A jar contains a gas and a few drops of water at absolute temperture `T_(1)`. The pressure in the jar is `830mm` of mercury. The temperature of the jar is reduced by `1%`. The saturation vapour pressures of water at the two temperatures are `30mm` of mercury and `25mm` of mercury. Calculate the new pressure in the jar.

Answer

A

Explanation

`P_1=830-30=800mm Hg, P_2 ?`
`V_1=V , V_2=V , T_1=T , T_2=T-0.01T=0.99T`
`(P_1V_1)/(T_1)=(P_2V_2)/(T_2) :. P_2=(P_1V_1)/(T_1)=(800xx0.099T)/T=792 mmHg`
`:. Total pressure in the jar =792+25=817mm Hg`