Tuesday, February 11, 2014

The capacity of human lungs is about 5.00 L of air with a deep breath. Assume the temperature of air inside human lungs is 35.0 C and the pressure...

Using the ideal gas law:


PV = nRT


where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant and T is the temperature.


In this case, for air: P = 1 atm, V = 5 l, T = 35 degrees Celsius = 308 K


and R = 0.0821 l atm / mole / K


Rearranging the equation, we get: n = PV/RT


substituting the values of all the parameters in this equation, we get:


n = PV/RT = (1 atm x 5 l)/(0.0821 l atm/mol/K x 308 K) = 0.19773 moles


Since there is 20.95% oxygen in the air, the content of oxygen in the lungs is:


Moles of oxygen = 20.95% of air


= 20.95/100 x 0.19773 moles = 0.04142 moles = 41.42 millimoles 


Thus, with the given conditions, the lungs will contain about 41.42 millimoles of oxygen.


Hope this helps.

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