Can a small amount of smoke be dense enough to stay in the air keeping its shape for a minute or so?
Or does it always dissipate quickly?
If not smoke, can anything else stay in the air for a minute while keeping its shape?
[Physics] Can smoke stay still in the air
airdiffusion
Related Solutions
You are right, these masks are almost useless as a protection against urban aerosols. With swine flu, there was a lot of discussion (example) that even the best masks cannot catch virus particles which are only 100 nm in size. The usual surgical masks are even less effective - they will hardly block anything smaller that 1 micron.
Now, urban aerosols have several size modes: most numerous are just 10-50 nm in size, although most of the mass will be in large 1-10 micron particles (this is the soot/dust that you can see).
The largest particles are blocked by the mask - but they are also filtered by your nose. The smallest particles - below 200 nm - that are considered much more dangerous because they can reach your lungs and even enter the bloodstream. The mask will also not help against nitrogen dioxide - the major component of urban pollution.
Punchline: The lift is directly proportional to the ambient pressure, and inversely proportional to the ambient (Kelvin) temperature.
Here's the derivation with some other related observations:
Let's say that the balloon has volume $V$. The gas inside has some temperature $T_\mathrm{in}$ and pressure $P_\mathrm{in}$. The gas outside has some temperature $T_\mathrm{out}$ and pressure $P_\mathrm{out}$. The gas on the inside of the balloon satisfies the ideal gas law: \begin{align} P_\mathrm{in}V = N_\mathrm{in}k T_\mathrm{in} \end{align} Where $N_\mathrm{in}$ is the number of gas molecules, and $k$ is Boltzmann's constant. If we let $\mu$ denote the mass of each gas molecule, then $\mu N/V = \rho_\mathrm{in}$ is the density of the gas inside the balloon, so we can rewrite the ideal gas law inside as \begin{align} P_\mathrm{in} = \frac{\rho_\mathrm{in}}{\mu}kT_\mathrm{in} \end{align} We can therefore write a similar expression for the gas outside the balloon: \begin{align} P_\mathrm{out} = \frac{\rho_\mathrm{out}}{\mu}kT_\mathrm{out} \end{align} If the gas pressure inside and the gas pressure outside are unequal, gas will flow through the opening in the balloon, and the pressures will equalize, so the inside and outside pressures are taken to be equal $P_\mathrm{in} = P_\mathrm{out}$. If we use this fact with the inside and outside ideal gas law statements, we obtain \begin{align} \rho_\mathrm{in}T_\mathrm{in} = \rho_\mathrm{out} T_\mathrm{out} \end{align} This show that if the temperature on the inside is greater then the temperature on the outside, then the density on the inside will be lower. This means that the balloon will float: here's why.
Archimedes' principle says that the bouyant force on an object in a fluid equals the weight of fluid displaced by the object. In this case, the balloon will displace a mass of fluid given by $\rho_\mathrm{out} V$ where $\rho_\mathrm{out}$ is the density of the ambient air. It follows that the buoyant force generated by the balloon is \begin{align} F_\mathrm{buoyant} = \rho_\mathrm{out} Vg \end{align} On the other hand, the weight of the air in the balloon is \begin{align} W_\mathrm{in} = \rho_\mathrm{in} Vg \end{align} So if the density inside is less than the density outside, then the buoyant force will be great than the weight, and the balloon will float. Note that we are assuming here that the balloon's material is light enough that its weight can be neglected.
What about if you want to support something else of mass $M$ with the balloon (like a camera)? Well in this case we want the buoyant force to counteract both the weight of the air, and the weight of the object. So we get the condition \begin{align} \rho_\mathrm{out} Vg \ge \rho_\mathrm{in} V g + Mg = (\rho_\mathrm{in} V + M)g \end{align} we can rewrite this as \begin{align} \rho_\mathrm{out} - \rho_\mathrm{in} \ge \frac{M}{V}. \end{align} Using the ideal gas law for the air inside and outside, we can write this as \begin{align} \frac{1}{T_\mathrm{out}} - \frac{1}{T_\mathrm{in}} \ge \frac{kM}{\mu P_\mathrm{out}V} \end{align} which gives \begin{align} \boxed{T_\mathrm{in} \ge \left(\frac{1}{T_\mathrm{out}}-\frac{kM}{\mu P_\mathrm{out}V}\right)^{-1}} \end{align} This allows you to compute the minimum temperature you need to keep the inside of the balloon at so that you can support the mass $M$ given the volume $V$ of the balloon, the molecular mass $\mu$ of the ambient air (which you can look up), and the outside temperature and pressure. You can also look up the value of Boltzmann's constant.
As for the differences in percent lift values you get when you change the ambient temperature and pressure, take the expression for the buoyant force written above, and note that using the ideal gas law for the outside air gives \begin{align} F_\mathrm{buoyant} = \frac{\mu Vg}{k}\frac{P_\mathrm{out}}{T_\mathrm{out}} \end{align} In other words, the buoyant force is directly proportional to the ambient pressure, and inversely proportional to the ambient temperature. Note that I have written the ideal gas law here in the Kelvin temperature scale. So if the pressure of the ambient air increases by a factor of two, so will the lift, while if the Kelvin temperature increases by a factor of two, then the lift will be halved.
Best Answer
I read this and think to myself "optimization problem". Firstly, you should know the following, which is the law of diffusion:
$$\frac{\partial \phi}{\partial t} = D\,\frac{\partial^2 \phi}{\partial x^2}$$
For clarity, $\phi$ is a function that represents the distribution of the concentration of the gas. It is a scalar function of 3 variables, which is to say that I could write it as $\phi(\vec{r})$, where $\vec{r}$ represents typical $x,y,z$ coordinates. If have an image in your mind of a cloud that is the shape of a snowman, that can be represented by that function, so can smoke rings or whatever you desire.
The clarity of the shape degrades over time, exactly per the above diffusion equation. Picture blurring an image in Photoshop. That is very similar to the process that happens.
The rate at which $\phi$ (your snowman) degrades in sharpness comes from the magnitude of $|d\phi/dt|$. This magnitude is proportional to the diffusion coefficient $D$ as well as that other derivative with respect to $dx^2$, but that term is representative of the sharpness itself, so we don't want to reduce that, we would rather reduce $D$. In order to reduce $D$, we need to first talk about mean free path and velocity of the gas molecules. I'll use this source and refer to the mean free path $\lambda$ (units of length) and average speed $\bar{c}$. In general D is proportional to those two.
$$D \propto \lambda \bar{c}$$
For a gas cloud the parameter $\lambda$ has mostly to do with the density of the gas, as well as some other things. Again, we would like to minimize $D$, but $\lambda$ might not have much design freedom. On the other hand, $\bar{c}$ could have great design freedom. This is also dependent on the temperature of the gas, but more specifically, the temperature is a measure of the kinetic energy of the molecules. I'll say fairly generally:
$$\frac{1}{2} m \bar{c}^2 = \frac{3}{2} k T$$
Never mind very much what $k$ is (it's just a physical constant), what matters is that this equation has temperature $T$ and $m$. I'm taking your question to be most likely concerned with normal air. That means that it is unlikely that we would have $T$ as a design variable. However, since you are not specifying the gas you are working with, it's possible we could choose that, and the selection of that gas determines $m$ which is a factor in determining $\bar{c}$ which is a factor in determining $D$, which determines the persistence of your cloud image.
Bottom line: Heavier gases will diffuse more slowly, meaning the image will persist longer.
An example of a high molecular weight gas is common refrigerant gases, like R-134a. If you released that into the air it will diffuse rather slowly compared to other examples. NOTE: don't do this, it would be dangerous and probably illegal.