The liquid in the open-tube manometer in Fig. 12.8a is mercury, y₁ = 3.00 cm, and y₂ = 7.00 cm. Atmospheric pressure is 980 millibars. What is (a) the absolute pressure at the bottom of the U-shaped tube; (b) the absolute pressure in the open tube at a depth of 4.00 cm below the free surface; (c) the absolute pressure of the gas in the container; (d) the gauge pressure of the gas in pascals?

The depth is${\mathrm{y}}_1=3\mathrm{cm}$ .

The depth is${\mathrm{y}}_2=7\mathrm{cm}$ .

The atmospheric pressure is${\mathrm{P}}_{\mathrm{atm}}=980\mathrm{mbar}$ .

The depth below the free surface is d=4 cm .

In this problem, the absolute pressure can be calculated by using the sum of gauge pressure in the tube and atmospheric pressure.

The relation of absolute pressure can be written as:

$$\begin{gathered} P_a=P_g+P_{alm} \\ {P}_a=pg{y}_2+P_{alm} \end{gathered}$$

Here, $P_g$is the gauge pressure,p is the density of mercury and g is the gravitational acceleration.

Substitute 980m bar for $P_{atm}$,13600kg/$m^3$for p ,7cm for $y_2$, and $9.80m/s^2$for g in the above relation.

$$\begin{gathered} P_a\left(13600\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(7\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right)+\left(980\mathrm{mbar}\times \frac{100\mathrm{Pa}}{1\mathrm{mbar}}\right) \\ {\mathrm{P}}_{\mathrm{a}}=107329.6\mathrm{Pa} \end{gathered}$$

Thus, the required absolute pressure is 107329.6 Pa

The relation of absolute pressure can be written as:

$$\begin{gathered} {P^{\text{'}}}_a={P^{\text{'}}}_g+P_{alm} \\ {P^{\text{'}}}_a=pg{y}_2+P_{alm} \end{gathered}$$

Here, ${P_a}^{\text{'}}$is the gauge pressure in open tube,p is the density of mercury and g is the gravitational acceleration.

Substitute 980 mbar for $P_{atm}$, 13600 $kg/m^3$for p ,4cm for d, and $9.80m/s^2$for g in the above relation.

$$\begin{gathered} {P^{\text{'}}}_a\left(13600\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(4\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right)+\left(980\mathrm{mbar}\times \frac{100\mathrm{Pa}}{1\mathrm{mbar}}\right) \\ {{\mathrm{P}}^{\text{'}}}_{\mathrm{a}}=103331.2\mathrm{Pa} \end{gathered}$$

Thus, the required absolute pressure is 103331.2Pa .

The relation of absolute pressure can be written as:

$$\begin{gathered} {P^{\text{'}\text{'}}}_a={P^{\text{'}}}_g+P_{alm} \\ {P^{\text{'}\text{'}}}_a=pg\left(y_2-y_1\right)+P_{alm} \end{gathered}$$

Here, ${P_g}^{\text{'}\text{'}}$is the gauge pressure of the gas.

Substitute 980mbar for $P_{atm}$,13600 $kg/m^3$for p ,3cm for $y_1$,7cm for $y_2$, and $980m/s^2$for g in the above relation.

$$\begin{gathered} {P^{\text{'}\text{'}}}_a\left(13600\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(7\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right)-3cm\times \frac{1m}{100cm}+\left(980\mathrm{mbar}\times \frac{100\mathrm{Pa}}{1\mathrm{mbar}}\right) \\ {{\mathrm{P}}^{\text{'}\text{'}}}_{\mathrm{a}}=103331.2\mathrm{Pa} \end{gathered}$$

Thus, the required absolute pressure is 103331.2Pa .

The relation of gauge pressure can be written as:

${\left(p_g\right)}_{gas}={p_a}^{\text{'}\text{'}}-p_{atm}$

Substitute 980mbar for $P_{atm}$, and 103331.2pa for ${P_a}^{\text{'}\text{'}}$in the above relation.

$$\begin{gathered} {\left(P_{g=}\right)}_{gas}=103331.2Pa-\left(980\mathrm{mbar}\times \frac{100\mathrm{Pa}}{1\mathrm{mbar}}\right) \\ {\left({\mathrm{P}}_{\mathrm{g}}\right)}_{gas}=5331.2\mathrm{Pa} \end{gathered}$$

Thus, the required absolute pressure is 5331.2 Pa .