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.

16.9816 psia

Explanation:

To calculate the absolute pressure in the bottom of tube we need to sum the atmosferic and gauge pressure.

And the gauge pressure is given by the contributions of columns of water () and mercury(), we can calculate the contribution of each column as:

(*)

where is the respective density, g gravity and h is height.

So we have all the data required to use the above equations (, 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 ( , and ). 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.

And pressure of water in mmHg is

The absolute pressure is:

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

Note: We need to use cm Hg for units coherence

Now passing from dyn/ to kN/ (or kPa) we need to consider that 1 dyn is kN and 1 is .  

Now passing kN/ to psia. We need to consider that 1 psia is 6.89476.

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