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A jar contains a gas and a few drops of ...

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

Text Solution

Verified by Experts

From the available data, `T_(1) = TK` , `T_(2) = (T-0.01T) = 0.99 TK`
`P_(1) = 830 - 30 = 800 mm`, `P_(2) = ?`
According to Charles' Law, `(P_(1))/(P_(2)) = (P_(2))/(T_(2))` or `P_(2) = (P_(1)T_(2))/(T_(1))`
`P_(2) = ((800 mm) xx (0.99 TK))/((TK)) = 794` mm
Total pressure inside the jar `= (792 + 25) = 817 mm`.
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