Question

A faulty barometer contains certain amount of air and saturated water vapour. It reads 74.0 cm when the atmospheric pressure is 76.0 cm of mercury and reads 72.10 cm when the atmospheric pressure is 74.0 cm of mercury. Saturation vapour pressure at the air temperature = 1.0 cm of mercury. Find the length of the barometer tube above the mercury level in the reservoir.

Answer 1

The length of the barometer tube is 71.1cm

Explanation:

Given :  

Let the Cross sectional area  be A.

Case:- 1

$V_{1}=(x-74) A$

Saturated vapour pressure = $1 \mathrm{cm} \mathrm{Hg}$

$P_{o}=76 \mathrm{cm} \mathrm{Hg}$

$P_{0}$ is  Atmospheric pressure,  

Height of a column of mercury $=74.0 \mathrm{cm}$

Let P be the pressure of air above the barometer.

Atmospheric pressure = Saturated vapor pressure + Air pressure above the level of mercury barometer + height of mercury column

$1+P+74=76$  

$75+P=76$

$x^{2} P=1 \mathrm{cm}$

$P=1 \mathrm{cm}$

Case :- 2

$P_{0}^{\prime}=74.0 \mathrm{cm} \mathrm{Hg}$

$P^{\prime}_{0}$ is Atmospheric pressure

Let the air pressure be P '.

$P^{\prime}+72 \cdot 10+1=76$

$P^{\prime}=76-73.10$

$P^{\prime}=2.9$

$V_{2}=(x-72.1) A$

Applying the law of Boyle, we get

$P V_{1}=P^{\prime} V_{2}$

$1 \times(x-74) A=2.9 \times(x-72.1) A$

$(x-74)=2.9 \times(x-72.1)$

$(x-74)=(2.9 x-209.09)$

$2.9 x-x=209.09-74$

$1.9 x=135.09$

$x=\frac{135.09}{1.9}$

$x=71.1 \mathrm{cm}$

Tube length$=71.1 \mathrm{cm}$