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(a) Calculate the number of molecules in a deep breath of air whose volume is \(2.25 \mathrm{~L}\) at body temperature, \(37{ }^{\circ} \mathrm{C},\) and a pressure of 735 torr. (b) The adult blue whale has a lung capacity of \(5.0 \times 10^{3} \mathrm{~L} .\) Calculate the mass of air (assume an average molar mass \(28.98 \mathrm{~g} / \mathrm{mol}\) ) contained in an adult blue whale's lungs at \(0.0{ }^{\circ} \mathrm{C}\) and 1.00 atm, assuming the air behaves ideally.

Short Answer

Expert verified
There are approximately \(5.48 \times 10^{22}\) molecules in a deep breath of air. The mass of air in an adult blue whale's lungs is approximately \(6.47 \times 10^3\) grams (or 6.47 kg).

Step by step solution

01

Calculate the amount of substance (moles) of air in the deep breath

First, we need to find the amount of substance (in moles) of the air in the deep breath. We will use the Ideal Gas Law, which states: \(PV=nRT \) Where: P - pressure V - volume n - the amount of substance (in moles) R - the ideal gas constant T - temperature The exercise gives us the pressure in torr, so we need to convert it to atmospheres since the value of R we will use is in L*atm/mol*K: \(1 \mathrm{~atm} = 760 \mathrm{~torr} \) Therefore, \(P = \frac{735 \mathrm{~torr}}{760 \frac{\mathrm{torr}}{\mathrm{atm}}} = 0.9671 \mathrm{~atm} \) We have the volume in liters, but we need the temperature in Kelvin: \(T = 37^{\circ} C + 273.15 = 310.15 \mathrm{~K} \) We will use the value of R in L*atm/mol*K: \(R = 0.08206 \mathrm{~L*atm/mol*K} \) Now we can solve the Ideal Gas Law for n:
02

Solve the Ideal Gas Law for n

\(n = \frac{PV}{RT} \) \(n = \frac{(0.9671 \mathrm{~atm})(2.25 \mathrm{~L})}{(0.08206 \mathrm{~L*atm/mol*K})(310.15 \mathrm{~K})} \) \(n \approx 0.09118 \mathrm{~mol} \)
03

Calculate the number of molecules using Avogadro's number

Finally, we can use Avogadro's number to find the number of molecules in this volume of air: Number of molecules = n x Avogadro's number Number of molecules = \(0.09118 \mathrm{~mol} * 6.022 \times 10^{23} \frac{\mathrm{molecules}}{\mathrm{mol}} \) Number of molecules \( \approx 5.48 \times 10^{22} \) molecules So, there are approximately \(5.48 \times 10^{22}\) molecules in a deep breath of air.
04

Calculate the amount of substance (moles) of air in the blue whale's lungs

In part (b), we need to find the mass of air in the blue whale's lungs. First, we will find the amount of substance (in moles) of the air in the whale's lungs. We have the volume, temperature, and pressure: \(V = 5.0 \times 10^3 \mathrm{~L} \) \(T = 0.0^{\circ} C + 273.15 = 273.15 \mathrm{~K} \) \(P = 1.00 \mathrm{~atm} \) Now we can solve the Ideal Gas Law for n:
05

Solve the Ideal Gas Law for n in the blue whale's lungs

\(n = \frac{PV}{RT} \) \(n = \frac{(1.00 \mathrm{~atm})(5.0 \times 10^3 \mathrm{~L})}{(0.08206 \mathrm{~L*atm/mol*K})(273.15 \mathrm{~K})} \) \(n \approx 223.2 \mathrm{~mol} \)
06

Calculate the mass of air in the blue whale's lungs

Now we can use the given average molar mass to find the mass of air in the blue whale's lungs: mass = n x average molar mass mass = \( (223.2 \mathrm{~mol})\times(28.98 \mathrm{~g/mol}) \) mass \( \approx 6.47 \times 10^3 \) g So, the mass of air in an adult blue whale's lungs is approximately \(6.47 \times 10^3\) grams (or 6.47 kg).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mole Calculations
In chemistry, a mole is like a bridge that connects the atomic world to our everyday world. When we talk about gases, like air, we often need to calculate how many moles, or units of molecules, are in a given volume. This is where the Ideal Gas Law comes into play. The Ideal Gas Law is a very useful tool that provides a relationship between pressure, volume, temperature, and the number of moles of a gas. To find the number of moles, we use the formula:\[ n = \frac{PV}{RT} \]Here, \(P\) is the pressure, \(V\) is the volume, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. It's important to ensure that all units are consistent when using this equation. For example, if using \(R = 0.08206\, \mathrm{L \cdot atm/mol \cdot K}\), pressure should be in atmospheres and volume in liters. By plugging in the known values, you can calculate the number of moles of gas. This calculation is crucial for further understanding how gases behave under different conditions.
Avogadro's Number
Avogadro's number is fundamental in chemistry, much like a star player is to a sports team. It allows us to convert between the mass of a substance and the number of individual atoms or molecules it contains. Avogadro's number is approximately \(6.022 \times 10^{23}\), and represents the number of atoms in exactly one mole of a substance.

When you've determined the moles of a substance using mole calculations, you can find the total number of molecules through multiplication with Avogadro's number:\[ \text{Number of molecules} = n \times 6.022 \times 10^{23} \]This calculation is essential when you want to understand how reactions occur on a molecular level. Consider a situation like calculating how many molecules are in a breath of air. By finding the moles first and then using Avogadro's number, you bridge the gap between macroscopic scales and individual molecular interactions. Knowing the exact number of molecules can be crucial for further chemical calculations and understanding reaction stoichiometry.
Molar Mass
To understand molar mass is to hold the key to converting between the mass of a substance and the amount (moles) of that substance. Molar mass, measured in grams per mole \(\text{g/mol}\), tells us how much one mole of a given substance weighs.
  • If you're dealing with a pure element, the molar mass can be directly taken from the periodic table, as it equals the atomic mass.
  • For compounds or mixtures such as air, which has an average molar mass, you calculate it by summing the atomic masses of all atoms in its chemical formula.
In practical scenarios, such as finding out the mass of air in a blue whale's lungs, you first calculate the number of moles of air present using the Ideal Gas Law. Once you have the moles, you use the molar mass to convert this into mass:\[ \text{mass} = n \times \text{molar mass} \]This conversion is indispensable for both laboratory experiments and industrial applications, where precise measurements of reactants and products are required.

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Most popular questions from this chapter

The temperature of a 5.00-L container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\). If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the root-mean-square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; (d) the total number of collisions of molecules with walls ner second.

A scuba diver's tank contains \(0.29 \mathrm{~kg}\) of \(\mathrm{O}_{2}\) compressed into a volume of 2.3 L. (a) Calculate the gas pressure inside the tank at \(9^{\circ} \mathrm{C} .\) (b) What volume would this oxygen occupy at \(26^{\circ} \mathrm{C}\) and 0.95 atm?

(a) What conditions are represented by the abbreviation STP? (b) What is the molar volume of an ideal gas at STP? (c) Room temperature is often assumed to be \(25^{\circ} \mathrm{C}\). Calculate the molar volume of an ideal gas at \(25^{\circ} \mathrm{C}\) and 1 atm pressure.

A plasma-screen TV contains thousands of tiny cells filled with a mixture of Xe, Ne, and He gases that emits light of specific wavelengths when a voltage is applied. A particular plasma cell, \(0.900 \mathrm{~mm} \times 0.300 \mathrm{~mm} \times 10.0 \mathrm{~mm},\) contains \(4 \%\) Xe in a 1: 1 Ne:He mixture at a total pressure of 500 torr. Calculate the number of Xe, Ne, and He atoms in the cell and state the assumptions you need to make in your calculation.

When a large evacuated flask is filled with argon gas, its mass increases by \(3.224 \mathrm{~g}\). When the same flask is again evacuated and then filled with a gas of unknown molar mass, the mass increase is 8.102 g. (a) Based on the molar mass of argon, estimate the molar mass of the unknown gas. (b) What assumptions did you make in arriving at your answer?

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