Problem 39 - Chemistry The Central Science

Question

(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.

Step-by-Step Solution

Answer

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 1: 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:

Step 2: Solve the Ideal Gas Law for n

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

Step 3: 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.

Step 4: 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:

Step 5: Solve the Ideal Gas Law for n in the blue whale's lungs

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

Step 6: 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).

Key Concepts

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.