Calculate the pressure at a height of 8000 m above sea-level if the atmospheric pressure is 101.3 kN/m² and temperature is 15°C at the sea-level assuming (i) air is incompressible, (ii) pressure variation follows adiabatic law, and (iii) pressure variation follows isothermal law. Take the density of air at the sea-level as equal to 1.285 kg/m³. Neglect variation of g with altitude.

Atmospheric Pressure at Altitude (3 Models)

Problem Statement

Given Data & Constants

Sea-level pressure, $P_0 = 101.3 \, \text{kN/m}^2 = 101300 \, \text{N/m}^2$
Sea-level temperature, $T_0 = 15^\circ\text{C} = 288.15 \, \text{K}$
Height, $z = 8000 \, \text{m}$
Sea-level air density, $\rho_0 = 1.285 \, \text{kg/m}^3$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$
Ratio of specific heats for air, $k = 1.4$ (Adiabatic index)
Gas constant for air, $R \approx 287 \, \text{J/kg·K}$ (Standard value)

Solution

(i) Air is Incompressible (Hydrostatic Law)

This model assumes air density is constant.

$$ P = P_0 - \rho_0 g z $$ $$ P = 101300 - (1.285 \times 9.81 \times 8000) $$ $$ P = 101300 - 100846.8 $$ $$ P = 453.2 \, \text{N/m}^2 $$

(ii) Pressure Variation Follows Isothermal Law

This model assumes air temperature is constant. The pressure decreases exponentially.

$$ P = P_0 e^{-(g z / (R T_0))} $$ $$ \text{First, calculate the exponent term:} $$ $$ \frac{g z}{R T_0} = \frac{9.81 \times 8000}{287 \times 288.15} = \frac{78480}{82708.05} \approx 0.94888 $$ $$ \text{Now, substitute into the equation:} $$ $$ P = 101300 \times e^{-0.94888} $$ $$ P \approx 101300 \times 0.38718 $$ $$ P \approx 39219.3 \, \text{N/m}^2 $$

(iii) Pressure Variation Follows Adiabatic Law

This model assumes no heat is exchanged with the surroundings. It accounts for both pressure and temperature changing with altitude.

$$ P = P_0 \left[1 - \frac{k-1}{k} \frac{g z}{R T_0}\right]^{\frac{k}{k-1}} $$ $$ \text{Calculate the terms inside the bracket:} $$ $$ \frac{k-1}{k} = \frac{1.4-1}{1.4} = \frac{0.4}{1.4} \approx 0.2857 $$ $$ \frac{g z}{R T_0} \approx 0.94888 \quad (\text{from isothermal calculation}) $$ $$ 1 - (0.2857 \times 0.94888) = 1 - 0.2711 = 0.7289 $$ $$ \text{Calculate the outer exponent:} $$ $$ \frac{k}{k-1} = \frac{1.4}{0.4} = 3.5 $$ $$ \text{Now, substitute into the main equation:} $$ $$ P = 101300 \times (0.7289)^{3.5} $$ $$ P \approx 101300 \times 0.3235 $$ $$ P \approx 32770.5 \, \text{N/m}^2 $$

Final Results:

Incompressible Law: $ P \approx 453 \, \text{N/m}^2 $ or $0.453 \, \text{kPa}$

Isothermal Law: $ P \approx 39219 \, \text{N/m}^2 $ or $39.22 \, \text{kPa}$

Adiabatic Law: $ P \approx 32771 \, \text{N/m}^2 $ or $32.77 \, \text{kPa}$

Explanation of the Models

(i) Incompressible Law: Assumes air has a constant density. This is highly inaccurate for large changes in altitude and predicts an impossibly low pressure, suggesting the atmosphere would end much sooner than it does.

(ii) Isothermal Law: Assumes constant temperature. This is more realistic as it allows density to decrease with pressure. It provides a better approximation than the incompressible model.

(iii) Adiabatic Law: Assumes no heat transfer. As a parcel of air rises, it expands and cools. This cooling effect makes the air denser than it would be in an isothermal atmosphere at the same altitude. Because the air column is denser, its weight is greater, and the pressure drops off more quickly with altitude. This is generally the most accurate of the three simple models for the Earth's troposphere.

Physical Meaning & Model Comparison

This problem clearly shows how the underlying physical assumptions dramatically change the calculated pressure at a high altitude like 8000 m (a typical cruising altitude for commercial aircraft).

The Incompressible model fails completely, predicting near-zero pressure.
The Isothermal model gives a pressure of 39.2 kPa.
The Adiabatic model, which accounts for the cooling of rising air, predicts the lowest pressure of 32.8 kPa.

The actual pressure at 8000 m in the International Standard Atmosphere (ISA) is approximately 35.6 kPa. Our adiabatic calculation is the closest, confirming that modeling the temperature change (lapse rate) is crucial for accurately predicting pressure at high altitudes.