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I'm trying to perform a physics problem, however, my issue is with the math. I want to sum each bit of contributing pressure felt at the base of a static gas column as additional "parcels of gas" are added to the top of the column.

I am working with the barometric formula, where the density of ideal gas varies with elevation following this relation: $$\tag{1} \rho (z) = \rho_o \cdot \exp\left(-\frac{Mg}{RT} dz\right)$$

The pressure at the base of a static gas column can be calculated using this relation: $$\tag{2} p=\rho gz$$

So, I think that the change in pressure with each incremental change in height of the gas column can be written mathematically as $$\tag{3} dp=\rho (z)gdz$$

which can then be integrated, I believe as so $$\tag{4} \int_{p_o}^p dp = g \int_{z_o}^z \rho (z) dz$$

$$\tag{5} p_z - p_o = g \int_{z_o}^z \rho_o \cdot e^{(-\frac{Mg}{RT}\ dz)}\ dz$$

I'm not sure if everything I have here is correct, but I do know that I am stumped on how to handle the integral on the rhs of eqn 5. The variables on the rhs of eqn 5: $\rho_o, M, g, R, T$ are all constants.

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    You are mistaken with the $dz$ inside the exponential. It should be something like $(z-z_0)$ where $z_0$ is the elevation at which the density is $\rho_0$. Use $e^{-a z}$ integrates to $\frac{-e^{-a z}}{a}$. – 2017-01-29
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    @user121049 when using $e^{-a z}$, does $z=(z-z_o)$? And then, I have $p_z-p_o=\rho_o \cdot \frac{-e^{-a (z-z_o)}}{a} \cdot (z-z_o)$? – 2017-01-29
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    You get $p(z)-p_0=\frac{\rho_0}{a}(1-e^{-a(z-z_0)})$. – 2017-01-30
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    @user121049 why does the function integrate to $1-e^{-a(z-z_0)}$? Also, looking back at Eqn 5, did we drop the gravity term $g$, i.e., should it be $p(z)-p_0=\frac{\rho_0 g}{a}(1-e^{-a(z-z_0)})$; with $a=\frac{Mg}{RT}$? If you like, put what you have as an answer with a bit of explanation on how to perform the integral. I'd be more than happy to accept as the answer. Thanks! – 2017-01-30

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If you replace Eq 5 with $p(z)-p_0=g\rho_0\int_{z_0}^{z}{e^{-a(z-z_0)}dz}$ as per comments. Then this integrates to $p(z)-p_0=-\frac{g\rho_0}{a}(e^{-a(z-z_0)})$ which has to be evaluated at $z$ and $z_0$. This gives $p(z)-p_0=\frac{g\rho_0}{a}(1-e^{-a(z-z_0)})$ because $e^{-a(z_0-z_0)}=1$.

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    and why does Eq 5 not include the $dz$ term from Eqn 2 (or Eqn 3)? I.e., why not written as $p(z)-p_0=g\rho_0\int_{z_0}^{z}{e^{-a(z-z_0)}} dz$? – 2017-02-01