Octane is a liquid component of gasoline. Given the following vapor pressures of octane at various temperatures, estimate the boiling point of octane in Leadville, Colorado, where the atmospheric pressure is 496 mmHg.
a. 400 mmHg at 104oC
b. 500 mmHg at 111oC
c. 600 mmHg at 117oC
d. 700 mmHg at 122oC
e. 760 mmHg at 125oC

110.8 ºC

Explanation:

To solve this problem we will make use of the Clausius-Clayperon equation:

lnP = - ΔHºvap/RT + C

where P is the pressure, ΔHºvap is the enthalpy of vaporization, R is the gas constant, T is the temperature, and C is a constant of integration.

Now this equation has a form y = mx + b where

y = lnP

x = 1/T

m = -ΔHºvap/R

Now we have to assume that ΔHºvap remains constant which is a good asumption given the narrow range of temperatures in the data ( 104-125) ºC

Thus what we have to do is find the equation of the best fit for this data using a  software as excel or your calculator.

T ( K)               1/T                  ln P

377               0.002653       5.9915

384              0.002604       6.2115

390              0.002564       6.3969

395              0.002532       6.5511

398              0.002513        6.6333

The best line has a fit:

y = -4609.5 x  + 18.218

with R² = 0.9998

Now that we have the equation of the line, we simply will substitute for a pressure of 496 mm in Leadville.

ln(496) = -4609.5(1/Tb) + 18.218

6.2066 = -4609.5(1/Tb) +18.218

⇒ 1/Tb = (18.218 - 6.2066)/4609.5 = 0.00261

Tb = 383.76 K  = (383.76 -273)K = 110.8 ºC

Notice we have touse up to 4 decimal places since rounding could lead to an erroneous answer ( i.e boiling temperature greater than 111, an impossibility given the data in the question). This is as a result of the value 496 mmHg so close to 500 mm Hg.

Perhaps that is the reason the question was flagged.

c

explanation: