Question

Assume that a person skiing high in the mountains at an altitude of h = 14200 ft takes in the same volume of air with each breath as she does while walking at sea level.

Required:
Determine the ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at sea level. Assume that the air composition (i.e. % of air that is oxygen) is the same at sea level as it is at 14200 ft.

Answer 1

The ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at sea level is 64.7 %

Step-by-step explanation:

Mass of air at sea level is given by;

`m_o = \rho_o V_o`

Mass of air at 14,200 ft altitude is given by;

`m_(14.2) = \rho _(14.2) V_(14.2)`

The ratio of the mass of oxygen at high altitude to that at sea level is given by;

`(m_(14.2))/(m_o) = (\rho _(14.2) V_(14.2))/(\rho _oV_(o))\\\\ Assume \ that \ the \ air \ composition \ (V_(14.2) = V_o)\\\\(m_(14.2))/(m_o) = (\rho _(14.2) )/(\rho _o)\\\\`

density of air at sea level,
`\rho _o = 0.002378 \ slug/ft^3`

density of air at 10,000 ft = 0.001756 slug/ft³

density of air at 14,200 ft = x

density of air at 15,000 ft, = 0.001496 slug/ft³

Interpolate between 10,000 ft and 15,000 ft

`(14,200 - 10,000)/(15,000-10,000) = (X - 0.001756)/(0.001496 -0.001756)\\\\ 0.84(-0.00026) = X - 0.001756\\\\-0.0002184 = X - 0.001756\\\\X = 0.001756 - 0.0002184\\\\ X = 0.001538 \ slug/ft^3`

The ratio of the mass of oxygen at 14,200 ft to that at sea level is given by;

`(m_(14.2))/(m_o) = (\rho _(14.2))/(\rho_o) \\\\(m_(14.2))/(m_o) =(0.001538)/(0.002378) = 0.647 = 64.7 \%`

Therefore, the ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at sea level is 64.7 %