Two vessels are labeled A and B. Vessel A contains \(\mathrm{NH}_{3}\) gas at \(70^{\circ} \mathrm{C},\) and vessel \(\mathrm{B}\) contains \(\mathrm{Ne}\) gas at the same temperature. If the average kinetic energy of \(\mathrm{NH}_{3}\) is \(7.1 \times 10^{-21} \mathrm{~J} / \mathrm{molecule},\) calculate the rootmean- square speed of Ne atoms.

When discussing the microscopic behavior of gases, a fundamental concept is the average kinetic energy of the gas molecules. It's intrinsically linked to the temperature of the gas and is quantitatively defined by the equation for kinetic energy, which in the context of gases is usually written as \( \frac {3}{2} k_{B} T \), where \( k_{B} \) is the Boltzmann constant and \( T \) is the temperature in Kelvin.

This equation shows that each degree increase in temperature leads to a proportional increase in the energy of gas molecules, explaining why hotter gases have faster-moving molecules. In the context of the provided exercise, by equating the average kinetic energies of two different gases at the same temperature, we can delve into the root-mean-square (rms) speed. The average kinetic energy of gas molecules can also be expressed as \( \frac{1}{2}m v_{rms}^2 \) for a single molecule, where \( m \) is the mass of the molecule, and \( v_{rms} \) is the root-mean-square speed of the gas molecules.

The Kelvin temperature scale is a cornerstone in the fields of physics and chemistry; it's an absolute temperature scale that sets zero at absolute zero, the point where molecular motion stops. On the Kelvin scale, 273.15 K is the equivalent to 0°C, the freezing point of water.

To work with any gas law or thermodynamic equation, temperatures must be converted to Kelvin because these equations rely on proportional changes in temperature, which are only consistent on an absolute scale like Kelvin. When you solve problems involving gases, and kinetic energy, like in the exercise, it's imperative to convert Celsius temperatures to Kelvin by using the formula \( T(K) = T(°C) + 273.15 \). This conversion ensures that the calculations accommodate the proper scale for thermal energy in the system.

Molar mass is another essential concept often coming up in chemistry and physics, particularly when dealing with gases. It represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol). For individual molecules, the molar mass will be a tiny value when expressed in kilograms per molecule, which is derived by dividing the mass in grams per mole by Avogadro's number \( (6.022 \times 10^{23} \text{ molecules/mol}) \).

In practice, molar mass allows us to switch between the mass of a collection of molecules and the number of molecules present. As seen in the exercise solution, when determining the rms speed of a gas molecule, knowing the molar mass is crucial. To perform a correct calculation, you use the molar masses of each gas (\( \text{NH}_3 \) and Ne in our case) to obtain the mass of a single molecule by adjusting for Avogadro's number, allowing you to equate kinetic energy or rms speeds across different substances.