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.
The boiling point of a gas is given by the point at which the vapor pressure
of the gas is equal to the prevailing atmospheric pressure.
Reasons:
The Clausius-Clayperon equation that can be used to estimate the vapor
pressure is presented as follows;
[tex]\dfrac{dP}{dT} =\dfrac{\Delta H}{T \cdot V}[/tex]
Which gives;
[tex]ln \left(P \right) = \mathbf{ -\dfrac{\Delta H_{vap}}{R \times T}+ C}[/tex]
Comparing the above equation to the equation of a straight line, y = m·x + c,
we have;
y = ㏑(p)
[tex]m = -\dfrac{\Delta H_{vap}}{R}[/tex]
[tex]x = \dfrac{1}{T}[/tex]
Using the given values, we have;
[tex]\begin{array}{|c|c|c|c}&\dfrac{1}{T} &ln(P)\\&&\\a&0.002651}&5.991465}\\b&0.002603&6.214608\\c&0.002563&6.39693\\d&0.002531&6.55108\\e&0.002512&6.633318\end{array}\right][/tex]
Plotting the above values on a graph and finding the line of best fir using MS Excel, we have;
[tex]ln \left(P \right) = -4603.4 \cdot \dfrac{1}{ T}+ 18.197[/tex]
When P = 496 mmHg, we get;
[tex]ln \left(496 \right) = -4603.4 \cdot \dfrac{1}{ T}+ 18.197[/tex]
[tex]T= \dfrac{4603.4 }{18.197 - ln \left(496 \right) } \approx 383.92[/tex]
T ≈ 383.92 K (110.77°C)
Similarly, given that the rate of change of pressure to temperature is
almost constant, using MS Excel, we have the pressure to temperature in
Celsius relationship of the given data presented as follows;
P = 17.197·T - 1399.4
Therefore, when the temperature is 496 mmHg, we get;
496 = 17.197·T - 1399.4
[tex]T = \dfrac{496 + 1399.4}{17.197} \approx 110[/tex]
Therefore;
The boiling point temperature of octane at 496 mmHg, T ≈ 110 °C
Learn more here:
The probable question options are;
A) 125 °C
B) 105 °C
C) 120 °C
D) 110 °C
E) 115° C
