Chapter 11, Problem 29PE

To determine

The gauge and absolute pressures in the balloon and the peanut jar.

Answer to Problem 29PE

For Balloon,

Gauge pressure, $P_g=5.00\text{cm}{\text{H}}_{\text{2}}\text{O}$

Absolute pressure, $P_{\text{abs}}=1.035\text{cm}{\text{H}}_{\text{2}}\text{O}$

For Jar,

Gauge pressure, $P_{\text{g}}=-50.0\text{mm}\text{Hg}$

Absolute pressure, $P_{\text{abs}}=708\text{mm}\text{Hg}$

Explanation of Solution

Given info:

The manometer along balloon and jar is shown below:

  

Height of the liquid inside the pipe, $h=0.05\text{m}$

Density of water, ${\rho }_w=1000{\text{kg/m}}^3$

Density of mercury, ${\rho }_{Hg}=13.6\times {10}^3{\text{kg/m}}^3$

Formula used:

Gauge pressure is defined as,

  $P_g=\rho gh$

Here,

  $\rho$ = density of the fluid

  $g$ = acceleration due to gravity

  $h$ = height or depth of the fluid

Absolute pressure is defined as,

  $P_{\text{abs}}=P_{\text{atm}}+P_{\text{gauge}}$

Here, $P_{\text{atm}}=1.01\times {10}^5\text{Pa}$.

Calculation:

For balloon,

Gauge pressure In the figure (b), it is seen that the pressure inside the balloon is greater than the atmospheric pressure. Since, water is filled inside the pipe to the height $h=0.05\text{m}$ and gauge pressure depends on the height if density of the fluid is constant.

Gauge pressure would be equal to the pressure of the water up to $h=5\text{cm}$. So,

  $P_{\text{g}}=5.0\text{cm}\text{of}{\text{H}}_2\text{O}$

Absolute pressure,

  $P_{\text{abs}}=P_{\text{atm}}+P_{\text{gauge}}$

  $P_{\text{abs}}=P_{\text{atm}}+{\rho }_{w}gh$

Substitute the value of $P_{\text{atm}}$, ${\rho }_w,g\text{and}h$ in the above equation, we get,

  $\begin{array}{l}{P}_{\text{abs}}=1.01\times {10}^5\text{Pa}+(1000{\text{kg/m}}^{\text{3}}\text{)}(9.8{\text{m/s}}^{\text{2}})(0.05\text{m})\text{}\\ P_{\text{abs}}=101490\text{Pa}\end{array}$

Pressure in terms of height can be written as,

  $\begin{array}{l}{P}_{\text{abs}}={\rho }_{w}gh\\ h=\frac{P_{abs}}{{\rho }_{w}g}\end{array}$

Now, substitute the value of ${\rho }_w$, $P_{abs}$, $g$ in the above equation.

  $\begin{array}{l}h=\frac{101490\text{Pa}}{(1000{\text{kg/m}}^{\text{3}})(9.8{\text{m/s}}^{\text{2}})}\\ h=10.35\text{m}\text{of}{\text{H}}_{\text{2}}\text{O}\\ h=1.035\times {10}^3\text{cm}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}$

For Jar,

Gauge pressure From the figure (c), it is seen that pressure inside the jar is lower than the atmospheric pressure. So, gauge pressure would be negative inside the jar.

Mercury is filled inside the pipe. So, gauge pressure would be equal to the pressure of mercury to the height of $0.05\text{m}$. Since pressure inside the jar is less than the atmospheric pressure, so the height of gauge pressure would be equal to negative height. Here, negative height shows the depth of the mercury from the top surface of the mercury inside the pipe.

Gauge pressure could be written as, $P_{\text{gauge}}=-0.05\text{m}=-50.0\text{mm}\text{.}$

Absolute pressure,

  $P_{\text{abs}}=P_{\text{atm}}+P_{\text{gauge}}$

  $P_{\text{abs}}=P_{\text{atm}}+{\rho }_{Hg}gh$

Substitute the value of $P_{\text{atm}}$, ${\rho }_{Hg},g\text{and}h$ in the above equation.

  $\begin{array}{l}{P}_{\text{abs}}=1.01\times {10}^5\text{Pa}+(13.6\times {10}^3{\text{kg/m}}^{\text{3}}\text{)}(9.8{\text{m/s}}^{\text{2}})(-0.05\text{m})\\ P_{\text{abs}}=101000\text{Pa}-6664\text{Pa}\\ P_{\text{abs}}=94336\text{Pa}\end{array}$

Pressure in terms of height of water can be written as,

  $\begin{array}{l}{P}_{\text{abs}}={\rho }_{Hg}gh\\ h=\frac{P_{abs}}{{\rho }_{\text{Hg}}g}\end{array}$

Now, substitute the value of ${\rho }_{Hg}$, $P_{abs}$, $g$ in the above equation,

  $\begin{array}{l}h=\frac{94336\text{Pa}}{(13.6\times {10}^3{\text{kg/m}}^{\text{3}})(9.8{\text{m/s}}^{\text{2}})}\\ h=0.708\text{m}\text{of}\text{Hg}\end{array}$

  $h=708\text{mm}\text{of}\text{Hg}$