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(D) For a gas in equilibrium under gravity,the change in pressure $dp$ for a small change in height $dh$ is given by $dp = -\rho g dh$.Using the ideal

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All of the above.

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(D) For a gas in equilibrium under gravity,the change in pressure $dp$ for a small change in height $dh$ is given by $dp = -\rho g dh$.Using the ideal gas law,$pV = nRT$,we can write $\rho = \frac{pM}{RT}$.Substituting this into the pressure equation: $dp = -\frac{pM}{RT} g dh$.Rearranging the terms: $\frac{dp}{p} = -\frac{Mg}{RT} dh$.Integrating from base $(h=0, p=p_0)$ to height $h$ $(p=p)$: $\int_{p_0}^{p} \frac{dp}{p} = -\int_{0}^{h} \frac{Mg}{RT} dh$.This yields $\ln(\frac{p}{p_0}) = -\frac{Mgh}{RT}$,which simplifies to $p = p_0 e^{-\frac{Mgh}{RT}}$.Since $\rho = \frac{pM}{RT}$,we have $\rho = \frac{p_0 M}{RT} e^{-\frac{Mgh}{RT}} = \rho_0 e^{-\frac{Mgh}{RT}}$.Thus,both pressure and density decrease exponentially with height. Therefore,all statements are correct.

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