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?
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.
- Q: Can you reduce the rate at which this blurring happens?
- A: Yes you can
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.
The expansion of a gas into the vacuum is not a diffusive process. Depending on the initial conditions (density and density gradient) the expansion is either described by the Euler/Navier-Stokes equation of fluid dynamics, or the Boltzmann equation of kinetic theory.
For the parameters that you mention the gas is sufficiently dense that most of it is in the regime of short mean free paths, and the expansion is described by the Euler equation (or the Navier-Stokes equation, if you would like to keep dissipative corrections). The Euler description breaks down near the dilute edge (and this is difficult to treat correctly), but this is not where most of the particles are.
If the initial density has a sharp edge, then the expansion takes the form of shock wave (in 1d, this is the Riemann problem). This is discussed in standard text books on fluid dynamics. If the initial density is smooth (say a Gaussian), then the expansion is approximately described by self-similar scaling solutions of the Euler equation.
Order of magnitude estimates can be obtained just from energy conservation. The initial energy is $E=\frac{3}{2}NT_0$ (for an ideal gas). The expansion cools the gas and converts this into kinetic energy,
$$
E=\int d^3x \frac{1}{2}\rho u^2.
$$
This means that after some initial acceleration phase the gas moves with mean velocity $u_{rms}=\sqrt{3T_0/m}$. This is not surprising, it says that the gas moves (approximately) with the speed of sound in the initial cloud. However, it is clearly different from a diffusive front, which slows down with time (the mean position goes as $\langle x^2\rangle^{1/2}\sim \sqrt{t}$).
Best Answer
This is covered in the standard convection-diffusion type of equation:
$$ \frac{\partial C}{\partial t} + \vec{u} \cdot \nabla C= D \nabla^2 C$$
Where $C$ is the smoke concentration, and $D$ is the diffusion coefficient of smoke.
While the air may be stagnant initially, it will come to move due to buoyancy effects, thus giving some non-zero air velocity $\vec{u}$, enhancing the spreading of the smoke. Smoke from the cigarette has a higher temperature than the surrounding, giving it lower density, which makes it rise. As it rises, it cools down, which also decrease the net force on the smoke parcel. At the same time hotter smoke from below hits the smoke that is more stagnant. This is typically what you see happening when the smoke comes directly from the cigarette.
The easiest way to let the smoke go away is opening a window. This will create draft, i. e. a high air velocity across the room, convecting all the smoke with it. This is why you might want to open a window when your house smells bad.