# Decay pressure altitude [closed]

Under the assumptions of isothermal layer and ideal gas, derive the equation of exponential decay of pressure with respect to altitude using the calculus method: cut a small piece of air of thickness equal to dz and base area equal to A as shown in Fig. and then integrate all the small pieces together from altitude $z_1$ to $z_2$ with corresponding pressure from $p_1$ to $p_2$. Eventually, one can derive the following $$p_2=p_1 e^{\frac{z_1-z_2}{h}}$$

[Hint: The figure indicate that the small pressure increment may be written as dp = −ρgdz. The state equation of ideal gas is assumed to hold: pV = nRT. The density ρ = n/V .]}

Extra Go through the derivation and find h as a function of R, T and g

Attempt

\begin{aligned} P(z+\Delta x)A -P(z)A&=\rho_{air}g \Delta z *g \\ \frac{P(z+\Delta z)-P(z)}{\Delta z }=\rho_{air} g \\ \frac{dP}{dz}=\rho_{air} g \end{aligned} From Diff Eq \begin{aligned} p(z)&=P_0 e^{\rho_{air}g*z} \\ &=p_{0}e^{\frac{M_{air} P_{abs}}{RT}*z} \end{aligned}

Cant see how $$p_2=p_1 e^{\frac{z_1-z_2}{h}}$$ was derive thinking diff assmptions were made

a hand drawn pic and free body diagram Give me a min

## closed as off-topic by Jon Custer, John Rennie, Gert, knzhou, user36790 Oct 21 '16 at 3:23

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• Have you tried using the ideal gas law yet? Think about what $h$ represents. Also, think about the bounds of pressure when you integrate. You may want to write out more steps. – HDE 226868 Oct 20 '16 at 18:12
• not sure . h is height?? think that I did use Ideal gas law. or h is the middle of the square height. IDK – Tiger Blood Oct 20 '16 at 18:14

You are given $dp = − \rho gdz$.
Your error, although you do not realise it, is that the density $\rho$ depends on the pressure.
Substitute for the density in the equation for $dp$ and do the integration.
You should now be able to call a constant in your pressure equation $h$ to get the required equation for pressure.