3367 2 33 10 5 9 9 10 5 2 36 10 6 5 185 10 3 6 37 10 7 2 677 The second iteration yields x1 3 …
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