Problem 120 - Chemistry The Molecular Science

Question

The relation between the average kinetic energy of a molecule, \(\frac{1}{2} m v^{2},\) and the absolute temperature is $$\frac{1}{2} m v^{2}=\frac{3}{2} k T$$ \(m\) is the mass of the molecule; \(v\) is its average velocity; \(k\) is \(1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} ; T\) is the absolute temperature. \(1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}\). Calculate the average velocity of a nitrogen dioxide molecule in the atmosphere at \(27.0^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The average velocity of an \( \mathrm{NO}_2 \) molecule at 27°C is approximately 403 m/s.

Step 1: Convert Temperature to Kelvin

The given temperature is in Celsius, so first convert it to Kelvin. The formula to convert Celsius to Kelvin is \( T(K) = T(^{\circ}C) + 273.15 \). Therefore, the temperature at \( 27.0^{\circ} \mathrm{C} \) in Kelvin is \( T(K) = 27.0 + 273.15 = 300.15 \ \mathrm{K} \).

Step 2: Analyze the Kinetic Energy Formula

The relationship between kinetic energy and temperature is given by \[ \frac{1}{2} m v^{2} = \frac{3}{2} k T \]. This can be rearranged to solve for \( v^{2} \): \[ v^{2} = \frac{3 k T}{m} \].

Step 3: Identify the Mass of the Molecule

Calculate the mass of a nitrogen dioxide molecule (\( \mathrm{NO}_2 \)). The molecular mass of \( \mathrm{NO}_2 \) is \( 14 + 16 \times 2 = 46 \ \mathrm{g/mol} \). Convert this to kilograms per molecule by dividing by Avogadro's number \( 6.022 \times 10^{23} \ \mathrm{mol}^{-1} \): \[ m = \frac{46 \times 10^{-3}}{6.022 \times 10^{23}} \ \mathrm{kg} = 7.64 \times 10^{-26} \ \mathrm{kg} \].

Step 4: Substitute Values into the Velocity Equation

Substitute \( k = 1.38 \times 10^{-23} \ \mathrm{J/K} \), \( T = 300.15 \ \mathrm{K} \), and \( m = 7.64 \times 10^{-26} \ \mathrm{kg} \) into the equation: \[ v^{2} = \frac{3 \times 1.38 \times 10^{-23} \times 300.15}{7.64 \times 10^{-26}} \].

Step 5: Calculate the Average Velocity

Perform the calculation: \[ v^{2} = \frac{(1.24269 \times 10^{-20})}{7.64 \times 10^{-26}} = 1.6262 \times 10^{5} \]. Take the square root to find the average velocity \( v \): \[ v = \sqrt{1.6262 \times 10^{5}} = 403.26 \ \mathrm{m/s} \].

Key Concepts

Kinetic Energy

The concept of kinetic energy is fundamental in understanding how molecules move and interact. Kinetic energy refers to the energy possessed by an object due to its motion. For molecules, this energy can be expressed in a formula that relates mass and velocity. The average kinetic energy of a molecule (\( \frac{1}{2} mv^2 \)) is directly linked to the molecular mass (\( m \)) and its velocity (\( v \)).
In the context of gases, kinetic energy plays a critical role in defining temperature and pressure. The equation for average kinetic energy can be rearranged to identify how velocity changes with temperature for a given mass. This relationship implies that at higher temperatures, molecules move faster because they have higher kinetic energies.

• Greater mass usually means slower speeds for the same energy.
• Lighter molecules will have higher velocities at the same energy level.

This interconnection illustrates why elements with different masses behave differently under identical conditions, emphasizing the importance of kinetic energy in molecular kinetics.

Absolute Temperature

The concept of absolute temperature is pivotal in understanding molecular behavior and energy distribution. Absolute temperature, measured in Kelvin (K), provides a scale where zero represents the complete absence of thermal energy, known as absolute zero.
In our example, we use the Kelvin scale because it directly links temperature to molecular kinetic energy. To convert from degrees Celsius to Kelvin, you add 273.15. For instance, \( 27\degree C \) becomes \( 300.15 K \), as demonstrated in the exercise.
This unit allows scientists to apply formulas consistently, as the kinetic energy-temperature relationship (\( \frac{1}{2} mv^2 = \frac{3}{2}kT \)) uses absolute temperatures. Absolute temperature thus provides a convenient baseline for understanding and predicting molecular movement and energy interactions, ensuring uniformity and precision in scientific calculations.

• Helps in comparing temperatures on a universally accepted scale.
• Ensures accurate calculations in thermodynamics and molecular physics.

Nitrogen Dioxide

Nitrogen dioxide (NO\(_2\)) is a significant molecule in numerous scientific discussions, primarily because of its atmospheric implications and kinetic properties. It is composed of one nitrogen atom and two oxygen atoms, making its molar mass 46 grams per mole. Understanding this mass is vital for calculating molecular kinetic energies and velocities.
To determine the mass of a single NO\(_2\) molecule in kilograms, it is necessary to use molecular mass and divide by Avogadro's number (\( 6.022 \times 10^{23} \)). This step transforms gram per mole to kilogram per molecule. For NO\(_2\), this means the molecule's mass is \( 7.64 \times 10^{-26} \; \text{kg} \).
In kinetic studies, especially those involving temperature and pressure influences in the atmosphere, knowing the mass and behavior of NO\(_2\) allows scientists to predict how this molecule will respond under varying environmental conditions. This understanding is crucial in both theoretical calculations and practical applications such as pollution tracking and air quality assessment.