Question

One-atmosphere pressure is equivalent to:

• A. 100 cm of water
• B. 76.4 cm of water
• C. 1036 cm of water
• D. 1036 cm of mercury

Hint:

A standard atmosphere supports a column of mercury of 76 cm but a column of water of about 10.3 meters. This is because water is about 13.6 times less dense than mercury. Remembering this ratio is a quick way to solve such problems.

Answer 1

The Correct Option is C

Step 1: Define one standard atmosphere (atm). One standard atmosphere is the pressure exerted by a 760 mm (or 76 cm) column of mercury at 0\(^{\circ}\)C. The pressure \(P\) is given by \(P = h \rho g\), where \(h\) is the height of the column, \(\rho\) is the density of the fluid, and \(g\) is the acceleration due to gravity.
Step 2: Set up the pressure equivalence equation. The pressure exerted by the mercury column must equal the pressure exerted by the water column. \[ P_{atm} = h_{water} \cdot \rho_{water} \cdot g = h_{mercury} \cdot \rho_{mercury} \cdot g \] \[ h_{water} \cdot \rho_{water} = h_{mercury} \cdot \rho_{mercury} \] Step 3: Solve for the height of the water column.
We have:
- \(h_{mercury} = 76\) cm
- Density of mercury, \(\rho_{mercury} \approx 13.6\) g/cm\(^3\)
- Density of water, \(\rho_{water} \approx 1.0\) g/cm\(^3\) \[ h_{water} = h_{mercury} \cdot \frac{\rho_{mercury}}{\rho_{water}} = 76 \text{ cm} \times \frac{13.6}{1.0} = 1033.6 \text{ cm} \] This value is approximately 1036 cm, which is a commonly used standard value in many texts and corresponds to the given option.