Water has a heat of vaporization of 44.01 KJ/mol and boils at 100 degrees Celcius at sea level. What is the boiling point of water in Celcius at the top of Mount Everest where the atmospheric pressur

asked by @alibethh1 • about 1 year ago • Chemistry • 5 pts

e is only 34% as strong as the pressure at sea level?

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1 answer

Hey there. Ok so for this question we are dealing with pressure, temperature and vaporization so that means we need to use the Clausius-Clapeyron Equation:

ln (P2/P1) = - delta Hvaporization/R x [(1/T2) - (1/T1)]

At 100 degrees Celsius we are dealing with normal pressure which is 760 mmHg or torr.

34% of 760 is 0.34 x 760 = 258.4

ln (258.4 torr/760 torr) =[ -44,010 J/mol / 8.314 J/mol x K] x [(1/T2) - (1/373.15 K)]

-1.07881 = -5293.48 x [(1/T2) - (1/373.15 K)]

0.000204 = [(1/T2) - (1/373.15 K)]

0.002884 = 1/T2

T2 = 346.8 K

Hope that helps.

answered by @jules • about 1 year ago