A given mass of gas occupies `919.0` mL of dry state at N.T.P. The mass when collect over water at `15^(@)C` and `0.987 "bar"` pressure occupies one litre volume. Calculate the vapour pressure of water at `15^(@)C`.

Answer

`0.005 "bar"`

Explanation

Step I. Vapour pressure of dry gas under experimently conditions.
`{:("N.T.P. conditions",,"Experimental conditions"),(V_(1)=919mL,,V_(2)=1000mL),(P_(1)=1.013"bar",,P_(2)=?),(T_(1)=273K,,T_(2)=15+273=288K):}`
`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `P_(2) = (P_(1)V_(1)T_(2))/(T_(1)V_(2)) = ((1.013 "bar") xx (919 mL) xx (288 K))/((273 K) xx (1000 mL)) = 0.982 "bar"`.
Step II. Pressure of water vapours (Aqueous tension)
Pressure of water vapours = Pressures of moist gas - Pressure of dry gas
`= 0.987 - 0.982 = 0.005 "bar"`