The normal airflow over the rocky mountains is west to east. The air loses much of its moisture content and is chilled as it climbes the western side of the mountains. When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a Chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressure p depends on altitude y according to$\mathit{p}\mathbf{=}{\mathit{p}}_{\mathbf{0}}\mathit{e}\mathit{x}\mathit{p}\mathbf{(}\mathbf{-}\mathbf{a}\mathbf{y}\mathbf{)}$,where${\mathit{p}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathit{a}\mathit{t}\mathit{m}$and$\mathit{a}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{16}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$. Also assume that the ratio of the molar specific heats is$\mathit{\gamma }\mathbf{=}\mathbf{4}\mathbf{/}\mathbf{3}$. A parcel of air with an initial temperature of$\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{°}\mathit{C}$descends adiabatically from${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{4267}\mathit{m}$to$\mathit{y}\mathbf{=}\mathbf{1567}\mathit{m}$

What is its temperature at the end of the descent?

• $\text{P}={\text{P}}_0{\text{e}}^{-ay}$
  
• ${\text{P}}_0=1\text{\hspace{0.17em}atm}$
  
• $\text{a}=1.16\times {10}^{-4}{\text{m}}^{-1}$
  
• $\gamma =\frac{4}{3}$
  
• ${\text{T}}_i=-5.00°\text{C}$
  

In case of adiabatic process,

$p{V}^{\gamma }=Constant$

Here P is the pressure, V is the volume and $\gamma$is the ratio specific heat capacity at constant pressure to the specific heat capacity at constant volume.

$\gamma =\frac{C_P}{C_V}$

Here $C_P$is specific heat capacity at constant pressure and $C_V$is specific heat capacity at constant volume.

First, find out the pressure at $y=4267\text{\hspace{0.17em}m}$usingthegiven formula

$p=p_0{\text{e}}^{-ay}$

Substitute$4267\text{\hspace{0.17em}m}$for y,$1.16\times {10}^{-4}{\text{m}}^{-1}$for and for$p_0$into the above equation,

$\begin{array}{c}{p}_i=1\times {\text{e}}^{-1.16\times {10}^{-4}\times 4267}\\ =0.6095\text{atm}\end{array}$

For ,$\text{y}=1567\text{m}$,usethegiven formula

$p=p_0{\text{e}}^{-ay}$

Substitute $1567\text{\hspace{0.17em}m}$for y,$1.16\times {10}^{-4}{\text{m}}^{-1}$for and$1atm$for$p_0$into the above equation,

$\begin{array}{c}{\text{P}}_f=1\times {\text{e}}^{-1.16\times {10}^{-4}\times 1567}\\ =0.8337\text{atm}\end{array}$

Now using the concept of adiabatic process,

$p_i{V}_i^{\gamma }=p_f{V}_f^{\gamma }$…… (1)

The ideal gas equation is given by,

$$\begin{gathered} pV=nRT \\ V\propto \frac{T}{P} \end{gathered}$$

Substitute$V\propto \frac{T}{P}$into the equation (1),

$\begin{array}{c}{p}_i{\left(\frac{T_i}{p_i}\right)}^{\gamma }=p_f{\left(\frac{T_f}{p_f}\right)}^{\gamma }\\ p_i^{1-\gamma }{T}_i^{\gamma }=p_f^{1-\gamma }{T}_f^{\gamma }\end{array}$

Substitute $0.6095atm$for $p_i$, $\frac{4}{3}$for $\gamma$, $(273-5)$for $T_i$, $0.8337atm$for $p_f$into the above equation,

$\begin{array}{c}{0.6095}^{1-\frac{4}{3}}\times {(273-5)}^{\frac{4}{3}}={0.8337}^{1-\frac{4}{3}}\times {\text{T}}_f^{\frac{4}{3}}\\ {\text{T}}_f=289.83\\ \approx 290\text{K}\\ =17°\text{C}\end{array}$

Therefore the temperature at the end of the descent is$17°\text{C}$