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
Text Solution
Verified by Experts
The correct Answer is:
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`
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