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Chapter 5: Problem 50

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
a. At sea level, there are approximately 0.245 moles of air in the lungs. b. 10 meters below water, there are approximately 0.482 moles of air in the lungs. c. At the top of Mount Everest, there are approximately 0.108 moles of air in the lungs.

Step by step solution

01

Solve for the number of moles (n) formula from the Ideal Gas Law equation

Starting with the Ideal Gas Law equation, solve for the number of moles (n) by dividing both sides by \(RT\). Here's the resulting formula: \[n = \frac{PV}{RT}\]
02

Calculate the number of moles at sea level

Plug in the values for sea level conditions (lung capacity = 6.0 L, T = 298 K, P = 1.00 atm) and gas constant (R = 0.0821 L∙atm/mol∙K) into the formula: \[n = \frac{(1.00\; \mathrm{atm})(6.0\; \mathrm{L})}{(0.0821\; \mathrm{L∙atm/mol∙K})(298\; \mathrm{K})}\] Calculate the value of n: \[n \approx 0.245 \mathrm{~moles}\]
03

Calculate the number of moles 10 meters below water

Plug in the values for 10 meters below water conditions (lung capacity = 6.0 L, T = 298 K, P = 1.97 atm) into the formula: \[n = \frac{(1.97\; \mathrm{atm})(6.0\; \mathrm{L})}{(0.0821\; \mathrm{L∙atm/mol∙K})(298\; \mathrm{K})}\] Calculate the value of n: \[n \approx 0.482\mathrm{~moles}\]
04

Calculate the number of moles at the top of Mount Everest

Plug in the values for Mount Everest conditions (lung capacity = 6.0 L, T = 200 K, P = 0.296 atm) into the formula: \[n = \frac{(0.296\; \mathrm{atm})(6.0\; \mathrm{L})}{(0.0821\; \mathrm{L∙atm/mol∙K})(200\; \mathrm{K})}\] Calculate the value of n: \[n \approx 0.108\mathrm{~moles}\] #Conclusion#: a. At sea level, there are approximately 0.245 moles of air in the lungs. b. 10 meters below water, there are approximately 0.482 moles of air in the lungs. c. At the top of Mount Everest, there are approximately 0.108 moles of air in the lungs.

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Most popular questions from this chapter

Freon- 12 is used as a refrigerant in central home air conditioners. The rate of effusion of Freon-12 to Freon-11 (molar mass \(=137.4 \mathrm{~g} / \mathrm{mol}\) ) is \(1.07: 1\). The formula of Freon-12 is one of the following: \(\mathrm{CF}_{4}, \mathrm{CF}_{3} \mathrm{Cl}, \mathrm{CF}_{2} \mathrm{Cl}_{2}, \mathrm{CFCl}_{3}\), or \(\mathrm{CCl}_{4} .\) Which formula is correct for Freon-12?

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