The volume of a given mass of a gas is 919 mL in dry state at STP. The same mass when collected over water at `15^(@)C` and 750 mm pressure occupies a volume of 1 L. Find out the vapor pressure of water at `15^(@)C`.

Explanation

Given , `V_(1) = 919 ` mL , `P_(1) = 760` mm , `T_(1) = 273 K , V_(2) = 1000 mL , T_(2) = 273 + 15 = 288` K
Pressure of the dry gas , `P_(2) = (P_(1) V_(1) T_(2))/(V_(2) T_(1)) = (760 mm xx 919 mL xx 288 K)/(1000 mL xx 273 K) = 736.8` mm
Vapour pressure of water = Pressure of the moist gas - Pressure of the dry gas = 750 - 736.8 = 13.2 mm