Let's solve the problem step by step as requested: --- ### **Step 1: Find the Atmospheric Pressure at the Boiling Point (75°C)** When water boils, its vapor pressure equals the atmospheric pressure. So, the atmospheric pressure is the vapor pressure of water at 75°C. **From steam tables or vapor pressure charts:** - At **75°C**, the **vapor pressure of water ≈ 0.39 atm**. **Convert atm to Pascals (Pa):** \[ 0.39~\text{atm} \times 1.013 \times 10^5~\text{Pa/atm} \approx 3.95 \times 10^4~\text{Pa} \] **Explanation:** At the boiling point, the atmospheric pressure equals the vapor pressure of water. By using standard vapor pressure tables, we find the pressure at 75°C. --- ### **Step 2: Write the Barometric Formula and Plug in the Values** The barometric formula is: \[ P = P_0 \exp(-cA) \] where \( P_0 \) is sea-level pressure (\( 1.01 \times 10^5 \) Pa), \( P \) is the pressure at altitude \( A \), and \( c = \frac{Mg}{RT} \). Given: - \( M = 0.030 \) kg/mol (30 g/mol) - \( g = 9.8 \) m/s² - \( R = 8.31 \) J/(mol·K) - \( T \) (assume average temperature between base and summit, let’s use the boiling point in Kelvin): \( T = 75 + 273 = 348 \) K Calculate \( c \): \[ c = \frac{Mg}{RT} = \frac{0.030 \times 9.8}{8.31 \times 348} \] \[ c \approx \frac{0.294}{2891.88} \approx 1.02 \times 10^{-4}~\text{m}^{-1} \] **Explanation:** This step uses the barometric formula and calculates the constant \( c \) with the given values. --- ### **Step 3: Solve for Altitude Using the Pressures** Rearrange the barometric formula to solve for \( A \): \[ \frac{P}{P_0} = \exp(-cA) \] \[ \ln\left(\frac{P}{P_0}\right) = -cA \] \[ A = -\frac{1}{c} \ln\left(\frac{P}{P_0}\right) \] Plug in the values: - \( P = 3.95 \times 10^4 \) Pa (from Step 1) - \( P_0 = 1.01 \times 10^5 \) Pa - \( c = 1.02 \times 10^{-4}~\text{m}^{-1} \) \[ A = -\frac{1}{1.02 \times 10^{-4}} \ln\left(\frac{3.95 \times 10^4}{1.01 \times 10^5}\right) \] \[ = -9803.9 \times \ln(0.391) \] \[ = -9803.9 \times (-0.939) \] \[ \approx 9199~\text{m} \] **Convert to feet:** \( 9199~\text{m} \times 3.281~\text{ft/m} \approx 30,190~\text{ft} \) **Explanation:** We rearrange the barometric formula to solve for altitude, plug in the known values, and calculate the answer. --- ## **Final Answers** ### (a) **Atmospheric pressure at the summit:** \[ \boxed{0.39~\text{atm} \approx 3.95 \times 10^4~\text{Pa}} \] ### (b) **Estimated altitude using this pressure:** \[ \boxed{9200~\text{m} \approx 30,200~\text{ft}} \] This is quite close to the actual altitude of Mt. McKinley (20,300 ft), but the difference can be attributed to the use of a rough average temperature in the barometric formula and the simplifications made.

The problem involves understanding how atmospheric pressure changes with altitude and how vapor pressure relates to boiling point temperature. First, we use the fact that at the boiling point (75°C), the vapor pressure of water equals the atmospheric pressure, allowing us to determine the pressure at the summit from standard vapor pressure tables. Next, we apply the barometric formula, which models how atmospheric pressure decreases exponentially with altitude, incorporating the ideal gas law parameters such as molar mass, gravity, gas constant, and temperature. By calculating the constant \( c \) and rearranging the formula, we solve for the altitude corresponding to the measured pressure, providing an estimate of the mountain's height based on the boiling point.