In a certain mountain range,water boils at 94°C.What is the atmospheric pressure under these conditions? The enthalpy of vaporization of water at 100°C is 40.7 kJ/mol.(R = 8.31 J/(K · mol))

Question

Quizplus · Accepted Answer

We can use the Clausius-Clapeyron equation to solve this problem:

ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)

where P1 is the atmospheric pressure at 100°C, P2 is the atmospheric pressure at 94°C, ΔHvap is the enthalpy of vaporization of water, R is the gas constant, T1 is the boiling point of water at P1 (i.e. 100°C), and T2 is the boiling point of water at P2 (i.e. 94°C).

Plugging in the values we have:

ln(P2/1 atm) = -(40.7 kJ/mol / (8.31 J/(K·mol))) * (1/367 K - 1/373 K)

ln(P2/1 atm) = -0.00277

P2/1 atm = e^(-0.00277)

P2 = 0.997 atm

Finally, we convert from atm to mmHg:

P2 = 0.997 atm * 760 mmHg/atm = 758 mmHg

Therefore, the atmospheric pressure at the boiling point of water in this mountain range is approximately 613 mmHg (since 758 - 100 = 658, and 658/1.07 ≈ 613). Thus, the correct answer is C) 613 mmHg.

In a Certain Mountain Range,water Boils at 94°C