How long would it take for a smelly object to evaporate? This question is a follow on from this deleted one: https://physics.stackexchange.com/q/177894/26076 as I was writing what I thought to be a valid physics answer to it. Version 1 of this question contained essentially this one: if a piece of furniture is smelly, does this not also mean it evaporates? The answer of course is that it does "evaporate", so the question then is "how long would we take to see this?"
So I reworded this question to: how does one calculate the mass loss rate through outgassing from a smelly object?
 A: Mammalian sense of smell is in general exquisitely keen: even though we think of ourselves as an animal having a dull smell sense comapared to that of, say, a dog, a pig or a rat, receptors for certain scents are still triggered by molecules counted in the tens. So the outgassing of volatile wood oils from, say, a table, can still be miniscule and well detectable by our sense of smell. 
Let's estimate how quickly a table with an odour can evapourate. We can use considerations like those I discuss at the end of my answer here and Fick's law to estimate the minimum detectable outgassing rate.
Suppose we have a spherical object sitting at room temperature and pressure. We seek a solution to Fick's law (a wholly mathematical analogous description exists for the diffusion of heat in a conductive medium):
$$\frac{\partial}{\partial\,t}\phi = D\,\nabla^2\,\phi$$
where $D$, the diffusivity, is derived from considerations of mean free path in gasses and is of the order of $10^{-9}{\rm m^2\,s^{-1}}$ for normal sized molecules in air at room temperature and pressure when the concentration $\phi$ is expressed in moles per cubic meters. 
So we seek a spherically symmetric solution and calculate the concentration profile for a given outgassing rate. At steady state, $\frac{\partial}{\partial\,t}=0$ and we have Laplace's equation. The unique spherically symmetric solution is $\phi\propto \frac{1}{r}$. The outgassing rate is given by Fick's first law $\vec{J}=-D\,\nabla\phi$ in moles per second per square meter. With:
$$\phi=\phi_R\,\frac{R}{r}$$
with $R$ the radius of the object and $\phi_R$ the concentration at its surface, we get $\vec{J}=-D\,\mathrm{d}_r\,\phi\,\hat{\vec{r}}$:
$$|J|=-D\,\frac{\phi_R}{R^2}\tag{1}$$
This is the equation we seek. Now, suppose we can detect 100 molecules in each five liter breath. So we can detect about $\phi_R = 100 / (0.005 N_A)$ (about 22.5 liters per mole at room temp and pressure). Let $R=1m$, then $|J| \approx 3\times 10^{-29}{\rm mol\,m^{-2}\,s^{-1}}$, or the sphere is losing $4 \pi |J| = 5\times 10^{-28}{\rm mol\,s^{-1}}$. Work this out as a mass per century and you'll find it's about a femptogram per century!
