This might be a bit more of an engineering question, but I'm calculating air density drop-off with altitude, and I'm having some problems calculating the pressure (I'll run through my method). This has been very useful in explaining, but the last bit lost me a little.
So we start with an ideal gas, then:
$$ pV = nRT $$ and using $$\frac{\rho V}{n} = M$$ where M is molar mass, you can calculate density to be:
$$ \rho = \frac{pM}{RT} $$ which implies a solution dependent only on pressure (p) and temperature (T).
Then define temperature using the Universal Standard Atmosphere lapse rate, $$T = T_0 – Lh$$ where L = 0.0065K/m and h is height in metres
Now at this point I'm a bit stuck. Wikipedia suggests the following equation:
$$p = p_0 \left(1 – \frac{L h}{T_0} \right)^\frac{g M}{R L}$$
Which, when calculated, provides values within a 5% tolerance – but where has it come from? I can't find any reference to its source or how it was derived. Can anyone help?
Best Answer
To derive it, you need to consider hydrostatic equilibrium for an air slab of thickness $dz$ at height $z$:
\begin{equation}P(z+dz)A-P(z)A=-mg\end{equation}
where $m$ is the mass of air with cross-section area $A$ and height $dz$. For $dz<<z$, we can write:
\begin{equation}P(z+dz)=P(z)+\dot{P}(z)dz\end{equation}
and substituting this in the first equation above finally gives:
\begin{equation}dP=-\rho g dz\end{equation} Using the definition of $\rho$ that you wrote in your post and solving the differential equation finally gives you the desired result.