JEE Main 2022 Physics question
The total internal energy of two mole monoatomic ideal gas at temperature T = 300 K will be J. (Given R = 8.31 J/mol.K)
Answer: 7479
Solution
U = n(3/2)RT = (3/2) J.
The total internal energy of two mole monoatomic ideal gas at temperature T = 300 K will be J. (Given R = 8.31 J/mol.K)
U = n(3/2)RT = (3/2) J.