Question

In: Chemistry

Ethyl alcohol has a vapor pressure of 20 mm Hg at 8 C and a normal...

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.

Solutions

Expert Solution

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 %


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