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The Boltzmann constant, denoted by the symbol \( k \), plays a crucial role in the world of thermodynamics and statistical mechanics. It is named after the physicist Ludwig Boltzmann and is predominantly used to bridge the gap between the macroscopic and microscopic worlds of physical systems.
This constant acts as a conversion factor between the microscopic energy of molecules and macroscopic temperature. In the context of gases, the Boltzmann constant allows us to relate the temperature of a gas to the kinetic energy of its molecules. It's the constant that appears in equations like the root mean square (rms) speed formula, helping us to calculate properties such as particle speed from thermal energy.
Here are some key points to understand about the Boltzmann constant:

• It has a fixed value of \( 1.38 \times 10^{-23} \) J/K (joules per kelvin).
• It is fundamental in the formula for the rms speed: \( v_{rms} = \sqrt{\frac{3kT}{m}} \). Here, \( k \) connects molecular motion to temperature.
• By canceling \( k \) in the rms speed equality equation, we see why it's essential for comparing different gases at various temperatures.

The root mean square speed is directly related to temperature, which means changes in temperature will affect how fast molecules within a gas move. Understanding this relationship helps explain why gases behave differently under varying thermal conditions.
The formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \) reveals how the rms speed (\(v_{rms}\)) increases with the temperature (\(T\)) of the gas. Here, we see:

• As temperature increases, the molecules have more kinetic energy, leading to a higher rms speed.
• A gas at lower temperatures will have molecules moving slower than at higher temperatures.
• This temperature dependency explains phenomena like diffusion rates, where hotter gases diffuse faster than colder ones.

In our exercise's context, to find the temperature at which hydrogen (\(\mathrm{H}_2\)) and oxygen (\(\mathrm{O}_2\)) have equal rms speeds, we exploit this temperature dependence. By equating their rms speeds and solving, we uncover the specific temperature for hydrogen that matches the condition for oxygen.

Molecular mass is a fundamental parameter that heavily influences the behavior of gases. It's the mass of a given molecule and plays a key role in determining physical properties like speed, particularly in the context of the kinetic theory of gases.
In the formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), you can see how molecular mass impacts speed:

• For a fixed temperature, a gas with a smaller molecular mass will have a higher rms speed than one with a larger molecular mass.
• Gases made up of lighter molecules, like \(\mathrm{H}_2\), are quicker to move compared to gases with heavier molecules like \(\mathrm{O}_2\).
• This is because the smaller mass allows the molecules to accelerate more for a given amount of energy.

In our exercise, understanding molecular masses of \(\mathrm{H}_2\) and \(\mathrm{O}_2\) is essential. They dictate how we calculate the temperature needed for matching rms speeds. Specifically, lighter \(\mathrm{H}_2\) requires less thermal energy to achieve a certain speed compared to heavier \(\mathrm{O}_2\), which is why the calculated temperature for \(\mathrm{H}_2\) is significantly lower.