How can i find the boilng point of a certain substance given temperature and vapor pressure?

Question

well we have this homework, and we are given a table with values of temperature and vapor pressure, and we have to find the boiling point of an object for example:

temperature vapor pressure
70 degrees celcius 542.5 mm Hg
80 degrees celcius 812.6 mm Hg

well iknow that for an object to boil it must equalize the atmospheric pressure 760 mm hg or 760 torr. i also know that 760 is given between those two pressures, and therefore the boiling point is between 70 and 80 degrees. what i don't know is the formula how to find the exact temperature between the two given temperature and pressures. please i need ur help and the derivation of the formula would be very much appreciated, and would prefereably be something from the boyle's or charles's law. thank you!!

Answer 1

You don't have enough information to get an exact answer. You would need antoine's coefficients to calculate a vapor pressure at a given temperature.

As a quick substitute, try graphing the temperatures and pressures you are given on a piece of graph paper or in Excel if you have access to it. Then look up an approximate temperature that coresponds to 760 mmHg.

Answer 2

use

dp/dT = DtrsH / T DtrsV
since DtrsV for a gas can be approximated as Vgas-Vliq and Vgas is so much larger than Vliq the final volume will be essentially Vgas = RT/p (assuming everything is in molar quantities)

dp / dT = p DtrsH / (T^2) nR
dp/p = DtrsH dT / (T^2) R (integrate this equation)
ln(p2/p1) = - (DtrsH / R)*(1/T2 - 1/T1)

putting the values in as above we get DtrsH.
Now that we know the value of DtrsH we put it back into the above equation with one of the set of values with the pressure of 760mmHg and solve for T2.

Example:
ln(760/542.5) = -(DtrsH /R) * ( 1/T2 - 1/343K)
rearrange and get T2.