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)792mm of Hg
(b)817mm of Hg
(c)800mm of Hg
(d)840mm of Hg

Answer

817

Explanation

Total pressure in the jar TK = Pressure of gas + Saturated vapour pressure of water at TK.
`:." "` Pressure of gas alone `(p_1)` at TK = 830 – 30 = 800 mm of Hg
Let pressure of gas alone at `(T -T/100)K = P_2`
Applying the formula
`P_1/T_1 = P_2/T_2`
We get, `800/T = P_2/((T - T/100))`
or `P_2 = 800 xx99/100 = 792 "m m of H g"`.
Saturated vapour pressure of water at
`(T - T/100 ) K =" 25 m m of H g"`
`:." "` Total pressure in the jar at this new temperature `"= 792 + 25 = 817 m m of H g" `