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$$\"3nH_`)p%GT%)F-F_pFi]l7$7$$\"3cG_`)p%GT%)F-F_pFj\\l7$7$FP$\"34.n:N>
[0:F-7$$\"3%***********fH\"*F-FT7$7$$\"3$))*********fH\"*F-FTFb]l7$7$F
A$\"3-(e#e>pV:$*F*7$$\"32?b^kC`F%*F-FE7$Fe_l7$FLFh^l7$7$F1F/7$$\"3q7dG
9d3())*F-F(7$F\\`l7$$\"37)3$feTAr(*F-$\"3/]\"pSTexG'F*7$7$$\"3[_aGd\"4
Xo*F-F6F``l7$7$$\"3g`aGd\"4Xo*F-F6Fb_l-%'COLOURG6&%$RGBG\"\"\"F0F0-%+A
XESLABELSG6$%\"tG%\"hG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "Relati
ve volume against time would be more difficult to plet - try it!" }
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "t100:=
1: t80:=1-10/7*h20^(3/2)+3/7*h20^(5/2):  (t100-t80)/t100*100;\n# time \+
to empty tha last 20% volume" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+O*
=&*3$!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "Q54:" }{MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 332 "restart: # speed of
 A+B->C reaction is proportional to the remaining amount of\nfirst rea
ctant, (a-y), in moles, and of the second reactant, (b-y),\nand a cons
tant k (a function of temperature, which we keep fixed),\nwhere y is t
he amount (in moles) of the resulting compound.\nThis explains the y'=
k(a-y)(b-y) equation. We solve it by:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "sol:=int(1/(a-y)/(b-y),y)=k*t+C;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$solG/,&*&,&%\"bG!\"\"%\"aG\"\"\"F*-%#lnG6#,&F+F*%\"y
GF,F,F,*&F(F*-F.6#,&F)F*F1F,F,F*,&*&%\"kGF,%\"tGF,F,%\"CGF," }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "where  C must be chosen to meet th
e obvious initial-value condition  y(0)=0, thus:" }{MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "solve(eval(sol,\{t=0,y=0\}
),C);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,&-%#lnG6#,$%\"aG!\"\"F+-
F'6#,$%\"bGF+\"\"\"F0,&F/F+F*F0F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 59 "Substituting this C back into the general solution yields: " }
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "y:=solv
e(eval(sol,\{C=%\}),y);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"yG**%\"
bG\"\"\"%\"aGF',&-%$expG6#,&*(%\"kGF'%\"tGF'F&F'!\"\"*(F/F'F0F'F(F'F'F
'F'F1F',&*&F*F'F(F'F'F&F1F1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "To
 get the solution for the a=b case (most common - why?), we compute th
e  b -> a limit:" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 65 "limit(y,b=a); # solution for a=b, as a limit of the a
<>b solution" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#**%\"kG\"\"\"%\"tGF%%
\"aG\"\"#,&F%F%*(F$F%F&F%F'F%F%!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 122 "Back to the   a-not-equal-to-b case: When  a>b,  y(t) must ten
d, in the  t -> infinity limit, to b  (why?), which is does:" }
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "assume(
a>b,k>0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "limit(y,t=infi
nity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%#b|irG" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 65 "assume(a<b):    # when  a<b, the same limit \+
must yield a, right?:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "li
mit(y,t=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%#a|irG" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 103 "Finally, building the  a=b soluti
on from the  y'=k.(a-y)^2 equation (getting the same answer as before:
" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "rest
art:  # solution for a=b, from the y'=k(a-y)^2 equation" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=int(1/(a-y)^2,y)=k*t+C;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/*&\"\"\"F',&%\"aGF'%\"yG!\"\"F
+,&*&%\"kGF'%\"tGF'F'%\"CGF'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 60 "solve(eval(sol,C=1/a),y);  # agrees with the previous answer" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#**%\"kG\"\"\"%\"tGF%%\"aG\"\"#,&*(F$F%
F&F%F'F%F%F%F%!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "limi
t(%,t=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%\"aG" }}}}{MARK "
47" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }