The total lung capacity of a typical adult is 5.0 L L . Approximately 20 % % of the air is oxygen.At sea level and at a body temperature of 37 ∘ C ∘C , how many oxygen molecules do the lungs contain at the end of a strong inhalation?

The number of mole of oxygen molecules contained in the lungs at the end of a strong inhalation is 0.04 mole

We'll begin by calculating the number of mole of air in the lungs.

Volume (V) = 5 L

Temperature (T) = 37 °C = 37 + 273 = 310 K

Pressure (P) = 1 atm

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?

PV = nRT

Divide both side by RT

n = PV / RT

n = (1 × 5) / (0.0821 × 310)

n = 0.196 mole Finally, we shall determine the number of mole of oxygen molecules.

Mole of air = 0.196 mole

Percentage of oxygen = 20% = 0.2

Mole of Oxygen molecules =?

Mole of Oxygen = percent × total mole

Mole of Oxygen = 0.04 mole

Learn more: link

There will be 0.040 mol of oxygen molecules containing at the end of strong inhalation.

Explanation:

Given

volume = 5.0 liter = 5.0 10^-3 m^3,  temperature = 37 degree celsius = 310 K,

Pressure p = 1 atm = 1.013 10^5 Pa.  

By using Boyle's law,

                        PV = nRT

where P represents the pressure,

          V represents the volume,

          R represents the gas constant,

          T represents the temperature

                      n = PV / RT

                         = (1.013 10^5) (5.0 10^-3) / (9.31) (310)

                      n = 0.20 mol

        N oxygen = 0.20 0.20 = 0.040 mol.

The graph is attached.

We first graph the point where his catch reached the surface, (35, 0).  Since it travels upward at a constant rate, the graph will be linear.  We also need to know where it starts (what depth it is at when he begins reeling it in).  We can use the formula d=rt as a template for our function.  d would be distance (in our case, depth), r is the rate (speed) and t is the amount of time.

To find how far the catch had to travel to reach the surface, we set up our equation as:
d = 0.1(35) 

This will tell us how much distance it traveled in 35 seconds.  0.1(35)=3.5, so the catch started 3.5m under water.  It then travels up at 0.1 m per second.