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 \(M_{\text {air }}=28.97 \mathrm{~g} / \mathrm{mol}\) 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 and density are half their sea-level values. What is the temperature at this altitude, in this model? c) Compare these results with the isothermal model of Chapter \(13 .\)

For a diatomic ideal gas, the adiabatic index (the ratio of specific heat capacities) is given by \(\gamma = \frac{7}{5}\). The equation relating pressure and density for an adiabatic process is: $$ p = K \rho^{\gamma} $$ where \(K\) is a constant of the process, \(\rho\) is the density, and \(\gamma\) is the adiabatic index. Using the equation of state for an ideal gas \(p = \rho R T\), where \(R\) is the specific gas constant, we can rewrite the equation above as: $$ p = K\left(\frac{p}{RT}\right)^{\gamma} $$ Now manipulate the equation to isolate temperature: $$ T = \frac{K^{\frac{1}{\gamma -1 }}p^{\frac{\gamma -1}{\gamma}}}{R^{\frac{\gamma -1}{\gamma}}} $$ To find the constant \(K\), we can use the given sea-level conditions, where \(p_0=101.0\,\text{kPa}\) and \(T_0 = (20.0 + 273.15)\,\text{K}\), and the specific gas constant for air \(R=\frac{8.314}{28.97} \mathrm{~J} / \mathrm{molK}\).

To find the altitude at which the air pressure and density are half their sea-level values, we can use the derived functions with \(p=\frac{p_0}{2}\). Then, we can calculate the corresponding temperature using the same derived function for temperature with respect to pressure. The altitude at which this occurs can be found using the hydrostatic equation: $$ \frac{dp}{dz}= -\rho g $$ with the isothermal assumption, $$ \frac{dp}{dz}= -\frac{pM_{\text {air }}g}{RT} $$ We can integrate this equation for the altitude \(z\) when \(p=\frac{p_0}{2}\): $$ z= \frac{RT}{M_{\text {air }}g} \ln \left(\frac{p_0}{p}\right) $$

For the isothermal model, we have that \(T(z)=T_0\) for all altitudes z, and we can use the same hydrostatic equation to derive the pressure: $$ p(z) = p_0 \exp \left( -\frac{M_{\text {air }}gz}{RT_0}\right) $$ Using this equation, we can calculate the altitude at which the pressure is half its sea-level value. By comparing these two models (adiabatic and isothermal), we can observe the differences in pressure and temperature variations with altitude.