Consider the diffusion of Bromine Gas within a tube filled with air.

After a prolonged period of time, is the colour composition uniform?

From my understanding, Bromine is a denser gas. Hence, on average, the Bromine gas is more concentrated at the bottom of the flask. There is always diffusion, even against gravity, so I think the colour would be relatively uniform. However, the bottom of the tube would still be slightly darker. ( May not be noticeable ).

I've read that the colour composition is entirely uniform. I can see where this comes from. I suspect we consider the particles to be too small to have any gravitational influence. However, if we were to have a longer tube would there be a slightly darker portion at the bottom of the tube? Do we consider the colour composition to be exactly uniform because the tube is too small to have any noticeable seperation of colour?

This problem has puzzled me a bit. Is my intuition right? That is, there is a slight seperation of colour, even though it may not be noticeable?


2 Answers 2


In a tall tube set upright, the density will change with height provided that the gas is not stirred. i.e. no convection occurs, so with that condition your intuition is correct. As the density (or pressure) changes so does the optical density ( colour). The relative density changes according to the Boltzmann distribution as

$$\displaystyle \frac{\rho}{\rho_0}=\exp(-gMh/(RT)) $$

where $\rho_o$ is the density at $ h=0$, where $h$ is the height in metres where you want to calculate the density, $g$, acceleration due to gravity(m/s$^2$), $M$, molecular weight (kg/mol), $R$, the gas constant (J/mol/K), and $T$ the temperature in Kelvin.


The equation above can be derived in an intuitive way from the ideal gas law and the fact that each molecule experiences a force under the effect of gravity as well as thermal motion. Using the ideal gas law is not a restriction as we can always choose a low concentration so that it accurately describes the gas/ vapour behaviour.

In the cylinder of gas of base area $A$ select at layer of thickness $\Delta Z$ and so a volume $V=A\Delta z$. The number of molecules $n$ in this volume is found using the ideal gas law ($pV=nk_BT$) and is

$$ \displaystyle n=\frac{p\,A}{k_BT}\Delta z$$

where $k_B$ is the Boltzmann constant. The force on a molecule is $f=mg$ so that there is a pressure change across layer $\Delta z$ of $\Delta p=p_{(z+\Delta z)}-p_z$. Equating forces (pressure is force/area) then $A\Delta p = -n\,mg$ (negative as the force acts downwards) then rearranging gives

$$\displaystyle \Delta p= -\frac{n\,mg}{A}=-\frac{p\,mg}{k_BT}\Delta z$$

and next make the $\Delta$ very small and in the limit into differentials so that

$$\displaystyle dp=-p\frac{mg}{k_BT}dz$$

which integrates to the equation above where molar units are used.

  • $\begingroup$ in theory this is true. But this is misleading. What is the density difference in, say, a 1m tall tube that fits in a typical lab? While, were the tube 1km tall, the difference might be detectable, is it detectable in anything that fits in a typical lab? $\endgroup$
    – matt_black
    Commented Jun 5, 2023 at 9:14
  • $\begingroup$ I agree in principle, the OP can calculate this for themselves what the difference can be. (In the lab its probable that convection would dominate even if utterly tiny differences in optical density could be measured. ) In the distant past I have done demonstrations using small air-agitated polystyrene spheres that can demonstrate the effect of gravity on gas density. $\endgroup$
    – porphyrin
    Commented Jun 5, 2023 at 10:31
  • $\begingroup$ You'd need a good glassworker to actually make it, but you could envisage a metre-long vacuum flask with the gas mix inside, to massively reduce thermal inputs that would drive convection. Silvered apart from a series of small windows through which to observe (e.g. shine a very weak laser that will be absorbed by the bromine) and measure the absorption as a function of height. Heater coils round the inner tube would provide even heating, and they could be switched off once it was all gaseous - or really on it having a non-zero vapour pressure $\endgroup$
    – Chris H
    Commented Jun 5, 2023 at 15:33

I will humbly disagree with the solution of my fellow trying to justify the opposite answer.

[OP] After a prolonged period of time, is the colour composition uniform?

Under the sole condition of diffusion, a solution made up of several substances can never achieve a uniform composition even at $t \rightarrow \infty$. Moreover, gravity does not influence the composition profile, it only ensures that the fluid does not "fall" by specifying an adequate pressure at every point.

We will consider for simplicity the diffusion evolution along the $z$-axis, so we can imagine that the $x$ and $y$ directions are of infinite lenght than that of the $z$ axis. In this scenario, the velocity field can only have the $z$ direction, and since air doesn't flow in the $x$ or $y$ direction, is only a function of $z$, i.e. $\mathbf{v}(z) = v_z(z)\mathbf{\hat{z}}$. The image is the following:

enter image description here

1. Mass Balance The continuity equation is \begin{align} \frac{\partial \rho}{\partial t} + \mathbf{\nabla} \cdot \mathbf{v} = 0 \tag{1} \end{align} but a prolonged period of time traduces into equating the time derivative to zero, that is, we want the solutions for $t \rightarrow \infty $ or the steady-state solutions. Eq. (1) says \begin{align} \mathbf{\nabla} \cdot \mathbf{v} = 0 \rightarrow \boxed{\frac{\mathrm{d}v_z}{\mathrm{d}z} = 0} \tag{2} \end{align} Thus, the velocity field at $t \rightarrow \infty $ is just a constant for every point. Here we need to make a decision:

  • If we consider $v_z \neq 0$, then we cannot stop at $z = L$, because the air will continuously go up. This can be circumvented by considering an ideal solution, of semi-infinite diffusion, where $L \rightarrow \infty$. This is an interesting solution often studied in transport phenomena but not our case.
  • If we want the tube of a finite size then $v_z = 0$, which is the condition that we will take.

2. Momentum Balance The equation of motion in the case for a newtonian fluid is \begin{align} \rho \left(\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \mathbf{\nabla}\mathbf{v}\right) &= -\mathbf{\nabla}P + \mu \nabla^2\mathbf{v} + \rho \mathbf{g} \tag{3} \\ \end{align} where $\mu$ is the viscosity of the air+bromine solution. In a prolonged period of time, employing Eq. (2), and remembering that the gravity points down in our coordinate system \begin{align} \rho(v_z \mathbf{\hat{z}}) \cdot \left(\underbrace{\frac{\mathrm{d}v_z}{\mathrm{d}z}}_{=0}\mathbf{\hat{z}}\right) &= -\frac{\mathrm{d}P}{\mathrm{d}z}\mathbf{\hat{z}} + \mu \underbrace{\frac{\mathrm{d}^2 v_z}{\mathrm{d}z^2}}_{=0}\mathbf{\hat{z}} + \rho (-g\mathbf{\hat{z}}) \\ 0 &= -\left(\frac{\mathrm{d}P}{\mathrm{d}z} + \rho g\right)\mathbf{\hat{z}} \rightarrow \boxed{\frac{\mathrm{d}P}{\mathrm{d}z} = -\rho g} \tag{4} \\ \end{align} Eq. (4) establishes the pressure profile in order to balance the weight of the fluid, that can be solved once stating a boundary condition. This is typically the pressure at a boundary, for example, if open to the atmosphere that would be $P(z=L) = P_\text{atm}$. This profile is a linear function and has its maximum at the base of the tube.

3. Mass Balance for Bromine The continuity equation for $\ce{Br2}$ under the absence of sources or sinks (like chemical reactions), and considering Fick's law to relate the flux and the concentration, is

$$ \frac{\partial C_\ce{Br2}}{\partial t} + \mathbf{v} \cdot \mathbf{\nabla}C_\ce{Br2} = D_\ce{Br2,air} \mathbf{\nabla}^2 C_\ce{Br2} \tag{5} $$

where $D_\ce{Br2,air}$ is the diffusion coefficient of $\ce{Br2}$ in air. Again, for steady state conditions Eq. (5) yields

\begin{align} (v_z \mathbf{\hat{z}}) \cdot \left(\frac{\mathrm{d}C_\ce{Br2}}{\mathrm{d}z}\mathbf{\hat{z}}\right) &= D_\ce{Br2,air} \frac{\mathrm{d}^2 C_\ce{Br2}}{\mathrm{d}z^2} \\ v_z \frac{\mathrm{d}C_\ce{Br2}}{\mathrm{d}z} &= D_\ce{Br2,air} \frac{\mathrm{d}^2 C_\ce{Br2}}{\mathrm{d}z^2} \rightarrow \boxed{\frac{\mathrm{d}^2 C_\ce{Br2}}{\mathrm{d}z^2} - \frac{v_z}{D_\ce{Br2,air}} \frac{\mathrm{d}C_\ce{Br2}}{\mathrm{d}z} = 0} \tag{6} \end{align} Eq. (6) establishes the concentration of bromine, for $t \rightarrow \infty$, which is independent of gravity, once two boundary conditions are stated. For our case of $v_z = 0$, Eq. (6) states a linear profile, with a maximum at the base of the tube, and a minimum at the top of the tube. Therefore we have proven that the concentration of $\ce{Br2}$ is not uniform in the tube.

4. Conclusion The lower the diffusion coefficient, the more difference of concentration there will be between the bottom and the top. This is the reason why we mix coffee with a spoon every morning, why we use a magnetic stirrer for titration, for the variables of interest to be independent of position. From a mathematical point of view, those actions are done to avoid this type of behaviour, where we have an undesired dependency of position in the concentration of a substance.

If $\ce{Br2}$ diffuses very easily in air, then we may consider that $C_\ce{Br2} \approx \text{constant}$ in the tube, because we may not be able to notice. However, in this scenario that we are studying, and strictly prohibiting to shake the tube, a difference of concentration and hence of colour is inevitable.

  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Chemistry Meta, or in Chemistry Chat. Comments continuing discussion may be removed. $\endgroup$
    – andselisk
    Commented Jun 7, 2023 at 17:37

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