The kinetic energy of translation of 20gm of oxygen at `47^(@)C` is (molecular wt. of oxygen is 32 gm/mol and R=8.3 J/mol/K)

The kinetic energy of translation of 20gm of oxygen at 47^@C is (molecular wt. of oxygen is 32gm/mol and R=8.3 J/mol/K)

Calculate the average molecular kinetic energy : per kg molecule of oxygen at 127 ∘ C, given that the molecular weight of oxygen is 32, R = 8.31 J m o l ^ -1 K − 1 , and Avogadros number N A ​ is 6.02×10 23 molecules mol −1 .

The kinetic energy for 14 g of nitrogen gas at 127^(@)C is nearly (mol. Mass of nitrogen = 28 and gas constant = 8.31 J/mol K)

The kinetic energy for 14 g of nitrogen gas at 127^(@)C is nearly (mol. Mass of nitrogen = 28 and gas constant = 8.31 J/mol K)

Find the kinetic energy of the molecules in 1 gm of NH_3 gas at 27^@C . Molecular mass of NH_3 is 17 gm/mole.

Calculate (a) the average kinetic energy of translation of an oxygen molecule at 27^(@) C (b) the total kinetic energy of an oxygen molecule at 27^(@)C (c ) the total kinetic energy in joule of one mole of oxygen at 27^(@) . Given Avogadro's number = 6.02 xx 10^(23) and Boltzmann's constant = 1.38 xx 10^(-23) J//(mol-K) .

Calculate (a) the average kinetic energy of translation of an oxygen molecule at 27^(@) C (b) the total kinetic energy of an oxygen molecule at 27^(@)C (c ) the total kinetic energy in joule of one mole of oxygen at 27^(@) . Given Avogadro's number = 6.02 xx 10^(23) and Boltzmann's constant = 1.38 xx 10^(-23) J//(mol-K) .