JEE Main202427 Jan 2024Evening ShiftPhysicsKinetic Theory of GasesActual
The total kinetic energy of 1 mole of oxygen at 27 ° C is : [Use universal gas constant ( R ) = 8 . 31 J mol - 1 K - 1 ]
Options
- A6845 . 5 J
- B5942 . 0 J
- C6232 . 5 J
- D5670 . 5 J
Correct answer
C. 6232 . 5 J
Step-by-step solution
The formula to calculate the kinetic energy can be written as K = f 2 n R T . . . 1 Since oxygen is a diatomic gas, the number of degrees of freedom is 5 . Thus, from equation (1), it follows that K = 5 2 × 1 × 8 . 31 × 300 J = 6232 . 5 J