Problem Statement
Calculate the pressure at a height of 7500 m above sea level if the atmospheric pressure is 10.143 N/cm² and temperature is 15°C at the sea-level, assuming (i) air is incompressible, (ii) pressure variation follows isothermal law, and (iii) pressure variation follows adiabatic law. Take the density of air at the sea-level as equal to 1.285 kg/m³. Neglect variation of g with altitude.
Given Data
- Height, \(Z = 7500 \, \text{m}\)
- Sea-level pressure, \(p_0 = 10.143 \, \text{N/cm}^2 = 101430 \, \text{N/m}^2\)
- Sea-level temperature, \(t_0 = 15^\circ\text{C} \implies T_0 = 273 + 15 = 288 \, \text{K}\)
- Sea-level air density, \(\rho_0 = 1.285 \, \text{kg/m}^3\)
- Adiabatic index for air, \(k = 1.4\)
Solution
(i) Pressure when Air is Incompressible (Hydrostatic Law)
This model assumes air density is constant. The pressure \(p\) is given by:
Substitute the given values:
Converting to N/cm²:
(ii) Pressure by Isothermal Law
This model assumes temperature is constant. The pressure \(p\) is given by:
Calculate the value of the exponent:
Now, calculate the final pressure:
Converting to N/cm²:
(iii) Pressure by Adiabatic Law
This model assumes no heat exchange (adiabatic process). The pressure \(p\) is given by:
Using \(p_0 = \rho_0 R T_0\), we can substitute for \(RT_0 = p_0 / \rho_0\).
Calculate the terms inside the bracket:
Now, substitute these into the main equation:
Converting to N/cm²:
(i) Incompressible (Hydrostatic) Law: \( \approx 0.69 \, \text{N/cm}^2 \)
(ii) Isothermal Law: \( \approx 3.99 \, \text{N/cm}^2 \)
(iii) Adiabatic Law: \( \approx 3.48 \, \text{N/cm}^2 \)
Explanation of Atmospheric Laws
This problem compares three models for atmospheric pressure variation:
1. Incompressible (Hydrostatic) Law: The simplest model, assuming constant air density. It is highly inaccurate for large altitudes as it doesn’t account for air being compressible.
2. Isothermal Law: A more realistic model that assumes constant temperature. As pressure decreases with altitude, density also decreases proportionally. This is a better approximation than the hydrostatic law.
3. Adiabatic Law: This is generally the most realistic of the three simple models for the troposphere (the lowest layer of the atmosphere). It assumes no heat is exchanged between air parcels as they rise and expand. In this process, the expanding air does work on its surroundings and cools down. This temperature drop causes the density to decrease less rapidly than in the isothermal case, leading to a different pressure profile.
Physical Meaning
The three laws give significantly different results, highlighting the importance of choosing the correct physical model:
- The Incompressible model predicts an extremely low pressure (\(0.69 \, \text{N/cm}^2\)) because it incorrectly assumes the air at 7500 m is as heavy as at sea level, drastically overestimating the weight of the air column.
- The Isothermal model gives a much higher pressure (\(3.99 \, \text{N/cm}^2\)) because it correctly accounts for the air becoming less dense.
- The Adiabatic model predicts the lowest pressure of the two realistic models (\(3.48 \, \text{N/cm}^2\)). This is because as air rises, it expands and cools. Colder air is denser than warmer air at the same pressure. The adiabatic model captures this cooling effect, resulting in a slightly denser (and heavier) air column on average compared to the constant-temperature isothermal model. However, the pressure is still much higher than the incompressible model. For real atmospheric conditions, the adiabatic model is often the most accurate of these three.
