Search any question & find its solution
Question:
Answered & Verified by Expert
An ideal gas of molar mass $\mathrm{M}$ is contained in a very tall vertical cylindrical column in the uniform gravitational field. Assuming the gas temperature to be $\mathrm{T}$, the height at which the centre of gravity of the gas is located is (R: universal gas constant)
Options:
Solution:
1573 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{RT}}{\mathrm{Mg}}$
$\because$ field is uniform COM and COG coincides.
$$
\text { So } \mathrm{y}_{\mathrm{cem}}=\frac{\int_{0}^{\infty} \mathrm{ydm}}{\int_{0}^{\infty} \mathrm{dm}}=\frac{\int_{0}^{\infty} \mathrm{y} \mathrm{Ady} \rho}{\int_{0}^{\infty} \mathrm{Ady} \rho}=\frac{\int_{0}^{\infty} \mathrm{y} \rho \mathrm{dy}}{\int_{0}^{\infty} \rho \mathrm{dy}}
$$
From Barometric formula :- $\rho=\rho_{0} \mathrm{e}^{-\mathrm{Mgy} / \mathrm{RT}}$
So $y_{\text {cen }}=\frac{\int_{0}^{\infty} y e^{-M g y / R T}}{\int_{0}^{\infty} e^{-M g y / R T}}=\frac{R T}{M g}$
$$
\text { So } \mathrm{y}_{\mathrm{cem}}=\frac{\int_{0}^{\infty} \mathrm{ydm}}{\int_{0}^{\infty} \mathrm{dm}}=\frac{\int_{0}^{\infty} \mathrm{y} \mathrm{Ady} \rho}{\int_{0}^{\infty} \mathrm{Ady} \rho}=\frac{\int_{0}^{\infty} \mathrm{y} \rho \mathrm{dy}}{\int_{0}^{\infty} \rho \mathrm{dy}}
$$
From Barometric formula :- $\rho=\rho_{0} \mathrm{e}^{-\mathrm{Mgy} / \mathrm{RT}}$
So $y_{\text {cen }}=\frac{\int_{0}^{\infty} y e^{-M g y / R T}}{\int_{0}^{\infty} e^{-M g y / R T}}=\frac{R T}{M g}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.