Problem 50 - Chemistry An Atoms First Approach

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 \((T=200 . \mathrm{K}, P=0.296 \mathrm{~atm})\).

Step-by-Step Solution

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 1: 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) = \(0.0821 \mathrm{~L~atm~K^{-1}~mol^{-1}}\)

Step 2: 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: \(n = \frac{(1.00 \mathrm{~atm})(6.0 \mathrm{~L})}{(0.0821 \mathrm{~L~atm~K^{-1}~mol^{-1}})(298 \mathrm{~K})}\)

Step 3: Calculate Number of Moles for Situation a

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

Step 4: Apply the Ideal Gas Law Formula for Situation b

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

Step 5: Calculate Number of Moles for Situation b

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

Step 6: Apply the Ideal Gas Law Formula for Situation c

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

Step 7: Calculate Number of Moles for Situation c

Now, we can calculate the number of moles of air in the lungs in situation c: \(n = \frac{(0.296)(6.0)}{(0.0821)(200)}\) \(n ≈ 0.108 \mathrm{~mol}\) 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.