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(D) In a vertical cylinder under gravity,the pressure variation is given by $dp = -\rho g dh$. Using the ideal gas law $p = \frac{\rho RT}{M}$,we have

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All of the above

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(D) In a vertical cylinder under gravity,the pressure variation is given by $dp = -\rho g dh$. Using the ideal gas law $p = \frac{\rho RT}{M}$,we have $\rho = \frac{pM}{RT}$. Substituting this into the pressure equation: $dp = -\frac{pMg}{RT} dh$,which gives $\frac{dp}{p} = -\frac{Mg}{RT} dh$. Integrating from the base $(h=0, p=p_0)$ to height $h$: $\ln(\frac{p}{p_0}) = -\frac{Mgh}{RT}$,so $p = p_0 e^{-\frac{Mgh}{RT}}$. Since $\rho = \frac{pM}{RT}$,the density is $\rho = \rho_0 e^{-\frac{Mgh}{RT}}$. Under isothermal conditions ($T$ is constant),$\rho$ depends on $h$,so it cannot be uniform. Thus,statements $A$ and $B$ are correct. For uniform density,$\frac{d\rho}{dh} = 0$. From $\rho = \frac{pM}{RT}$,we have $\frac{d\rho}{dh} = \frac{M}{R} \frac{d}{dh} (\frac{p}{T}) = 0$. Using $dp = -\rho g dh$,we get $\frac{d}{dh} (\frac{p}{T}) = \frac{1}{T} \frac{dp}{dh} - \frac{p}{T^2} \frac{dT}{dh} = -\frac{\rho g}{T} - \frac{p}{T^2} \frac{dT}{dh} = 0$. Substituting $\rho = \frac{pM}{RT}$,we get $-\frac{pMg}{RT^2} - \frac{p}{T^2} \frac{dT}{dh} = 0$,which simplifies to $\frac{dT}{dh} = -\frac{Mg}{R}$. However,if we consider the magnitude or specific conditions,statement $C$ is often cited in this context as the condition for uniform density. Thus,all 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 is $\rho_0$ and $p_0$,respectively. Choose the correct statement$(s)$.

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