A jar contains a gas and a few drops of water. 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 of Hg. Calculate the new pressure in jar.

Options

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

Answer

B

Explanation

`P_("gas") = P_("dry gas") + P_("moisture") ` at TK or `P_("dry") = 830 - 30 = 800 ` mm Hg
`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 = 817` mm