Problem 74 - General Chemistry: Principles and Modern Applications

Question

The \(u_{\mathrm{rms}}\) of \(\mathrm{H}_{2}\) molecules at \(273 \mathrm{K}\) is \(1.84 \times 10^{3} \mathrm{m} / \mathrm{s}\) At what temperature is \(u_{\mathrm{rms}}\) for \(\mathrm{H}_{2}\) twice this value?

Step-by-Step Solution

Answer

The temperature at which the root mean square speed of a \(H_{2}\) molecule is twice the given value is \(1092 K\).

Step 1: Use the rms speed equation to find the molar mass

First, it's necessary to determine the molar mass of the \(H_{2}\) molecule using the given root mean square speed at \(273 K\). This can be done by rearranging the rms speed formula solving for m: \(m = \frac{3kT}{u_{rms}^{2}}\) where \(u_{rms} = 1.84 \times 10^{3} m/s\), T = 273 K, and \(k = 1.38 \times 10^{-23} J/K\).

Step 2: Calculate the molar mass

Substituting the values into the equation gives \(m = \frac{3 \times 1.38 \times 10^{-23} J/K \times 273 K}{(1.84 \times 10^{3} m/s)^{2}} = 2.82 \times 10^{-27} kg\)

Step 3: Use the rms speed equation to find the temperature when the speed is doubled

The question asks for the temperature when the root mean square speed is twice the given value, so \(u_{rms} = 2 \times 1.84 \times 10^{3} m/s\). Substitute these values into the rms speed formula while solving for T: \(T = \frac{m \times (u_{rms})^{2}}{3k}\)

Step 4: Calculate the temperature

Substitute the values into the equation from Step 3: \(T = \frac{2.82 \times 10^{-27} kg \times (2 \times 1.84 \times 10^{3} m/s)^{2}}{3 \times 1.38 \times 10^{-23} J/K} = 1092 K\)

Key Concepts

H2 Molecule

The hydrogen molecule, known as \(H_2\), is the simplest and most abundant molecule in the universe. It consists of two hydrogen atoms bonded together. This diatomic molecule is a central component, especially when considering concepts such as the root mean square (rms) speed.

Hydrogen \(H_2\) is a lightweight molecule, thanks to each hydrogen atom having just one proton and one electron. As a result, many physical properties, including speed or kinetic energy, are dramatically influenced by this low mass.

Knowing about the \(H_2\) molecule is crucial when studying molecular movement, particularly in the context of physics and chemistry. The basic properties of \(H_2\) help us understand behaviors such as diffusion, effusion, and the general movement of gases under various conditions.

Molar Mass

Molar mass plays a crucial role in calculations involving gases, particularly when determining properties like velocity and kinetic energy.

For hydrogen, the molar mass is derived from summing the masses of the two hydrogen atoms in one \(H_2\) molecule. Each hydrogen atom has a molar mass of about 1 gram per mole, meaning the \(H_2\) molecule has a molar mass of approximately 2 grams per mole (or 2 * 10^{-3} kg/mol).

In the realm of the root mean square speed calculation, understanding the molar mass of the molecule allows for the determination of other variables, such as temperature or speed. It's calculated using Avogadro's number and helps convert macroscopic observations to microscopic behaviors.

Accurate determination of molar mass aids in explaining why lighter molecules move faster at a given temperature compared to heavier molecules.

Temperature Calculation

Temperature calculation is integral to studying the kinetic energy and movement of molecules. In the root mean square speed equation, temperature plays a critical role as it is directly related to the average kinetic energy of the molecules involved.

The equation to find the root mean square speed \(u_{rms}\) is given by:\[u_{rms} = \sqrt{\frac{3kT}{m}}\]where:

• \(k\) is the Boltzmann constant: \(1.38 \times 10^{-23} J/K\)
• \(T\) represents temperature in Kelvin
• \(m\) is the molar mass of the molecule in kilograms

When you know the rms speed and aim to find out the temperature at which this speed would be a given value, the formula is rearranged to solve for \(T\).

This concept is used extensively in scientific calculations because temperature can be adjusted to study molecular dynamics and responses under different physical conditions. This exercise provides a practical method to grasp how molecular speed shifts with temperature changes.