Combined Gas Law, 11-4-19

On one of my old cars I want to increase the pressure in my copper radiator and raise the boiling point of the coolant. Using water as the coolant, what temperature would my coolant boil with a 16 psi radiator cap instead of my current 15 psi cap? I wanted the answer in customary English units of which I'm most familiar. I found this Combined Gas Law calculator online....

https://www.calcprofi.com/combined-g...alculator.html

I made a program for my Ti74. (I don't always have access to wifi and my phone is 'dumb') The program yields similar results to the online calculator. If I'm understanding and calculating correctly, going from a 15 psi cap at a 212 deg f boiling point to a 16 psi cap should yield a boiling temp of 257 deg f. I'm no math expert or programmer so check the results for yourself. But I think I'll give a 16 psi cap a try.

Code:

1100 PRINT "Combined Gas Law":PAUSE .8
1105 PRINT "Enter 0 For":PAUSE .8
1107 PRINT "Desired Quanity":PAUSE .8
1110 PRINT A;"> V1,in3 ";:ACCEPT NULL(A),A:V1=A
1120 PRINT B;"> P1,psi ";:ACCEPT NULL(B),B:P1=B
1130 PRINT C;"> T1,deg f ";:ACCEPT NULL(C),C:T1=C+459.67
1140 PRINT D;"> V2,in3 ";:ACCEPT NULL(D),D:V2=D
1150 PRINT E;"> P2,psi ";:ACCEPT NULL(E),E:P2=E
1160 PRINT F;"> T2,deg f ";:ACCEPT NULL(F),F:T2=F+459.67
1170 IF A=0 THEN X=(V2*P2*T1)/(P1*T2):X$="V1,in3=":A=X:GOTO 1240
1180 IF B=0 THEN X=(V2*P2*T1)/(V1*T2):X$="P1,psi=":B=X:GOTO 1240
1190 IF C=0 THEN X=(V1*P1*T2)/(V2*P2):X=X-459.67:X$="T1,deg f=":C=X:GOTO 1240
1200 IF D=0 THEN X=(V1*P1*T2)/(P2*T1):X$="V2,in3=":D=X:GOTO 1240
1210 IF E=0 THEN X=(V1*P1*T2)/(V2*T1):X$="P2,psi=":E=X:GOTO 1240
1220 IF F=0 THEN X=(V2*P2*T1)/(V1*P1):X=X-459.67:X$="T2,deg f=":F=X:GOTO 1240
1230 PRINT "error..":PAUSE:GOTO 1100
1240 ANS=X:PRINT "ans ";X$;X:PAUSE:GOTO 1100

For my radiator:

V1=1 in3 [enter]

P1=15 psi [enter]

T1=212 deg f [enter]

V2=1 in3 [enter]

P2=16 psi [enter]

T2=0 [enter]

T2=256.778 deg f

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