# In a column of fluid does density vary?

In a column of fluid does density vary?

• Hi Karan, and welcome to Physics Stack Exchange! We prefer that you only post one question per post. Since Kyle pointed out in an earlier comment that your first question had already been answered here, I removed it and the last question. Feel free to repost question #3 separately (just make sure you check that it doesn't duplicate something). – David Z Oct 8 '14 at 19:28
• Related question that might shed light here (but isn't an exact duplicate) physics.stackexchange.com/questions/139346/… – tpg2114 Oct 8 '14 at 19:33
• The pressure certainly does if you're in a gravity field and not in free fall. The density may or may not change depending on the fluid. For water the density changes very little with pressure. – DanielSank Oct 8 '14 at 19:44

Let's investigate as follows. Imagine that the column is in the $z$-direction and has cross-sectional area $A$. Let $z=0$ at the ground. Consider a small, vertical "piece" of the column between height $z$ and $z+\Delta z$. The force pushing up on the bottom of this piece due to the air below it is the pressure $P(z)$ at the bottom times the cross sectional area $A$ of the piece.
\begin{align} F_\mathrm{top} = P(z)A \end{align} Similarly, the force pushing down on top due to the air above is \begin{align} F_\mathrm{bottom} = -P(z+\Delta z)A. \end{align} If the column is in equilibrium, these forces along with the weight of the piece will sum to zero -- they will balance. Since the mass of the piece is it's volume times its density, and since $\Delta z$ is taken small, the density can be taken to be the density $\rho(z)$ at its bottom which gives a mass $A\Delta z \,\rho(z)$. Therefore, it's weight is \begin{align} W = -A\Delta z\rho(z) g \end{align} Summing up all the forces and setting them to zero, namely \begin{align} F_\mathrm{top} + F_\mathrm{bottom} + W = 0 \end{align} gives \begin{align} P(z)A-P(z+\Delta z)A-A\Delta z\rho(z) g = 0 \end{align} which can be rearranged to give \begin{align} \frac{P(z+\Delta z) - P(z)}{\Delta z} = -\rho(z) g \end{align} If one takes the limit $\Delta z \to 0$, then the left hand side becomes $P'(z)$, namely the derivative of the pressure with respect to $z$, so we obtain a relationship between the derivative of the pressure and the density \begin{align} P'(z) = -\rho(z) g \tag{$\star$} \end{align} On the other hand, the ideal gas Law tell us that \begin{align} P(z) = \frac{\rho(z)}{m}kT \end{align} where $m$ is the mass per molecule, $k$ is Boltzmann's constant, and $T$ is the temperature of the gas. Combining this with $(\star)$ gives the following differential equation for the density: \begin{align} \rho'(z) = -\frac{mg}{kT} \rho(z) \end{align} and therefore, the density as a function of height $z$ above the ground is \begin{align} \boxed{\rho(z) = \rho(0) e^{-\frac{mg}{kT} z}} \end{align} In other words, in this simplified model, the density of an ideal gas decreases exponentially with height.