Question

If nitrogen gas molecules goes straight up from the surface of the earth with its rms speed at $0^{\circ }\text{C}$ and there is no collision with other molecules, then it will rise to an approximate height of:
(Given, The acceleration due to gravity at a comparable height $h$ above earth's surface is given by $g^{\prime }=\frac{h}{1+\frac{h}{R}}$)

• A. $8\text{km}$
• B. $12\text{km}$
• C. $12\text{m}$
• D. $8\text{m}$

Answer 1

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

We know that,
$v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.3\times 273}{28\times {10}^{-3}}}$
Molecular weight of $N_2$ is,
$M=28\text{g}=28\times {10}^{-3}\text{kg}$
$\therefore v_{\text{rms}}\simeq 493\text{m/s}$
Now applying the mechanical energy conservation for molecule when it reaches maximum height $h$,
$\text{Increase in Grav P.E}=\text{decrease in K.E}$
$\Rightarrow \frac{mgh}{1+\frac{h}{R}}=\frac{1}{2}m{v}_{\text{rms}}^2$
$\Rightarrow \frac{2gh}{1+\frac{h}{R}}=v_{rms}^2$
$\Rightarrow \frac{2\times 9.8\times h}{1+\frac{h}{6400\times {10}^3}}={493}^2$
Solving this we get,
$h\approx 1200\text{m}\approx 12\text{km}$