Find the KE per kg of nitrogen molecule at `127^@C` (molecular weight of nitrogen = 28, R = `8320 J//K mole 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 .

Find the KE:- per molecule of nitrogen at N.T.P.(molecular weight of N_2 = 28, Normal pressure = 76 cm of mercury, density of mercury 1.36 gm//cm^2 , g = 980 cm//s^2 , Avogadro's number 6.023 xx 10^23 , R = 8.314 xx 10^7 erg//mol^k , molecular weight of nitrogen = 31).

0btain the mean free path of nitrogen molecule at 0^@C and 1.0 atm pressure. The molecular diameter of nitrogen is 324 pm (assume that the gas is ideal).

Calculate the average molecular K.E:- per molecule of oxygen at 127^@C given that molecular weight =32, R = 8.31 J mol^-1 K^-1 , N_A = 6.02 xx 10^23 , T =127^@C

Compare the R.M.S. velocity of hydrogen molecules at 400 K with R.M.S. velocity of oxygen molecules at 900 K. (Molecular weight of hydrogen = 2, molecular weight of oxygen = 32)

Calculate kinetic energy of 10 gm of Argon molecule at 127^@C . (R = 8320 J//K mole K, Atomic weight of Argon = 40)

At what temperature will nitrogen molecules have same R.M.S. speed as oxygen molecules at 400 K (molecular weight of oxygen = 32, molecular weight of nitrogen = 28)

Total random kinetic energy of the molecules in 1 mole of a gas at a temperature of 300 K is (R = 2 cal/mole °C)