Speed of sound in gaseous medium Why does the speed of sound decrease as the density of the medium increases? I know why this happens mathematically, but I want to know what happens at the molecular level that results in higher speed in solid. Could somebody explain that phenomenon, taking into account density and elasticity?
 A: There is both a rigorous and intuitive explanation for the dependence of the speed of propagation of mechanical waves and the density of the medium.
Intuitively: You know that sound waves are longitudinal waves, resulting from a collective oscillation of the medium's molecules. When sound's propagating, it compresses and re-expands the air molecules, as one group of molecules is compressed, the previous group re-expands itself(after having passed on its excitation energy), so you can imagine that if you have very dense air already, you very much have exhausted the possibility of air molecules to compress-re-expand themselves. This in turn reduces the speed of propagation.
More rigorously: Module of compressibility of a material is defined as $$K = -\frac{\Delta p}{\Delta V/V} $$ interpreted as constraint of compression over deformation imposed by compression, and the inverse of $K$ is called the coefficient of compressibility $\kappa$.
Now when calculating the speed of propagation of a mechanical wave in a medium, two very important quantities have to be computed, yielding the speed of propagation: 

First: Deformation of the medium. For a longitudinal wave like sound, during each passage of the wave, each collective piece of the gas is deformed and displaces along its axis (without its cross section being modified to simplify), so: $$\frac{\Delta V}{V} = \frac{\Delta l}{l} $$
Remembering the coefficient of compressibility $\kappa$ : 
$$\frac{\Delta V}{V} = -\kappa p $$
In our case (considered here) $$\frac{\Delta l}{l}=-\kappa p$$
Second: Acceleration. Each element of volume $dV$ of the gas is accelerated during the passage of the wave, so for any case, you always set out the fundamental principle of dynamics (considering the longitudinal deformation of a portion $dx$ of the gas to $dx+d\epsilon$): 
$$\frac{\partial F}{\partial x}dx=dm \frac{\partial^2\epsilon}{\partial t^2}$$
From here it is clear that $dm$ is replaced with $dV \rho$, using $dV$ expression in first step of deformation, you reach the equation of propagation of waves:
$$\frac{\partial^2 \epsilon}{\partial x^2}=\kappa \rho \frac{\partial^2 \epsilon}{\partial t^2} $$ 
From which we know the coefficient of the right hand side is $\frac{1}{c²}$, with c the speed of propagation. To conclude, you should always find out how a change of density, will modify the deformation of the medium caused by the passage of the wave. 
$$c_{gas}=\sqrt{\frac{1}{\kappa \rho}}=\sqrt{\frac{\gamma nRT}{\rho V}} $$
Where $\gamma$ is a constant depending on the nature of the gas, and $n$ number of moles in the gas, the second expression is obtained using the law of ideal gases: 
$$PV = nRT$$
and $1/\kappa$ replaced by $\gamma p$
A: When two gas molecules collide their momentum will have to be conserved. Now when one molecule, with a velocity in the general direction of the propagation direction of the sound, collides head on with another molecule with the same mass, then the velocity of the other molecule will be the same as the velocity of the initial molecule travelling in the direction of the sound wave. However most collision will not be head on, but under some angle. In that case both molecules will have a velocity component in the direction of the propagation direction of the sound, but due to conservation of momentum their velocity will have to be lower then the single initial molecule.
Like Ignacio Vergara Kausel mentioned in his answer when the density increases the mean free path of a molecule decreases. And shorter mean free path means more collisions and therefore a higher frequency that molecules collide under an angle, lowering the mean velocity and thus the speed of sound.
