Mount Everest rises to a height of 8.850 × 10³ m above sea level.At this height,the atmospheric pressure is 231 mm Hg.At what temperature (in °C)will water boil at the summit of Mount Everest? The vapor pressure of water at 373 K is 760.0 mm Hg.(Δ_vapH° for H₂O = 40.7 kJ/mol and R = 8.314 J/K ⋅ mol)

Question

Quizplus · Accepted Answer

First, we need to calculate the atmospheric pressure at the summit of Mount Everest using the equation:

P = P₀e^(-Mgh/RT)

Where P₀ is the atmospheric pressure at sea level (760 mm Hg), M is the molar mass of air (0.02897 kg/mol), g is the acceleration due to gravity (9.81 m/s²), h is the height above sea level (8.850 × 10^3 m), R is the gas constant (8.314 J/K⋅mol), and T is the temperature in Kelvin.

Plugging in the values, we get:

P = (760 mm Hg) × exp[-(0.02897 kg/mol) × (9.81 m/s²) × (8.850 × 10^3 m) / (8.314 J/K⋅mol × 288.15 K)]

P = 231 mm Hg

Now we can use the Clausius-Clapeyron equation to determine the boiling point of water at this pressure:

ln(P₀/P) = ΔH_vap/R × (1/T - 1/T₀)

Where P₀ is the vapor pressure of water at 373 K (760 mm Hg), T₀ is 373 K, ΔH_vap is the enthalpy of vaporization of water (40.7 kJ/mol), and R is the gas constant (8.314 J/K⋅mol).

Solving for T, we get:

T = ΔH_vap/R × (1/ln(P₀/P) + 1/T₀)^-1

Plugging in the values, we get:

T = (40.7 kJ/mol)/(8.314 J/K⋅mol) × (1/ln(760 mm Hg/231 mm Hg) + 1/373 K)^-1

T = 69.0 °C

Therefore, the boiling point of water at the summit of Mount Everest is 69.0 °C, which is choice B.

Mount Everest Rises to a Height of 8