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 .

Options

A. 792 mm of Hg
B. 817 mm Hg Hg
C. 800 mm of Hg
D. 840 mm of Hg

Answer

B

Explanation

`p_(gas)=P _("dry gas")+P_("moisture")(at T K ) `
OR
` P_("dry")=830-30=800`
`Now at T_(2) =0.99 T_(1),`
At constant volume ,`(P_(1))/(T_(1))=(p_(2))/(T_(2))`
`P _("dry")=(800xx0.99T)/(T) =792 mm`
`therefore P_(gas)=P_("dry") + P_("moisture ")=792 +25 `
`=817 mm`