Question

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 300 and 25 mm of Hg Calculate the new pressure in the jar .
A. 792 mm of Hg
B. 817 mm of Hg
C. 800 mm of Hg
D. 840 mm of Hg

Answer 1

Correct Answer - b
(b) `P_("gas")=P_("dry gas")+P_("moisture") "at " TK`
or `P_("dry")=830-30=800`
Now at `T_(2)=0.99T_(1),`
at constant volume `(P_(1))/(T_(1))=(P_(2))/(T_(2))`
`P_("dry")=(800xx0.99T)/(T)=792 mm`
`:. " " P_("gas")=P_("dry")+P_("moisture")`
`=792+25=817 mm`