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 of Hg
C. 800 mm of Hg
D. 840 mm of Hg

Answer

B

Explanation

`P_("gas") = P_("dry gas")+ P_("moisture ")` at TK
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") = ( 800 xx 0.99 T )/( T) = 792 mm`
`:. P_("gas") = P_("dry") + P_("moisture") `
`=792+ 25=817mm`