Question

1. atmospheric pressure decreases approximately exponentially as elevation increases. if the continuous rate of decrease is determined to be 12% per kilometer and the pressure at sea level is 751.0 mmhg, what is the pressure at 8.3 kilometers above sea level?

Explanation:

Step1: Identify the exponential - decay formula

The formula for exponential decay is $P = P_0e^{-rt}$, where $P_0$ is the initial amount, $r$ is the rate of decay, and $t$ is the time (or in this case, the elevation). Here, $P_0 = 751.0$ mmHg, $r=0.12$ per kilometer, and $t = 8.3$ kilometers.

Step2: Substitute the values into the formula

$P=751.0\times e^{-0.12\times8.3}$.
First, calculate the exponent: $-0.12\times8.3=- 0.996$.
Then, find the value of $e^{-0.996}$. Using a calculator, $e^{-0.996}\approx0.369$.

Step3: Calculate the final pressure

$P = 751.0\times0.369$.
$P\approx277.1$ mmHg.

Answer:

$277.1$ mmHg