Question

If nitrogen gas molecules goes straight up with its rms speed at $0^{\circ }$ from the surface of the earth and there are no collision with other molecules, then it will rise to an approximate height of
(Take $g=10m/s^2$)

• A. $60\text{km}$
• B. $40\text{km}$
• C. $35\text{km}$
• D. $12\text{km}$

Answer 1

The correct option is D $12\text{km}$

Given,
Temperature, $T=0^{\circ }\text{C}=273\text{K}$
We know that, $V_{rms}=\sqrt{\frac{3RT}{M}}\therefore V_{rms}=\sqrt{\frac{3\times 8.3\times 273}{28\times {10}^{-3}}}\left[\text{molar mass of nitrogen = 28 g}\right]=493\text{m/s}$
Let us suppose, it rises to maximum height $h$. At maximum height, molecule will stop instantly and all its kinetic energy will be converted into potential energy.
$\therefore$ Increase in PE = Decrease in KE
$\Rightarrow mgh=\frac{1}{2}m{V}_{rms}^2\Rightarrow h=\frac{V_{rms}^2}{2g}=\frac{{493}^2}{2\times 10}=12152.45\text{m}\Rightarrow h\approx 12\text{km}$