Question

Total lung capacity of a typical adult is approximately 5.0 l. approximately 20% of the air is oxygen. you may want to review (pages 360 - 366) . part a at sea level and at an average body temperature of 37∘c, how many moles of oxygen do the lungs contain at the end of an inflation?

Answer 1

To calculate the number of moles of oxygen in the lungs, use the ideal gas law with the given lung capacity of 5.0 liters, knowing that 20% of the air is oxygen, at a pressure of 1 atm and body temperature of 37°C.

Step-by-step explanation:

The student's question concerns the calculation of the number of moles of oxygen contained in the lungs at the end of an inflation at sea level and body temperature, which is a Chemistry question related to the ideal gas law. Given that the total lung capacity is approximately 5.0 liters and 20% of the air is oxygen, we can estimate the volume of oxygen in the lungs. 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 gas constant, and T is the temperature, we can calculate the number of moles of oxygen.

At sea level, the pressure is 1 atm and the temperature is 37°C or 310 K (Kelvin). The volume of oxygen in the lungs is 20% of 5.0 L, which equals 1.0 L. The molar volume of an ideal gas at standard temperature and pressure is 22.4 L. By rearranging the ideal gas law and inserting the values, we calculate the number of moles of oxygen as n = (PV)/(RT).

Answer 2

n = 0.039 moles

Step-by-step explanation:

Given that as we know that air content for oxygen are 20%, so as same for lungs which is equals to 1 liter.

For mole (n) calculation we use

pV = nRT

we know that

10^5 Pa is the pressure(p) at sea level

1 liter or 1000cm^3 or 0.001m^3 is the volume(V) of air

8.314 is the value of R which is molar gas constant

For this equation we need to have temperature(T) in Kelvin (273+37)

273 is the temperature at zero in kelvins + 37 normal human body temperature

Re arranging the equation

n = pV/RT

substituting the values

n=10^5 x 0.001 / (8.314 * (273 + 37))

n = 0.039 moles