Question

(a) A 75.0-kg man floats in freshwater with 3.00% of his volume above water when his lungs are empty, and 5.00% of his volume above water when his lungs are full. Calculate the volume of air he inhales-called his lung capacity-in liters. (b) Does this lung volume seem reasonable?

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(a)\textbf{(a)} Here the mass of the man is

m=75.0kg\begin{align*} m = 75.0\,\mathrm{kg} \end{align*}

Case-I:\textbf{Case-I:} The fraction of the volume of the man above water when his lungs are empty is

f=3%=0.03\begin{align*} f' = 3\,\% = 0.03 \end{align*}

Therefore the fraction submerged is

f=97%=0.97\begin{align*} f = 97\,\% = 0.97 \end{align*}

Hence from the equation 11.41, the defination of fraction submerged is

f=ρmanρwarer\begin{align*} f & = \frac{\rho_{\text{man}}}{\rho_{\text{warer}}} \\ \end{align*}

Hence, we get

ρman=fρwater=0.97×1.000×103kgm3=0.97×103kgm3\begin{align*} \rho_{\text{man}} & = f\, \rho_{\text{water}}\\ & = 0.97 \times\,1.000\times 10^3\,\mathrm{kg\,m^{-3}}\\ & = 0.97\times 10^3\,\mathrm{kg\,m^{-3}} \end{align*}

Hence, the volume of the body of the man is

V1=mρman=75.0kg0.97×103kgm3=77.32×103m3\begin{align*} V_1 & = \frac{m}{\rho_{\text{man}}} \\ & = \frac{75.0\,\mathrm{kg}}{0.97\times 10^3\,\mathrm{kg\,m^{-3}}}\\ & = 77.32\times 10^{-3}\,\mathrm{m^{3}}\tag{1} \end{align*}

Case-II:\textbf{Case-II:} The fraction of the volume of the man above water when his lungs are full with air is

f=5%=0.05\begin{align*} f' = 5\,\% = 0.05 \end{align*}

Therefore the fraction submerged is

f=95%=0.95\begin{align*} f = 95\,\% = 0.95 \end{align*}

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