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I have solved the following exercise but the answer I get is different from the one stated in the book I am using: I can't see what I am doing wrong so I would be grateful if someone pointed it out to me, thanks.

"Consider a box of volume $1.5L$ full of nitrogen gas which exerts a pressure of $3 atm$ on the walls of the box. The translational kinetic energy of the nitrogen molecules of the gas is $6.42\cdot 10^{-28}J$. Find the number of nitrogen molecules contained in the box"

My solution:

Perfect gas law ($n=$ number of moles, $N=$number of molecules) $$PV=nRT\Rightarrow n=\frac{PV}{RT}\Rightarrow N=\frac{PV}{RT}N_A\overset{R=k_bN_A}{=}\frac{PV}{k_bT}\overset{K=\frac{3}{2}k_bT}{=}\frac{3PV}{2K}=\frac{3\cdot(303975Pa)(0.0015m^3)}{2\cdot6.44\cdot 10^{-28}J}=1.06\cdot 10^{30}$$ but the book says the correct answer is $1.06\cdot 10^{23}$.

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I guess there is a typo in the book, most likely in the number for translational kinetic energy. If you try to find temperature from the provided number, it will be about $3\cdot10^{-5}$ K - even if you manage to reach such temperature, nitrogen would be solid. They probably meant to have $6.42\cdot10^{-21}$ J for translational kinetic energy which corresponds to $310$ K and then it is clear that you should have some fraction of a mole (remember, one mole at normal conditions takes up 22.4 liters).

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  • $\begingroup$ Thank you very much. $\endgroup$ – lorenzo Apr 2 at 17:51
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    $\begingroup$ I'd guess that the original question was in cgs units and that the energy was in ergs (1 erg is $10^-7$ J), and someone changed the units at some point but forgot to change the value. $\endgroup$ – PhillS Apr 2 at 18:40
  • $\begingroup$ @PhillS That seemed to me to make sense because the difference in powers is 7, but then I realized that the difference is in the wrong direction: $10^{-28}$ erg = $10^{-35}$ J. $\endgroup$ – Viking Apr 5 at 13:30

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