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Chapter 5: Problem 114

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

Expert verified
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

01

Write down the Ideal Gas Law equation

First, write down the ideal gas law equation: \(PV = nRT\)
02

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: \(n = \frac{PV}{RT}\)
03

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 \(n_a \approx 0.244~mol\).
04

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 \(n_b \approx 0.479~mol\).
05

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

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