Q. A jar contains a gas and few drops of water at $T\ K.$ The pressure in the jar is $830\ mm$ of mercury. The temperature of jar is reduced by 1%. The saturated vapour pressure of water at the two temperatures are $30\ mm$ and $25\ mm$ of mercury. Then the new pressure in the jar will be
A) $917\ mm$ of $Hg$
B) $717\ mm$ of $Hg$
C) $817\ mm$ of $Hg$
D) None of these
Explanation
(c) At $TK,$ pressure of gas $(P)$ in the jar
= Total pressure $–$ saturated vapour pressure
$\Rightarrow P = (830 - 30) = 800mm\ of\ Hg$
New temperature $T^{'} = \left( T - \frac{T}{100} \right) = \frac{99T}{100}$
Using Charle’s law $\ \frac{P}{T} = \frac{P^{'}}{T^{'}} \Rightarrow P^{'} = \frac{PT^{'}}{T}$
$= \frac{800 \times 99T}{100T} = 792mm\ of\ Hg$
Saturated vapour pressure at $T^{'} = 25\ mm\ of\ Hg$
$\therefore$ Total pressure in the jar
$=$ Actual pressure of gas $+$ Saturated vapour pressure
$= 792 + 25 = 817\ mm\ of\ Hg$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.