A jar contains a gas and a few drops of water at `TK` The pressure in the jar is `830 mm` of Hg The temperature of the jar is reduced by `1%` The vapour pressure of water at two temperatures are 30 and 25 mm og Hg Calculate the new pressure in the jar .

Explanation

At `T K,P_(gas) =P_("dry gas") + P_("moisture")`
`:. P_("dry gas") = 830 -30 =800 mm`
Now at new temperature `T_(1) =T - (T)/(100) = 0.99T`
Since `V_(1) =V_(2), (P)/(T) =const`
or `(P_(1))/(T_(1)) = (P_(2))/(T_(2))`
`:. P_("dry gas") = (800 xx 0.99T)/(T) = 792 mm`
`:. P_(gas) = P_(dry gas) + P_(moisture)`
`= 792 + 25 =817 mm pf Hg` .