The average kinetic energy of one molecule of an ideal gas at 27^(@)C and 1 atm pressure is:

The average kinetic energy of one molecule of an ideal gas at 27^(@)C

1.	The average kinetic energy of one molecule of an ideal gas at 27^(@)C and 1 atm pressure is:
Answer» `900 cal K^(-1) mol^(-1)` `6.21 xx 10^(-21) J K^(-1) "MOLECULE"^(-1)` `336.7 J K^(-1) "molecule"^(-1)` `3741.3 J K^(-1) mol^(-1)` Solution :`KE = (3)/(2) (R)/(N) T` `= (3)/(2) xx (8.314)/(6.023 xx 10^(23)) xx 300` `= 6.21 xx 10^(-21) J K^(-1) "molecule"^(-1)`