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!
