# Confusion about number of molecules in a gas [closed]

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 $$1.06\cdot 10^{23}$$.

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).
• 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. – PhillS Apr 2 at 18:40
• @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. – Viking Apr 5 at 13:30