The barometric pressure at sea level is 760 mm of mercury while that on a mountain top is 735 mm. If the density of air is assumed constant at 1.2 kg/m³, what is the elevation of the mountain top?

Mountain Elevation from Pressure Difference

Problem Statement

The barometric pressure at sea level is 760 mm of mercury while that on a mountain top is 735 mm. If the density of air is assumed constant at 1.2 kg/m³, what is the elevation of the mountain top?

Given Data

  • Pressure head at sea level, \(h_0 = 760 \, \text{mm of Hg}\)
  • Pressure head at mountain top, \(h = 735 \, \text{mm of Hg}\)
  • Density of air, \(\rho_{air} = 1.2 \, \text{kg/m}^3\)
  • Specific gravity of mercury, \(S_{Hg} = 13.6\)
  • Density of water, \(\rho_w = 1000 \, \text{kg/m}^3\)

Solution

1. Define Density of Mercury

$$ \rho_{Hg} = S_{Hg} \times \rho_w $$ $$ \rho_{Hg} = 13.6 \times 1000 $$ $$ \rho_{Hg} = 13600 \, \text{kg/m}^3 $$

2. Calculate Pressure at Sea Level (\(p_0\))

Convert the pressure head from mm of Hg to N/m².

$$ p_0 = \rho_{Hg} \cdot g \cdot h_0 $$ $$ p_0 = 13600 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times (760/1000) \, \text{m} $$ $$ p_0 = 101396.16 \, \text{N/m}^2 $$

3. Calculate Pressure at Mountain Top (\(p\))

$$ p = \rho_{Hg} \cdot g \cdot h $$ $$ p = 13600 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times (735/1000) \, \text{m} $$ $$ p = 98060.76 \, \text{N/m}^2 $$

4. Calculate the Elevation of the Mountain

The pressure difference is due to the weight of the column of air. Assuming constant air density:

$$ p_0 - p = \rho_{air} \cdot g \cdot Z $$ $$ Z = \frac{p_0 - p}{\rho_{air} \cdot g} $$ $$ Z = \frac{101396.16 - 98060.76}{1.2 \times 9.81} $$ $$ Z = \frac{3335.4}{11.772} $$ $$ Z \approx 283.33 \, \text{m} $$
Final Result:

The elevation of the mountain top is approximately \( Z \approx 283.3 \, \text{m} \).

Explanation of the Principle

Atmospheric pressure is caused by the weight of the column of air above a certain point. As one goes up in altitude (e.g., climbing a mountain), the height of the air column above decreases, and therefore, the atmospheric pressure also decreases.

For relatively small changes in elevation, the density of air can be assumed to be constant. In this case, the relationship between the pressure drop (\(p_0 - p\)) and the change in elevation (\(Z\)) is linear, following the basic hydrostatic equation: $$ \Delta p = \rho g Z $$ This problem uses this simplified model to estimate the mountain's height based on the measured difference in barometric pressure.

Physical Meaning

The result shows that a drop in barometric pressure of 25 mm of mercury (from 760 mm to 735 mm) corresponds to an elevation gain of approximately 283.3 metres under the assumed conditions.

This principle is the basis for how an altimeter works in an aircraft or for a hiker. An altimeter is essentially a sensitive barometer (an instrument that measures air pressure). By measuring the change in atmospheric pressure, it can calculate and display the corresponding change in altitude. This calculation is a simplified example of how such a device functions.

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