Pressure with Two Liquids, Hg and Water. An open test tube at 293 K is filled at the bottom with 12.1 cm of Hg, and 5.6 cm of water is placed above the Hg. Calculate the pressure at the bottom of the test tube if the atmospheric pressure is 756 mm Hg. Use a density of 13.55 g/cm3 for Hg and 0.998 g/cm3 for water. Give the answer in terms of dyn/cm2, psia, and kN/m2. See Appendix A.1 for conversion factors.

And the gauge pressure is given by the contributions of columns of water ([tex]P_{w}[/tex]) and mercury([tex]P_{Hg}[/tex]), we can calculate the contribution of each column as:

[tex]P= \rho g h[/tex] (*)

where [tex]\rho[/tex] is the respective density, g gravity and h is height.

So we have all the data required to use the above equations ([tex]P_{atm}[/tex], height and density of each column) we only need to be carefully with the units.

For simplicity we can to express all pressure contributions in mmHg ( [tex] P_{atm} [/tex], [tex]P_{w}[/tex] and [tex]P_{Hg}[/tex]). Note that the units "x" mmHg means the pressure at the bottom of a column of mercury of "x" mm high. For example, in this case we have a column 12.1 cm of Hg, that is a column of 121 mmHg (passing from cm to mm only requires multiply by 10) pressure exerted by that column is 121 mmHg.

Now pressure of 5.6 cm (56 mm) of water would be 56 mm of water, but it is not the same that mmHg, since the density of water is lower, the pressure exerted by 1 mm of water is lower than the exerted by 1 mm of Hg. The conversion between mmHg and mm of water is given by the relation between the densities.

[tex]mmHg=\frac{\rho_w*mmH_2O}{\rho_{Hg}}[/tex]

[tex]mmHg=\frac{0.998*mmH_2O}{13.55}=0.0737 mmH_2O[/tex]

And pressure of water in mmHg is

[tex]0.0737*56=4.1246 mmHg[/tex]

The absolute pressure is:

[tex]P_{abs}=P_{atm}+P_G= 756 + 121 + 4.1246 = 881.1246 mmHg = 88.11246cmHg[/tex]

To pass to dyn/cm^2 units we need to use the equation (*)

[tex]P= \rho g h = 13.55 \frac{g}{cm^3} * 980.665 \frac{cm}{s^2} * 88.11246 cmHg = 1170839.28 \frac{g}{cm s^2} = 1170839.28 \frac{dyn}{cm^2} [/tex]

Note: We need to use cm Hg for units coherence

Now passing from dyn/[tex]cm^2[/tex] to kN/[tex]m^2[/tex] (or kPa) we need to consider that 1 dyn is [tex]10^{-8}[/tex] kN and 1 [tex]cm^2[/tex] is [tex]10^{-4} m^2[/tex].

[tex]1170839.28 \frac{dyn}{cm^2} * \frac{10^{-8}kN}{1 dyn}*\frac{cm^2}{10^{-4}m^2}=117.083928kN/m^2[/tex]

Now passing kN/[tex]m^2[/tex] to psia. We need to consider that 1 psia is 6.89476.

[tex]117.083928kN/m^2*\frac{1psia}{6.89476kN/m^2}=16.9816 psia[/tex]

mickymike92 mickymike92 · Answer 2 · 2022-03-03T14:23:40+01:00

The pressure at the bottom of the tube, absolute pressure, is given as follows:

P = 881.124 mmHg
P = 1.175 * 10⁶ dyn/cm²
P = 17.04 psia
P = 117.5 kN/m²

What is pressure?

Pressure is defined as perpendicular force per unit area.

Pressure = force/area

The pressure due to fluids is calculated using the formula:

Pressure = hρg

Where;

h is height
hρ is density of the fluid
g is acceleration due to gravity

The pressure at the bottom is known as absolute pressure.

Absolute pressure = Atmospheric pressure + Gauge pressure

Atmospheric pressure = 756 mm Hg

Gauge pressure = Pressure due mercury + pressure due to water

Pressure due to mercury = 12.1 cmHg = 121 mmHg

Pressure due to water = 5.6 cmH₂O = 56 mmH₂O

1 mmH₂O = 0.998/13.55 mmHg = 0.07365 mmHg

56 mmH₂O = 0.07365 mmHg * 56 = 4.124 mmHg

Gauge pressure = 121 mmHg + 4.124 mmHg = 125.124 mmHg

Absolute pressure = 756 mm Hg + 125.124 mmHg

Absolute pressure = 881.124 mmHg

Converting to dyn/cm²

1 mmHg = 1333.22 dyn/cm²

881.124 mmHg = 1.175 * 10⁶ dyn/cm²

Converting to psia:

1 mmHg = 0.0193368

881.124 mmHg = 17.04 psia

Converting to kN/m²:

1 mmHg = 0.133322 kN/m²

881.124 mmHg = 117.5 kN/m²

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