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

$\begin{array}{l}\text{Given:}\\ \text{Let the CSA be A.}\\ \text{Case 1:}\\ V_1=(x-74)A\\ \text{SVP = 1 cm Hg}\\ \text{Atmospheric pressure,}{P}_o=76\text{cm Hg}\\ \text{Mercury column height = 74}\text{.0 cm}\\ \text{Let}\mathit{\text{P}}\text{be the air pressure above the barometer. Then,}\\ \text{Atmospheric pressure}=\text{SVP + Air pressure above the barometer mercury level + Mercury column height}\\ \\ \text{⇒1+}\mathit{\text{P}}\text{+74 =76}\\ \Rightarrow \text{P = 1 cm}\\ \text{Case 2:}\\ \text{Atmospheric pressure,}{P}_0\text{'}=74.0\text{cm Hg}\\ \text{Let}\mathit{\text{P}}\text{'}\text{}\text{be the air pressure}\text{. Then,}\\ P\text{'}+72.10+1=76\\ \Rightarrow P\text{'}=0.9\\ V_2=(x-72.1)A\\ \text{Applying Boyle's law, we get}\\ P{V}_1\text{=}P\text{'}{V}_2\\ \text{⇒1}\times (x-74)A=0.9\times (x-72.1)A\\ \Rightarrow x=91.1\text{cm}\\ \text{Length of the tube = 91}\text{.1 cm}\\ \end{array}$