Physics quiz question: air pressure in Krubera Cave

Physics 205A Quiz 5, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 9.19(a)

The entrance to Krubera Cave[*] in the Abkhazia region of Georgia is 2,256 m above sea level. It is the deepest known cave, explored to a depth of 2,197 m from its entrance. Assuming that the gravitational constant g and the density of air do not vary with elevation, the air pressure at the deepest explored point in Krubera Cave is __________ the air pressure at sea level.
(A) less than.
(B) equal to.
(C) greater than.
(D) (Not enough information is given.)

[*] Source: wki.pe/Krubera_Cave.

Correct answer (highlight to unhide): (A)

For static fluids, the energy density relation between pressure and changes in elevation is given by:

0 = ∆P + ρ·g·∆y,

0 = (P_{deepest point} – P_{sea level}) + ρ·g·(y_{deepest point} – y_{sea level}),

where the air pressure at sea level is 101.3 kPa = 1.013×10⁵ Pa, such that:

P_{deepest point} = P_{sea level} – ρ·g·(y_{deepest point} – y_{sea level}).

Since the entrance to Krubera Cave is 2,256 m above sea level, and the deepest explored point is 2,197 m below the entrance, the deepest explored point in Krubera Cave is 2,256 m – 2,197 m = 59 m above sea level, making the quantity (y_{deepest point} – y_{sea level}) positive, and thus the pressure in the deepest explored point in Krubera Cave is slightly less than the air pressure at sea level (with the above assumptions).

Sections 70854, 70855, 73320
Exam code: quiz05LuF7
(A) : 28 students
(B) : 6 students
(C) : 27 students
(D) : 0 students

Success level: 46%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.43