Question

At different elevations, the boiling point of water changes because of the differences in air pressure. The air pressure in bar at different elevations is given by: P = (1 − 2.25577×10^-5 h)^5.25588 where h is the elevation above sea level in metres. Compare the boiling point of water at h = 175 m and at h = 5364 m. The heat of vapourization and boiling point of water at 1 bar are 40.66 kJ mol–1 and 100ºC.

Answer 1

Pressure at 175m

P = (1 − 2.25577×10^-5 h)^ 5.25588

   = (1 − 2.25577×10^-5 *175 m)5.25588 = 0.98 bar

Pressure ar sea level = 1.01 bar

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

or, ln (0.98/1.01) = 40.66*10^3 J/mol/ 8.314 J/K/mol [1/373-1/T2]

or, T2 = 372.54 K =99.14oC

-----------------------------------------------------

Pressure at 5364m

P = (1 − 2.25577×10^-5 h)^ 5.25588

   = (1 − 2.25577×10^-5 *5364 m)5.25588 = 0.51 bar

Pressure ar sea level = 1.01 bar

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

or, ln (0.51/1.01) = 40.66*10^3 J/mol/ 8.314 J/K/mol [1/373-1/T2]

or, T2 = 354.52 K =81.5oC