Question

Two vessels convain saturated air, one at a temperature of 20°C and the other at 10°C. What quantity of water is precipitated upon mixing the two equal masses of air if the volume of each vessel is 1m3? It may be assumed that the change in saturated vapour pressure in this temperature interval is approximately proportional to the temperature and that the saturated vapour pressure is equal to 9 mm of mercury at 10°C and 17mm of mercury at 20°C. Neglect the loss of heat resulting from heat exchange with the walls of the vessel during the mixing. Assume the specific heat of saturated air to be independent of temperature

Answer 1

Answer: There is no precipitation.

Explanation:

From the equality of the masses of air and the equality of their specific heats it follows that, after the mixing, the temperature of the mixture will be 15°C. From the proportionality of the saturated vapour pressure to the temperature it follows that the vapour pressure at 15°C will equal 13 mm of mercury. The absolute humidity of saturated air at 10, 15, and 20°C will be 9.2, 13.0, and 16.8 g/m3, respectively. The excess quantity of water vapour in the air at 15°C will be equal to (9.2 + 16.8) — 2 x 13.0 = 0.