Molecular Speeds: Mean, RMS, and Most Probable

Hello! Welcome back to our study of the Kinetic Theory of Gases.

In our last lesson, we built a microscopic model of an ideal gas and used it to derive a powerful expression for pressure: $P = \frac{1}{3} nm ⟨ v^{2} ⟩$ . We concluded by noting the similarity between the resulting equation $P V = \frac{2}{3} N ⟨ K . E . ⟩_{a v g}$ and the empirical Ideal Gas Law.

Today, we will connect these two equations to uncover one of the most profound insights of physics: the direct relationship between temperature and molecular motion. This will allow us to define and calculate three distinct, statistically important measures of molecular speed: the root-mean-square speed ( $v_{r m s}$ ), the average speed ( $v_{a v g}$ ), and the most probable speed ( $v_{m p}$ ). Mastering these concepts is crucial for solving a wide range of problems in thermodynamics, especially at the JEE Advanced level.

1. Temperature, Kinetic Energy, and RMS Speed

Let's begin by equating the two expressions for the product $P V$ :

From the Kinetic Theory (last lesson):
$P V = \frac{1}{3} N m ⟨ v^{2} ⟩$
where $N$ is the total number of molecules, $m$ is the mass of one molecule, and $⟨ v^{2} ⟩$ is the mean square speed.
From the Ideal Gas Law (empirical):
$P V = N k_{B} T$
where $k_{B}$ is the Boltzmann constant ( $1.38 \times 1 0^{- 23}$ J/K) and $T$ is the absolute temperature in Kelvin.

By setting these two equations equal to each other, we bridge the microscopic and macroscopic worlds:
$\frac{1}{3} N m ⟨ v^{2} ⟩ = N k_{B} T$

Canceling $N$ from both sides and rearranging, we get:
$\frac{1}{3} m ⟨ v^{2} ⟩ = k_{B} T$
$\frac{2}{3} (\frac{1}{2} m ⟨ v^{2} ⟩) = k_{B} T$

This leads to a cornerstone result:
$⟨ K . E . ⟩_{t r an s} = \frac{1}{2} m ⟨ v^{2} ⟩ = \frac{3}{2} k_{B} T$

This equation states that the absolute temperature of an ideal gas is directly proportional to the average translational kinetic energy of its molecules. This is the kinetic interpretation of temperature.

With this relationship, we can now define and calculate the root-mean-square (RMS) speed. As the name suggests, it's the square root of the mean of the squares of the molecular speeds, $v_{r m s} = ⟨ v^{2} ⟩$ .

From our energy-temperature relation, we can solve for $⟨ v^{2} ⟩$ :
$⟨ v^{2} ⟩ = \frac{3 k _{B} T}{m}$
Taking the square root gives us the formula for RMS speed:
$v_{r m s} = \frac{3 k _{B} T}{m}$

For practical problem-solving, it's often more convenient to work with molar mass ( $M$ ) and the universal gas constant ( $R$ ). Recalling that $M = m \cdot N_{A}$ and $R = k_{B} \cdot N_{A}$ (where $N_{A}$ is Avogadro's number), we can substitute $m = M / N_{A}$ and $k_{B} = R / N_{A}$ into the equation:
$v_{r m s} = \frac{3 ( R / N _{A} ) T}{M / N _{A}} = \frac{3 R T}{M}$

This is the most common form you will use.

A crucial note on units: In the formula $v_{r m s} = \frac{3 R T}{M}$ , to get the speed in m/s, you must use SI units consistently:

$R = 8.314$ J/(mol·K)
$T$ in Kelvin (K)
$M$ in kilograms per mole (kg/mol). (e.g., for O₂, $M$ is 0.032 kg/mol, not 32 g/mol).

2. A Spectrum of Speeds: Average and Most Probable

In a gas, molecules don't all travel at the same speed. Due to constant collisions, their speeds vary, following a statistical pattern known as the Maxwell-Boltzmann distribution. This distribution shows that while very slow and very fast molecules are rare, there is a well-defined peak and a spread of speeds.

Maxwell-Boltzmann Distribution of Molecular Speeds — This graph shows the distribution of molecular speeds for Oxygen at 300 K. The x-axis is speed and the y-axis represents the fraction of molecules at that speed. Notice the curve is not symmetric. The peak represents the most probable speed ($v_{mp}$), while the RMS speed ($v_{rms}$) is further to the right.

From this distribution, we define two other important speeds:

Most Probable Speed ( $v_{m p}$ ): The speed possessed by the maximum number of molecules. This corresponds to the peak of the Maxwell-Boltzmann distribution curve.
Average Speed ( $v_{a v g}$ ): The arithmetic mean of the speeds of all molecules.

Deriving the formulas for these speeds requires calculus on the Maxwell-Boltzmann distribution function, which is beyond the scope of this lesson. However, the results are essential.

The following resource provides a concise summary of the formulas for all three speeds, which are fundamental for the JEE.

KINETIC THEORY OF GASES AND THERMODYNAMICS - SelfStudys

This PDF document, designed for JEE preparation, clearly defines the three molecular speeds and presents their final formulas. It also shows the ratio between them, a useful fact to remember.

Please read Section 5, 'TYPES OF SPEED' (pages 4-5 of the PDF). Focus on the definitions and the boxed formulas for v_rms, v_av, and v_mps. Also, note the relationship between them discussed at the bottom.

Let's summarize the key formulas from that reading:

Speed Type	Formula (in terms of T and M)	Formula (in terms of T and m)
Most Probable	$v_{m p} = \frac{2 R T}{M}$	$v_{m p} = \frac{2 k _{B} T}{m}$
Average (Mean)	$v_{a v g} = \frac{8 R T}{π M}$	$v_{a v g} = \frac{8 k _{B} T}{π m}$
Root-Mean-Square	$v_{r m s} = \frac{3 R T}{M}$	$v_{r m s} = \frac{3 k _{B} T}{m}$

A Deeper Look: Deriving the Most Probable Speed

Given your background in mathematics, you might appreciate seeing how $v_{m p}$ is derived. It involves finding the value of $v$ that maximizes the Maxwell-Boltzmann probability distribution function, $f (v)$ , by taking its derivative with respect to $v$ and setting it to zero.

The video below provides a clear, step-by-step derivation using the product rule.

Most Probable Molecular Speed

This short video demonstrates the calculus-based derivation of the most probable speed by differentiating the Maxwell-Boltzmann distribution. This provides a deeper justification for the formula.

Watch from the beginning to 05:33. Follow the logic of finding the peak of a function by setting its derivative to zero. Pay attention to how the product rule is applied to the distribution function to find the expression for the most probable velocity.

3. Comparing the Speeds

By examining the formulas, we can compare the magnitudes of the three speeds:

$v_{m p} = 2 \frac{R T}{M} \approx 1.414 \frac{R T}{M}$
$v_{a v g} = \frac{8}{π} \frac{R T}{M} \approx 1.596 \frac{R T}{M}$
$v_{r m s} = 3 \frac{R T}{M} \approx 1.732 \frac{R T}{M}$

This consistently gives the relationship:
$v_{m p} < v_{a v g} < v_{r m s}$

The RMS speed is the largest because the squaring process gives greater weight to the faster-moving molecules in the long "tail" of the Maxwell-Boltzmann distribution.

The ratio between them is a useful fact for quick comparisons:
$v_{m p} : v_{a v g} : v_{r m s} \approx 1 : 1.128 : 1.224$

Test your understanding!

At what temperature will the root-mean-square speed of oxygen ( $O_{2}$ ) molecules be equal to the root-mean-square speed of hydrogen ( $H_{2}$ ) molecules at 300 K?

(Molar mass of $O_{2} \approx 32$ g/mol; Molar mass of $H_{2} \approx 2$ g/mol)

Show answer

Let's denote the properties of oxygen with subscript $O_{2}$ and hydrogen with $H_{2}$ .
We are given the condition: $(v_{r m s})_{O_{2}} = (v_{r m s})_{H_{2}}$ .

Using the formula $v_{r m s} = \frac{3 R T}{M}$ , we can write:
$\frac{3 R T _{O_{2}}}{M _{O_{2}}} = \frac{3 R T _{H_{2}}}{M _{H_{2}}}$
Squaring both sides and canceling the $3 R$ term gives:
$\frac{T _{O_{2}}}{M _{O_{2}}} = \frac{T _{H_{2}}}{M _{H_{2}}}$
We need to find $T_{O_{2}}$ . Rearranging the formula:
$T_{O_{2}} = T_{H_{2}} (\frac{M _{O_{2}}}{M _{H_{2}}})$
We are given $T_{H_{2}} = 300$ K. We can use the molar masses in g/mol since their ratio will be dimensionless:
$T_{O_{2}} = 300 K \times (\frac{32 g/mol}{2 g/mol}) = 300 K \times 16$
$T_{O_{2}} = 4800 K$
So, oxygen molecules would need to be at a much higher temperature (4800 K) to have the same RMS speed as the much lighter hydrogen molecules at 300 K. This is a classic JEE-style problem.

Conclusion

In this lesson, we have quantified the motion of gas molecules, moving from the abstract concept of average kinetic energy to concrete, calculable speeds.

Key Takeaways:

Temperature is a direct measure of the average translational kinetic energy of gas molecules: $⟨ K . E . ⟩_{t r an s} = \frac{3}{2} k_{B} T$ .
Molecules in a gas have a distribution of speeds, not a single speed.
We use three statistical measures to characterize this motion: most probable speed ( $v_{m p}$ ), average speed ( $v_{a v g}$ ), and root-mean-square speed ( $v_{r m s}$ ).
All three speeds are proportional to $T / M$ , meaning they increase with temperature and decrease with molar mass.
The speeds are always ordered as: $v_{m p} < v_{a v g} < v_{r m s}$ .

Preview of the Next Lesson:
We saw that the average translational kinetic energy is $\frac{3}{2} k_{B} T$ . But what about other forms of energy, like rotation or vibration? In the next lesson, we will explore the Equipartition of Energy Theorem. This powerful principle states that energy is shared equally among all available degrees of freedom. This will allow us to calculate the total internal energy and, consequently, the molar specific heats ( $C_{V}$ and $C_{P}$ ) for monatomic, diatomic, and polyatomic gases.

Recommended resources

1. Temperature, Kinetic Energy, and RMS Speed

2. A Spectrum of Speeds: Average and Most Probable

A Deeper Look: Deriving the Most Probable Speed

3. Comparing the Speeds

Conclusion