Consider a gas of particles with mass m in a gravitational field g. The number density n(h) at height h follows the Boltzmann distribution. What is the ratio of the number of particles in the height range 0 to H to the number of particles in the range H to infinity?

A. ( mgHkT) - 1 — The number of particles is the integral of the density function over the height. Integrating n(h) from 0 to Hand H to infinity gives the ratio (1 - exp(-mgH/kT)) / exp(-mgH/kT), which simplifies to exp(mgH/kT) - 1.

JEE Advanced · PhysicsHard

$Consider a gas of particles with mass m in a gravitational field g . The number density n (h) = n_{0} exp (- m g h / k T) . What is the ratio of the number of particles in the height range 0 to H to the number of particles in the range H to \infty ?$

A. $exp (\frac{m g H}{k T}) - 1$
B. $1 - exp (- \frac{m g H}{k T})$
C. $exp (- \frac{m g H}{k T})$
D. $exp (\frac{m g H}{k T})$

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Correct answer: A — $exp (\frac{m g H}{k T}) - 1$

Solution

Number N = \int n (h) d h . Ratio is \frac{\int _{0}^{H} e ^{- m g h / k T} d h}{\int _{H}^{\infty} e ^{- m g h / k T} d h} = \frac{1 - e ^{- m g H / k T}}{e ^{- m g H / k T}} = e^{m g H / k T} - 1 .

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Consider a gas of particles with mass m in a gravitational field g. The number density n(h)=n0​exp(−mgh/kT). What is the ratio of the number of particles in the height range 0 to H to the number of particles in the range H to ∞?

Solution

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More Kinetic Theory of Gases practice questions

$Consider a gas of particles with mass m in a gravitational field g . The number density n (h) = n_{0} exp (- m g h / k T) . What is the ratio of the number of particles in the height range 0 to H to the number of particles in the range H to \infty ?$