Answer:
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.
