Chapter 10, Problem 19P

a.

To determine

To Find: The absolute pressure in the tank.

Answer to Problem 19P

The absolute pressure in the tank tis $4.74\times {10}^5\text{Pa}$ .

Explanation of Solution

Given:

The atmospheric pressure ${\mathrm{P}}_{\mathrm{o}}=1040\text{m}\text{bar =1}\text{.040}\times {\text{10}}^5\text{Pa}$

The height of the mercury in the tube $\mathrm{h}=28.0\text{cm=0}\text{.28}\text{m}$

The density of the mercury $\mathrm{\rho }=13600{\text{kg/m}}^{\text{3}}$

Formula Used:

Write the expression for the absolute pressure.

  $\mathrm{P}={\mathrm{P}}_{\mathrm{o}}+\mathrm{\rho }\mathrm{g}\mathrm{h}$

Here, $\mathrm{\rho }$ is the density of the liquid,

  $\mathrm{h}$ is the height of the mercury in the manometer

  ${\mathrm{P}}_{\mathrm{o}}$ is the atmospheric pressure.

Calculation:

Substitute the values in the above expression.

  $\begin{array}{c}\mathrm{P}=1.014\times {10}^5+\left(13600\times 9.8\times .28\right)\\ =4.74\times {10}^5\text{Pa}\end{array}$

Conclusion:

The absolute pressure in the tank is $4.74\times {10}^5\text{Pa}$ .

b.

To determine

To Find: The absolute pressure in the tank.

Answer to Problem 19P

The pressure in the tank is $\mathrm{P}=0.958\times {10}^5\text{Pa}$ .

Explanation of Solution

Given:

The atmospheric pressure ${\mathrm{P}}_{\mathrm{o}}=1040\text{m}\text{bar =1}\text{.040}\times {\text{10}}^5\text{Pa}$

The height of the mercury in the tube $\mathrm{h}=4.2\text{cm=0}\text{.042}\text{m}$

The density of the mercury $\mathrm{\rho }=13600{\text{kg/m}}^{\text{3}}$

Formula Used:

Write the expression for the absolute pressure.

  $\mathrm{P}={\mathrm{P}}_{\mathrm{o}}+\mathrm{\rho }\mathrm{g}\mathrm{h}$

Here, $\mathrm{\rho }$ is the density of the liquid,

  $\mathrm{h}$ is the height of the mercury in the manometer

  ${\mathrm{P}}_{\mathrm{o}}$ is the atmospheric pressure.

Calculation:

Substitute the values in the above expression.

  $\begin{array}{c}\mathrm{P}=1.014\times {10}^5-\left(13600\times 9.8\times .042\right)\\ =0.958\times {10}^5\text{Pa}\end{array}$

Conclusion:

The pressure in the tank is $\mathrm{P}=0.958\times {10}^5\text{Pa}$ .