Animated Solution for Chemistry - States of Matter: If the distribution of molecular speeds of a gas is as per the figure shown below, then the ratio of the most probable, the average and the roots mean square speeds, respectively, is

Welcome to a fascinating twist on a classic kinetic theory problem! This question is a brilliant example of why blindly memorizing formulas can lead you straight into a trap. Let's break down the physics and the statistics behind this distribution curve.

When we study the Kinetic Theory of Gases, we spend a lot of time analyzing the Maxwell-Boltzmann distribution. That standard curve is famously right-skewed (it has a long tail extending to the right). Because of that skewness, the three characteristic speeds follow a strict order: the most probable speed ($v_{mp}$) is the lowest, followed by the average speed ($v_{avg}$), and the root mean square speed ($v_{rms}$) is the highest.

The standard ratio for a Maxwell-Boltzmann distribution is:

If you just glanced at the options, you might be tempted to immediately select Option (C). But wait! The question explicitly provides a graph, and we must respect the data given to us.

Look closely at the figure provided in the problem. Unlike the Maxwell-Boltzmann curve, this graph is perfectly symmetrical, resembling a classic bell curve or normal distribution.

In statistics, the symmetry of a distribution dictates the relationship between its central tendencies:

1. Mode ($v_{mp}$): This is the peak of the curve, representing the speed possessed by the maximum fraction of molecules. 2. Mean ($v_{avg}$): This is the center of mass of the distribution.

For any perfectly symmetrical distribution, the peak is exactly in the middle, which means the Mode and the Mean coincide perfectly. Therefore, for this specific gas:

This immediately tells us that the ratio of the first two speeds must be $1:1$.

Now, what about the root mean square speed ($v_{rms}$)? Does it also equal the average speed?

To answer this, we look at the mathematical definition of variance (${\sigma }^2$) for any set of varying data:

Since $⟨v^2⟩$ is simply $(v_{rms}{)}^2$ and $⟨v⟩$ is $v_{avg}$, we can rewrite this as:

Because the molecules in the gas do not all move at the exact same speed (the graph has a visible spread or width), the variance must be strictly positive (${\sigma }^2>0$). Therefore:

This is a universal mathematical truth: for any distribution with a spread, the RMS value is always greater than the simple average.

Bringing it all together, we have established the following relationship for the given symmetrical distribution:

Now, let's evaluate the given options: - (A) $1:1:1$ implies $v_{avg}=v_{rms}$, which is impossible since there is a spread in speeds. - (B) $1:1:1.224$ implies $v_{mp}=v_{avg}<v_{rms}$. This perfectly matches our derived condition! - (C) $1:1.128:1.224$ implies $v_{mp}<v_{avg}<v_{rms}$, which is only true for a right-skewed distribution. - (D) $1:1.128:1$ implies $v_{rms}<v_{avg}$, which violates the mathematical property of variance.

Thus, the correct ratio is $1:1:1.224$, making Option (B) the correct answer. Always trust the graph over memorized formulas!