(II) An open-tube mercury manometer measures the pressure in an oxygen tank. When the atmospheric pressure is 1040 m bar, what is the absolute pressure (in Pa) in the tank if the height of the mercury in the open tube is (a) 18.5cm higher, (b) 5.6cm lower than the mercury in the tube connected to the tank? See Fig. 10–7a.

In a manometer, the change in pressure at a place is determined by the height of the reading in the curved open and straight end.

The atmospheric pressure is P_atm = 1040m bar.

The mercury at the open end is h = 18.5 cm higher in the first case.

The mercury at the open end is h' = 5.6cm lower in the first case.

The density of mercury is ρ = 13.6 x 10³ kg/m³.

Here, g =9.8 m/s² is the acceleration due to gravity.

The atmospheric pressure is calculated as,

\begin{aligned}{P_{atm}}=1040\;{\rm{mbar}}\times\frac{{{{10}^{-3}}\;{\rm{bar}}}}{{1\;{\rm{mbar}}}}\\{P_{atm}}=1040\times{10^{-3}}\;{\rm{bar}}\\{P_{atm}}=1040\times{10^{-3}}\;{\rm{bar}}\times\frac{{1\;{\rm{Pa}}}}{{{{10}^{-5}}\;{\rm{bar}}}}\\{P_{atm}}=1040\times{10^2}\;{\rm{Pa}}\end{aligned}

The absolute pressure at higher than the normal height is,

\begin{aligned}P'={P_{atm}}+h rho g\\=\left({1040\times{{10}^2}\;{\rm{Pa}}}\right)+\left({18.5\times{{10}^{-2}}\;{\rm{m}}}\right)\times\left({13.6\times{{10}^3}\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}}\right)\times\left({9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}\right)\\\approx129\times{10^3}\;{\rm{Pa}}\end{aligned}

Hence, the absolute pressure is 129 x 10³ Pa.

The absolute pressure at lower than the normal height is,

P’’ = p_atm – h’ρg

On plugging the values in the above relation.

Hence, the absolute pressure is 9.6 x 10⁴ Pa.