Question

How many particles are present in a closed container if the energy it contains is 95577.88J, and the diatomic oxygen gas is moving at a velocity of 51.28m/s?

Answer 1

Solution:

Energy and number of particle relation is given by as

$E=\frac{nM_Av^2}{2N_A}$

$n=\frac{2E N_A}{M_Av^2}$

$n=\frac{2\times95577.88\times 6.023\times 10^{23}}{32\times(51.28)^2}$

$n=13.68\times10^{23}$