Atmospheric pressure as a hydrostatic head: Which of the following is the closest equivalent of 1 atm in terms of fluid column height?

Introduction / Context:Engineering calculations often require converting standard atmospheric pressure into an equivalent height of a liquid column. Knowing these equivalences helps when reading manometers, sizing barometric condensers, or interpreting pressure heads in piping systems.

Given Data / Assumptions:

• 1 atm ≈ 101.325 kPa.
• Equivalent heads: about 10.33 m of water, 33.9–34.0 ft of water, and 760 mm (76 cm) of mercury at standard conditions.
• Densities and gravity at standard conditions are implied.

Concept / Approach:The hydrostatic relation is ΔP = ρ g h. For a given pressure, required height h is inversely proportional to liquid density ρ. Thus, a denser fluid like mercury needs a much shorter column than water to balance atmospheric pressure. Recognized “rules of thumb” are typically used in quick selections.

Step-by-Step Solution:

Verification / Alternative check:Using ΔP = ρ g h with ρ_H2O ≈ 1000 kg/m^3 and g ≈ 9.81 m/s^2 yields h ≈ 10.33 m for 101.325 kPa. Converting 10.33 m to feet gives ~33.9 ft.

Why Other Options Are Wrong:

• 13.6 cm Hg confuses mercury’s specific gravity with the required head.
• 1 m H2O is too small; it does not balance 1 atm.
• 13.6 m Hg is far too large.
• (Extra reference) 760 mm Hg is correct but was provided as an additional option; the best match among the listed head choices is 34 ft of H2O.

Common Pitfalls:Mixing up 13.6 (SG of Hg) with 76 cm; ignoring unit conversions between metric and imperial; forgetting that heads scale inversely with density.

Final Answer:34 ft of H2O