Question

Atmospheric pressure The pressure of Earth’s atmosphere at sea level is approximately 1000 millibars and decreases exponentially with elevation. At an elevation of 30,000 ft (approximately the altitude of Mt. Everest), the pressure is one-third the sea-level pressure. At what elevation is the pressure half the sea-level pressure? At what elevation is it 1% of the sea-level pressure?

Accepted Answer

Identify the given information: the atmospheric pressure at sea level is $$P_0 = 1000$ millibars, and the pressure decreases exponentially with elevation h$. This means the pressure P(h$$)$$ can be modeled as P(h$$) = $$P_0$$ $$e^{kh}$$, where $$k is a $$constant to be determined and $$h is $$the elevation in feet. Use the information at 30,000 ft to find the constant $$k. At$$ $$h = 30000$$, the pressure is one-third of sea-level pressure, so $$P(30000) = \frac{1}{3} P_0$. Substitute into the model: $$\frac{1}{3} $$P_0$$ = $$P_0$$ $$e^{k 	imes 30000}. $$Simplify to get $$\frac{1}{3} = e$$^{30000k}$$. Solve for k$ by taking the natural logarithm of both sides: $$\ln\left(\frac{1}{3}ight) = 30000k$$, which gives k$$ = \frac{\ln\left(\frac{1}{3}ight)}{30000}$$. Find the elevation h$$_{1/2}$$ where the pressure is half the sea-level pressure: P(h$$_{1/2}) = \frac{1}{2} $$P_0. $$Substitute into the model: $$\frac{1}{2} P_0$ = P_0$ e$$^{k h_{1/2}$$}$$, which simplifies to $$\frac{1}{2} = e$$^{k h_{1/2}$$}. $$Take the natural logarithm to solve for $$h_{1/2}$$: $$\ln\left(\frac{1}{2}ight) = k h_{1/2}$$, so $$h_{1/2} = \frac{\ln\left(\frac{1}{2}ight)}{k}. $$Similarly, find the elevation $$h_{0.01}$$ where the pressure is 1% of sea-level pressure: $$P(h_{0.01}) = 0.01 P_0$. Substitute into the model: 0.01$$ = $$e^{k h_{0.01}$$}$$. Take the natural logarithm: $$\ln(0.01) = k h_{0.01}$$, so h$$_{0.01} = \frac{\ln(0.01)}{k}$$.

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Key Concepts

Exponential Decay

Pressure as a Function of Elevation

Solving for Variables in Exponential Equations