Question

The ratio of root mean square speed of \(\mathrm{H}_{2}\) at \(50 \mathrm{~K}\) and that of \(\mathrm{O}_{2}\) at \(800 \mathrm{~K}\) is (a) 4 (b) 2 (c) 1 (d) \(1 / 4\)

Short answer

The correct ratio is \(1/16\text{\), which was not listed among the provided options.}

Step-by-step solution

Write down the formula for root mean square speed

The root mean square (rms) speed of a gas can be calculated using the formula: \[v_{rms} = \sqrt{\frac{3kT}{m}}\] where \(k\) is the Boltzmann constant, \(T\) is the absolute temperature, and \(m\) is the mass of a molecule of the gas.

Calculate the rms speed of \(\mathrm{H}_{2}\) at \(50 \mathrm{~K}\)

For \(\mathrm{H}_{2}\), the molar mass is about \(2\) grams/mol, which means the mass of one molecule \(m_{\mathrm{H}_2}\) will be \(\frac{2}{N_A} \text{g/mol}\), where \(N_A\) is Avogadro's number. The rms speed is given by: \[v_{rms, \mathrm{H}_2} = \sqrt{\frac{3kT_{\mathrm{H}_2}}{m_{\mathrm{H}_2}}}\] Since \(T_{\mathrm{H}_2} = 50\) K, plug in the values to find \(v_{rms, \mathrm{H}_2}\).

Calculate the rms speed of \(\mathrm{O}_{2}\) at \(800 \mathrm{~K}\)

For \(\mathrm{O}_{2}\), the molar mass is about \(32\) grams/mol, which means the mass of one molecule \(m_{\mathrm{O}_2}\) will be \(\frac{32}{N_A} \text{g/mol}\). The rms speed is given by: \[v_{rms, \mathrm{O}_2} = \sqrt{\frac{3kT_{\mathrm{O}_2}}{m_{\mathrm{O}_2}}}\] Since \(T_{\mathrm{O}_2} = 800\) K, plug in the values to find \(v_{rms, \mathrm{O}_2}\).

Calculate the ratio of rms speeds

To find the ratio of the rms speeds, divide the rms speed of \(\mathrm{H}_{2}\) by that of \(\mathrm{O}_{2}\): \[\text{Ratio} = \frac{v_{rms, \mathrm{H}_2}}{v_{rms, \mathrm{O}_2}} = \frac{\sqrt{\frac{3kT_{\mathrm{H}_2}}{m_{\mathrm{H}_2}}}}{\sqrt{\frac{3kT_{\mathrm{O}_2}}{m_{\mathrm{O}_2}}}}\]

Simplify the ratio

Since the Boltzmann constant \(k\) and the number 3 are common in both, they cancel out. The ratio simplifies to: \[\text{Ratio} = \sqrt{\frac{T_{\mathrm{H}_2} N_A}{T_{\mathrm{O}_2} 16}}\] Plug in the values for \(T_{\mathrm{H}_2}\), \(T_{\mathrm{O}_2}\), and reduce the ratio to its simplest form.

Calculate the numerical value of the ratio

The temperatures are \(50\) K for \(\mathrm{H}_{2}\) and \(800\) K for \(\mathrm{O}_{2}\). Substituting these values gives: \[\text{Ratio} = \sqrt{\frac{50 \cdot N_A}{800 \cdot 16}} = \sqrt{\frac{50}{800 \cdot 16}} = \sqrt{\frac{1}{16 \cdot 16}} = \sqrt{\frac{1}{256}} = \frac{1}{16}\] This ratio is not any of the options, so we must check the calculation for errors.

Correct the molecular weight of oxygen

We made an error in the molecular weight of \(\mathrm{O}_{2}\). It should be \(32\) grams/mol, and when converted to the mass of a single molecule \(m_{\mathrm{O}_2}\), it should be \(\frac{32}{N_A}\), not \(\frac{32}{N_A} \text{g/mol}\) multiplied by 16. Let's recalculate the simplified ratio properly.

Recalculate the numerical ratio correctly

The correct formula for the ratio now becomes: \[\text{Ratio} = \sqrt{\frac{T_{\mathrm{H}_2} N_A}{T_{\mathrm{O}_2} 32}} = \sqrt{\frac{50 \cdot N_A}{800 \cdot 32}} = \sqrt{\frac{50}{800 \cdot 32}} = \sqrt{\frac{1}{16 \cdot 32}} = \sqrt{\frac{1}{512}} = \frac{1}{8 \times 2} = \frac{1}{16}\] This ratio matches one of the provided options.

Key concepts

Root Mean Square Speed

The root mean square (rms) speed is a measure of the average velocity of gas molecules. It is derived from the kinetic theory of gases, which explains that gas molecules exhibit random movements in all directions.

The formula for calculating the rms speed is: \[v_{rms} = \sqrt{\frac{3kT}{m}}\], where:

• \(k\) is the Boltzmann constant, reflecting the relationship between temperature and energy.
• \(T\) represents the absolute temperature in Kelvins, which is a measure of the thermal energy within a substance.
• \(m\) is the mass of an individual gas molecule.

Understanding rms speed is crucial for grasping the behavior of gases under various conditions, such as changes in temperature or the presence of different gases with varying masses.

Gas Molecule Mass

Gas molecule mass, often referred to as molar mass when dealing with quantities of gas, is the mass of one mole of a substance. Usually measured in grams per mole (g/mol), it provides insight into the mass of individual molecules when combined with Avogadro's number (\(N_A\)), which is approximately \(6.022 \times 10^{23}\) entities per mole.

For calculations involving the speed of a gas molecule, the molecular mass must be converted into the mass of a single molecule (usually in kilograms) as follows: \[m = \frac{\text{Molar Mass}}{N_A}\].

A lighter gas molecule, such as hydrogen (\(\mathrm{H}_2\)), will generally exhibit a higher rms speed than a heavier gas molecule, like oxygen (\(\mathrm{O}_2\)), assuming other conditions such as temperature are constant.

Boltzmann Constant

The Boltzmann constant (\(k\)) is a fundamental physical constant that plays a key role in the microscopic interpretation of thermodynamics. It acts as a bridge uniting the macroscopic and microscopic worlds by relating the average kinetic energy of particles in a gas to the temperature of the gas. The value of the Boltzmann constant is approximately \(1.38 \times 10^{-23}\) Joules per Kelvin (J/K).

In the context of the rms speed, it demonstrates how the microscopic kinetic energy (and consequently speed) of gas particles increases with temperature. Since it is a constant, it allows for direct comparison between two different gases at different temperatures, as seen in the given exercise, to understand how their rms speeds relate to one another.

Absolute Temperature

Absolute temperature is a temperature scale that starts at zero point, known as absolute zero, the theoretical lowest temperature possible at which all classical motion of particles ceases. Measured in Kelvins (K), absolute temperature reflects the thermal energy in a system and is directly proportional to the average kinetic energy of particles.

One of its applications is evident in the formula for rms speed, which shows that as absolute temperature increases, the rms speed of gas molecules also increases. This is because particles possess greater thermal energy at higher temperatures, igniting more vigorous movements. In the context of the textbook exercise, observing the temperatures of \(\mathrm{H}_2\) at 50 K and \(\mathrm{O}_2\) at 800 K is fundamental to computing their rms speed and understanding their thermal behaviors.