MCAT Physics Question — Atmospheric Pressure
 by
 Apr 23, 2017
 MCAT Question of the Day
 Reviewed By: Liz Flagge
What is the atmospheric pressure at the top of Mount Everest, which stands approximately 9000 meters above sea level? (Assume an average air density = 0.8 kg/m^{3} and g = 10 m/s^{2})
 7200 Pa
 29000 Pa
 72,000 Pa
 101,000 Pa
Click for Explanation
This question asks the examinee to determine the pressure due to the atmosphere at the top of Mount Everest. To answer this question, you must determine the difference in air pressure between sea level (101,000 Pa) and the top of Everest. The best method to calculate the pressure differential is to apply the gauge pressure formula, P_{gauge} = ρgh, where P_{gauge }is the pressure due to the atmospheric column of height h, ρ is the density of the fluid, and g is the gravitational acceleration constant. The difference between the atmospheric pressure at sea level and the gauge pressure applying the equation above yields the atmospheric pressure at the top of Mt. Everest, as follows:
P_{Everest }= P_{Sea Level }– P_{gauge}
P_{Everest }= 101,000 Pa– ρgh
P_{Everest }= 101,000 Pa– (0.8 kg/m^{3})(10 m/s^{2})(9000 m)
P_{Everest }=29,000 Pa
Thus, the correct answer is 29,000 Pa, which is answer choice B.
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