Q. A horizontal uniform glass tube of 100 cm length sealed at both ends contains 10 cm mercury column in the middle. The temperature and pressure of air on either side of mercury column are respectively 31$℃$ and 76 cm of mercury. If the air column at one end is kept at 0$℃$ and the other end at 273$℃$ , the pressure of air which is at 0$℃$ is (in cm of Hg)
Explanation
(c) On keeping the temperature of the ends of tube at 0$℃$ and 273$℃$.
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Applying ideal gas equation
$\frac{p_{1}V_{1}}{T_{1}} = \frac{p_{2}V_{2}}{T_{2}} = \frac{p_{3}V_{3}}{T_{3}}$
$\frac{76 \times 45}{(273 + 31)} = \frac{p_{2} \times l}{(273 + 0)} = \frac{p_{3}(90 - l)}{273 + 273}$
$\frac{76 \times 45}{304} = \frac{p_{2} \times l}{273} = \frac{p_{3}(90 - l)}{546}$
I II III
From II and III
$\frac{p_{2} \times l}{273} = \frac{p_{3}(90 - l)}{546}$
(Mercury column is at rest, so pressure
difference $p_{2} - p_{3} = 0\ \Rightarrow p_{2} = p_{3})$
$\therefore$ $\frac{p_{2} \times l}{273} = \frac{p_{2}(90 - l)}{546}$
$\Rightarrow \ \ \ \ \ \ \ 2l = 90 - l\ \Rightarrow l = 30$ cm
From I and II
$\frac{76 \times 45}{304} = \frac{p_{2} \times 30}{273}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p_{2} = \frac{76 \times 45 \times 273}{30 \times 304}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p_{2} = 102.4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.