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

Answer:

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

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;

$\frac{m_{14.2}}{m_o} = \frac{\rho _{14.2} V_{14.2}}{\rho _oV_{o}}\\\\ Assume \ that \ the \ air \ composition \ (V_{14.2} = V_o)\\\\\frac{m_{14.2}}{m_o} = \frac{\rho _{14.2} }{\rho _o}$

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

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

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

$\frac{m_{14.2}}{m_o} = \frac{\rho _{14.2}}{\rho_o} \\\\\frac{m_{14.2}}{m_o} =\frac{0.001538}{0.002378} = 0.647 = 64.7 \%$

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 %