Calculate average molecular kinetic energy:- per kilogram of oxygen at `27^@C` (R = 8320 J/k mole K, Avogadro s number `6.02 xx 10^27` molecule/K mole.

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

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 .

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

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

The average translational kinetic energy and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21 xx 10^(-21) J and 484 m/s respectively. Then the corresponding values at 600^@K are

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).

The kinetic energy of 1 kg of oxygen at 300 K is 1.356 xx 10^6 J . Find kinetic energy of 4 kg of oxygen at 400 K.

Two vessels A and B are identical. A has 1 g hydrogen at 0^(@)C and B has 1 g oxygen at 0^(@)C . Vessel A contains x molecules and B contains y molecules. The average kinetic energy per molecules in A is 'n' times the average kinetic energy per molecule in B. The value of 'n' is

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)