Calculate the boiling point of water if the atmospheric pressure is 767 mmHg.

Thanks!

the B.P. of water is 100.0ºC at 760 mmHg

the "correction factor" is ... 0.045 ºC/mmHg

this works for a "normal" range of atmospheric pressures

To calculate the boiling point of water at a given atmospheric pressure, you can use the Clausius-Clapeyron equation. The equation relates the boiling point of a substance to its vapor pressure and the heat of vaporization.

The Clausius-Clapeyron equation is given by:

ln(P2/P1) = (-ΔHvap/R)(1/T2 - 1/T1)

where P1 and P2 are the initial and final pressures, T1 and T2 are the initial and final temperatures, ΔHvap is the heat of vaporization of the substance, and R is the ideal gas constant.

For water, the heat of vaporization (ΔHvap) is approximately 40.7 kJ/mol, and R is 8.314 J/(mol·K).

Given that the atmospheric pressure is 767 mmHg, you need to convert it to its equivalent SI unit, which is Pascals (Pa). 1 mmHg = 133.3224 Pa.

So, the atmospheric pressure is:
767 mmHg * 133.3224 Pa/mmHg = 102266.56 Pa

Assuming that the normal boiling point of water is 100°C (373.15 K) at atmospheric pressure (1 atm), we can plug the values into the Clausius-Clapeyron equation to solve for the boiling point at 102266.56 Pa:

ln(102266.56/101325) = (-40700/8.314)(1/T2 - 1/373.15)

Solving for T2, the boiling point at 767 mmHg, we get:

T2 = 1 / (1/373.15 - (8.314/40700) * ln(102266.56/101325))

Calculating this expression will give you the boiling point of water at the given pressure of 767 mmHg.

Please note that this is an approximation, as the heat of vaporization can vary slightly with temperature.