91影视

• $\text{P}={\text{P}}_0{\text{e}}^{鈭�ay}$
  
• ${\text{P}}_0=1\text{鈥塧迟尘}$
  
• $\text{a}=1.16脳{10}^{鈭�4}{\text{m}}^{鈭�1}$
  
• $纬=\frac{4}{3}$
  
• ${\text{T}}_i=鈭�5.00掳\text{C}$
  

In case of adiabatic process,

$p{V}^{纬}=Constant$

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

$纬=\frac{C_P}{C_V}$

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 $y=4267\text{鈥尘}$usingthegiven formula

$p=p_0{\text{e}}^{鈭�ay}$

Substitute$4267\text{鈥尘}$for y,$1.16脳{10}^{鈭�4}{\text{m}}^{鈭�1}$for and for$p_0$into the above equation,

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

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

$p=p_0{\text{e}}^{鈭�ay}$

Substitute $1567\text{鈥尘}$for y,$1.16脳{10}^{鈭�4}{\text{m}}^{鈭�1}$for and$1\text{鈥夆赌�}atm$for$p_0$into the above equation,

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

Now using the concept of adiabatic process,

$p_i{V}_i^{纬}=p_f{V}_f^{纬}$鈥︹�� (1)

The ideal gas equation is given by,

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

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

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

Substitute $0.6095\text{鈥�}atm$for $p_i$, $\frac{4}{3}$for $纬$, $(273鈭�5)$for $T_i$, $0.8337\text{鈥�}atm$for $p_f$into the above equation,

$\begin{array}{c}{0.6095}^{1鈭�\frac{4}{3}}脳{(273鈭�5)}^{\frac{4}{3}}={0.8337}^{1鈭�\frac{4}{3}}脳{\text{T}}_f^{\frac{4}{3}}\\ {\text{T}}_f=289.83\\ 鈮�290\text{K}\\ =17掳\text{C}\end{array}$

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