(a) What fraction of H2 gas at sea level and T=300 K has sufficient speed to escape from the ea

(a) What fraction of H2 gas at sea level and T=300 K has...

(a) What fraction of $\mathrm{H}_2$ gas at sea level and $T=300 \mathrm{~K}$ has sufficient speed to escape from the earth's gravitational field? (You may assume an ideal gas. Leave your answer in integral form.) (b) Now imagine an $\mathrm{H}_2$ molecule in the upper atmosphere with a speed equal to the earth's escape velocity. Assume that the remaining atmosphere above the molecule has thickness $d=100 \mathrm{~km}$, and that the earth's entire atmosphere is isothermal and homogeneous with mean number density $n=$ $2.5 \times 10^{25} / \mathrm{m}^3$ (not a very realistic atmosphere). Using simple arguments, estimate the average time needed for the molecule to escape. Assume all collisions are elastic, and that the total atmospheric height is small compared with the earth's radius. Some useful numbers: $$ \begin{aligned} M_{\text {earth }} & =6 \times 10^{24} \mathrm{~kg}, \\ R_{\text {earth }} & =6.4 \times 10^3 \mathrm{~km} . \end{aligned} $$

(a) What fraction of $\mathrm{H}_2$ gas at sea level and $T=300 \mathrm{~K}$ has sufficient speed to escape from the earth's gravitational field? (You may assume an ideal gas. Leave your answer in integral form.)
(b) Now imagine an $\mathrm{H}_2$ molecule in the upper atmosphere with a speed equal to the earth's escape velocity. Assume that the remaining atmosphere above the molecule has thickness $d=100 \mathrm{~km}$, and that the earth's entire atmosphere is isothermal and homogeneous with mean number density $n=$ $2.5 \times 10^{25} / \mathrm{m}^3$ (not a very realistic atmosphere).

Using simple arguments, estimate the average time needed for the molecule to escape. Assume all collisions are elastic, and that the total atmospheric height is small compared with the earth's radius.
Some useful numbers:
$$
\begin{aligned}
M_{\text {earth }} & =6 \times 10^{24} \mathrm{~kg}, \\
R_{\text {earth }} & =6.4 \times 10^3 \mathrm{~km} .
\end{aligned}
$$

Key Concepts

Maxwell?Boltzmann Distribution

This distribution provides the probability that a molecule in an ideal gas has a given speed at a particular temperature. It is central to calculating the fraction of particles in the high?velocity tail that exceed any specific threshold speed, such as the escape velocity required to overcome a planet's gravitational pull.

Escape Velocity

Escape velocity is the minimum speed an object or molecule must have to break free from the gravitational field of a celestial body without further propulsion. It is derived from equating a particle’s kinetic energy to the gravitational potential energy and is crucial for understanding which gas particles can naturally leave an atmosphere.

Integration over the Speed Distribution

To determine the fraction of molecules with speeds above a particular threshold (like the escape velocity), one integrates the Maxwell?Boltzmann speed distribution from that threshold to infinity. This integral quantifies the proportion of particles that have sufficient kinetic energy to escape the gravitational field.

Mean Free Path and Collision Dynamics

In kinetic theory, the mean free path is the average distance a molecule travels between successive collisions. Coupled with the collision frequency, these concepts are used to estimate transport properties, including how long it takes for a molecule to traverse a defined region of an atmosphere and potentially escape, which is important in understanding molecular escape times in a collisional environment.

Transcript

00:01 So we have the root mean squared speed of a gas molecule equaling the square root of three times the ideal gas constant multiplied by the temperature divided by m, the molar mass of the gas molecule.

00:17 And so we can then say that the root means squared speed for the hydrogen molecule would be equalling then the square root of three multiplied by the ideal gas speed.

00:32 As constant 8 .3145 multiplied by the temperature of 293 kelvin, 20 degrees celsius converted into kelvin.

00:48 And this would be divided by the molar mass of hydrogen, which is going to be 1 .008 grams per mole times 2.

00:57 So we have 2 .016 times 10 to the negative third kilograms per mole.

01:11 Units for the ideal gas constant 8 .3145 units would be joules per mole kelvin.

01:24 And so we find that then this is going to be equal to 1 ,9004.

01:31 Meters per second.

01:37 And then for part b, we know that the escape velocity on the surface of the earth is nearly six times larger than the rms speed than the rms speed of hydrogen at 20 degrees celsius.

01:57 So we can say velocity of the escape is going to be six times the root mean squared speed for hydrogen.

02:07 And so we can say that then the hydrogen molecule cannot escape.

02:26 And so given this, we're going to find the rms speed for oxygen.

02:39 This would be the same formula.

02:42 3 multiplied by 8 .3145, multiplied by 293 kelvin, and this would all be divided by, this would be 31 .998 times 10 to the name...

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