https://r2.erweima.ai/mathbot/image/172481961805775y7cebw.jpg

To solve the problem given in the image, let's break it down step by step.

### Problem Statement:

The problem provides the following information:
- The atmosphere is considered isothermal at 15°C, and follows the equation of state for an ideal gas, \( P = \rho R T \).
- We need to calculate the pressure in bars and the density in \( \text{kg/m}^3 \) at:
  1. 2500 meters above sea level.
  2. 8700 meters above sea level.

It also provides that the pressure and density at sea level are 1 bar and 1.19 \( \text{kg/m}^3 \) respectively, using the following expression:
\[ P = P_0 e^{-\frac{g h}{R T}} \]

Where:
- \( P_0 \) is the pressure at sea level.
- \( g \) is the acceleration due to gravity \( \approx 9.81 \, \text{m/s}^2 \).
- \( h \) is the height above sea level.
- \( R \) is the specific gas constant for air \( \approx 287 \, \text{J/kg·K} \).
- \( T \) is the absolute temperature in Kelvin.

### Solution:

#### Convert Temperature:
- Temperature in Celsius: 15°C
- Convert to Kelvin: \( T = 15 + 273.15 = 288.15 \, \text{K} \)

#### Pressure and Density Calculations:

1. **At 2500 meters:**

   \[
   P = 1 \, \text{bar} \times e^{-\frac{9.81 \times 2500}{287 \times 288.15}}
   \]

   - Compute the exponent:

   \[
   -\frac{9.81 \times 2500}{287 \times 288.15} \approx -0.297
   \]

   - Now, calculate the pressure:

   \[
   P = 1 \times e^{-0.297} \approx 1 \times 0.743 \, \text{bar} \approx 0.743 \, \text{bar}
   \]

   - For density:
   \[
   \rho = \frac{P}{RT} = \frac{0.743 \times 10^5}{287 \times 288.15} \approx 0.883 \, \text{kg/m}^3
   \]

2. **At 8700 meters:**

   \[
   P = 1 \, \text{bar} \times e^{-\frac{9.81 \times 8700}{287 \times 288.15}}
   \]

   - Compute the exponent:

   \[
  

Learn how to calculate pressure and density at different altitudes above sea level using the ideal gas law and exponential decay formula. This problem involves solving for atmospheric conditions given specific altitudes and sea level data. Suitable for advanced high school students.

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