Please help me with this equation: Temperature in Farenheit=-3200/
log(pressure in psi/1700) -459.7. Use pressure of 14.7. Answer I think
should be 212. I can't get it to come out. I know log(a/b)= loga-logb,
but I'm screwing up somewhere. HELP!!!! Thanks Robert 3rrotec

Hi!!

As you said it is log(a/b)= log(a)-log(b), where log(x) is the natural
logarithm (base e) ---> 8probably here are your problem, may be you
are using logarithm base 10).

Note that the given formula is a result of a non-linear regression
analysis, see "Boiling Point Of Water As A Function Of Pressure":
http://www.nlreg.com/boil.htm

According to documentation on this site "the boiling point of water
decreases as the pressure in the vessel containing the water
decreases. "Clapeyron's equation? shows that the boiling point is
related to pressure according to the following function:
Temperature = b / log(Pressure/a) - 459.7 
Where Temperature is in degrees Fahrenheit (the 459.7 constant
converts degrees Fahrenheit to degrees Rankine -- relative to absolute
zero), Pressure is the pressure in the vessel in pounds per square
inch, and a and b are parameters whose values are to be determined (by
a regression analysis program)."
http://www.nlreg.com/NLREG.pdf

The data for the boiling example was collected by the author's son for
a science project. So the result could not be completely accurate but
a very good aproximation.

With the above in mind we can do the calculations for boiling point:
Farenheit = -3200 / log(pressure in psi/1700) - 459.7 =
          = -3200 / log(14.7/1700) - 459.7 = 
          = -3200 / [log(14.7) - log(1700)] - 459.7 =
          = -3200 / [2.68 - 7.44] - 459.7 =
          = -3200 / [-4.75] - 459.7 =
          = 673.7 - 459.7 = 
          = 214°F
This is a good aproximation.

A more accurate formula is the following one:
Boiling point[°F] = 49.161 * log(in. Hg) + 44.932 

From the above we can derive a formula to use with PSI:
Boiling point[°F] = 49.161 * log(2.036027*PSI) + 44.932 =
                  = 49.161 * [log(2.036027) + log(PSI)] + 44.932 =
                  = 49.161 * 0.711 + 49.161 * log(PSI) + 44.932 =
                  = 34.953 +  49.161 * log(PSI) + 44.932 =
                  = 79.885 + 49.161 * log(PSI) 

Then for 14.7 PSI you have:
Boiling point[°F] = 79.885 + 49.161 * log(14.7) =
                  = 79.885 + 49.161 * 2.688 =
                  = 79.885 + 132.145 =
                  = 212.03°F

More precisely, since 1 atm is equal to 14.69595 PSI, then:
Boiling point[°F] = 79.885 + 49.161 * log(14.69595) =
                  = 79.885 + 132.123 =
                  = 212.008°F

For reference on the last proposed formula see:
"Boiling Point of Water Calculator":
'In the tables, the following equations have been used:
pressure (in. Hg) = 29.921* (1-6.8753*0.000001 * altitude, ft.)^5.2559
boiling point = 49.161 * Ln (in. Hg) + 44.932 '
http://www.biggreenegg.com/boilingPoint.htm

"CALIBRATING THERMOMETERS IN BOILING WATER: Boiling Point /
Atmospheric Pressure / Altitude Tables" by O. Peter Snyder, Jr., Ph.D.
:
http://www.hi-tm.com/Documents/Calib-boil.html


Search strategy:
"Clapeyron's equation" boiling pressure
"Clapeyron's equation" temperature fahrenheit

Hope that this helps you. Feel free to request for a clarification if you need it.

Best regards,
livioflores-ga

Yep, I was using base 10. Thanks for fast responce! 3rrotec Robert Williams