1. The problem statement, all variables and given/known data Hello. I have a small question... How does one correctly "modify" an equation to work in different units? The equation below gives the boiling temperature of isopropanol as a function of pressure: T = B / (A - log10P) - C Where T is in kelvin, P is in bar, and A= 4.57795 B= 1221.423 C= -87.474 Obtain an equation that gives the boiling temperature in degrees F, as a function of ln(P), with P in psi. It must be in the form T = B' / (A' - ln(P)) - C', where A', B', C' are different numerical values from those given above. 2. Relevant equations log10x = ln x / ln (10) ln(ab) = ln(a) + ln(b) 1 bar = 14.5 psi 1 atm = 1.013 bar = 14.7 psi Kelvin to degrees F: ((T - 273.15) * 1.8) + 32 3. The attempt at a solution I worked it out, but when I try a "test" value of 1 atm (14.7 psi, 1.013 bar) the temperature values do not match after conversion! If overall, the temperature T is in kelvin, then change the RHS into degrees F. T = ([B / (A - ln(P*14.5 psi/bar)/ln(10)) - C] - 273.15)*1.8 + 32 T = ([B / ((A - ln(P) + ln(14.5))/ln(10)) - C] - 273.15)*1.8 + 32 I now multiply the B/A fraction by ln(10)/ln(10), and factor in the 1.8. T = ([1.8*ln(10)B / (ln(10)A - ln(P) - ln(14.5)) - 1.8C] - 1.8*273.15) + 32 T = [1.8*ln(10)B / (ln(10)A - ln(P) - ln(14.5)) - 1.8(C + 273.15)] + 32 Now, I define 1.8*ln(10)B = B' ln(10)A - ln(14.5) = A' 1.8(C + 273.15) + 32 = C' So now I have T = B' / (A' - ln(P)) - C' A' = 7.866970777 B' = 5062.374706 C' = 302.2168 (I accounted for the negative so that it works in the equation written as is) But when I test a pressure of 1 atm (1.013 bar = 14.7 psi), I get a temperature of 355K. This should correspond to a degrees F of around 180, but I'm getting around 600 when I use my new formula with pressure 14.7 psi. Any help?