E Rank the following gases \(\begin{array}{llll}\text { NO } & \text { Ar } & \text { N }_{2} & \text { N }_{2} \mathrm{O}_{5}\end{array}\) in order of (a) increasing speed of effusion through a tiny opening. (b) increasing time of effusion.

Understanding Graham's Law of Effusion is pivotal for comprehending how different gases will effuse, or escape, through a small opening. The law states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass. In simpler terms, lighter gases will effuse more rapidly than heavier gases.

When comparing two gases, Graham's Law can be mathematically expressed as:
\[\begin{equation}\frac{Rate of effusion of Gas 1}{Rate of effusion of Gas 2} = \sqrt{\frac{Molar mass of Gas 2}{Molar mass of Gas 1}}\end{equation}\]
This equation illustrates that if you know the molar mass of two gases, you can determine how much faster one gas will effuse compared to the other. Therefore, the lesser the molar mass, the faster the effusion rate, which plays directly into the solution of our exercise. This principle also explains why a balloon filled with helium gas deflates faster than one filled with oxygen; helium has a much lower molar mass compared to oxygen.

Molar Mass is a fundamental concept in chemistry and is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It corresponds to the atomic or molecular mass in grams. Knowing the molar mass of each gas is crucial for applying Graham's Law of Effusion.

To calculate the molar mass of a compound, sum the molar masses of all the atoms present in the molecule. For example, the molar mass of water (H2O) is the sum of the molar masses of two hydrogens and one oxygen.
\[\begin{equation}2 \times H (1.01 g/mol) + O (16.00 g/mol) = 18.02 g/mol\end{equation}\]
In the given exercise, calculating the molar masses of each gas correctly is the first critical step. Without these calculations, we would be unable to apply Graham's Law to predict effusion rate or time.

The concept of Effusion Speed is directly tied to how fast a gas can pass through a tiny opening into a vacuum. It's a measure of how quickly a gas escapes and is influenced by the kinetic energy of the gas particles. Higher temperatures often result in higher effusion speeds because the particles have more energy and move faster. However, the molecular weight of the gas also significantly affects the effusion speed, as outlined by Graham's Law.

Considering our exercise, when we've determined the molar mass of each gas, we use this information to calculate the relative effusion speeds using the inverse relation of molar mass to effusion rate. It’s also important to note that the 'speed' in this context refers to the rate at which gas particles pass through the opening, not their velocity. Because of this inverse relation, we can conclude that the gas with the highest molar mass will effuse the slowest and the one with the lowest molar mass will effuse the fastest.