Question

The vapor pressure of water is 0.2561 psi at 60F and 0.3629 psi at 70F.

A. Estimate it's heat of vaporization over the temp range (Ans: 49.2 kJ)

B. One day, the weatherman reported the temperature and relative humidity to be 70F and 75% respectively. Use your result from A to estimate the dew point on that day (the temp at which water vapor condenses into liquid).

Answer 1

(A) Given temperature, T1 = 60 F = 288.7 K, vapor pressure, P1 = 0.2561 psi

temperature, T2 = 70 F = 294.26 K, vapor pressure, P2 = 0.3629 psi

Applying Vant hoff equation

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

=> ln(0.3629 psi / 0.2561 psi) =  (Hv / 8.314 J.K-1mol-1) x [1/288.7K - 1/294.26K]

=> 0.34856 = (Hv / 8.314 J.K-1mol-1) x [5.56K / 288.7K x 294.26K]

=> Hv = (0.34856 x 8.314 J.K-1mol-1 x 288.7K x 294.26K) / 5.56K = 44.3 KJ (answer)

(B) Given ambient temperature = 70 F = 294.26 K

RH = 75 %

Dew point temperature can be calculated from the following formulae