Problem 61 - University Physics with Modern Physics

Question

As noted in the text, the speed distribution of molecules in the Earth's atmosphere has a significant impact on its composition. a) What is the average speed of a nitrogen molecule in the atmosphere, at a temperature of \(18.0^{\circ} \mathrm{C}\) and a (partial) pressure of \(78.8 \mathrm{kPa} ?\) b) What is the average speed of a hydrogen molecule at the same temperature and pressure?

Step-by-Step Solution

Answer

(Please provide your answer in m/s) Answer: The average speed of a nitrogen molecule in the atmosphere, at a temperature of 18.0°C and a (partial) pressure of 78.8 kPa is approximately 515.4 m/s. The average speed of a hydrogen molecule at the same temperature and pressure is approximately 1827.1 m/s.

Step 1: Convert temperature to Kelvin

To convert the temperature from Celsius to Kelvin, use the formula: \(T(K) = T(^\circ C) + 273.15\) For this problem, we have: \(T(K) = 18.0^\circ C + 273.15 = 291.15 \ K\)

Step 2: Calculate the mass of nitrogen and hydrogen molecules

To find the mass of one molecule of each gas, we can use the molar mass and the Avogadro constant: \(m_1 = \dfrac{M_1}{N_A}\) Here, \(m_1\) is the mass of a single molecule, \(M_1\) is the molar mass, and \(N_A\) is the Avogadro constant (\(6.022 \times 10^{23}\ \text{mol}^{-1}\)). For nitrogen (N2) and hydrogen (H2) we have the following molar masses: - Nitrogen (N2): \(M_1 = 2 \times 14.01\ \text{g/mol} = 28.02\ \text{g/mol}\) - Hydrogen (H2): \(M_1 = 2 \times 1.01\ \text{g/mol} = 2.02\ \text{g/mol}\) Now, we can calculate the mass of a single molecule (in kilograms) for each gas: - Nitrogen: \(m_{N2} = \dfrac{28.02\ \text{g/mol}}{6.022 \times 10^{23}\ \text{mol}^{-1}} = 4.65 \times 10^{-26}\ \text{kg}\) - Hydrogen: \(m_{H2} = \dfrac{2.02\ \text{g/mol}}{6.022 \times 10^{23}\ \text{mol}^{-1}} = 3.36 \times 10^{-27}\ \text{kg}\)

Step 3: Calculate the average speed of nitrogen and hydrogen molecules

Now, we can use the formula for the average speed of a gas molecule and substitute the values we have: \(u_{avg} = \sqrt{\dfrac{8kT}{\pi m}}\) For nitrogen (N2) and hydrogen (H2) we get: - Nitrogen: \(u_{avg_{N2}} = \sqrt{\dfrac{8 \times 1.38 \times 10^{-23}\ \text{J/K} \times 291.15\ \text{K}}{\pi \times 4.65 \times 10^{-26}\ \text{kg}}} = 515.4\ \text{m/s}\) - Hydrogen: \(u_{avg_{H2}} = \sqrt{\dfrac{8 \times 1.38 \times 10^{-23}\ \text{J/K} \times 291.15\ \text{K}}{\pi \times 3.36 \times 10^{-27}\ \text{kg}}} = 1827.1\ \text{m/s}\)

Step 4: Write the final answers

The average speed of a nitrogen molecule in the atmosphere, at a temperature of \(18.0^{\circ} \mathrm{C}\) and a (partial) pressure of \(78.8\ \mathrm{kPa}\) is approximately \(515.4\ \text{m/s}\). The average speed of a hydrogen molecule at the same temperature and pressure is approximately \(1827.1\ \text{m/s}\).

Key Concepts

Molar Mass

Molar mass is a crucial concept in chemistry, referring to the weight of one mole of a substance. It is expressed in grams per mole (g/mol) and is directly related to the mass of individual atoms or molecules. In our textbook exercise, understanding molar mass allows us to compare different gases, like nitrogen and hydrogen.

Nitrogen, with a chemical symbol N₂, has a molar mass of 28.02 g/mol, calculated by doubling the atomic mass of nitrogen, as it is diatomic. Hydrogen's diatomic form, H₂, is much lighter, having a molar mass of only 2.02 g/mol. Why is this important? It helps us determine the mass of an individual gas molecule when combined with Avogadro's constant, which is necessary to calculate the average molecular speed.

Avogadro's Constant

Avogadro's constant, denoted as N_A, is a fundamental number in chemistry that represents the quantity of atoms or molecules in one mole of a substance. This constant, approximately 6.022 x 10²³ mol⁻¹, is named after the italian scientist Amedeo Avogadro. It serves as a bridge between the macroscopic scale of substances we handle everyday and the microscopic scale of atoms and molecules.

In our exercise, we use this constant to find the mass of one nitrogen or hydrogen molecule from their respective molar masses. By dividing the molar mass by Avogadro's constant, we convert from the macroscale (grams per mole) to the microscale (kilograms per molecule), which is required to further investigate their behaviors as individual gas molecules.

Kinetic Molecular Theory

Kinetic Molecular Theory (KMT) gives us a framework to understand the behavior of gases at the molecular level. It explains the properties of gases in terms of the energy their molecules contain and how they move. Some fundamental points of KMT include the constant, random motion of gas molecules and the negligible volume of the molecules compared to the space they occupy.

Moreover, KMT tells us that the temperature of a gas is a measure of the average kinetic energy of its molecules. Consequently, at a given temperature, lighter gas molecules, like hydrogen, move faster on average than heavier ones, like nitrogen. This concept explains our findings in the exercise, where despite the same temperature, hydrogen molecules have a significantly higher average speed.

Gas Molecule Velocity Calculation

To calculate the average velocity of gas molecules, we use an equation derived from the Kinetic Molecular Theory. The equation is
\( u_{avg} = \sqrt{\dfrac{8kT}{\pi m}} \)
, where

• \(u_{avg}\) is the average speed,
• k is the Boltzmann's constant,
• T is the temperature in Kelvin,
• m is the mass of a single molecule.

This equation embodies the direct relationship between energy (as temperature) and the motion of molecules. In this problem, upon substituting in the calculated mass of nitrogen and hydrogen molecules, we found their average speeds, considering the temperature in Kelvin. The lighter hydrogen molecules travel at a greater speed due to their smaller mass, demonstrating an inverse relationship between mass and average velocity.