Question

Chapter 13 examined the variation of pressure with altitude in the Earth's atmosphere, assuming constant temperature-a model known as the isothermal atmosphere. A better approximation is to treat the pressure variations with altitude as adiabatic. Assume that air can be treated as a diatomic ideal gas with effective molar mass \(M_{\text {air }}=28.97 \mathrm{~g} / \mathrm{mol}\). Note that in reality many complexities of the Earth's atmosphere cannot be modeled as an adiabatic ideal gas, for example, the thermosphere. a) Find the air pressure and temperature of the atmosphere as functions of altitude. Let the pressure at sea level be \(p_{0}=101.0 \mathrm{kPa}\) and the temperature at sea level be \(20.0^{\circ} \mathrm{C}\). b) Determine the altitude at which the air pressure is half its sea level value and the temperature at this altitude. Also determine the altitude at which the air density is half its sea level value and the temperature at that altitude. c) Compare these results with the isothermal model of Chapter 13 .

Short answer

What is the temperature at this altitude? Answer: At an altitude of approximately 5431.22 meters above sea level, the air pressure becomes half of its value at sea level. The temperature at this altitude is about 236.97 K.

Step-by-step solution

Define given constants and variables

Let's write down given values and variables: \(p_0 = 101.0\) kPa - pressure at sea level \(T_0 = 20.0 + 273.15\) K - temperature at sea level (converted to Kelvin) \(M_{\text {air }} = 28.97\) g/mol - molar mass of air \(R = 8.314\) J/mol·K - universal gas constant \(g = 9.81\) m/s² - gravity acceleration \(\gamma = 1.4\) - adiabatic index for diatomic ideal gas \(h\) - altitude above sea level

Derive pressure formula

We know that the adiabatic equation for an ideal gas is \(P \cdot V^{\gamma} = \text{constant}\). This can be expressed in terms of pressure, temperature, and altitude as: $$ P = p_0 \left( \frac{T}{T_0} \right)^{\frac{\gamma}{\gamma-1}} $$

Derive temperature formula

The temperature as a function of altitude can be derived from the adiabatic lapse rate formula: $$ T = T_0 - \frac{g \cdot M_{\text {air }}}{R \cdot \gamma} \cdot h $$

Calculate pressure and temperature at half sea level pressure

We'll find the altitude at which the air pressure is half its sea level value and the temperature at this altitude. Using the pressure formula, we can write: $$ \frac{1}{2}p_0 = p_0 \left(\frac{T}{T_0}\right)^{\frac{\gamma}{\gamma-1}} $$ Solving for T, we get: $$ T = T_0 \left(\frac{1}{2}\right)^{\frac{\gamma-1}{\gamma}} $$ Now substituting the given values, $$ T = 293.15 \left(\frac{1}{2}\right)^{\frac{1.4-1}{1.4}} \approx 236.97 \, K $$ Now we can use the temperature formula to find the altitude at this temperature: $$ h = \frac{T_0 - T}{\frac{g \cdot M_{\text {air }}}{R \cdot \gamma}} $$ $$ h \approx 5431.22 \, \text{m} $$

Calculate pressure and temperature at half sea level density

We'll find the altitude at which the air density is half its sea level value and the temperature at this altitude. The density as a function of altitude can be derived from the ideal gas law, knowing that density decreases with increasing altitude: $$ \text{Density} = \rho = \frac{P \cdot M_{\text {air }}}{RT} $$ Now we set the density at half the sea level value, so that: $$ \frac{1}{2} \frac{p_0 \cdot M_{\text {air }}}{R \cdot T_0} = \frac{P \cdot M_{\text {air }}}{RT} $$ Solving for P, we get: $$ P = \frac{1}{2}p_0 \frac{T}{T_0} $$ Substituting this result into the temperature formula, we can solve for altitude: $$ h = \frac{\frac{1}{2}T_0 - T}{\frac{g \cdot M_{\text {air }}}{R \cdot \gamma}} $$ $$ h \approx 5003.86 \, \text{m} $$ The temperature at this altitude can be calculated using the temperature formula: $$ T = 293.15 - \frac{9.81 \cdot 28.97}{8.314 \cdot 1.4} \cdot 5003.86 \approx 249.67 \, K $$

Compare with isothermal model

In the isothermal model, the pressure and temperature follow an exponential decay with altitude. However, in the adiabatic model, the temperature linearly decreases with increasing altitude, resulting in different pressure and temperature profiles. The adiabatic model gives a more accurate representation of the Earth's atmosphere compared to the isothermal model, which could overestimate the pressure in higher altitudes and underestimate it at lower altitudes.