Imagine designing an experiment in which the presence of a gas is determined by simply listening to the gas with your ear. The human ear can detect pressures as low as 2 x 10^-5 N*m^-2. Assuming that the eardrum has an area of roughly 1 mm^2, what is the minimum collisional rate that can be detected by ear? Assume that the gas of interest is N2 at 298 K.

Question

Chemistry

Amelia

20 January, 22:25

Imagine designing an experiment in which the presence of a gas is determined by simply listening to the gas with your ear. The human ear can detect pressures as low as 2 x 10^-5 N*m^-2. Assuming that the eardrum has an area of roughly 1 mm^2, what is the minimum collisional rate that can be detected by ear? Assume that the gas of interest is N2 at 298 K.

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Douglas Barton · Answer 1

Pressure = Force/Area

so,

Force = Pressure x Area

Force = (2x 10⁻⁵) N/M² x (1 x (10⁻³) ² M²

Force = 2 x 10⁻¹¹N

as we know,

Force = mass x acceleration (F=m. a)

a = F/m

a = (2 x 10⁻¹¹N) / 28

g since 1 N=1. kg. m. s⁻²

a = (2 x 10-11kg. m. s⁻²) / (28 x 10⁻³kg)

a = 5.6 x 10-7 m. s⁻²

thus minimum collision rate that can be detected is 5.6 x 10-7 m. s⁻²