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?
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}"
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