Can doubling the particle density in a fusion reactor increase energy output?

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jjson775
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Homework Statement
A fusion reactor has achieved an ion temperature of 4 x 10^8 K, an ion density of 2 x 10^13 /cm^3 and a confinement time of 1.4 s. Calculate the amount of energy stored in the plasma of the reactor. The plasma volume of the reactor is about 50 m^3.
Relevant Equations
E = 3/2 NKbT
E= 3/2 (1.381x10^-23)(4x10^8)(2x10^19)50
E= 8.29 x 10^6 Wrong
 
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A plasma contains ions and also electrons (such that the plasma is electrically neutral overall). Maybe the thermal energy of the electrons should be included. But no information is given regarding the number of electrons per ion.
 
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jjson775 said:
E= 8.29 x 10^6 Wrong
Two obvious mistakes are:
1) missing units;
2) answer needs rounding to 1 significant figure to match the data.

And, as noted by TSny, electrons are also present. Assuming the plasma is an isotope of hydrogen (typically deuterium), there will be equal numbers of ions and electrons. The particle density is then 2 x 2 x 10¹³ /cm³ which doubles your answer.

[Edit: However, I've seen 'ions' used as a general term for charged particles. So 2 x 10¹³ /cm³ might include both the nuclei and electrons. Or it might be just the nuclei. You may have to try both.]

Note that treating the plasma as an ideal gas is a crude aproximation. Plasma physics is very complex. (I don't claim to understand it.) If you are studying ideal gases at an introductory level, your method (plus the above corrections) should be adequate. But if you are at a more advanced level and studying fusion/fusion-reactors, a more sophisticated analysis may be needed.

I don’t see why the confinement time is given, unless needed for a subsequent question.
 
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Doubling the particle density yields the right answer so the electrons had to be considered. I am very aware that using the formula for an ideal gas is very crude. I had also decided that confinement time is irrelevant. I am an old guy self studying modern physics from a textbook and am in a chapter on fission and fusion. Thanks for your help.
 
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