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The equation below gives the boiling temperature of isopropanol as a function of pressure: T =(...

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 the parameters A, B and C are 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. Hint: the equation is of the form T = (B' / (A'-lnP) ) - C' but the constants A', B' and C' have different values from those given above.

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