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 wa

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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 wa

ter 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.

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arsen [2.9K] · Answer 1 · 2026-02-23T17:04:34+03:00

Answer:

16.9816 psia

$117.083928 kN/m^2$

Explanation:

To determine the absolute pressure at the tube's bottom, we must add atmospheric pressure to gauge pressure.

$P_{abs}=P_{atm}+P_G$

Gauge pressure accounts for contributions from water columns ( $P_{w}$ ) and mercury ( $P_{Hg}$ ), allowing us to derive the contribution of each as:

$P= \rho g h$ (*)

where $\rho$ represents density, g is gravity, and h is height.

With all necessary data available to apply the above equations ( $P_{atm}$ , height, and density of each fluid), we must be diligent about unit consistency.

For clarity, we can express all pressure contributions in mmHg ( $P_{atm}$ , $P_{w}$ , and $P_{Hg}$ ). The units "x" mmHg indicate the pressure at the bottom of a mercury column that is "x" mm tall. In this case, a 12.1 cm Hg column equals 121 mmHg (conversion from cm to mm requires multiplying by 10) showing the pressure exerted is 121 mmHg.

For a water pressure at 5.6 cm (56 mm), it equals 56 mm of water. However, this differs from mmHg because water's density is less than mercury's, resulting in 1 mm of water exerting less pressure than 1 mm of Hg. The conversion between mmHg and mm of water relies on their densities.

$mmHg=\frac{\rho_w*mmH_2O}{\rho_{Hg}}$