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.