Question

Problem 2. (40 pts) Water vapor pressure at room temperature.

a. (5) The molar heat of boiling water is +40.66 kJ/mol (= deltaHº) at 373 K and 1 atm. Find the entropy of boiling.

b. (10 pts) Assume that deltaH and deltaS stay constant at their values of deltaHº and Sº which are determined at 373 K, respectively, and do not change as a function of temperature. Show that deltaGº₂₉₃ at 20 ºC ( = 293 K) and at 1 atm is about 8720 J. Should water spontaneously boil at 20 ºC and 1 atm; why?

c. (20 pts) Now find the equilibrium pressure at which water boils at 20 ºC = 293 K. Explicitly what is this pressure in atm? You can use the following thermodynamic cycle similar to the ones which we have used in class. (R = 8.314 J K^-1mol^-1)

d. (5 pts) If the water vapor pressure at 20 ºC is 0.04 atm, will it rain? Why?

Answer 1

a. entropy of boiling delta S = molar heat of boiling/boiling point

= 40.66/373 = 0.109 kJ/K.mol

b. delta Go = delta Ho - TdeltaSo

= 40.66 - 293 x 0.109

= 8.723 kJ/mol

As the deltaGo is +ve, it means water will not spontaneously boil at 20 oC and 1 atm.

c. Using Claussius-Clapeyron equation,

ln(P2/P1) = deltaHvap/R[1/T1-1/T2]

we have,

P1 = unknown

P2 = 1 atm

T1 = 20 oC = 293 K

T2 = 373 K

deltaHvap = 40.66 kJ/mol

R = 8.314 J/K.mol

Feed values,

ln(1/P1) = 40660/8.314[1/293 - 1/373]

P1 = 0.03 atm

So the equilibrium pressure would be 0.03 atm for water to boil at 20 oC.

d. If water vapor pressure at 20 oC is 0.04 atm, it will rain. To have a rain we require the vapor pressure of water about 2% of atmospheric pressure which here is 0.02 atm. Since the vapor pressure of exceeds the required value for rain it will rain at 20 oC and 0.04 atm vapor pressure.