Question

A box contains about 5.54 1021 hydrogen

Answer 1

Answer:$E=33.71\ J$

Explanation:

Given

No of atoms of hydrogen $N=5.54\times 10^{21}$

Temperature of room $T=21^{\circ}\approx 294\ K$

Thermal energy of the atoms is given by

$E=\frac{3}{2}NkT$

where k=boltzmann constant

$E=\frac{3}{2}\times 5.54\times 10^{21}\times 1.38\times 10^{-23}\times 294$

$E=33.71\ J$

Hence the energy of the atoms is $33.71\ J$