Problem 15 - Chemistry The Central Science In Si Units

Question

The average concentration of carbon monoxide in air in a city in 2007 was \(3.0 \mathrm{ppm} .\) Calculate the number of \(\mathrm{CO}\) molecules in \(1.0 \mathrm{~L}\) of this air at a pressure of \(100 \mathrm{kPa}\) and a temperature of \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

There are approximately \(7.32 \times 10^{16}\) CO molecules in 1.0 L of air.

Step 1: Convert the Temperature to Kelvin

The temperature in Celsius needs to be converted to Kelvin. The formula to convert Celsius to Kelvin is: \\[ T(K) = T(℃) + 273.15 \]Substituting the given temperature: \\[ T(K) = 25 + 273.15 = 298.15 \, \text{K} \]

Step 2: Use the Ideal Gas Law to Find moles of Air

The ideal gas law is given by \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature. Rearrange to solve for \( n \): \\[ n = \frac{PV}{RT} \]Substitute \( P = 100 \text{kPa} = 100,000 \, \text{Pa} \), \( V = 1.0 \, \text{L} = 0.001 \, \text{m}^3 \), \( R = 8.314 \, \text{J/mol} \cdot \text{K} \), and \( T = 298.15 \, \text{K} \): \\[ n = \frac{100,000 \times 0.001}{8.314 \times 298.15} \approx 0.0405 \, \text{mol} \]

Step 3: Calculate the Total Number of Molecules in the Air Sample

The number of molecules can be determined using Avogadro's number, which is \( 6.022 \times 10^{23} \, \text{molecules/mol} \). Multiply by the number of moles calculated: \\[ \text{Total molecules} = 0.0405 \times 6.022 \times 10^{23} \approx 2.44 \times 10^{22} \, \text{molecules} \]

Step 4: Calculate the Number of CO Molecules

The concentration of CO in air is given as \(3.0 \, \text{ppm}\), which means there are 3 parts per million parts of air. Calculate the number of CO molecules as follows: \\[ \text{CO molecules} = \frac{3}{10^6} \times 2.44 \times 10^{22} \approx 7.32 \times 10^{16} \]

Key Concepts

Avogadro's Number

In chemistry, Avogadro's Number is a fundamental constant used to describe the number of particles, which can be atoms or molecules, in one mole of any substance. This number is approximately \( 6.022 \times 10^{23} \) particles per mole. It serves as a crucial link between the macroscopic world (which we can see and measure) and the microscopic world (the tiny atoms and molecules that we can't see with our eyes).

Avogadro's Number allows us to convert moles of a substance to the actual number of molecules. For example, if we have 1 mole of carbon monoxide (CO), it means we have \( 6.022 \times 10^{23} \) molecules of CO. This concept is often used when calculating the number of particles in a given sample and helps scientists and students understand and predict the behavior of gases under different conditions. By multiplying the number of moles by Avogadro's Number, we can find out exactly how many molecules we have.

Molecular Concentration

Molecular Concentration is a measure of the amount of a particular substance in a given volume. For gases like carbon monoxide (CO), concentrations are often expressed in parts per million (ppm). This unit indicates how many parts of a particular gas are in one million parts of air.

Concentration in ppm is calculated using a simple ratio. For instance, a CO concentration of \(3.0 \) ppm means there are 3 molecules of CO per one million molecules of air. This is critical for understanding pollution levels and the potential impact on health because even small concentrations of gases like CO can be significant.

• To find the number of specific molecules in a sample, use the formula: \(\frac{\text{concentration in ppm}}{10^6} \times \text{Total number of molecules}\).
• This calculation gives insight into how molecular concentrations impact everyday life, especially in assessing air quality.

Temperature Conversion

Temperature Conversion is essential in many gas law calculations because temperature must be in an absolute scale, Kelvin (K), when using the Ideal Gas Law. In Kelvin, the temperature starts at absolute zero, which is the point at which all molecular motion ceases.

The formula to convert Celsius to Kelvin is straightforward:

• \( T(K) = T(℃) + 273.15 \)

This formula ensures that calculations involving temperature are universally consistent and not dependent on relative scales like Celsius. For example, when working with temperatures, starting with \(25^{\circ} \text{C} \), conversion to Kelvin gives \( 298.15 \text{ K} \).

• Using Kelvin is crucial in solving gas law problems to avoid errors associated with negative temperatures in Celsius.
• This universal scale helps in uniformity and better understanding across different scientific and real-life applications.