Problem 37 - Chemistry The Central Science

Question

(a) Calculate the number of molecules in a deep breath of air whose volume is 2.25 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

(a) There are approximately \(5.90 \times 10^{22} \ molecules\) in a deep breath of air. (b) The mass of air contained in an adult blue whale's lungs is approximately \(6.51 \times 10^3 \, g\).

Step 1: Write down the Ideal Gas Law formula

The Ideal Gas Law formula is given by: \(PV = nRT\) Where: P = pressure (in atm) V = volume (in L) n = number of moles R = Ideal Gas Constant (R = 0.0821 L atm mol^{-1} K^{-1}) T = temperature (in K)

Step 2: Perform part a of the exercise

To find the number of molecules in a deep breath, we first need to find the number of moles using the Ideal Gas Law formula. Then, we can multiply the number of moles with the Avogadro's number to find the number of molecules. Given: Volume (V) = 2.25 L Temperature (T) = 37 °C = 37 + 273.15 = 310.15 K Pressure (P) = 735 torr; we need to convert it to atm: \(P = \frac{735 \ torr}{760 \ torr/atm} = 0.967 \ atm\) The Ideal Gas Law formula becomes: \(n = \frac{PV}{RT}\) Plug in the values: \(n = \frac{(0.967 \ atm)(2.25 \ L)}{(0.0821 \ L \ atm \ mol^{-1} K^{-1})(310.15 \ K)}\) \(n \approx 0.098 \, mol\) Now, multiply the number of moles by Avogadro's number to find the number of molecules: Number of molecules = \(n \times N_A\) Number of molecules = \((0.098 \, mol)(6.022 \times 10^{23} \ molecules/mol)\) Number of molecules \( ≈ 5.90 \times 10^{22} \ molecules\)

Step 3: Perform part b of the exercise

First, we will find the number of moles of air in the adult blue whale's lungs. Given: Volume (V) = 5.0 x 10^3 L Temperature (T) = 0.0 °C = 0 + 273.15 = 273.15 K Pressure (P) = 1.00 atm Use the Ideal Gas Law formula to find the moles of air: \(n = \frac{(1.00 \ atm)(5.0 \times 10^3 \ L)}{(0.0821 \ L \ atm \ mol^{-1} K^{-1})(273.15 \ K)}\) \(n \approx 224.6 \, mol\) Now, convert the moles of air into mass using the given average molar mass: Mass of air = \(n \times M\) Mass of air = \((224.6 \, mol)(28.98 \, g/mol)\) Mass of air \( ≈ 6.51 \times 10^3 \, g\) #Answer#: (a) There are approximately \(5.90 \times 10^{22} \ molecules\) in a deep breath of air. (b) The mass of air contained in an adult blue whale's lungs is approximately \(6.51 \times 10^3 \, g\).

Key Concepts

Moles Calculation

Understanding how to calculate moles is an essential skill when working with the Ideal Gas Law. Moles represent the number of particles, such as atoms or molecules, in a given substance. The symbol for moles is 'n'. In the context of the Ideal Gas Law formula, \(PV = nRT\), 'n' is what we solve for to get the number of moles of gas in a particular system.

To find 'n', we rearrange the Ideal Gas Law formula to \(n = \frac{PV}{RT}\). By inserting the values for pressure, volume, temperature, and the ideal gas constant (R), we can compute the moles directly.

For example, with a volume of 2.25 L, a pressure of 0.967 atm, and a temperature of 310.15 K, the moles of air in a deep breath can be calculated as follows:

• Convert the units if necessary (e.g., pressure needs to be in atm).
• Insert the values into \(n = \frac{(0.967 \, \text{atm})(2.25 \, \text{L})}{(0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1})(310.15 \, \text{K})}\).
• The result (approximately 0.098 mol) gives the moles of air present.

Avogadro's Number

Avogadro's number is a constant that links the microscopic world of atoms and molecules to the macroscopic world that we can measure. It is defined as the number of particles in one mole of a substance and is valued at approximately \(6.022 \times 10^{23}\).

This constant is crucial for converting moles to actual numbers of molecules or atoms. In practical terms, it helps us understand quantities on a human scale, like the number of molecules in a breath of air.

In a typical calculation, once you've used the Ideal Gas Law to find the moles (e.g., 0.098 mol), you multiply this by Avogadro's number. The mathematics look like this:

• Number of molecules = number of moles \(\times\) Avogadro's number.
• Number of molecules \(\approx (0.098 \, \text{mol})(6.022 \times 10^{23} \, \text{molecules/mol})\).
• The result, approximately \(5.90 \times 10^{22}\) molecules, gives a tangible perspective on how many molecules are in a breath of air.

Molar Mass Conversion

Understanding molar mass conversion is vital for turning information about gases into different forms, such as mass. The molar mass of a substance tells you how much one mole of it weighs. In chemistry, it's usually expressed in grams per mole (g/mol).

In the example of an adult blue whale's lungs, using the Ideal Gas Law gives us the moles of air (about 224.6 mol). We need to find the mass, which requires multiplying by the average molar mass of air. Here's how it's done:

• Given that the average molar mass of air is 28.98 g/mol, we multiply the moles by this value.
• Mass of air \( = 224.6 \, \text{mol} \times 28.98 \, \text{g/mol}\).
• This results in a mass of approximately \(6.51 \times 10^3 \, \text{g}\), representing the actual weight of the air in the whale's lungs at the specified conditions.