Chapter 4.5, Problem 85E

Atmospheric Pressure Atmospheric pressure P (in kilopascals, kPa) at altitude h (in kilometers, km) is governed by the formula

$\ln \left(\frac{P}{P_0}\right)=-\frac{h}{k}$

where k = 7 and P₀ = 100 kPa are constants.

1. (a) Solve the equation for P.
2. (b) Use part (a) to find the pressure P at an altitude of 4 km.

To determine

To find: The Atmospheric pressure $P$ from the equation $\ln \left(\frac{P}{P_0}\right)=-\frac{h}{k}$.

Answer to Problem 85E

The Atmospheric pressure is $P=P_0{e}^{-\frac{h}{k}}$ from the equation $\ln \left(\frac{P}{P_0}\right)=-\frac{h}{k}$.

Explanation of Solution

Given:

The formula for Atmospheric pressure $P$(in kilopascal, kPa) at altitude $h$(in kilometers, km) is,

$\ln \left(\frac{P}{P_0}\right)=-\frac{h}{k}$

Calculation:

The given formula for Atmospheric pressure $P$(in kilopascal, kPa) at altitude $h$(in kilometers, km) is,

$\ln \left(\frac{P}{P_0}\right)=-\frac{h}{k}$

Simplify the given equation by raising $e$ to each side,

$\begin{array}{c}{e}^{\ln \left(\frac{P}{P_0}\right)}=e^{-\frac{h}{k}}\\ \frac{P}{P_0}=e^{-\frac{h}{k}}\left(\because \ln e^x=x\right)\\ P=P_0{e}^{-\frac{h}{k}}\end{array}$

Thus, the Atmospheric pressure is $P=P_0{e}^{-\frac{h}{k}}$ from the equation $\ln \left(\frac{P}{P_0}\right)=-\frac{h}{k}$.

To determine

To find: The pressure $P$ at an altitude of $4$ km.

Answer to Problem 85E

The value of Atmospheric pressure at an altitude of $4\text{km}$ is $P=56.47$.

Explanation of Solution

Given:

From part (a) the value of atmospheric pressure is,

$P=P_0{e}^{-\frac{h}{k}}$

Altitude $h=4$.

And constant values are $k=7$ and $p_0=100\text{kPa}$.

Calculation:

The given value of atmospheric pressure is,

$P=P_0{e}^{-\frac{h}{k}}$

Substitute $7$ for $k$, $100$ for $P_0$ and $4$ for $h$ in the above equation,

$\begin{array}{c}P=100\left(e^{-\frac{4}{7}}\right)\\ =100\left(e^{-0.57143}\right)\\ =100\left(0.5647\right)\\ =56.47\end{array}$

Thus, the value of Atmospheric pressure at an altitude of $4\text{km}$ is $P=56.47$.