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

In conclusion, the number of moles of air in the lungs is approximately: a. 0.245 mol at sea level. b. 0.482 mol at 10 m below water. c. 0.108 mol at the top of Mount Everest.

Step-by-step solution

Extract Given Values

From the exercise, we are given the following values: 1. Lung capacity (V) = 6.0 L 2. For situation a: T = 298 K, P = 1.00 atm 3. For situation b: T = 298 K, P = 1.97 atm 4. For situation c: T = 200 K, P = 0.296 atm 5. The gas constant (R) = $0.0821\mathrm{L}\mathrm{atm}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$

Apply the Ideal Gas Law Formula for Situation a

The Ideal Gas Law formula is PV = nRT. We will rearrange the formula to solve for n, which gives: $n=\frac{PV}{RT}$. Now we will plug in the given values for situation a: $n=\frac{(1.00\mathrm{atm})(6.0\mathrm{L})}{(0.0821\mathrm{L}\mathrm{atm}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1})(298\mathrm{K})}$

Calculate Number of Moles for Situation a

Now, we can calculate the number of moles of air in the lungs in situation a: $n=\frac{(1.00)(6.0)}{(0.0821)(298)}$ $n\approx 0.245\mathrm{mol}$

Apply the Ideal Gas Law Formula for Situation b

Now we will plug in the given values for situation b: $n=\frac{(1.97\mathrm{atm})(6.0\mathrm{L})}{(0.0821\mathrm{L}\mathrm{atm}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1})(298\mathrm{K})}$

Calculate Number of Moles for Situation b

Now, we can calculate the number of moles of air in the lungs in situation b: $n=\frac{(1.97)(6.0)}{(0.0821)(298)}$ $n\approx 0.482\mathrm{mol}$

Apply the Ideal Gas Law Formula for Situation c

Now we will plug in the given values for situation c: $n=\frac{(0.296\mathrm{atm})(6.0\mathrm{L})}{(0.0821\mathrm{L}\mathrm{atm}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1})(200\mathrm{K})}$

Calculate Number of Moles for Situation c

Now, we can calculate the number of moles of air in the lungs in situation c: $n=\frac{(0.296)(6.0)}{(0.0821)(200)}$ $n\approx 0.108\mathrm{mol}$ In conclusion, the number of moles of air in the lungs is approximately: a. 0.245 mol at sea level. b. 0.482 mol at 10 m below water. c. 0.108 mol at the top of Mount Everest.