Question

The average lung capacity of a human is \(6.0 \mathrm{~L}\). How many moles of air are in your lungs when you are in the following situations? a. At sea level \((T=298 \mathrm{~K}, P=1.00 \mathrm{~atm})\). b. \(10 . \mathrm{m}\) below water \((T=298 \mathrm{~K}, P=1.97 \mathrm{~atm})\). c. At the top of Mount Everest \((T=200 . \mathrm{K}, P=0.296 \mathrm{~atm})\).

Short answer

The number of moles of air in the lungs is approximately: a. 0.244 moles at sea level b. 0.479 moles 10 meters below water c. 0.108 moles at the top of Mount Everest

Step-by-step solution

Write down the Ideal Gas Law equation

First, write down the ideal gas law equation: \(PV = nRT\)

Rearrange the equation to solve for n

Next, rearrange the ideal gas law equation to solve for the number of moles (n). Divide both sides of the equation by RT, so that: \(n = \frac{PV}{RT}\)

Calculate the number of moles in situation a

In situation a, at sea level, the temperature T is given as \(298~K\), the pressure P is given as \(1.00~atm\), and the volume V is \(6.0~L\). The ideal gas constant R in the units of L, atm, K, and mol is \(0.0821~\frac{L \cdot atm}{mol \cdot K}\). Plug these values into the equation \(n = \frac{PV}{RT}\) and calculate the number of moles of air in the lungs: \(n_a = \frac{(1.00~atm)(6.0~L)}{(0.0821~\frac{L \cdot atm}{mol \cdot K})(298~K)}\) Calculating this, we find that \(n_a \approx 0.244~mol\).

Calculate the number of moles in situation b

In situation b, 10 meters below water, the temperature T is given as \(298~K\), the pressure P is given as \(1.97~atm\), and the volume V is \(6.0~L\). Plug these values into the equation \(n = \frac{PV}{RT}\) and calculate the number of moles of air in the lungs: \(n_b = \frac{(1.97~atm)(6.0~L)}{(0.0821~\frac{L \cdot atm}{mol \cdot K})(298~K)}\) Calculating this, we find that \(n_b \approx 0.479~mol\).

Calculate the number of moles in situation c

In situation c, at the top of Mount Everest, the temperature T is given as \(200 K\), the pressure P is given as \(0.296~atm\), and the volume V is \(6.0~L\). Plug these values into the equation \(n = \frac{PV}{RT}\) and calculate the number of moles of air in the lungs: \(n_c = \frac{(0.296~atm)(6.0~L)}{(0.0821~\frac{L \cdot atm}{mol \cdot K})(200~K)}\) Calculating this, we find that \(n_c \approx 0.108~mol\). So, the number of moles of air in the lungs is approximately: a. 0.244 moles at sea level b. 0.479 moles 10 meters below water c. 0.108 moles at the top of Mount Everest