Problem 37 - 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 of \(28.98 \mathrm{~g} / \mathrm{mol}\) ) contained in an adult blue whale's lungs at \(0.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\), assuming the air behaves ideally.

Step-by-Step Solution

Answer

In short, (a) there are approximately \(5.61 \times 10^{22}\) molecules in a deep breath of air with a volume of \(2.25 \mathrm{~L}\) at body temperature and a pressure of 735 torr, and (b) the mass of air contained in an adult blue whale's lungs, with a lung capacity of \(5.0 \times 10^{3} \mathrm{~L}\), at \(0.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\), is approximately 6477.8 g.

Step 1: Part (a): Find the number of moles

Given: Volume (V) = 2.25 L, temperature (T) = 37°C = 310.15 K (converting to Kelvin), pressure (P) = 735 torr = 735/760 atm (converting to atm), and gas constant (R) = 0.0821 L atm / mol K. First, we need to find the number of moles (n) using the ideal gas law equation: PV = nRT Rearranging the equation to solve for "n": n = PV / RT Now plug in the given values and calculate "n": n = (735/760 atm × 2.25 L) / (0.0821 L atm / mol K × 310.15 K) n ≈ 0.0932 mol

Step 2: Part (a): Calculate the number of molecules

Now, we will use Avogadro's number (6.022 × 10^23 molecules/mol) to calculate the number of molecules in the deep breath of air: Number of molecules = Number of moles × Avogadro's number Number of molecules = 0.0932 mol × 6.022 × 10^23 molecules/mol Number of molecules ≈ 5.61 × 10^22 molecules

Step 3: Part (b): Find the number of moles

Given: Volume (V) = 5.0 x 10^3 L, temperature (T) = 0.0°C = 273.15 K (converting to Kelvin), pressure (P) = 1.00 atm, and gas constant (R) = 0.0821 L atm / mol K. First, we need to find the number of moles (n) using the ideal gas law equation: PV = nRT Rearranging the equation to solve for "n": n = PV / RT Now plug in the given values and calculate "n": n = (1.00 atm × 5.0 x 10^3 L) / (0.0821 L atm / mol K × 273.15 K) n ≈ 223.5981 mol

Step 4: Part (b): Calculate the mass of air

To find the mass of air in an adult blue whale's lungs, we need to multiply the number of moles by the average molar mass of air (28.98 g/mol): Mass of air = Number of moles × Average molar mass Mass of air = 223.5981 mol × 28.98 g/mol Mass of air ≈ 6477.8 g So, the mass of air contained in an adult blue whale's lungs is approximately 6477.8 g.

Key Concepts

Avogadro's Number

One of the fundamental concepts in chemistry is Avogadro's number, which is incredibly important for understanding the behavior of gases. Avogadro's number is the constant used to describe the number of particles, usually atoms or molecules, in one mole of a substance. This number is approximately equal to 6.022 × 10^{23}, a massive number which speaks to how small particles like atoms and molecules really are.

In the context of gases and the ideal gas law, Avogadro's number allows us to connect the macroscopic world of measurements to the microscopic world of molecules. For example, when we breathe in air, we're inhaling a large number of molecules. By using the ideal gas law and knowing the volume, temperature, and pressure of the air, we can calculate the number of moles present. Then, multiplying by Avogadro's number, we arrive at the number of molecules in a sample.

• 6.022 × 10^23 molecules/mol.
• Helps relate observable quantities to atomic scale.
• Essential for converting moles to particles or vice versa.

Understanding Avogadro's number means appreciating just how many particles make up even the tiniest samples of substances we encounter every day.

Molar Mass

Molar mass is another key concept in chemistry, especially when using the ideal gas law. Defined as the mass of one mole of a substance, it is typically expressed in grams per mole (g/mol). Molar mass gives us a bridge between the amount of substance in moles and its physical mass.

For gases, molar mass is used to calculate the mass of the gas when its volume, temperature, and pressure are known. In the blue whale lung example, we used the average molar mass of air, 28.98 g/mol, to find out the weight of air a massive creature like the blue whale can hold.

Key steps for using molar mass:

• Calculate moles using the ideal gas law: \(n = \frac{PV}{RT}\).
• Multiply the moles by the molar mass to find the total mass: \(\text{mass} = n \times \text{molar mass}\).

Recognizing how to handle molar mass allows us to convert the abstract ideas of quantity and mass into practical, real-world applications.

Gas Constant

The gas constant is the keystone of the ideal gas law, symbolized as \(R\). It allows us to use the ideal gas law: \(PV = nRT\) to connect different physical properties such as pressure, volume, and temperature to calculate the number of moles of a gas.

The gas constant \(R\) takes the form of \(0.0821 \ \text{L atm / mol K}\) when working with pressure in atmospheres, volume in liters, and temperature in Kelvin. This specific value for \(R\) ensures that when all parameters are provided in these units, we can precisely calculate relationships between them using the ideal gas law.

Why Gas Constant Matters:

• The value of \(R\) depends on the units used, ensuring calculations remain consistent.
• Facilitates direct calculations between macro-level measurements (like volume and pressure) and chemical moles.
• By rearranging equations, \(R\) helps solve for missing terms such as moles, pressure, or volume.

In the physics of gases, fully understanding how to appropriately apply the gas constant ensures accurate and reliable scientific and practical outcomes.