According to [1], the plasma produced by lightning has a temperature ~ 28,000 K, and an electron density ~ $10^{24} / \rm m^3$. What is the density of the plasma in $\rm kg / m^3$?
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$\begingroup$ Don't you just multiply the number density of electrons by the electron rest mass? $\endgroup$ – honeste_vivere Jun 9 '17 at 13:09
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$\begingroup$ @honeste_vivere the mass of the plasma is mainly due to the nuclei (since electrons are ~1800x lighter), so one needs to figure out the density of those, in the actual plasma, i.e., not the air before the lightning strikes. $\endgroup$ – Ondřej Čertík Jun 11 '17 at 14:46
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1$\begingroup$ If you consider the lightning plasma being a proper plasma, is it quasineutral, i.e. electron density = ion density (if we assume single ionized particles). So you would need to take a look at the composition of air and which of those components is ionized at 28,000 K $\approx$ 2.4 eV. $\endgroup$ – Alf Jun 14 '17 at 10:38
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According to [2], equation (10), the average density in the lightning stroke at 24,000K is $\left({\rho/\rho_0}\right)_{\rm avg} = 0.1$. The $\rho_0$ is given before as $\rho_0 = 1.29 \times 10^{-3}\ {\rm g/cm^3} = 1.29\ {\rm kg / m^3}$.
So the average density of the lightning plasma is 0.129 kg / $\rm m^3$ according to [2].
[2] Uman, M. A., Orville, R. E., & Salanave, L. E. (1964). The Density, Pressure, and Particle Distribution in a Lightning Stroke near Peak Temperature. Journal of the Atmospheric Sciences. http://doi.org/10.1175/1520-0469(1964)021<0306:TDPAPD>2.0.CO;2
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$\begingroup$ Note that $\rho_{0}$ is the density of air before the lightning strike... $\endgroup$ – Jon Custer Jun 9 '17 at 0:25
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$\begingroup$ @JonCuster yes, that's my understanding of the article. $\endgroup$ – Ondřej Čertík Jun 11 '17 at 14:47
