<p>The <em>free</em> electrons in an ionised gas are included in the calculation of the mean molecular weight <span class="math-container">$\mu$</span> because they must be present in the gas (it can't be net positive) and they contribute equally (per particle) to the gas pressure. They are not included in <span class="math-container">$\mu$</span> when the gas is neutral because <em>bound</em> electrons are not treated as separate particles from the point of view of calculating the gas pressure and they contribute almost no mass to the gas.</p>
<p><strong>Details</strong></p>
<p>The &quot;mean molecular weight&quot; <span class="math-container">$\mu$</span> is used in calculations to relate pressure to density (and temperature) - the so-called &quot;equation of state&quot;. An example would be the perfect gas law, which can be written
<span class="math-container">$$ P = n k_B T\ , $$</span>
where <span class="math-container">$P$</span> is pressure, <span class="math-container">$T$</span> is temperature and <span class="math-container">$n$</span> is the nuber density of particles - i.e. the number of particles per cubic metre.</p>
<p>Note that <span class="math-container">$n$</span> includes ALL isolated, free particles - it does not matter whether they are protons, electrons, ions, atoms or molecules; all free particles contribute equally to the kinetic gas pressure. But note that a single atom, consisting of a nucleus with electrons is treated as a <em>single</em> free particle for the purposes of working out the gas pressure.</p>
<p>However, we often wish to express pressure in terms of mass density, since if we know the mass and size of the gaseous object, then we have a direct measure of that rather than the particle number density.</p>
<p>To relate mass density to number density, we need to know something about the composition of the gas and its ionisation state and that is where <span class="math-container">$\mu$</span> comes in. It is alternatively defined as <em>the mean number of atomic mass units per free particle in the gas</em>, and with that definition it is clear that mass density <span class="math-container">$\rho$</span> is related to <span class="math-container">$n$</span> by <span class="math-container">$\rho =  \mu m_u n$</span>, where <span class="math-container">$m_u$</span> is an atomic mass unit and is very close to the mass of a proton. The perfect gas law then becomes
<span class="math-container">$$ P = \frac{\rho}{\mu m_u} k_B T\ . $$</span></p>
<p>In an ionised gas, free electrons MUST be included in this calculation because, as I said, they make just as much contribution to the gas pressure as do any other free particles in the gas. Thus for every ionised hydrogen atom there is 1 electron and 1 proton and therefore 2 particles with about 1 atomic mass unit between them. Hence <span class="math-container">$\mu = 1/2$</span>. The mass of the electron is so small (about <span class="math-container">$0.00055 m_u$</span>) that it is neglected in approximate calculations. On the other hand a neutral hydrogen atom also has mass of 1 atomic mass unit, but now consists of only one free particle and so neutral hydrogen gas would have <span class="math-container">$\mu = 1/1 = 1$</span>.</p>

CC BY-SA 4.0
2023-12-30T13:32:56.573Z
2023-12-30T13:38:55.237Z
2023-12-30T13:38:55.237Z
ProfRob
<p>Electrons are so light that their mass can be treated as 0 for the purposes of this calculation. So for atomic hydrogen, the mass of each particle is 1+0 = 1. The electron is tightly bound to the proton, so these don't behave like separate particles.</p>
<p>Similarly for molecular Hydrogen, <span class="math-container">$H_2$</span>, the two atoms are bound into a single molecule, and the average mass is 2.</p>
<p>If the hydrogen is fully ionized, then half the particles have mass 1 (the protons) and half the particles have mass 0 (the electrons), so the average mass per particle is 1/2.  Remember that the electrons don't disappear when hydrogen is ionized, they are still there.  We care about the free electron because it is a component of the plasma.</p>
<p>We could do this calculation for Helium too.  Neutral atoms have an average mass of 4.  If you fully ionize helium you get three particles, one nucleus with a mass of 4 and two electrons.  Hence the average mass is 4/3 or 1.33. Most larger atoms have roughly equal numbers of protons and neutrons, so when you fully ionize an atom with <span class="math-container">$n$</span> protons, you get <span class="math-container">$n+1$</span> particles, one with a mass of about <span class="math-container">$2n$</span> and <span class="math-container">$n$</span> with a mass of 0, so the average mass is about <span class="math-container">$2n/(n+1)\approx 2$</span>,</p>
<p>(<a href="https://faculty.fiu.edu/%7Evanhamme/ast3213/mu.pdf" rel="nofollow noreferrer">source</a>)</p>

CC BY-SA 4.0
2023-12-30T11:03:08.527Z
2023-12-30T11:11:43.283Z
2023-12-30T11:11:43.283Z
James K
<p>I am confused about the calculation of mean molecular weights, especially for ionized particles. From my understanding, the mean molecular weight <span class="math-container">$\mu$</span> is the average mass per the number of particles:</p>
<p><span class="math-container">$$ \mu = \frac{\langle m \rangle}{m_p}$$</span></p>
<p>Resources typically explain it as follows:</p>
<p>For neutral H, <span class="math-container">$\mu = 1$</span> since there is one nucleus. For fully ionized H, since the mass of H is 1 amu that is shared between a nucleus and a free electron, <span class="math-container">$\mu = 1/2$</span>. But isn't a fully ionized hydrogen <span class="math-container">$H^+$</span>, where it does <strong>not</strong> have an electron? So shouldn't <span class="math-container">$\mu_{H, ionized} = 1$</span> since there is only a nucleus? Why are we not considering the electron present in neutral H for neutral H mean mol. weight calculations? Why do we care about the free electron? I'm thoroughly confused.</p>

<star><mass>
Neutral vs ionized mean molecular weights
justauser
# Neutral vs ionized mean molecular weights

(Asked by: username_0 on 2023-12-30)

I am confused about the calculation of mean molecular weights, especially for ionized particles. From my understanding, the mean molecular weight $\mu$ is the average mass per the number of particles:

$$ \mu = \frac{\langle m \rangle}{m\_p}$$

Resources typically explain it as follows:

For neutral H, $\mu = 1$ since there is one nucleus. For fully ionized H, since the mass of H is 1 amu that is shared between a nucleus and a free electron, $\mu = 1/2$. But isn't a fully ionized hydrogen $H^+$, where it does **not** have an electron? So shouldn't $\mu\_{H, ionized} = 1$ since there is only a nucleus? Why are we not considering the electron present in neutral H for neutral H mean mol. weight calculations? Why do we care about the free electron? I'm thoroughly confused.

**Tags:** star, mass


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## 2 answers

### Answer by username_1 on 2023-12-30. Score: 4 | Accepted answer

The *free* electrons in an ionised gas are included in the calculation of the mean molecular weight $\mu$ because they must be present in the gas (it can't be net positive) and they contribute equally (per particle) to the gas pressure. They are not included in $\mu$ when the gas is neutral because *bound* electrons are not treated as separate particles from the point of view of calculating the gas pressure and they contribute almost no mass to the gas.

**Details**

The "mean molecular weight" $\mu$ is used in calculations to relate pressure to density (and temperature) - the so-called "equation of state". An example would be the perfect gas law, which can be written
$$ P = n k\_B T\ , $$
where $P$ is pressure, $T$ is temperature and $n$ is the nuber density of particles - i.e. the number of particles per cubic metre.

Note that $n$ includes ALL isolated, free particles - it does not matter whether they are protons, electrons, ions, atoms or molecules; all free particles contribute equally to the kinetic gas pressure. But note that a single atom, consisting of a nucleus with electrons is treated as a *single* free particle for the purposes of working out the gas pressure.

However, we often wish to express pressure in terms of mass density, since if we know the mass and size of the gaseous object, then we have a direct measure of that rather than the particle number density.

To relate mass density to number density, we need to know something about the composition of the gas and its ionisation state and that is where $\mu$ comes in. It is alternatively defined as *the mean number of atomic mass units per free particle in the gas*, and with that definition it is clear that mass density $\rho$ is related to $n$ by $\rho = \mu m\_u n$, where $m\_u$ is an atomic mass unit and is very close to the mass of a proton. The perfect gas law then becomes
$$ P = \frac{\rho}{\mu m\_u} k\_B T\ . $$

In an ionised gas, free electrons MUST be included in this calculation because, as I said, they make just as much contribution to the gas pressure as do any other free particles in the gas. Thus for every ionised hydrogen atom there is 1 electron and 1 proton and therefore 2 particles with about 1 atomic mass unit between them. Hence $\mu = 1/2$. The mass of the electron is so small (about $0.00055 m\_u$) that it is neglected in approximate calculations. On the other hand a neutral hydrogen atom also has mass of 1 atomic mass unit, but now consists of only one free particle and so neutral hydrogen gas would have $\mu = 1/1 = 1$.


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### Answer by username_2 on 2023-12-30. Score: 4

Electrons are so light that their mass can be treated as 0 for the purposes of this calculation. So for atomic hydrogen, the mass of each particle is 1+0 = 1. The electron is tightly bound to the proton, so these don't behave like separate particles.

Similarly for molecular Hydrogen, $H\_2$, the two atoms are bound into a single molecule, and the average mass is 2.

If the hydrogen is fully ionized, then half the particles have mass 1 (the protons) and half the particles have mass 0 (the electrons), so the average mass per particle is 1/2. Remember that the electrons don't disappear when hydrogen is ionized, they are still there. We care about the free electron because it is a component of the plasma.

We could do this calculation for Helium too. Neutral atoms have an average mass of 4. If you fully ionize helium you get three particles, one nucleus with a mass of 4 and two electrons. Hence the average mass is 4/3 or 1.33. Most larger atoms have roughly equal numbers of protons and neutrons, so when you fully ionize an atom with $n$ protons, you get $n+1$ particles, one with a mass of about $2n$ and $n$ with a mass of 0, so the average mass is about $2n/(n+1)\approx 2$,

([source](https://faculty.fiu.edu/%7Evanhamme/ast3213/mu.pdf))


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