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Chapter 6: Problem 119

A nitrogen molecule ( \(\mathrm{N}_{2}\) ) having the average kinetic energy at \(300 \mathrm{K}\) is released from Earth's surface to travel upward. If the molecule could move upward without colliding with other molecules, then how high would it go before coming to rest? Give your answer in kilometers. [Hint: When the molecule comes to rest, the potential energy of the molecule will be \(m g h\) where \(m\) is the molecular mass in kilograms, \(g=9.81 \mathrm{m} \mathrm{s}^{-2}\) is the acceleration due to gravity, and \(h\) is the height, in meters, above Earth's surface.]

Short Answer

Expert verified
The nitrogen molecule would reach an elevation of approximately 90.5 kilometers above Earth's surface before coming to rest.

Step by step solution

01

Calculate Initial Kinetic Energy of Nitrogen Molecule

First, calculate the initial kinetic energy of the nitrogen molecule. The average kinetic energy at \(300K\) can be given by \(kT\), where \(k = 1.38 x 10^{-23} J/K\) is the Boltzmann constant and \(T = 300K\) is temperature. The kinetic energy is then calculated by multiplying the given temperature with the Boltzmann constant, i.e., \(Ek_{initial} = kT = 1.38 x 10^{-23} J/K x 300 K = 4.14 x 10^{-21} J\).
02

Calculate Molecular Mass

As nitrogen (N) has an atomic mass of approximately 14 atomic mass units (a.m.u), the molecular mass (as \(\mathrm{N}_{2}\)) is \(2x14 = 28 a.m.u\). Convert this to kilograms by realizing that \(1 a.m.u = 1.66 x 10^{-27} Kg\), so \(m = 28 a.m.u * 1.66 x 10^{-27} Kg/a.m.u = 4.65 x 10^{-26} Kg\).
03

Calculate Final Potential Energy

When the nitrogen molecule comes to rest at height \(h\), it will have potential energy given by \(mgh = Ek_{initial}\), where \(g\) is acceleration due to gravity \(9.81 m/s^{2}\).
04

Solve for Height

Now, solve the previous equation for \(h\). Rearranging the equation with respect to \(h\), we have \(h = Ek_{Initial}/(mg) = 4.14 x 10^{-21} J / (4.65 x 10^{-26} Kg x 9.81 m/s^{2}) = 90473 m\).

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