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I need to find the boiling point (in degrees centigrade) of ethanol on a day when the atmospheric pressure is $780\text{ torr}$.

I know that the molar mass of ethanol is $46.07\text{ g/mol}$ but I have no idea how to continue. Can you tell me what equations should be used or how is it solved, please?

Thanks!

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    $\begingroup$ Antoine Equation is fitted to experimental data if you just need a number. $\endgroup$ – MaxW Jul 3 '18 at 0:28
  • $\begingroup$ @MaxW I appreciate that but I don't know how to find the unique solution. It seems we need some more data... right? Because I have the answer, and it's unique. With your link I don't understand where the solution is. $\endgroup$ – manooooh Jul 3 '18 at 0:41
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    $\begingroup$ Huh? $$P = 10^{A-\frac{B}{C-T}}$$ is the solution. A,B,and C are constants. You can solve for T knowing P, or P knowing T. $\endgroup$ – MaxW Jul 3 '18 at 1:11
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    $\begingroup$ Care, @MaxW, the equation is $$P = 10^{A-\frac{B}{C{\bf{\color{red}+}}T}},$$ with a plus :). Cheers $\endgroup$ – manooooh Jul 3 '18 at 2:55
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You could use the Clausius Clapeyron equation:

$$ \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta H_\mathrm{v}}{\Delta V \cdot T}$$

Since the change in pressure is rather small, it is fair to assume the molar enthalpy of evaporation to be constant and use the value given for standard conditions, which should be easy to obtain.

The other simplification you can make is to assume that the volume of the liquid is very small compared to the volume of the gas:

$$ V_\mathrm{l} << V_\mathrm{g}$$ so that $$\Delta V = V_\mathrm{g}-V_\mathrm{l} \sim V_\mathrm{g}$$

Now by using the ideal gas law and expressing volume as molar quantity,

$$v=\frac{p}{RT}$$

you get the ODE

$$ \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta H_\mathrm{v}\cdot p}{T^2}$$

from whose solution you get a nice equation to solve your problem.

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A good estimate of the boiling point of a material at 760 mm Hg can be made from other data, using a nomograph at the back of the Aldrich catalog (under Tables and Graphs) and online at https://www.sigmaaldrich.com/chemistry/solvents/learning-center/nomograph.html.

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