Problem 37 - Chemistry The Central Science In Si Units

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 \(97.99 \mathrm{kPa} .(\mathbf{b})\) 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 \(101.33 \mathrm{kPa}\), assuming the air behaves ideally.

Step-by-Step Solution

Answer

(a) About \(5.16 \times 10^{22}\) molecules; (b) About 6.49 kg of air.

Step 1: Understand Ideal Gas Law

To solve these types of problems, we need to use the Ideal Gas Law, which is given by the equation: \[ PV = nRT \] where \( P \) is pressure in pascals (Pa), \( V \) is volume in cubic meters (m³), \( n \) is the number of moles, \( R \) is the universal gas constant (8.314 J/mol·K), and \( T \) is temperature in Kelvin (K).

Step 2: Convert Units for Part (a)

For part (a), convert the volume from liters to cubic meters: \( V = 2.25 \text{ L} = 2.25 \times 10^{-3} \text{ m}^3 \).Convert the temperature from Celsius to Kelvin: \( T = 37 + 273.15 = 310.15 \text{ K} \).Convert the pressure from kPa to Pa: \( P = 97.99 \text{ kPa} = 97990 \text{ Pa} \).

Step 3: Calculate the Number of Moles for Part (a)

Rearrange the Ideal Gas Law to solve for \( n \): \[ n = \frac{PV}{RT} \]Substitute the values: \( n = \frac{97990 \times 2.25 \times 10^{-3}}{8.314 \times 310.15} \). This gives \( n \approx 0.0857 \text{ moles} \).

Step 4: Calculate the Number of Molecules for Part (a)

Use Avogadro's number to find the number of molecules: \[ N = n \times 6.022 \times 10^{23} \]Substitute the value of \( n \) from Step 3: \( N = 0.0857 \times 6.022 \times 10^{23} \), which results in \( N \approx 5.16 \times 10^{22} \) molecules.

Step 5: Convert Units for Part (b)

For part (b), convert the lung capacity from liters to cubic meters: \( V = 5.0 \times 10^3 \text{ L} = 5.0 \text{ m}^3 \).Convert the temperature from Celsius to Kelvin: \( T = 0 + 273.15 = 273.15 \text{ K} \).Convert the pressure from kPa to Pa: \( P = 101.33 \text{ kPa} = 101330 \text{ Pa} \).

Step 6: Calculate the Number of Moles for Part (b)

Rearrange the Ideal Gas Law to solve for \( n \): \[ n = \frac{PV}{RT} \]Substitute the values: \( n = \frac{101330 \times 5.0}{8.314 \times 273.15} \). This gives \( n \approx 223.97 \text{ moles} \).

Step 7: Calculate the Mass of Air for Part (b)

The mass can be calculated using the number of moles and the molar mass: \[ \text{mass} = n \times \text{molar mass} \]Substitute the values: \( \text{mass} = 223.97 \times 28.98 \), which results in \( \text{mass} \approx 6488.23 \text{ g} = 6.49 \text{ kg} \).

Key Concepts

Mole calculations

Mole calculations are a fundamental part of chemistry used to determine the number of particles in a given quantity of substance. In our exercise, moles (\(n\)) were calculated using the Ideal Gas Law formula: \[n = \frac{PV}{RT}\].
This formula highlights the relationship between pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and moles (\(n\)) of a gas.
Written out, to find moles, we:

• Multiply the pressure (\(P\)) of the gas by its volume (\(V\)).
• Divide by the product of the universal gas constant (\(R = 8.314\) J/mol·K) and the temperature (\(T\).) in Kelvin.

It's important to ensure all units are consistent with the equation—pressure in pascals, volume in cubic meters, and temperature in Kelvin.
For example, if you calculated that there are 0.0857 moles of gas in a deep breath of air, it describes a portion of air containing that number of mole quantities of particles.

Avogadro's number

Avogadro's number is crucial for converting between moles and molecules or atoms. It's defined as \(6.022 \times 10^{23}\), the number of atoms, ions, or molecules in one mole of a substance.
This number allows chemists to convert between the macroscopic scale that can be measured directly (grams, liters) and the microscopic scale of atoms and molecules.
So, to get the total number of molecules from the moles calculated, the equation is straightforward:

• Using \(N = n \times 6.022 \times 10^{23}\), where \(n\) is the number of moles, you'll accurately find the microscopic representation in numbers of particles.

For part (a) of your exercise, after finding 0.0857 moles of air, you would multiply by Avogadro's number to find there are approximately \(5.16 \times 10^{22}\) molecules.

Unit conversion in chemistry

Unit conversion plays a critical role in chemistry for maintaining consistency and ensuring correct calculations.
In many chemical processes and calculations, converting units is essential as different measurements may be initially given in non-standard units. Here’s a breakdown of common conversions:

• Volume: Convert from liters to cubic meters by using \(1 \text{L} = 0.001 \text{m}^3\).
• Pressure: Convert from kilopascals (\(\text{kPa}\)) to pascals (\(\text{Pa}\)) by multiplying by 1,000 (1 \(\text{kPa}\) = 1,000 \(\text{Pa}\)).
• Temperature: Convert from Celsius to Kelvin by adding 273.15 to the Celsius temperature.

These conversions ensure that when applying the Ideal Gas Law, everything fits together neatly, ensuring the integrity of calculations.

Gas laws applications

The Ideal Gas Law is a powerful tool for solving many practical and theoretical gas-related questions. It describes how gases behave under various conditions of temperature, pressure, and volume and is expressed as \(PV = nRT\).
In real-world applications, understanding gas laws allows predictions and calculations for various situations, such as:

• Calculating the amount of gas in a container, like lung capacity for living organisms.
• Understanding changes in conditions, like how temperature changes affect pressure at constant volume.
• Predicting outcomes in industrial processes, automotive engines, or weather patterns.

In part (b) of the exercise, using the Ideal Gas Law allowed calculating not just the number of moles but also the corresponding mass of air, enabling comprehensive analysis of gas content both in everyday situations and complex systems.