Answer
a. $P=100e^{-h/7}$
b. 56.47 kPa
Work Step by Step
a. $\displaystyle \ln(\frac{P}{P_{0}})=-\frac{h}{k}\qquad$ .. apply $e^{(..)}$ to both sides
$\displaystyle \frac{P}{P_{0}}=e^{-h/k}\qquad$ .. $/\times P_{0}$
$P=P_{0}e^{-h/k}$
$P=100e^{-h/7}$
b. when h=4
$ P=100e^{-4/7}\approx$56.47 kPa
