# Maxwell Distribution of Speed

The Maxwell-Boltzmann distribution describes the distribution of speeds of particles in a gas at a given temperature. Specifically, it gives the probability that a particle in a gas will have a certain speed.

The distribution depends on three variables: the mass of the particles, the temperature of the gas, and the universal gas constant. The distribution is a bell-shaped curve that peaks at a speed called most probable speed. The probability density function of Maxwell-Boltzmann distribution is \begin{align} f(v)=\left(\frac{m}{2\pi kT}\right)^{3/2}\,e^{-\frac{mv^2}{2kT}}, \end{align} where $m$ is the mass of the particles, $k$ is the Boltzmann constant and $T$ is the temperature of the gas.

Let $m$ be mass of a gas molecule. The molar mass ($M$) of a gas is mass of the Avogadro number ($N_A=6.023\times{10}^{23}$) of gas molecules \begin{align} M=mN_A. \end{align}

The Boltzmann's constant is the ratio of gas constant $R$ and Avogadro number i.e., \begin{align} k=\frac{R}{N_A} \end{align}

## Root Mean Square (RMS) Speed

The root mean square (RMS) speed is the square root of the average of the squared speeds of all the particles in a gas. The RMS speed of molecules of a gas kept at absolute temperature $T$ is given by \begin{align} v_{rms}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3RT}{M}}, \end{align} where $m$ is the mass of a molecule and $M$ is the molar mass.

The RMS speed depends on the temperature and the mass of the particles. At higher temperatures, the particles in a gas have higher RMS speeds, and lighter particles have higher RMS speeds than heavier particles at the same temperature. The RMS speed is also related to pressure and kinetic energy.

## Average Speed

The average speed of gas molecules is the arithmetic mean of the speeds of all the particles in the gas. It is given by \begin{align} \bar{v}=\sqrt{\frac{8kT}{\pi m}}=\sqrt{\frac{8RT}{\pi M}} \end{align} The average speed is always less than the rms speed.

## Most Probable Speed

The most probable speed of gas molecules is the speed at which the highest number of gas molecules in a gas sample have that particular speed. In other words, it is the speed at which the Maxwell-Boltzmann speed distribution of a gas has its maximum value. The most probable speed is given by the formula \begin{align} v_p=\sqrt{\frac{2kT}{m}} \end{align}

The most probable speed of gas molecules is lower than the average speed and the root mean square speed of gas molecules. The ratio of the most probable speed to the average speed is given by: \begin{align} \frac{v_p}{\bar{v}} = \sqrt{\frac{2}{\pi}} < 1. \end{align}

## Problems from IIT JEE

Problem (IIT JEE 1982): Two different gases at the same temperature have equal rms velocities. (True/False)

Solution: The rms velocity of a gas, with molecular mass $m$ at temperature $T$, is given by \begin{align} v_\text{rms}=\sqrt{3kT/m}.\nonumber \end{align} Thus, the two gases of different molecular masses will have different $v_\text{rms}$ at the same temperature.

Problem (IIT JEE 1987): The rms speed of oxygen molecules O2 at a certain temperature $T$ (degree absolute) is $v$. If the temperature is doubled and oxygen gas dissociates into atomic oxygen, the rms speed remains unchanged.

Solution: The rms speed of a gas at temperature $T$ is given by \begin{align} v_\text{rms}=\sqrt{\frac{3RT}{M}}, \end{align} where $M$ is the molecular mass of the gas. When temperature is raised to $T^\prime=2T$, oxygen molecule dissociates into two atoms with atomic mass $M^\prime=M/2$. Substitute in above equation to get \begin{align} v_\text{rms}^\prime&=\sqrt{\frac{3RT^\prime}{M^\prime}}\\ &=\sqrt{\frac{3R(2T)}{M/2}}\\ &=2\sqrt{\frac{3RT}{M}}\\ &=2v_\text{rms}.\nonumber \end{align}

Problem (IIT JEE 1998): Let $\bar{v}$, $v_\text{rms}$ and $v_\text{p}$ respectively denote the mean speed, root mean square speed and most probable speed of the molecules in an ideal monatomic gas at absolute temperature $T$. The mass of the molecule is $m$. Then,

1. no molecule can have speed greater than $\sqrt{2}v_\text{rms}$.
2. no molecule can have speed less than $v_\text{p}/\sqrt{2}$.
3. $v_\text{p}<\bar{v} 4. the average kinetic energy of a molecule is$\frac{3}{4}m v_\text{p}^2$. Solution: The molecules in a gas travel with different speeds. The number of molecules having speed between$v$and$v+\mathrm{d}v$varies with$v$as shown in the figure. This variation is given by the Maxwell's speed distribution law. This law gives expressions for the most probable speed ($v_\text{p}$), average speed$\bar{v}$, and rms speed$v_\text{rms}as \begin{align} v_\text{p}&=\sqrt{{2kT}/{m}},\nonumber\\ \bar{v}&=\sqrt{{8kT}/{(\pi m)}},\nonumber\\ v_\text{rms}&=\sqrt{{3kT}/{m}}.\nonumber \end{align} Note that the speed distribution is asymmetric aboutv_p$. It expands more towards the right of$v_p$, which makes$\bar{v} > v_p\$ (see figure). The average kinetic energy of monatomic gas molecule is given by \begin{align} K=\frac{3}{2}kT=\tfrac{3}{4}mv_\text{p}^2.\nonumber \end{align}

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