Problem 85 - Algebra and Trigonometry

Question

Atmospheric pressure \(P\) (in kilo-pascals, 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 \(P_{0}=100 \mathrm{kPa}\) are constants. (a) Solve the equation for \(P .\) (b) Use part (a) to find the pressure \(P\) at an altitude of 4 \(\mathrm{km}\) .

Step-by-Step Solution

Answer

(a) \( P = 100 \cdot e^{-\frac{h}{7}} \); (b) At 4 km, \( P \approx 56.48 \ \mathrm{kPa} \).

Step 1: Isolate \(P\) in the equation

We start with the formula \( \ln \left(\frac{P}{P_{0}}\right)=-\frac{h}{k} \). Our goal is to solve for \( P \). To do this, we need to get rid of the natural logarithm \( \ln \). We use the property that if \( \ln(x) = a \), then \( x = e^{a} \). Thus, we have: \( \frac{P}{P_{0}} = e^{-\frac{h}{k}} \).

Step 2: Solve for \(P\)

To solve for \( P \), multiply both sides of the equation \( \frac{P}{P_{0}} = e^{-\frac{h}{k}} \) by \( P_{0} \) to isolate \( P \): \( P = P_{0} \cdot e^{-\frac{h}{k}} \). This equation gives \( P \) in terms of \( h \), \( P_{0} \), and \( k \).

Step 3: Substitute Constants for Specific Altitude

Now that we have the formula \( P = P_{0} \cdot e^{-\frac{h}{k}} \), we can substitute the given values: \( k = 7 \), \( P_{0} = 100 \ \mathrm{kPa} \), and \( h = 4 \ \mathrm{km} \). This becomes \( P = 100 \cdot e^{-\frac{4}{7}} \).

Step 4: Calculate the Exponent

Calculate \(-\frac{4}{7}\) to find the exponent in the expression: \(-\frac{4}{7} \approx -0.5714\).

Step 5: Compute the Atmospheric Pressure \(P\)

Calculate \( e^{-0.5714} \) using a calculator or a mathematical software to find the approximate value of the exponential expression. \( e^{-0.5714} \approx 0.5648 \). Then multiply by 100: \( P = 100 \times 0.5648 \approx 56.48 \ \mathrm{kPa} \).

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a fundamental mathematical concept that helps to simplify exponential problems. It is defined as the logarithm to the base \( e \), where \( e \approx 2.71828 \). This special number, \( e \), is an irrational constant used frequently in mathematics, especially in calculus and complex analysis.
The natural logarithm has several important properties:

• It converts multiplication into addition: \( \ln(ab) = \ln(a) + \ln(b) \).
• It turns division into subtraction: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \).
• It deals efficiently with powers: \( \ln(a^b) = b \cdot \ln(a) \).

In the context of atmospheric pressure and altitude, the formula given \( \ln\left(\frac{P}{P_0}\right) = -\frac{h}{k} \) uses the natural logarithm to relate pressure \( P \) to altitude \( h \). It allows us to solve for \( P \) by isolating it from within the logarithmic expression. We convert it into an exponential form to easily handle the calculations.

Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of the air above us in the Earth's atmosphere, and it decreases with increasing altitude. Measured in kilo-pascals (kPa), atmospheric pressure at sea level is usually around 101.3 kPa, which corresponds to \( P_0 = 100 \text{ kPa} \) in the exercise.
Atmospheric pressure:

• Is influenced by altitude: The higher you go, the less air there is above you, resulting in lower pressure.
• Affects weather and temperature: Variations in pressure can cause changes in weather patterns.
• Is essential for life: It influences breathing and the distribution of gases in our environment.

In our exercise, the atmospheric pressure \( P \) at a specific altitude is derived from an exponential decay model, explaining how pressure diminishes as height increases. The formula used helps us predict pressure at various altitudes by considering the decay factor \( k \).

Altitude and Pressure Relationship

The relationship between altitude and pressure is characterized by a noticeable decrease in atmospheric pressure as altitude increases. This inverse relationship is mathematically described using the exponential function \( P = P_0 \cdot e^{-\frac{h}{k}} \). Here, \( h \) represents altitude, and \( k \) is a constant that dictates the rate of change.
The relationship can be interpreted as follows:

• The exponential term \( e^{-\frac{h}{k}} \) reveals how pressure decreases with altitude, making it an exponential decay situation where pressure diminishes rapidly as we rise.
• The constant \( k = 7 \) in this scenario controls the rate at which pressure decreases. A larger value of \( k \) implies a more gradual decrease.
• This understanding allows us to predict and calculate atmospheric pressure at different altitudes by substituting the specific values into the formula.

When solving the exercise, by using \( k = 7 \) and substituting \( h = 4 \text{ km} \), we computed the pressure \( P \) at 4 km. The formula reflects the response of pressure change with altitude due to the exponential nature of the Earth’s atmosphere and its pressure system.