Antoine equation and unit conversion. The equation below allows the calculation

Question

Antoine equation and unit conversion. The equation below allows the calculation of the boiling point of a pure substance as a function of pressure. A - log1o where Tis in Kelvin, P is in bar. The parameters A, B and C are specific to a pure substance. In the case of isopropanol, the parameters are A - 4.57795, B 1221.423 C-87.474 Determine the form of the equation that provides the boiling point in , as a function of the In P, with P in psi. The equation is of the form C" A'- InP where the constants A',B'and C' have different values than those given above.

Explanation / Answer

T(K) = [B/(A - log P(bar))] - C

As we know that

T(K) = (T(°F) - 32) * (5/9) + 273.15

T(K) = 0.5555 T(°F) - 17.7777 + 273.15

T(K) = 0.5555 T(°F) + 255.3723

P(bar) = 14.504 P (psi)

log P (bar) = log (14.504 P (psi))

Convert log to natural log

log P (bar) = log (14.504 P (psi)) = [ln (14.504 P (psi))]/[2.303]

= (ln 14.504)/2.303 + ( ln P) / 2.303

log P (bar) = 1.16127 + 0.43421 ln P (psi)

Put the values in given equation

0.5555 T(°F) + 255.3723 = [B/(A - 1.16127 - 0.43421 ln P (psi))] - C

T(°F) = [(B/0.5555)/(A - 1.16127 - 0.43421 ln P (psi))] - (C/0.5555) - (255.3723/0.5555)

Put the values of A, B and C

T(°F) = [(1221.423/0.5555)/(4.57795 - 1.16127 - 0.43421 ln P (psi))] - (-87.474/0.5555) - (255.3723/0.5555)

T(°F) = [(2198.781)/(3.41668 - 0.43421 ln P (psi))] - 302.2471

T(°F) = [(2198.781/0.43421)/(3.41668/0.43421 - ln P (psi))] - 302.2471

T(°F) = [(5063.865)/(7.86872 - ln P (psi))] - 302.2471

T(°F) = [(B')/(A' - ln P (psi))] - C'

Where

A' = 7.86872

B' = 5063.865

C' = 302.2471

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