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(D) $1$. In a vertical column of gas,the pressure $p$ decreases as height $h$ increases due to the weight of the gas column above it. Thus,$\frac{dp}{

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All of the above.

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(D) $1$. In a vertical column of gas,the pressure $p$ decreases as height $h$ increases due to the weight of the gas column above it. Thus,$\frac{dp}{dh} = -\rho g$ is the correct hydrostatic equilibrium equation for a gas.$2$. From the ideal gas law,$pV = nRT$,we can write $p = \frac{n}{V} RT$. Since the number of moles $n = \frac{m}{M}$ (where $m$ is mass and $M$ is molar mass),we have $p = \frac{m}{VM} RT$. Given that density $\rho = \frac{m}{V}$,we get $p = \rho \frac{RT}{M}$.$3$. Since pressure $p$ depends on the density $\rho$ and height $h$,and the weight of the gas column acts downwards,the pressure must decrease as we move up the cylinder.$4$. Therefore,all the given statements are correct.

$A$ very tall vertical cylinder is filled with a gas of molar mass $M$ under isothermal conditions at temperature $T$. The density and pressure of the gas at the base of the container are $\rho_0$ and $p_0$,respectively. Choose the correct statement$(s)$.

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