# I do not understand this " rate of diffusion of gases is inversely proportional to the square root of their molar masses provided that pressure and temperature remain constant. ?

##### 1 Answer
Aug 24, 2017

DEPENDENCE OF RATE OF EFFUSION ON TEMPERATURE

Consider the root-mean-square speed of a gas following Boltzmann statistics (we could have chosen the most probable speed or average speed; it doesn't matter).

${v}_{R M S} = \sqrt{\frac{3 R T}{M}}$

where $R$ and $T$ are known from the ideal gas law, and $M$ is the molar mass in $\text{kg/mol}$.

Two gases with speeds $v$ have effusion rates $z \propto v$, so that:

${z}_{A} / {z}_{B} = {v}_{A} / {v}_{B}$,

Thus, two gases at the same temperature will have effusion rates given by:

$\textcolor{b l u e}{{z}_{A} / {z}_{B}} = \frac{\sqrt{\frac{3 R T}{M} _ A}}{\sqrt{\frac{3 R T}{M} _ B}}$

$= \textcolor{b l u e}{\sqrt{{M}_{B} / {M}_{A}}}$

Or, $z \propto \frac{1}{\sqrt{M}}$.

And this is known as Graham's Law of effusion. This holds true as long as the gases are held under the same pressure and temperature.

At constant temperature and pressure, the rates of effusion themselves also remain the same.

The temperature dependence is clear from the derivation. However, the pressure dependence is implicit at the moment. Let's see how this comes in.

DEPENDENCE OF RATE OF EFFUSION ON PRESSURE

One way to look at this is to use the ideal gas law $P V = n R T$ to get $R T = \frac{P V}{n}$, so that...

$\underline{{v}_{R M S} = \sqrt{\frac{3 P V}{n M}}}$

So, in this manner, gas speed (and therefore effusion rate) is a function of pressure, and pressure therefore matters.

Alternatively, consider that

$P = \frac{F}{A}$

where $F$ is the force the gas imparts onto a surface and $A$ is the surface area that it collides with.

The effusion can be modeled by a "collision" with a pinhole:

And one would have used the Maxwell-Boltzmann speed distribution to set up the following integral:

$P = {\int}_{0}^{2 \pi} {\int}_{0}^{\pi} {\int}_{0}^{\infty} 2 m v \cos \theta \rho {\left(\frac{m}{2 \pi {k}_{B} T}\right)}^{3 / 2} {v}^{3} {e}^{- m {v}^{2} / 2 {k}_{B} T} \cos \theta \sin \theta \mathrm{dv} d \theta d \phi$

$= \left[. . .\right]$

$= \frac{4}{3} \pi m \rho {\left(\frac{m}{2 \pi {k}_{B} T}\right)}^{3 / 2} {\overbrace{{\int}_{0}^{\infty} {v}^{4} {e}^{- m {v}^{2} / 2 {k}_{B} T} \mathrm{dv}}}^{\frac{3 {\pi}^{1 / 2}}{8} {\left(\frac{2 {k}_{B} T}{m}\right)}^{5 / 2}}$,

where $\rho = \frac{N}{V}$ is the number density, i.e. the number of particles per unit volume. Note that $N {k}_{B} = n R$, where ${k}_{B}$ is the Boltzmann constant.

With some algebraic simplification, the result is:

$P = \frac{N}{V} {k}_{B} T = \underline{\frac{n}{V} R T}$

which is what we expect, as we just got $P V = n R T$, the ideal gas law.

Therefore, if the pressure changes for fixed mols $n$ of gas at constant temperature $T$, the gas volume would change. That would consequently change the surface area $A$ that it "collides" with, which would therefore alter its effusion rate.

As a result, the rate of effusion also depends on whether the pressure was constant. Therefore, at constant pressure AND temperature, it is ONLY dependent on gas speed, and consequently, molar mass.