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Chapter 8: 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
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

01

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) = \(0.0821 \mathrm{~L~atm~K^{-1}~mol^{-1}}\)
02

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: \(n = \frac{(1.00 \mathrm{~atm})(6.0 \mathrm{~L})}{(0.0821 \mathrm{~L~atm~K^{-1}~mol^{-1}})(298 \mathrm{~K})}\)
03

Calculate Number of Moles for Situation a

Now, we can calculate the number of moles of air in the lungs in situation a: \(n = \frac{(1.00)(6.0)}{(0.0821)(298)}\) \(n ≈ 0.245 \mathrm{~mol}\)
04

Apply the Ideal Gas Law Formula for Situation b

Now we will plug in the given values for situation b: \(n = \frac{(1.97 \mathrm{~atm})(6.0 \mathrm{~L})}{(0.0821 \mathrm{~L~atm~K^{-1}~mol^{-1}})(298 \mathrm{~K})}\)
05

Calculate Number of Moles for Situation b

Now, we can calculate the number of moles of air in the lungs in situation b: \(n = \frac{(1.97)(6.0)}{(0.0821)(298)}\) \(n ≈ 0.482 \mathrm{~mol}\)
06

Apply the Ideal Gas Law Formula for Situation c

Now we will plug in the given values for situation c: \(n = \frac{(0.296 \mathrm{~atm})(6.0 \mathrm{~L})}{(0.0821 \mathrm{~L~atm~K^{-1}~mol^{-1}})(200 \mathrm{~K})}\)
07

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.

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Most popular questions from this chapter

Ideal gas particles are assumed to be volumeless and to neither attract nor repel each other. Why are these assumptions crucial to the validity of Dalton's law of partial pressures?

Which of the following statements is(are) true? a. If the number of moles of a gas is doubled, the volume will double, assuming the pressure and temperature of the gas remain constant. b. If the temperature of a gas increases from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) the volume of the gas would double, assuming that the pressure and the number of moles of gas remain constant. c. The device that measures atmospheric pressure is called a barometer. d. If the volume of a gas decreases by one half, then the pressure would double, assuming that the number of moles and the temperature of the gas remain constant.

You have a helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is \(79.0 \%\) nitrogen and \(21.0 \%\) oxygen by volume. The "lift" of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C} ?\) Explain. b. Calculate the temperature of the air required for the hotair balloon to provide the same lift as the helium balloon at \(1.00\) atm and \(25^{\circ} \mathrm{C}\). Assume atmospheric conditions are \(1.00\) atm and \(25^{\circ} \mathrm{C}\).

We state that the ideal gas law tends to hold best at low pressures and high temperatures. Show how the van der Waals equation simplifies to the ideal gas law under these conditions.

Nitrogen gas \(\left(\mathrm{N}_{2}\right)\) reacts with hydrogen gas \(\left(\mathrm{H}_{2}\right)\) to form ammonia gas \(\left(\mathrm{NH}_{3}\right) .\) You have nitrogen and hydrogen gases in a \(15.0\)-\(\mathrm{L}\) container fitted with a movable piston (the piston allows the container volume to change so as to keep the pressure constant inside the container). Initially the partial pressure of each reactant gas is \(1.00\) atm. Assume the temperature is constant and that the reaction goes to completion. a. Calculate the partial pressure of ammonia in the container after the reaction has reached completion. b. Calculate the volume of the container after the reaction has reached completion.
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