Question

Ethyl alcohol has a vapor pressure of 20 mm Hg at 8 C and a normal boiling point of 78.4 C.
Estimate the vapor pressure at 45 C using (a) the Antoine equation
(b) the Clausius–Clapeyron equation and the two given data points; and (c) linear interpolation
between the two given points. Taking the first estimate to be correct, calculate the percentage error associated with the second and third estimates.

Answer 1

Vapor pressure of ethyl alcohol at 8 oC = 20mm Hg

Normal boiling point of ethyl alcohol = 78.4 oC

Vapor pressure of ethyl alcohol at 45 oC = ?

(a). Antoine equation is given as:

log P = A - B/(C+T)

where, for ethyl alcohol:

A = 8.20417 , B = 1642.89 , C=230.3

At T= 45 oC

log P = 8.20417-1642.89 / (230.3 + 45)

log P = 2.24

P = 173.78 mm Hg

(b). Causius Clapeyron equation is given as:

ln (P1/P2) = -[ Hvap /R] * (1/T1 - 1/T2)

Taking,

P1 = 760 mmHg , T1 = 78.4 oC = 351.4 K

P2 = 20 mmHg , T2 = 8 oC = 281 K

ln (760 / 20) = - [Hvap / 8.314 ] (1/ 351.4 - 1/ 281)

3.64 = - [Hvap/ 8.314] * (-7.13 *10^-4)

Hvap = (3.64 * 8.314) / (7.13 *10^-4)

= 4.24 *10^4 J/mol

Again, at T = 45 oC = 318 K

ln (P1/P2) = - [Hvap / R] * (1/ T1 - 1/ T2)

ln (760/P2) = - [4.24 *10^4 / 8.314 ] * (1/ 351.4 - 1/ 318)

ln (760/P2) = 1.50

P2 = 169.64 mm Hg

(c). Linear interpolation formula is given as:

P = Po + (T - To) [(P1 - Po) / (T1 - To)]

P = 20 + (318 - 281) [(760 - 20) / (351.4 - 281)]

= 20 + 37 * (740 / 70.4)

= 20 + 37 * 10.51

P = 408.87 mmHg

Percentage error with the second estimate = (| (173.78 - 169.64) | / 173.78 )* 100

= 2.38 %

Percentage error with the third estimate = (| (173.78 - 408.87) | / 173.78 )* 100

= 135.28 %